3.601 \(\int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=534 \[ \frac {b (A b-a B) \cot ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2+b^2\right ) (a \cot (c+d x)+b)^2}+\frac {b \left (-7 a^3 B+11 a^2 A b+a b^2 B+3 A b^3\right ) \sqrt {\cot (c+d x)}}{4 a^2 d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}-\frac {\left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {\left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}-\frac {\left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {\left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {\sqrt {b} \left (-15 a^5 B+35 a^4 A b+18 a^3 b^2 B+6 a^2 A b^3+a b^4 B+3 A b^5\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{5/2} d \left (a^2+b^2\right )^3} \]

[Out]

1/2*b*(A*b-B*a)*cot(d*x+c)^(3/2)/a/(a^2+b^2)/d/(b+a*cot(d*x+c))^2+1/2*(3*a^2*b*(A-B)-b^3*(A-B)-a^3*(A+B)+3*a*b
^2*(A+B))*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)+1/2*(3*a^2*b*(A-B)-b^3*(A-B)-a^3*(A+B)+3*a
*b^2*(A+B))*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)-1/4*(a^3*(A-B)-3*a*b^2*(A-B)+3*a^2*b*(A+B
)-b^3*(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*(a^3*(A-B)-3*a*b^2*(A-B)+3*a^
2*b*(A+B)-b^3*(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*(35*A*a^4*b+6*A*a^2*b
^3+3*A*b^5-15*B*a^5+18*B*a^3*b^2+B*a*b^4)*arctan(a^(1/2)*cot(d*x+c)^(1/2)/b^(1/2))*b^(1/2)/a^(5/2)/(a^2+b^2)^3
/d+1/4*b*(11*A*a^2*b+3*A*b^3-7*B*a^3+B*a*b^2)*cot(d*x+c)^(1/2)/a^2/(a^2+b^2)^2/d/(b+a*cot(d*x+c))

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Rubi [A]  time = 1.37, antiderivative size = 534, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3581, 3605, 3645, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac {b (A b-a B) \cot ^{\frac {3}{2}}(c+d x)}{2 a d \left (a^2+b^2\right ) (a \cot (c+d x)+b)^2}+\frac {b \left (11 a^2 A b-7 a^3 B+a b^2 B+3 A b^3\right ) \sqrt {\cot (c+d x)}}{4 a^2 d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}-\frac {\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {\left (3 a^2 b (A+B)+a^3 (A-B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}-\frac {\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {\left (3 a^2 b (A-B)+a^3 (-(A+B))+3 a b^2 (A+B)-b^3 (A-B)\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}-\frac {\sqrt {b} \left (6 a^2 A b^3+35 a^4 A b+18 a^3 b^2 B-15 a^5 B+a b^4 B+3 A b^5\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{5/2} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[(Sqrt[Cot[c + d*x]]*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^3,x]

[Out]

-(((3*a^2*b*(A - B) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(A + B))*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sq
rt[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))*ArcTan[1 + Sqrt[2]*
Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) - (Sqrt[b]*(35*a^4*A*b + 6*a^2*A*b^3 + 3*A*b^5 - 15*a^5*B + 18*
a^3*b^2*B + a*b^4*B)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(4*a^(5/2)*(a^2 + b^2)^3*d) + (b*(A*b - a*B
)*Cot[c + d*x]^(3/2))/(2*a*(a^2 + b^2)*d*(b + a*Cot[c + d*x])^2) + (b*(11*a^2*A*b + 3*A*b^3 - 7*a^3*B + a*b^2*
B)*Sqrt[Cot[c + d*x]])/(4*a^2*(a^2 + b^2)^2*d*(b + a*Cot[c + d*x])) - ((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*
b*(A + B) - b^3*(A + B))*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[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))*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x
]])/(2*Sqrt[2]*(a^2 + b^2)^3*d)

Rule 63

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 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 617

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 628

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 1162

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 1165

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 1168

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 3534

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 3581

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 3605

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 3634

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 3645

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*T
an[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 3653

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*} \int \frac {\sqrt {\cot (c+d x)} (A+B \tan (c+d x))}{(a+b \tan (c+d x))^3} \, dx &=\int \frac {\cot ^{\frac {5}{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 {3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}-\frac {\int \frac {\sqrt {\cot (c+d x)} \left (-\frac {3}{2} b (A b-a B)+2 a (A b-a B) \cot (c+d x)-\frac {1}{2} \left (4 a^2 A+3 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 {3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cot (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}{4} b \left (11 a^2 A b+3 A b^3-7 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+3 a^2 A b^2+3 A b^4+9 a^3 b B+a b^3 B\right ) \cot ^2(c+d x)}{\sqrt {\cot (c+d x)} (b+a \cot (c+d x))} \, dx}{2 a^2 \left (a^2+b^2\right )^2}\\ &=\frac {b (A b-a B) \cot ^{\frac {3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac {\int \frac {-2 a^2 \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right )+2 a^2 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{2 a^2 \left (a^2+b^2\right )^3}+\frac {\left (b \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+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^2 \left (a^2+b^2\right )^3}\\ &=\frac {b (A b-a B) \cot ^{\frac {3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}+\frac {\operatorname {Subst}\left (\int \frac {2 a^2 \left (3 a^2 A b-A b^3-a^3 B+3 a b^2 B\right )-2 a^2 \left (a^3 A-3 a A b^2+3 a^2 b B-b^3 B\right ) x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{a^2 \left (a^2+b^2\right )^3 d}+\frac {\left (b \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {-x} (b-a x)} \, dx,x,-\cot (c+d x)\right )}{8 a^2 \left (a^2+b^2\right )^3 d}\\ &=\frac {b (A b-a B) \cot ^{\frac {3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}-\frac {\left (b \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right )\right ) \operatorname {Subst}\left (\int \frac {1}{b+a x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{4 a^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 ) \operatorname {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 ) \operatorname {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 {\sqrt {b} \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{5/2} \left (a^2+b^2\right )^3 d}+\frac {b (A b-a B) \cot ^{\frac {3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cot (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 ) \operatorname {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 (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname {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 ) \operatorname {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 ) \operatorname {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 {\sqrt {b} \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{5/2} \left (a^2+b^2\right )^3 d}+\frac {b (A b-a B) \cot ^{\frac {3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (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 ) \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 ) \operatorname {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 (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \operatorname {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 (3 a^2 b (A-B)-b^3 (A-B)-a^3 (A+B)+3 a b^2 (A+B)\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\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 ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}-\frac {\sqrt {b} \left (35 a^4 A b+6 a^2 A b^3+3 A b^5-15 a^5 B+18 a^3 b^2 B+a b^4 B\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{5/2} \left (a^2+b^2\right )^3 d}+\frac {b (A b-a B) \cot ^{\frac {3}{2}}(c+d x)}{2 a \left (a^2+b^2\right ) d (b+a \cot (c+d x))^2}+\frac {b \left (11 a^2 A b+3 A b^3-7 a^3 B+a b^2 B\right ) \sqrt {\cot (c+d x)}}{4 a^2 \left (a^2+b^2\right )^2 d (b+a \cot (c+d x))}-\frac {\left (a^3 (A-B)-3 a b^2 (A-B)+3 a^2 b (A+B)-b^3 (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 ) \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*}

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Mathematica [A]  time = 6.39, size = 566, normalized size = 1.06 \[ \frac {2 \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {b (A b-a B) \sqrt {\tan (c+d x)}}{4 a \left (a^2+b^2\right ) (a+b \tan (c+d x))^2}+\frac {b \left (a^2 (-B)+2 a A b+b^2 B\right ) \sqrt {\tan (c+d x)}}{2 a \left (a^2+b^2\right )^2 (a+b \tan (c+d x))}+\frac {3 b (A b-a B) \left (\frac {\tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{a^{3/2} \sqrt {b}}+\frac {\sqrt {\tan (c+d x)}}{a (a+b \tan (c+d x))}\right )}{8 a \left (a^2+b^2\right )}+\frac {\sqrt {b} \left (a^2 (-B)+2 a A b+b^2 B\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{2 a^{3/2} \left (a^2+b^2\right )^2}+\frac {\left (-\left (a^3 (A+B)\right )+3 a^2 b (A-B)+3 a b^2 (A+B)-b^3 (A-B)\right ) \left (\sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )}{4 \left (a^2+b^2\right )^3}+\frac {\sqrt {b} \left (a^3 (-B)+3 a^2 A b+3 a b^2 B-A b^3\right ) \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {\tan (c+d x)}}{\sqrt {a}}\right )}{\sqrt {a} \left (a^2+b^2\right )^3}-\frac {\left (a^3 (A-B)+3 a^2 b (A+B)-3 a b^2 (A-B)-b^3 (A+B)\right ) \left (\sqrt {2} \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )-\sqrt {2} \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )}{8 \left (a^2+b^2\right )^3}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(Sqrt[Cot[c + d*x]]*(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]]*(((3*a^2*b*(A - B) - b^3*(A - B) - a^3*(A + B) + 3*a*b^2*(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) + (Sqrt[b]*(2*a*A*b - a^2*B + b^2*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(2*a^(3/2)*(a^2 + b^2)^
2) + (Sqrt[b]*(3*a^2*A*b - A*b^3 - a^3*B + 3*a*b^2*B)*ArcTan[(Sqrt[b]*Sqrt[Tan[c + d*x]])/Sqrt[a]])/(Sqrt[a]*(
a^2 + b^2)^3) - ((a^3*(A - B) - 3*a*b^2*(A - B) + 3*a^2*b*(A + B) - b^3*(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
) + (b*(A*b - a*B)*Sqrt[Tan[c + d*x]])/(4*a*(a^2 + b^2)*(a + b*Tan[c + d*x])^2) + (b*(2*a*A*b - a^2*B + b^2*B)
*Sqrt[Tan[c + d*x]])/(2*a*(a^2 + b^2)^2*(a + b*Tan[c + d*x])) + (3*b*(A*b - a*B)*(ArcTan[(Sqrt[b]*Sqrt[Tan[c +
d*x]])/Sqrt[a]]/(a^(3/2)*Sqrt[b]) + Sqrt[Tan[c + d*x]]/(a*(a + b*Tan[c + d*x]))))/(8*a*(a^2 + b^2))))/d

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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {\cot \left (d x + c\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(cot(d*x + c))/(b*tan(d*x + c) + a)^3, x)

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maple [C]  time = 8.88, size = 100811, normalized size = 188.78 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x)

[Out]

result too large to display

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maxima [A]  time = 0.61, size = 558, normalized size = 1.04 \[ \frac {\frac {{\left (15 \, B a^{5} b - 35 \, A a^{4} b^{2} - 18 \, B a^{3} b^{3} - 6 \, A a^{2} b^{4} - B a b^{5} - 3 \, A b^{6}\right )} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{8} + 3 \, a^{6} b^{2} + 3 \, a^{4} b^{4} + a^{2} 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 {7 \, B a^{3} b^{2} - 11 \, A a^{2} b^{3} - B a b^{4} - 3 \, A b^{5}}{\sqrt {\tan \left (d x + c\right )}} + \frac {9 \, B a^{4} b - 13 \, A a^{3} b^{2} + B a^{2} b^{3} - 5 \, A a b^{4}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{a^{6} b^{2} + 2 \, a^{4} b^{4} + a^{2} b^{6} + \frac {2 \, {\left (a^{7} b + 2 \, a^{5} b^{3} + a^{3} b^{5}\right )}}{\tan \left (d x + c\right )} + \frac {a^{8} + 2 \, a^{6} b^{2} + a^{4} b^{4}}{\tan \left (d x + c\right )^{2}}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/4*((15*B*a^5*b - 35*A*a^4*b^2 - 18*B*a^3*b^3 - 6*A*a^2*b^4 - B*a*b^5 - 3*A*b^6)*arctan(a/(sqrt(a*b)*sqrt(tan
(d*x + c))))/((a^8 + 3*a^6*b^2 + 3*a^4*b^4 + a^2*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)))) - sqr
t(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) - ((7*B*a^3*b^2 - 11*A*a^2*b^3 - B*a*b^4 -
3*A*b^5)/sqrt(tan(d*x + c)) + (9*B*a^4*b - 13*A*a^3*b^2 + B*a^2*b^3 - 5*A*a*b^4)/tan(d*x + c)^(3/2))/(a^6*b^2
+ 2*a^4*b^4 + a^2*b^6 + 2*(a^7*b + 2*a^5*b^3 + a^3*b^5)/tan(d*x + c) + (a^8 + 2*a^6*b^2 + a^4*b^4)/tan(d*x + c
)^2))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cot(c + d*x)^(1/2)*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^3,x)

[Out]

int((cot(c + d*x)^(1/2)*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x))^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (A + B \tan {\left (c + d x \right )}\right ) \sqrt {\cot {\left (c + d x \right )}}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**(1/2)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c))**3,x)

[Out]

Integral((A + B*tan(c + d*x))*sqrt(cot(c + d*x))/(a + b*tan(c + d*x))**3, x)

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