\(\int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx\) [600]
Optimal result
Integrand size = 33, antiderivative size = 601 \[
\int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {b^{3/2} \left (63 a^4 A b+46 a^2 A b^3+15 A b^5-35 a^5 B-6 a^3 b^2 B-3 a b^4 B\right ) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{7/2} \left (a^2+b^2\right )^3 d}-\frac {\left (8 a^4 A+31 a^2 A b^2+15 A b^4-11 a^3 b B-3 a b^3 B\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b (A b-a B) \cot ^{\frac {5}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (13 a^2 A b+5 A b^3-9 a^3 B-a b^2 B\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}
\]
[Out]
1/4*b^(3/2)*(63*A*a^4*b+46*A*a^2*b^3+15*A*b^5-35*B*a^5-6*B*a^3*b^2-3*B*a*b^4)*arctan(a^(1/2)*cot(d*x+c)^(1/2)/
b^(1/2))/a^(7/2)/(a^2+b^2)^3/d+1/2*b*(A*b-B*a)*cot(d*x+c)^(5/2)/a/(a^2+b^2)/d/(b+a*cot(d*x+c))^2+1/4*b*(13*A*a
^2*b+5*A*b^3-9*B*a^3-B*a*b^2)*cot(d*x+c)^(3/2)/a^2/(a^2+b^2)^2/d/(b+a*cot(d*x+c))+1/2*(a^3*(A-B)-3*a*b^2*(A-B)
+3*a^2*b*(A+B)-b^3*(A+B))*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)+1/2*(a^3*(A-B)-3*a*b^2*(A-
B)+3*a^2*b*(A+B)-b^3*(A+B))*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)+1/4*(3*a^2*b*(A-B)-b^3*(A
-B)-a^3*(A+B)+3*a*b^2*(A+B))*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)-1/4*(3*a^2*b*(A-B
)-b^3*(A-B)-a^3*(A+B)+3*a*b^2*(A+B))*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)-1/4*(8*A*
a^4+31*A*a^2*b^2+15*A*b^4-11*B*a^3*b-3*B*a*b^3)*cot(d*x+c)^(1/2)/a^3/(a^2+b^2)^2/d
Rubi [A] (verified)
Time = 2.15 (sec) , antiderivative size = 601, normalized size of antiderivative = 1.00, number of
steps used = 18, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3662, 3686, 3726, 3728,
3734, 3615, 1182, 1176, 631, 210, 1179, 642, 3715, 65, 211} \[
\int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\frac {b (A b-a B) \cot ^{\frac {5}{2}}(c+d x)}{2 a d \left (a^2+b^2\right ) (a \cot (c+d x)+b)^2}-\frac {\left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {\left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \arctan \left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {b \left (-9 a^3 B+13 a^2 A b-a b^2 B+5 A b^3\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}+\frac {\left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}-\frac {\left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}-\frac {\left (8 a^4 A-11 a^3 b B+31 a^2 A b^2-3 a b^3 B+15 A b^4\right ) \sqrt {\cot (c+d x)}}{4 a^3 d \left (a^2+b^2\right )^2}+\frac {b^{3/2} \left (-35 a^5 B+63 a^4 A b-6 a^3 b^2 B+46 a^2 A b^3-3 a b^4 B+15 A b^5\right ) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{7/2} d \left (a^2+b^2\right )^3}
\]
[In]
Int[(Cot[c + d*x]^(3/2)*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^3,x]
[Out]
-(((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sq
rt[2]*(a^2 + b^2)^3*d)) + ((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*ArcTan[1 + Sqrt[2]*
Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + (b^(3/2)*(63*a^4*A*b + 46*a^2*A*b^3 + 15*A*b^5 - 35*a^5*B - 6
*a^3*b^2*B - 3*a*b^4*B)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(4*a^(7/2)*(a^2 + b^2)^3*d) - ((8*a^4*A
+ 31*a^2*A*b^2 + 15*A*b^4 - 11*a^3*b*B - 3*a*b^3*B)*Sqrt[Cot[c + d*x]])/(4*a^3*(a^2 + b^2)^2*d) + (b*(A*b - a*
B)*Cot[c + d*x]^(5/2))/(2*a*(a^2 + b^2)*d*(b + a*Cot[c + d*x])^2) + (b*(13*a^2*A*b + 5*A*b^3 - 9*a^3*B - a*b^2
*B)*Cot[c + d*x]^(3/2))/(4*a^2*(a^2 + b^2)^2*d*(b + a*Cot[c + d*x])) + ((3*a^2*b*(A - B) - b^3*(A - B) - a^3*(
A + B) + 3*a*b^2*(A + B))*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) - ((
3*a^2*b*(A - B) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*
x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d)
Rule 65
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]
Rule 210
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^(-1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])
], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])
Rule 211
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 631
Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[a*(c/b^2)]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + 2*c*(x/b)], x] /; RationalQ[q] && (EqQ[q^2, 1] || !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]
Rule 642
Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[d*(Log[RemoveContent[a + b*x +
c*x^2, x]]/b), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]
Rule 1176
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[2*(d/e), 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]
Rule 1179
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[-2*(d/e), 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]
Rule 1182
Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[(-a)*c]
Rule 3615
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]
Rule 3662
Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]
Rule 3686
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*c - a*d)*(B*c - A*d)*(a + b*Tan[e + f*x])^(m - 1)*((c + d*Tan[e
+ f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m -
2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a*A*d*(b*d*(m - 1) - a*c*(n + 1)) + (b*B*c - (A*b + a*B)*d)*(b*c*(m - 1)
+ a*d*(n + 1)) - d*((a*A - b*B)*(b*c - a*d) + (A*b + a*B)*(a*c + b*d))*(n + 1)*Tan[e + f*x] - b*(d*(A*b*c + a
*B*c - a*A*d)*(m + n) - b*B*(c^2*(m - 1) - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f
, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 1] && LtQ[n, -1] && (Inte
gerQ[m] || IntegersQ[2*m, 2*n])
Rule 3715
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_) + (C_.)*
tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[A/f, Subst[Int[(a + b*x)^m*(c + d*x)^n, x], x, Tan[e + f*x]], x]
/; FreeQ[{a, b, c, d, e, f, A, C, m, n}, x] && EqQ[A, C]
Rule 3726
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(A*d^2 + c*(c*C - B*d))*(a + b*Ta
n[e + f*x])^m*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 + d^2))), x] - Dist[1/(d*(n + 1)*(c^2 + d^2)), I
nt[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1)*Simp[A*d*(b*d*m - a*c*(n + 1)) + (c*C - B*d)*(b*c
*m + a*d*(n + 1)) - d*(n + 1)*((A - C)*(b*c - a*d) + B*(a*c + b*d))*Tan[e + f*x] - b*(d*(B*c - A*d)*(m + n + 1
) - C*(c^2*m - d^2*(n + 1)))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c -
a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && LtQ[n, -1]
Rule 3728
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*
tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[C*(a + b*Tan[e + f*x])^m*((c + d
*Tan[e + f*x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c +
d*Tan[e + f*x])^n*Simp[a*A*d*(m + n + 1) - C*(b*c*m + a*d*(n + 1)) + d*(A*b + a*B - b*C)*(m + n + 1)*Tan[e + f
*x] - (C*m*(b*c - a*d) - b*B*d*(m + n + 1))*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}
, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && GtQ[m, 0] && !(IGtQ[n, 0] && ( !Intege
rQ[m] || (EqQ[c, 0] && NeQ[a, 0])))
Rule 3734
Int[(((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (
f_.)*(x_)]^2))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/(a^2 + b^2), Int[(c + d*Tan[e + f*
x])^n*Simp[b*B + a*(A - C) + (a*B - b*(A - C))*Tan[e + f*x], x], x], x] + Dist[(A*b^2 - a*b*B + a^2*C)/(a^2 +
b^2), Int[(c + d*Tan[e + f*x])^n*((1 + Tan[e + f*x]^2)/(a + b*Tan[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e,
f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && !GtQ[n, 0] && !LeQ[n, -
1]
Rubi steps \begin{align*}
\text {integral}& = \int \frac {\cot ^{\frac {7}{2}}(c+d x) (B+A \cot (c+d x))}{(b+a \cot (c+d x))^3} \, dx \\ & = \frac {b (A b-a B) \cot ^{\frac {5}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac {\int \frac {\cot ^{\frac {3}{2}}(c+d x) \left (-\frac {5}{2} b (A b-a B)+2 a (A b-a B) \cot (c+d x)-\frac {1}{2} \left (4 a^2 A+5 A b^2-a b B\right ) \cot ^2(c+d x)\right )}{(b+a \cot (c+d x))^2} \, dx}{2 a \left (a^2+b^2\right )} \\ & = \frac {b (A b-a B) \cot ^{\frac {5}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (13 a^2 A b+5 A b^3-9 a^3 B-a b^2 B\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac {\int \frac {\sqrt {\cot (c+d x)} \left (\frac {3}{4} b \left (13 a^2 A b+5 A b^3-9 a^3 B-a b^2 B\right )-2 a^2 \left (2 a A b-a^2 B+b^2 B\right ) \cot (c+d x)+\frac {1}{4} \left (8 a^4 A+31 a^2 A b^2+15 A b^4-11 a^3 b B-3 a b^3 B\right ) \cot ^2(c+d x)\right )}{b+a \cot (c+d x)} \, dx}{2 a^2 \left (a^2+b^2\right )^2} \\ & = -\frac {\left (8 a^4 A+31 a^2 A b^2+15 A b^4-11 a^3 b B-3 a b^3 B\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b (A b-a B) \cot ^{\frac {5}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (13 a^2 A b+5 A b^3-9 a^3 B-a b^2 B\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}-\frac {\int \frac {\frac {1}{8} b \left (8 a^4 A+31 a^2 A b^2+15 A b^4-11 a^3 b B-3 a b^3 B\right )+a^3 \left (a^2 A-A b^2+2 a b B\right ) \cot (c+d x)+\frac {1}{8} \left (24 a^4 A b+31 a^2 A b^3+15 A b^5-8 a^5 B-3 a^3 b^2 B-3 a b^4 B\right ) \cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{a^3 \left (a^2+b^2\right )^2} \\ & = -\frac {\left (8 a^4 A+31 a^2 A b^2+15 A b^4-11 a^3 b B-3 a b^3 B\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b (A b-a B) \cot ^{\frac {5}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (13 a^2 A b+5 A b^3-9 a^3 B-a b^2 B\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}-\frac {\int \frac {a^3 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right )+a^3 \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{a^3 \left (a^2+b^2\right )^3}-\frac {\left (b^2 \left (63 a^4 A b+46 a^2 A b^3+15 A b^5-35 a^5 B-6 a^3 b^2 B-3 a b^4 B\right )\right ) \int \frac {1+\cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{8 a^3 \left (a^2+b^2\right )^3} \\ & = -\frac {\left (8 a^4 A+31 a^2 A b^2+15 A b^4-11 a^3 b B-3 a b^3 B\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b (A b-a B) \cot ^{\frac {5}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (13 a^2 A b+5 A b^3-9 a^3 B-a b^2 B\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}-\frac {2 \text {Subst}\left (\int \frac {-a^3 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right )-a^3 \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^3 \left (a^2+b^2\right )^3 d}-\frac {\left (b^2 \left (63 a^4 A b+46 a^2 A b^3+15 A b^5-35 a^5 B-6 a^3 b^2 B-3 a b^4 B\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{8 a^3 \left (a^2+b^2\right )^3 d} \\ & = -\frac {\left (8 a^4 A+31 a^2 A b^2+15 A b^4-11 a^3 b B-3 a b^3 B\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b (A b-a B) \cot ^{\frac {5}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (13 a^2 A b+5 A b^3-9 a^3 B-a b^2 B\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac {\left (b^2 \left (63 a^4 A b+46 a^2 A b^3+15 A b^5-35 a^5 B-6 a^3 b^2 B-3 a b^4 B\right )\right ) \text {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 a^3 \left (a^2+b^2\right )^3 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{\left (a^2+b^2\right )^3 d} \\ & = \frac {b^{3/2} \left (63 a^4 A b+46 a^2 A b^3+15 A b^5-35 a^5 B-6 a^3 b^2 B-3 a b^4 B\right ) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{7/2} \left (a^2+b^2\right )^3 d}-\frac {\left (8 a^4 A+31 a^2 A b^2+15 A b^4-11 a^3 b B-3 a b^3 B\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b (A b-a B) \cot ^{\frac {5}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (13 a^2 A b+5 A b^3-9 a^3 B-a b^2 B\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \text {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \left (a^2+b^2\right )^3 d} \\ & = \frac {b^{3/2} \left (63 a^4 A b+46 a^2 A b^3+15 A b^5-35 a^5 B-6 a^3 b^2 B-3 a b^4 B\right ) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{7/2} \left (a^2+b^2\right )^3 d}-\frac {\left (8 a^4 A+31 a^2 A b^2+15 A b^4-11 a^3 b B-3 a b^3 B\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b (A b-a B) \cot ^{\frac {5}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (13 a^2 A b+5 A b^3-9 a^3 B-a b^2 B\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d} \\ & = -\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \arctan \left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \arctan \left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {b^{3/2} \left (63 a^4 A b+46 a^2 A b^3+15 A b^5-35 a^5 B-6 a^3 b^2 B-3 a b^4 B\right ) \arctan \left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{7/2} \left (a^2+b^2\right )^3 d}-\frac {\left (8 a^4 A+31 a^2 A b^2+15 A b^4-11 a^3 b B-3 a b^3 B\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b (A b-a B) \cot ^{\frac {5}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (13 a^2 A b+5 A b^3-9 a^3 B-a b^2 B\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} \left (a^2+b^2\right )^3 d} \\
\end{align*}
Mathematica [A] (verified)
Time = 6.56 (sec) , antiderivative size = 617, normalized size of antiderivative = 1.03
\[
\int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\frac {2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (A+B)\right ) \left (\sqrt {2} \arctan \left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\sqrt {2} \arctan \left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )\right )}{4 \left (a^2+b^2\right )^3}-\frac {3 b^{3/2} (A b-a B) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{8 a^{7/2} \left (a^2+b^2\right )}-\frac {b^{3/2} \left (3 a^2 A b+A b^3-2 a^3 B\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{2 a^{7/2} \left (a^2+b^2\right )^2}-\frac {b^{3/2} \left (6 a^4 A b+3 a^2 A b^3+A b^5-3 a^5 B+a^3 b^2 B\right ) \arctan \left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{a^{7/2} \left (a^2+b^2\right )^3}+\frac {\left (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \left (\sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )\right )}{8 \left (a^2+b^2\right )^3}-\frac {A}{a^3 \sqrt {\tan (c+d x)}}-\frac {b^2 (A b-a B) \sqrt {\tan (c+d x)}}{4 a^2 \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}-\frac {3 b^2 (A b-a B) \sqrt {\tan (c+d x)}}{8 a^3 \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac {b^2 \left (3 a^2 A b+A b^3-2 a^3 B\right ) \sqrt {\tan (c+d x)}}{2 a^3 \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}\right )}{d}
\]
[In]
Integrate[(Cot[c + d*x]^(3/2)*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^3,x]
[Out]
(2*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(A + B))*(Sq
rt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]))/(4*(a^2 + b^2)
^3) - (3*b^(3/2)*(A*b - a*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(8*a^(7/2)*(a^2 + b^2)) - (b^(3/2)*
(3*a^2*A*b + A*b^3 - 2*a^3*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(2*a^(7/2)*(a^2 + b^2)^2) - (b^(3/
2)*(6*a^4*A*b + 3*a^2*A*b^3 + A*b^5 - 3*a^5*B + a^3*b^2*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(a^(7
/2)*(a^2 + b^2)^3) + ((3*a^2*b*(A - B) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*(Sqrt[2]*Log[1 - Sqrt[2]
*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]]))/(8*(a^2 + b
^2)^3) - A/(a^3*Sqrt[Tan[c + d*x]]) - (b^2*(A*b - a*B)*Sqrt[Tan[c + d*x]])/(4*a^2*(a^2 + b^2)*(a + b*Tan[c + d
*x])^2) - (3*b^2*(A*b - a*B)*Sqrt[Tan[c + d*x]])/(8*a^3*(a^2 + b^2)*(a + b*Tan[c + d*x])) - (b^2*(3*a^2*A*b +
A*b^3 - 2*a^3*B)*Sqrt[Tan[c + d*x]])/(2*a^3*(a^2 + b^2)^2*(a + b*Tan[c + d*x]))))/d
Maple [B] (verified)
Leaf count of result is larger than twice the leaf count of optimal. \(4580\) vs. \(2(549)=1098\).
Time = 0.42 (sec) , antiderivative size = 4581, normalized size of antiderivative =
7.62
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method | result | size |
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derivativedivides |
\(\text {Expression too large to display}\) |
\(4581\) |
default |
\(\text {Expression too large to display}\) |
\(4581\) |
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[In]
int(cot(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
[Out]
-1/4/d*(1/tan(d*x+c))^(3/2)*tan(d*x+c)*(15*A*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(5/2)*b^9+15*A*
(a*b)^(1/2)*tan(d*x+c)^2*b^8+24*A*(a*b)^(1/2)*a^6*b^2+24*A*(a*b)^(1/2)*a^4*b^4+8*A*(a*b)^(1/2)*a^2*b^6+8*A*(a*
b)^(1/2)*a^8+39*A*(a*b)^(1/2)*tan(d*x+c)^2*a^4*b^4+46*A*(a*b)^(1/2)*tan(d*x+c)^2*a^2*b^6-11*B*(a*b)^(1/2)*tan(
d*x+c)^2*a^5*b^3-14*B*(a*b)^(1/2)*tan(d*x+c)^2*a^3*b^5-3*B*(a*b)^(1/2)*tan(d*x+c)^2*a*b^7+65*A*(a*b)^(1/2)*tan
(d*x+c)*a^5*b^3+74*A*(a*b)^(1/2)*tan(d*x+c)*a^3*b^5+25*A*(a*b)^(1/2)*tan(d*x+c)*a*b^7+8*A*(a*b)^(1/2)*a^6*b^2*
tan(d*x+c)^2+63*A*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(1/2)*a^6*b^3+46*A*arctan(b*tan(d*x+c)^(1/
2)/(a*b)^(1/2))*tan(d*x+c)^(1/2)*a^4*b^5+15*A*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(1/2)*a^2*b^7-
13*B*(a*b)^(1/2)*tan(d*x+c)*a^6*b^2-18*B*(a*b)^(1/2)*tan(d*x+c)*a^4*b^4-5*B*(a*b)^(1/2)*tan(d*x+c)*a^2*b^6-35*
B*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(1/2)*a^7*b^2-6*B*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*t
an(d*x+c)^(1/2)*a^5*b^4-3*B*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(1/2)*a^3*b^6+16*A*(a*b)^(1/2)*a
^7*b*tan(d*x+c)-70*B*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(3/2)*a^6*b^3-12*B*arctan(b*tan(d*x+c)^
(1/2)/(a*b)^(1/2))*tan(d*x+c)^(3/2)*a^4*b^5-6*B*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(3/2)*a^2*b^
7+63*A*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(5/2)*a^4*b^5+46*A*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1
/2))*tan(d*x+c)^(5/2)*a^2*b^7-35*B*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(5/2)*a^5*b^4-6*B*arctan(
b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(5/2)*a^3*b^6-3*B*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)
^(5/2)*a*b^8+126*A*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(3/2)*a^5*b^4+92*A*arctan(b*tan(d*x+c)^(1
/2)/(a*b)^(1/2))*tan(d*x+c)^(3/2)*a^3*b^6+30*A*arctan(b*tan(d*x+c)^(1/2)/(a*b)^(1/2))*tan(d*x+c)^(3/2)*a*b^8-2
*A*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^5*b^3-3*A*2^(1/2)*(a*b)^(1/2)*ln
(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*tan(d*x+c)^(1/2)*a^6*b^2+3*
B*2^(1/2)*(a*b)^(1/2)*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*tan
(d*x+c)^(1/2)*a^6*b^2+6*B*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^7*b+6*B*2^
(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^6*b^2-2*B*2^(1/2)*(a*b)^(1/2)*arctan(1
+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^5*b^3+6*B*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2)
)*tan(d*x+c)^(1/2)*a^7*b-2*B*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^8+2*A*
2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^8+2*A*2^(1/2)*(a*b)^(1/2)*arctan(-1+
2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^8+A*2^(1/2)*(a*b)^(1/2)*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-
1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*tan(d*x+c)^(1/2)*a^8-B*2^(1/2)*(a*b)^(1/2)*ln(-(1+2^(1/2)*tan(d*x+
c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*tan(d*x+c)^(1/2)*a^8-2*B*2^(1/2)*(a*b)^(1/2)*arc
tan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^8+6*A*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2
))*tan(d*x+c)^(5/2)*a^5*b^3-6*A*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(5/2)*a^4*b
^4-2*A*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(5/2)*a^3*b^5+A*2^(1/2)*(a*b)^(1/2)*
ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*tan(d*x+c)^(5/2)*a^6*b^2-
3*A*2^(1/2)*(a*b)^(1/2)*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*t
an(d*x+c)^(5/2)*a^4*b^4-B*2^(1/2)*(a*b)^(1/2)*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^
(1/2)-tan(d*x+c)-1))*tan(d*x+c)^(5/2)*a^6*b^2+3*B*2^(1/2)*(a*b)^(1/2)*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+
c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*tan(d*x+c)^(5/2)*a^4*b^4-2*B*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)
*tan(d*x+c)^(1/2))*tan(d*x+c)^(5/2)*a^6*b^2+6*B*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x
+c)^(5/2)*a^5*b^3+6*B*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(5/2)*a^4*b^4-2*B*2^(1
/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(5/2)*a^3*b^5-2*B*2^(1/2)*(a*b)^(1/2)*arctan(-1+
2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(5/2)*a^6*b^2+6*B*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))
*tan(d*x+c)^(5/2)*a^5*b^3+6*B*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(5/2)*a^4*b^4
+2*A*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(5/2)*a^6*b^2+6*A*2^(1/2)*(a*b)^(1/2)*a
rctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(5/2)*a^5*b^3-6*A*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)
^(1/2))*tan(d*x+c)^(5/2)*a^4*b^4-2*A*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(5/2)*a
^3*b^5+3*A*2^(1/2)*(a*b)^(1/2)*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c
)-1))*tan(d*x+c)^(5/2)*a^5*b^3-A*2^(1/2)*(a*b)^(1/2)*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(
d*x+c)^(1/2)-tan(d*x+c)-1))*tan(d*x+c)^(5/2)*a^3*b^5+3*A*2^(1/2)*(a*b)^(1/2)*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+t
an(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*tan(d*x+c)^(1/2)*a^7*b-A*2^(1/2)*(a*b)^(1/2)*ln(-(1+2^(1/2
)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*tan(d*x+c)^(1/2)*a^5*b^3+6*A*2^(1/2)*(
a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^7*b-6*A*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*
tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^6*b^2-2*A*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+
c)^(1/2)*a^5*b^3+6*A*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^7*b-6*A*2^(1/2
)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^6*b^2-4*A*2^(1/2)*(a*b)^(1/2)*arctan(1+2^
(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^4*b^4+4*A*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*t
an(d*x+c)^(3/2)*a^7*b+12*A*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^6*b^2-12
*A*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^5*b^3-4*A*2^(1/2)*(a*b)^(1/2)*ar
ctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^4*b^4+2*A*2^(1/2)*(a*b)^(1/2)*ln(-(2^(1/2)*tan(d*x+c)^(1/
2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*tan(d*x+c)^(3/2)*a^7*b-6*A*2^(1/2)*(a*b)^(1/2)*ln(-(
2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*tan(d*x+c)^(3/2)*a^5*b^3-2*B*2
^(1/2)*(a*b)^(1/2)*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*tan(d*
x+c)^(3/2)*a^7*b+6*B*2^(1/2)*(a*b)^(1/2)*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)
-tan(d*x+c)-1))*tan(d*x+c)^(3/2)*a^5*b^3+3*B*2^(1/2)*(a*b)^(1/2)*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(
1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*tan(d*x+c)^(1/2)*a^7*b-B*2^(1/2)*(a*b)^(1/2)*ln(-(2^(1/2)*tan(d*x+c)^(
1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*tan(d*x+c)^(1/2)*a^5*b^3-2*B*2^(1/2)*(a*b)^(1/2)*l
n(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*tan(d*x+c)^(3/2)*a^4*b^4+1
2*B*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^5*b^3-4*B*2^(1/2)*(a*b)^(1/2)*ar
ctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^4*b^4-4*B*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)
^(1/2))*tan(d*x+c)^(3/2)*a^7*b+12*B*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a
^6*b^2+12*B*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^5*b^3-4*B*2^(1/2)*(a*b)
^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^4*b^4+6*B*2^(1/2)*(a*b)^(1/2)*ln(-(2^(1/2)*tan(d
*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*tan(d*x+c)^(3/2)*a^6*b^2-4*B*2^(1/2)*(a*b)^
(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^7*b+12*B*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(
d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^6*b^2-2*B*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^
(5/2)*a^3*b^5+3*B*2^(1/2)*(a*b)^(1/2)*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+
tan(d*x+c)))*tan(d*x+c)^(5/2)*a^5*b^3-B*2^(1/2)*(a*b)^(1/2)*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(
1/2)*tan(d*x+c)^(1/2)+tan(d*x+c)))*tan(d*x+c)^(5/2)*a^3*b^5+6*A*2^(1/2)*(a*b)^(1/2)*ln(-(1+2^(1/2)*tan(d*x+c)^
(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*tan(d*x+c)^(3/2)*a^6*b^2-2*A*2^(1/2)*(a*b)^(1/2)*ln
(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))*tan(d*x+c)^(3/2)*a^4*b^4+4*
A*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^7*b+12*A*2^(1/2)*(a*b)^(1/2)*arcta
n(1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(3/2)*a^6*b^2-12*A*2^(1/2)*(a*b)^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1
/2))*tan(d*x+c)^(3/2)*a^5*b^3+2*A*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(5/2)*a^6
*b^2+6*B*2^(1/2)*(a*b)^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^6*b^2-2*B*2^(1/2)*(a*b)^(1
/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))*tan(d*x+c)^(1/2)*a^5*b^3)/(a^2+b^2)^3/a^3/(a+b*tan(d*x+c))^2/(a*b)^(1/
2)
Fricas [B] (verification not implemented)
Leaf count of result is larger than twice the leaf count of optimal. 8979 vs. \(2 (544) = 1088\).
Time = 187.83 (sec) , antiderivative size = 17988, normalized size of antiderivative = 29.93
\[
\int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\text {Too large to display}
\]
[In]
integrate(cot(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="fricas")
[Out]
Too large to include
Sympy [F]
\[
\int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \cot ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx
\]
[In]
integrate(cot(d*x+c)**(3/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**3,x)
[Out]
Integral((A + B*tan(c + d*x))*cot(c + d*x)**(3/2)/(a + b*tan(c + d*x))**3, x)
Maxima [A] (verification not implemented)
none
Time = 0.32 (sec) , antiderivative size = 577, normalized size of antiderivative = 0.96
\[
\int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=-\frac {\frac {{\left (35 \, B a^{5} b^{2} - 63 \, A a^{4} b^{3} + 6 \, B a^{3} b^{4} - 46 \, A a^{2} b^{5} + 3 \, B a b^{6} - 15 \, A b^{7}\right )} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{9} + 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + a^{3} b^{6}\right )} \sqrt {a b}} - \frac {2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} + 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} - {\left (A + B\right )} b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left ({\left (A - B\right )} a^{3} + 3 \, {\left (A + B\right )} a^{2} b - 3 \, {\left (A - B\right )} a b^{2} - {\left (A + B\right )} b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left ({\left (A + B\right )} a^{3} - 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} + {\left (A - B\right )} b^{3}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left ({\left (A + B\right )} a^{3} - 3 \, {\left (A - B\right )} a^{2} b - 3 \, {\left (A + B\right )} a b^{2} + {\left (A - B\right )} b^{3}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {\frac {11 \, B a^{3} b^{3} - 15 \, A a^{2} b^{4} + 3 \, B a b^{5} - 7 \, A b^{6}}{\sqrt {\tan \left (d x + c\right )}} + \frac {13 \, B a^{4} b^{2} - 17 \, A a^{3} b^{3} + 5 \, B a^{2} b^{4} - 9 \, A a b^{5}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{a^{7} b^{2} + 2 \, a^{5} b^{4} + a^{3} b^{6} + \frac {2 \, {\left (a^{8} b + 2 \, a^{6} b^{3} + a^{4} b^{5}\right )}}{\tan \left (d x + c\right )} + \frac {a^{9} + 2 \, a^{7} b^{2} + a^{5} b^{4}}{\tan \left (d x + c\right )^{2}}} + \frac {8 \, A}{a^{3} \sqrt {\tan \left (d x + c\right )}}}{4 \, d}
\]
[In]
integrate(cot(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="maxima")
[Out]
-1/4*((35*B*a^5*b^2 - 63*A*a^4*b^3 + 6*B*a^3*b^4 - 46*A*a^2*b^5 + 3*B*a*b^6 - 15*A*b^7)*arctan(a/(sqrt(a*b)*sq
rt(tan(d*x + c))))/((a^9 + 3*a^7*b^2 + 3*a^5*b^4 + a^3*b^6)*sqrt(a*b)) - (2*sqrt(2)*((A - B)*a^3 + 3*(A + B)*a
^2*b - 3*(A - B)*a*b^2 - (A + B)*b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*((A - B
)*a^3 + 3*(A + B)*a^2*b - 3*(A - B)*a*b^2 - (A + B)*b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c))))
+ sqrt(2)*((A + B)*a^3 - 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 + (A - B)*b^3)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/
tan(d*x + c) + 1) - sqrt(2)*((A + B)*a^3 - 3*(A - B)*a^2*b - 3*(A + B)*a*b^2 + (A - B)*b^3)*log(-sqrt(2)/sqrt(
tan(d*x + c)) + 1/tan(d*x + c) + 1))/(a^6 + 3*a^4*b^2 + 3*a^2*b^4 + b^6) - ((11*B*a^3*b^3 - 15*A*a^2*b^4 + 3*B
*a*b^5 - 7*A*b^6)/sqrt(tan(d*x + c)) + (13*B*a^4*b^2 - 17*A*a^3*b^3 + 5*B*a^2*b^4 - 9*A*a*b^5)/tan(d*x + c)^(3
/2))/(a^7*b^2 + 2*a^5*b^4 + a^3*b^6 + 2*(a^8*b + 2*a^6*b^3 + a^4*b^5)/tan(d*x + c) + (a^9 + 2*a^7*b^2 + a^5*b^
4)/tan(d*x + c)^2) + 8*A/(a^3*sqrt(tan(d*x + c))))/d
Giac [F]
\[
\int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\int { \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{3}} \,d x }
\]
[In]
integrate(cot(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="giac")
[Out]
integrate((B*tan(d*x + c) + A)*cot(d*x + c)^(3/2)/(b*tan(d*x + c) + a)^3, x)
Mupad [F(-1)]
Timed out. \[
\int \frac {\cot ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx=\text {Hanged}
\]
[In]
int((cot(c + d*x)^(3/2)*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^3,x)
[Out]
\text{Hanged}