3.834 \(\int \frac {\cot ^{\frac {3}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx\)

Optimal. Leaf size=444 \[ \frac {b^2 \left (13 a^2+5 b^2\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}+\frac {b^2 \cot ^{\frac {5}{2}}(c+d x)}{2 a d \left (a^2+b^2\right ) (a \cot (c+d x)+b)^2}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {b^{5/2} \left (63 a^4+46 a^2 b^2+15 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{7/2} d \left (a^2+b^2\right )^3}-\frac {\left (8 a^4+31 a^2 b^2+15 b^4\right ) \sqrt {\cot (c+d x)}}{4 a^3 d \left (a^2+b^2\right )^2} \]

[Out]

1/4*b^(5/2)*(63*a^4+46*a^2*b^2+15*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^2*
cot(d*x+c)^(5/2)/a/(a^2+b^2)/d/(b+a*cot(d*x+c))^2+1/4*b^2*(13*a^2+5*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-b)*(a^2+4*a*b+b^2)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)+1/2*(a-b)*(
a^2+4*a*b+b^2)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/(a^2+b^2)^3/d*2^(1/2)-1/4*(a+b)*(a^2-4*a*b+b^2)*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+b)*(a^2-4*a*b+b^2)*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^4+31*a^2*b^2+15*b^4)*cot(d*x+c)^(1/2)/a^3/(a^2+b^2)^2/d

________________________________________________________________________________________

Rubi [A]  time = 1.21, antiderivative size = 444, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 15, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.652, Rules used = {3673, 3565, 3645, 3647, 3653, 3534, 1168, 1162, 617, 204, 1165, 628, 3634, 63, 205} \[ \frac {b^2 \left (13 a^2+5 b^2\right ) \cot ^{\frac {3}{2}}(c+d x)}{4 a^2 d \left (a^2+b^2\right )^2 (a \cot (c+d x)+b)}+\frac {b^2 \cot ^{\frac {5}{2}}(c+d x)}{2 a d \left (a^2+b^2\right ) (a \cot (c+d x)+b)^2}-\frac {\left (31 a^2 b^2+8 a^4+15 b^4\right ) \sqrt {\cot (c+d x)}}{4 a^3 d \left (a^2+b^2\right )^2}-\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {(a+b) \left (a^2-4 a b+b^2\right ) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d \left (a^2+b^2\right )^3}+\frac {b^{5/2} \left (46 a^2 b^2+63 a^4+15 b^4\right ) \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{4 a^{7/2} d \left (a^2+b^2\right )^3}-\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d \left (a^2+b^2\right )^3}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d \left (a^2+b^2\right )^3} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^(3/2)/(a + b*Tan[c + d*x])^3,x]

[Out]

-(((a - b)*(a^2 + 4*a*b + b^2)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d)) + ((a - b)*(
a^2 + 4*a*b + b^2)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*(a^2 + b^2)^3*d) + (b^(5/2)*(63*a^4 + 46*a
^2*b^2 + 15*b^4)*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/(4*a^(7/2)*(a^2 + b^2)^3*d) - ((8*a^4 + 31*a^2*
b^2 + 15*b^4)*Sqrt[Cot[c + d*x]])/(4*a^3*(a^2 + b^2)^2*d) + (b^2*Cot[c + d*x]^(5/2))/(2*a*(a^2 + b^2)*d*(b + a
*Cot[c + d*x])^2) + (b^2*(13*a^2 + 5*b^2)*Cot[c + d*x]^(3/2))/(4*a^2*(a^2 + b^2)^2*d*(b + a*Cot[c + d*x])) - (
(a + b)*(a^2 - 4*a*b + b^2)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*(a^2 + b^2)^3*d) +
((a + b)*(a^2 - 4*a*b + b^2)*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 3565

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)^2*(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 + d^2)), x] - D
ist[1/(d*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^(m - 3)*(c + d*Tan[e + f*x])^(n + 1)*Simp[a^2*d*(b*d*(
m - 2) - a*c*(n + 1)) + b*(b*c - 2*a*d)*(b*c*(m - 2) + a*d*(n + 1)) - d*(n + 1)*(3*a^2*b*c - b^3*c - a^3*d + 3
*a*b^2*d)*Tan[e + f*x] - b*(a*d*(2*b*c - a*d)*(m + n - 1) - b^2*(c^2*(m - 2) - d^2*(n + 1)))*Tan[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && Gt
Q[m, 2] && LtQ[n, -1] && IntegerQ[2*m]

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 3647

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 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]

Rule 3673

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int \frac {\cot ^{\frac {3}{2}}(c+d x)}{(a+b \tan (c+d x))^3} \, dx &=\int \frac {\cot ^{\frac {9}{2}}(c+d x)}{(b+a \cot (c+d x))^3} \, dx\\ &=\frac {b^2 \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 b^2}{2}+2 a b \cot (c+d x)-\frac {1}{2} \left (4 a^2+5 b^2\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^2 \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^2 \left (13 a^2+5 b^2\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^2 \left (13 a^2+5 b^2\right )-4 a^3 b \cot (c+d x)+\frac {1}{4} \left (8 a^4+31 a^2 b^2+15 b^4\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+31 a^2 b^2+15 b^4\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b^2 \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^2 \left (13 a^2+5 b^2\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+31 a^2 b^2+15 b^4\right )+a^3 \left (a^2-b^2\right ) \cot (c+d x)+\frac {1}{8} b \left (24 a^4+31 a^2 b^2+15 b^4\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+31 a^2 b^2+15 b^4\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b^2 \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^2 \left (13 a^2+5 b^2\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^4 \left (a^2-3 b^2\right )+a^3 b \left (3 a^2-b^2\right ) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx}{a^3 \left (a^2+b^2\right )^3}-\frac {\left (b^3 \left (63 a^4+46 a^2 b^2+15 b^4\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+31 a^2 b^2+15 b^4\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b^2 \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^2 \left (13 a^2+5 b^2\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 \operatorname {Subst}\left (\int \frac {-a^4 \left (a^2-3 b^2\right )-a^3 b \left (3 a^2-b^2\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^3 \left (63 a^4+46 a^2 b^2+15 b^4\right )\right ) \operatorname {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+31 a^2 b^2+15 b^4\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b^2 \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^2 \left (13 a^2+5 b^2\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 ((a+b) \left (a^2-4 a b+b^2\right )\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-b) \left (a^2+4 a b+b^2\right )\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 (b^3 \left (63 a^4+46 a^2 b^2+15 b^4\right )\right ) \operatorname {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 {b^{5/2} \left (63 a^4+46 a^2 b^2+15 b^4\right ) \tan ^{-1}\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+31 a^2 b^2+15 b^4\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b^2 \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^2 \left (13 a^2+5 b^2\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 ((a+b) \left (a^2-4 a b+b^2\right )\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+b) \left (a^2-4 a b+b^2\right )\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-b) \left (a^2+4 a b+b^2\right )\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-b) \left (a^2+4 a b+b^2\right )\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 {b^{5/2} \left (63 a^4+46 a^2 b^2+15 b^4\right ) \tan ^{-1}\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+31 a^2 b^2+15 b^4\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b^2 \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^2 \left (13 a^2+5 b^2\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 {(a+b) \left (a^2-4 a b+b^2\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 {(a+b) \left (a^2-4 a b+b^2\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-b) \left (a^2+4 a b+b^2\right )\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 ((a-b) \left (a^2+4 a b+b^2\right )\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 {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {(a-b) \left (a^2+4 a b+b^2\right ) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} \left (a^2+b^2\right )^3 d}+\frac {b^{5/2} \left (63 a^4+46 a^2 b^2+15 b^4\right ) \tan ^{-1}\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+31 a^2 b^2+15 b^4\right ) \sqrt {\cot (c+d x)}}{4 a^3 \left (a^2+b^2\right )^2 d}+\frac {b^2 \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^2 \left (13 a^2+5 b^2\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 {(a+b) \left (a^2-4 a b+b^2\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 {(a+b) \left (a^2-4 a b+b^2\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 [C]  time = 6.23, size = 530, normalized size = 1.19 \[ -\frac {\frac {4 a^2 \cot ^{\frac {11}{2}}(c+d x) \, _2F_1\left (2,\frac {11}{2};\frac {13}{2};-\frac {a \cot (c+d x)}{b}\right )}{11 b \left (a^2+b^2\right )^2}+\frac {2 b \left (3 a^2-b^2\right ) \left (-7 \cot ^{\frac {3}{2}}(c+d x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\cot ^2(c+d x)\right )-3 \cot ^{\frac {7}{2}}(c+d x)+7 \cot ^{\frac {3}{2}}(c+d x)\right )}{21 \left (a^2+b^2\right )^3}-\frac {2 a \left (a^2-3 b^2\right ) \cot ^{\frac {9}{2}}(c+d x)}{9 \left (a^2+b^2\right )^3}+\frac {a \left (a^2-3 b^2\right ) \left (40 \cot ^{\frac {9}{2}}(c+d x)-72 \cot ^{\frac {5}{2}}(c+d x)+360 \sqrt {\cot (c+d x)}+45 \left (\sqrt {2} \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-\sqrt {2} \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+2 \left (\sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-\sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )\right )\right )}{180 \left (a^2+b^2\right )^3}+\frac {2 a^2 \cot ^{\frac {11}{2}}(c+d x) \, _2F_1\left (3,\frac {11}{2};\frac {13}{2};-\frac {a \cot (c+d x)}{b}\right )}{11 b^3 \left (a^2+b^2\right )}+\frac {2 b \left (a^2-3 b^2\right ) \left (15 \cot ^{\frac {7}{2}}(c+d x)-7 b \left (\frac {3 \cot ^{\frac {5}{2}}(c+d x)}{a}-\frac {5 b \left (\frac {\cot ^{\frac {3}{2}}(c+d x)}{a}-\frac {3 b \left (\frac {\sqrt {\cot (c+d x)}}{a}-\frac {\sqrt {b} \tan ^{-1}\left (\frac {\sqrt {a} \sqrt {\cot (c+d x)}}{\sqrt {b}}\right )}{a^{3/2}}\right )}{a}\right )}{a}\right )\right )}{105 \left (a^2+b^2\right )^3}}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^(3/2)/(a + b*Tan[c + d*x])^3,x]

[Out]

-(((-2*a*(a^2 - 3*b^2)*Cot[c + d*x]^(9/2))/(9*(a^2 + b^2)^3) + (2*b*(a^2 - 3*b^2)*(15*Cot[c + d*x]^(7/2) - 7*b
*((3*Cot[c + d*x]^(5/2))/a - (5*b*((-3*b*(-((Sqrt[b]*ArcTan[(Sqrt[a]*Sqrt[Cot[c + d*x]])/Sqrt[b]])/a^(3/2)) +
Sqrt[Cot[c + d*x]]/a))/a + Cot[c + d*x]^(3/2)/a))/a)))/(105*(a^2 + b^2)^3) + (2*b*(3*a^2 - b^2)*(7*Cot[c + d*x
]^(3/2) - 3*Cot[c + d*x]^(7/2) - 7*Cot[c + d*x]^(3/2)*Hypergeometric2F1[3/4, 1, 7/4, -Cot[c + d*x]^2]))/(21*(a
^2 + b^2)^3) + (4*a^2*Cot[c + d*x]^(11/2)*Hypergeometric2F1[2, 11/2, 13/2, -((a*Cot[c + d*x])/b)])/(11*b*(a^2
+ b^2)^2) + (2*a^2*Cot[c + d*x]^(11/2)*Hypergeometric2F1[3, 11/2, 13/2, -((a*Cot[c + d*x])/b)])/(11*b^3*(a^2 +
b^2)) + (a*(a^2 - 3*b^2)*(360*Sqrt[Cot[c + d*x]] - 72*Cot[c + d*x]^(5/2) + 40*Cot[c + d*x]^(9/2) + 45*(2*(Sqr
t[2]*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]) + Sqrt[2]*Log[1
- Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])))/(
180*(a^2 + b^2)^3))/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)^(3/2)/(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 \[ \int \frac {\cot \left (d x + c\right )^{\frac {3}{2}}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 5.88, size = 78085, normalized size = 175.87 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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maxima [A]  time = 0.94, size = 439, normalized size = 0.99 \[ \frac {\frac {{\left (63 \, a^{4} b^{3} + 46 \, a^{2} b^{5} + 15 \, b^{7}\right )} \arctan \left (\frac {a}{\sqrt {a b} \sqrt {\tan \left (d x + c\right )}}\right )}{{\left (a^{9} + 3 \, a^{7} b^{2} + 3 \, a^{5} b^{4} + a^{3} b^{6}\right )} \sqrt {a b}} + \frac {2 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (a^{3} + 3 \, a^{2} b - 3 \, a b^{2} - b^{3}\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left (a^{3} - 3 \, a^{2} b - 3 \, a b^{2} + b^{3}\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )}{a^{6} + 3 \, a^{4} b^{2} + 3 \, a^{2} b^{4} + b^{6}} - \frac {\frac {15 \, a^{2} b^{4} + 7 \, b^{6}}{\sqrt {\tan \left (d x + c\right )}} + \frac {17 \, a^{3} b^{3} + 9 \, a b^{5}}{\tan \left (d x + c\right )^{\frac {3}{2}}}}{a^{7} b^{2} + 2 \, a^{5} b^{4} + a^{3} b^{6} + \frac {2 \, {\left (a^{8} b + 2 \, a^{6} b^{3} + a^{4} b^{5}\right )}}{\tan \left (d x + c\right )} + \frac {a^{9} + 2 \, a^{7} b^{2} + a^{5} b^{4}}{\tan \left (d x + c\right )^{2}}} - \frac {8}{a^{3} \sqrt {\tan \left (d x + c\right )}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*((63*a^4*b^3 + 46*a^2*b^5 + 15*b^7)*arctan(a/(sqrt(a*b)*sqrt(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^3 + 3*a^2*b - 3*a*b^2 - b^3)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d
*x + c)))) + 2*sqrt(2)*(a^3 + 3*a^2*b - 3*a*b^2 - b^3)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) +
sqrt(2)*(a^3 - 3*a^2*b - 3*a*b^2 + b^3)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - sqrt(2)*(a^3 -
3*a^2*b - 3*a*b^2 + 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) - ((15*a^2*b^4 + 7*b^6)/sqrt(tan(d*x + c)) + (17*a^3*b^3 + 9*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^3*sqrt(tan(d*x + c))))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\mathrm {cot}\left (c+d\,x\right )}^{3/2}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\cot ^{\frac {3}{2}}{\left (c + d x \right )}}{\left (a + b \tan {\left (c + d x \right )}\right )^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(c + d*x)**(3/2)/(a + b*tan(c + d*x))**3, x)

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