3.785 \(\int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=184 \[ -\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {1}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {i}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \]

[Out]

(-1/8+1/8*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2
)/a^(5/2)/d+1/4/a^2/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2)+1/5/d/cot(d*x+c)^(5/2)/(a+I*a*tan(d*x+c))^(5/2
)+1/6*I/a/d/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.36, antiderivative size = 184, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {4241, 3547, 3546, 3544, 205} \[ \frac {1}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {i}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/(Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((-1/8 + I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt
[Tan[c + d*x]])/(a^(5/2)*d) + 1/(5*d*Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(5/2)) + (I/6)/(a*d*Cot[c + d*x
]^(3/2)*(a + I*a*Tan[c + d*x])^(3/2)) + 1/(4*a^2*d*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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 3544

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*a*b)/f, Subst[Int[1/(a*c - b*d - 2*a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]

Rule 3546

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*b*f*m), x] - Dist[(a*c - b*d)/(2*b^2), Int[(a + b*Tan[e
+ f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &&
EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && LeQ[m, -2^(-1)]

Rule 3547

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2*f*m*(b*c - a*d)), x] + Dist[1/(2*a), Int[(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && Eq
Q[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n + 1, 0] && LtQ[m, -1]

Rule 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps

\begin {align*} \int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{(a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\tan ^{\frac {3}{2}}(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx}{2 a}\\ &=\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}-\frac {\left (i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}} \, dx}{4 a^2}\\ &=\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {1}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3}\\ &=\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {1}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (i \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{-i a-2 a^2 x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{4 a d}\\ &=-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{a^{5/2} d}+\frac {1}{5 d \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {i}{6 a d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}+\frac {1}{4 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 1.25, size = 171, normalized size = 0.93 \[ -\frac {i e^{-6 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {\cot (c+d x)} \left (-4 e^{2 i (c+d x)}-16 e^{4 i (c+d x)}+17 e^{6 i (c+d x)}-15 e^{5 i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )+3\right )}{60 \sqrt {2} a^3 d} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((-1/60*I)*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(3 - 4*E^((2*I)*(c + d*x)) - 16*E^((4*I)*(c
+ d*x)) + 17*E^((6*I)*(c + d*x)) - 15*E^((5*I)*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*
x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Sqrt[Cot[c + d*x]])/(Sqrt[2]*a^3*d*E^((6*I)*(c + d*x)))

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fricas [B]  time = 0.91, size = 366, normalized size = 1.99 \[ -\frac {{\left (30 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left ({\left (\sqrt {2} {\left (8 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} + 4 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 30 \, a^{3} d \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left ({\left (\sqrt {2} {\left (-8 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {-\frac {i}{8 \, a^{5} d^{2}}} + 4 i \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} {\left (-17 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 16 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/120*(30*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log((sqrt(2)*(8*I*a^3*d*e^(2*I*d*x + 2*I*c) - 8*I*
a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/8
*I/(a^5*d^2)) + 4*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 30*a^3*d*sqrt(-1/8*I/(a^5*d^2))*e^(5*I*d*x + 5*I*c)
*log((sqrt(2)*(-8*I*a^3*d*e^(2*I*d*x + 2*I*c) + 8*I*a^3*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*
x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt(-1/8*I/(a^5*d^2)) + 4*I*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) -
sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*(-17*I*
e^(6*I*d*x + 6*I*c) + 16*I*e^(4*I*d*x + 4*I*c) + 4*I*e^(2*I*d*x + 2*I*c) - 3*I))*e^(-5*I*d*x - 5*I*c)/(a^3*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate(1/((I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^(3/2)), x)

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maple [B]  time = 3.32, size = 660, normalized size = 3.59 \[ \frac {\left (\frac {1}{120}+\frac {i}{120}\right ) \left (60 i \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right ) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )-28 i \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-30 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {2}+60 \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right ) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )-40 i \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right )+28 \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-15 i \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {2}\, \sin \left (d x +c \right )-40 \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\right )-30 \left (\cos ^{2}\left (d x +c \right )\right ) \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {2}+13 i \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )-45 \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {2}\, \cos \left (d x +c \right )+40 i \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \cos \left (d x +c \right )-13 \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )+40 \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \cos \left (d x +c \right )+15 \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {2}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{d \left (4 i \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 \left (\cos ^{3}\left (d x +c \right )\right )-i \sin \left (d x +c \right )-3 \cos \left (d x +c \right )\right ) \sin \left (d x +c \right )^{2} \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )^{\frac {3}{2}} a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

(1/120+1/120*I)/d*(60*I*cos(d*x+c)^2*sin(d*x+c)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))
*2^(1/2)-28*I*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-30*I*cos(d*x+c)*sin(d*x+c)*arctan((1/
2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+60*2^(1/2)*cos(d*x+c)^3*arctan((1/2+1/2*I)*((-1+c
os(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))-40*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3+28*((-1+cos(d*x+c))
/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-15*I*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))
*2^(1/2)*sin(d*x+c)-40*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-30*cos(d*x+c)^2*arctan((1/2+1/2*I)*((-1
+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+13*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-45*arctan((
1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)*cos(d*x+c)+40*I*cos(d*x+c)*((-1+cos(d*x+c))/sin
(d*x+c))^(1/2)-13*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+40*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+
c)+15*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2))*cos(d*x+c)^2*(a*(I*sin(d*x+c)+co
s(d*x+c))/cos(d*x+c))^(1/2)/(4*I*cos(d*x+c)^2*sin(d*x+c)+4*cos(d*x+c)^3-I*sin(d*x+c)-3*cos(d*x+c))/sin(d*x+c)^
2/((-1+cos(d*x+c))/sin(d*x+c))^(1/2)/(cos(d*x+c)/sin(d*x+c))^(3/2)/a^3

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cot(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

int(1/(cot(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(5/2)), x)

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sympy [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(1/cot(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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