3.8.34 \(\int \frac {(a+i a \tan (c+d x))^3}{\sqrt {\cot (c+d x)}} \, dx\) [734]

Optimal. Leaf size=106 \[ \frac {8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {8 a^3}{5 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {8 i a^3}{d \sqrt {\cot (c+d x)}}-\frac {2 \left (i a^3+a^3 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)} \]

[Out]

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

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Rubi [A]
time = 0.16, antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {3754, 3634, 3672, 3610, 3614, 214} \begin {gather*} -\frac {8 a^3}{5 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (a^3 \cot (c+d x)+i a^3\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {8 i a^3}{d \sqrt {\cot (c+d x)}}+\frac {8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 214

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

Rule 3610

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

Rule 3614

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

Rule 3634

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-a^2)*(b*c - a*d)*(a + b*Tan[e + f*x])^(m - 2)*((c + d*Tan[e + f*x])^(n + 1)/(d*f*(b*c + a*d)*(n + 1))), x]
+ Dist[a/(d*(b*c + a*d)*(n + 1)), Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*(b*c*(
m - 2) - a*d*(m - 2*n - 4)) + (a*b*c*(m - 2) + b^2*d*(n + 1) - a^2*d*(m + n - 1))*Tan[e + f*x], x], 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] && GtQ[m, 1] && Lt
Q[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3672

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(A*b - a*B)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2
+ b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rule 3754

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 {(a+i a \tan (c+d x))^3}{\sqrt {\cot (c+d x)}} \, dx &=\int \frac {(i a+a \cot (c+d x))^3}{\cot ^{\frac {7}{2}}(c+d x)} \, dx\\ &=-\frac {2 \left (i a^3+a^3 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int \frac {(i a+a \cot (c+d x)) \left (-6 i a^2-4 a^2 \cot (c+d x)\right )}{\cot ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {8 a^3}{5 d \cot ^{\frac {3}{2}}(c+d x)}-\frac {2 \left (i a^3+a^3 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int \frac {-10 i a^3-10 a^3 \cot (c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {8 a^3}{5 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {8 i a^3}{d \sqrt {\cot (c+d x)}}-\frac {2 \left (i a^3+a^3 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {2}{5} \int \frac {-10 a^3+10 i a^3 \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=-\frac {8 a^3}{5 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {8 i a^3}{d \sqrt {\cot (c+d x)}}-\frac {2 \left (i a^3+a^3 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}-\frac {\left (80 a^6\right ) \text {Subst}\left (\int \frac {1}{10 a^3+10 i a^3 x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {8 \sqrt [4]{-1} a^3 \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {8 a^3}{5 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {8 i a^3}{d \sqrt {\cot (c+d x)}}-\frac {2 \left (i a^3+a^3 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]
time = 2.53, size = 164, normalized size = 1.55 \begin {gather*} \frac {a^3 e^{-3 i c} \sqrt {\cot (c+d x)} (\cos (3 (c+d x))+i \sin (3 (c+d x))) \left (\sec ^3(c+d x) (-5 \cos (c+d x)+5 \cos (3 (c+d x))+17 i \sin (c+d x)+21 i \sin (3 (c+d x)))-80 \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right ) \sqrt {i \tan (c+d x)}\right )}{10 d (\cos (d x)+i \sin (d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*Sqrt[Cot[c + d*x]]*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)])*(Sec[c + d*x]^3*(-5*Cos[c + d*x] + 5*Cos[3*(c
+ d*x)] + (17*I)*Sin[c + d*x] + (21*I)*Sin[3*(c + d*x)]) - 80*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^(
(2*I)*(c + d*x)))]]*Sqrt[I*Tan[c + d*x]]))/(10*d*E^((3*I)*c)*(Cos[d*x] + I*Sin[d*x])^3)

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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 29.32, size = 520, normalized size = 4.91

method	result	size
default	\(\frac {a^{3} \left (-1+\cos \left (d x +c \right )\right ) \left (20 i \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \EllipticPi \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-20 i \EllipticF \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \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 )+20 \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {\cos \left (d x +c \right )-1+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{2}\left (d x +c \right )\right ) \EllipticPi \left (\sqrt {-\frac {\cos \left (d x +c \right )-1-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+21 i \sqrt {2}\, \left (\cos ^{3}\left (d x +c \right )\right )-21 i \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right )-5 \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-i \cos \left (d x +c \right ) \sqrt {2}+5 \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+i \sqrt {2}\right ) \left (\cos \left (d x +c \right )+1\right )^{2} \sqrt {2}}{5 d \sqrt {\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \cos \left (d x +c \right )^{2} \sin \left (d x +c \right )^{4}}\)	\(520\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a^3/d*(-1+cos(d*x+c))*(20*I*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2)
)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/
sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)-20*I*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(
1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+
c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)+20*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+
1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(cos(d*x+
c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)+21*I*2^(1/2)*cos(d*x+c)^3-21*I*2^(1/2)*cos(d*x+c)^2
-5*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-I*2^(1/2)*cos(d*x+c)+5*2^(1/2)*cos(d*x+c)*sin(d*x+c)+I*2^(1/2))*(cos(d*x+c)
+1)^2/(cos(d*x+c)/sin(d*x+c))^(1/2)/cos(d*x+c)^2/sin(d*x+c)^4*2^(1/2)

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Maxima [A]
time = 0.50, size = 161, normalized size = 1.52 \begin {gather*} \frac {5 \, {\left (\left (2 i - 2\right ) \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \left (2 i - 2\right ) \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \left (i + 1\right ) \, \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \left (i + 1\right ) \, \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} - 2 \, {\left (i \, a^{3} + \frac {5 \, a^{3}}{\tan \left (d x + c\right )} - \frac {20 i \, a^{3}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac {5}{2}}}{5 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*(5*((2*I - 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + (2*I - 2)*sqrt(2)*arctan(-1/2
*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) - (I + 1)*sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) +
1) + (I + 1)*sqrt(2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^3 - 2*(I*a^3 + 5*a^3/tan(d*x +
c) - 20*I*a^3/tan(d*x + c)^2)*tan(d*x + c)^(5/2))/d

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 390 vs. \(2 (86) = 172\).
time = 0.47, size = 390, normalized size = 3.68 \begin {gather*} -\frac {5 \, \sqrt {\frac {64 i \, a^{6}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {64 i \, a^{6}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3}}\right ) - 5 \, \sqrt {\frac {64 i \, a^{6}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac {{\left (8 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {64 i \, a^{6}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{3}}\right ) - 16 \, {\left (13 \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + 6 \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 11 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - 8 \, a^{3}\right )} \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}}}{20 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/20*(5*sqrt(64*I*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*lo
g(1/4*(8*I*a^3*e^(2*I*d*x + 2*I*c) + sqrt(64*I*a^6/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt((I*e^(2*I*d*x + 2*I*c
) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c)/a^3) - 5*sqrt(64*I*a^6/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3
*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(8*I*a^3*e^(2*I*d*x + 2*I*c) - sqrt(64*I*a^6/d^2
)*(d*e^(2*I*d*x + 2*I*c) - d)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))*e^(-2*I*d*x - 2*I*c
)/a^3) - 16*(13*a^3*e^(6*I*d*x + 6*I*c) + 6*a^3*e^(4*I*d*x + 4*I*c) - 11*a^3*e^(2*I*d*x + 2*I*c) - 8*a^3)*sqrt
((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*
d*e^(2*I*d*x + 2*I*c) + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} - i a^{3} \left (\int \frac {i}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {3 \tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \frac {\tan ^{3}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {3 i \tan ^{2}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-I*a**3*(Integral(I/sqrt(cot(c + d*x)), x) + Integral(-3*tan(c + d*x)/sqrt(cot(c + d*x)), x) + Integral(tan(c
+ d*x)**3/sqrt(cot(c + d*x)), x) + Integral(-3*I*tan(c + d*x)**2/sqrt(cot(c + d*x)), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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