3.98 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\)
Optimal. Leaf size=184 \[ \frac{11 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a^2 \cot ^2(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{i a^2 \cot (c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i a \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d} \]
[Out]
(11*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (2*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c
+ d*x]]/(Sqrt[2]*Sqrt[a])])/d - ((I/2)*a^2*Cot[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (a^2*Cot[c + d*x]^2
)/(2*d*Sqrt[a + I*a*Tan[c + d*x]]) - (((5*I)/4)*a*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d
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Rubi [A] time = 0.605892, antiderivative size = 184, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used =
{3553, 3596, 3598, 3600, 3480, 206, 3599, 63, 208} \[ \frac{11 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{a^2 \cot ^2(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{i a^2 \cot (c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i a \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^(3/2),x]
[Out]
(11*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (2*Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c
+ d*x]]/(Sqrt[2]*Sqrt[a])])/d - ((I/2)*a^2*Cot[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (a^2*Cot[c + d*x]^2
)/(2*d*Sqrt[a + I*a*Tan[c + d*x]]) - (((5*I)/4)*a*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d
Rule 3553
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(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] /; Fr
eeQ[{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] && LtQ[
n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])
Rule 3596
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((a*A + b*B)*(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*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*Tan[e + f*x], x], x], x
] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && !GtQ[n,
0]
Rule 3598
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]
Rule 3600
Int[(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[(A*b + a*B)/(b*c + a*d), Int[(a + b*Tan[e + f*x])^m, x], x] - Dist[(B*c
- A*d)/(b*c + a*d), Int[((a + b*Tan[e + f*x])^m*(a - b*Tan[e + f*x]))/(c + d*Tan[e + f*x]), x], x] /; FreeQ[{
a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
Rule 3480
Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 3599
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*B)/f, Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x
]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && EqQ[A*b + a*B,
0]
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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac{a^2 \cot ^2(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{1}{2} \int \frac{\cot ^2(c+d x) \left (-\frac{7 i a^2}{2}+\frac{9}{2} a^2 \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=-\frac{i a^2 \cot (c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \cot ^2(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{\int \cot ^2(c+d x) \sqrt{a+i a \tan (c+d x)} \left (-\frac{5 i a^3}{2}+\frac{3}{2} a^3 \tan (c+d x)\right ) \, dx}{2 a^2}\\ &=-\frac{i a^2 \cot (c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \cot ^2(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i a \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{\int \cot (c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{11 a^4}{4}+\frac{5}{4} i a^4 \tan (c+d x)\right ) \, dx}{2 a^3}\\ &=-\frac{i a^2 \cot (c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \cot ^2(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i a \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{11}{8} \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)} \, dx-(2 i a) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{i a^2 \cot (c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \cot ^2(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i a \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}-\frac{\left (11 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}-\frac{\left (4 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{i a^2 \cot (c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \cot ^2(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i a \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}+\frac{(11 i a) \operatorname{Subst}\left (\int \frac{1}{i-\frac{i x^2}{a}} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{4 d}\\ &=\frac{11 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{4 d}-\frac{2 \sqrt{2} a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{i a^2 \cot (c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{a^2 \cot ^2(c+d x)}{2 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i a \cot (c+d x) \sqrt{a+i a \tan (c+d x)}}{4 d}\\ \end{align*}
Mathematica [A] time = 1.84419, size = 190, normalized size = 1.03 \[ -\frac{a e^{-\frac{1}{2} i (2 c+3 d x)} \sqrt{1+e^{2 i (c+d x)}} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (\cos \left (\frac{d x}{2}\right )+i \sin \left (\frac{d x}{2}\right )\right ) \left (16 \sinh ^{-1}\left (e^{i (c+d x)}\right )-11 \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{1+e^{2 i (c+d x)}}}\right )+\sqrt{1+e^{2 i (c+d x)}} (2 \cot (c+d x)+5 i) \csc (c+d x)\right )}{4 \sqrt{2} d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^(3/2),x]
[Out]
-(a*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*(16*ArcSinh[E^(I*(c
+ d*x))] - 11*Sqrt[2]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)*(c + d*x))]] + Sqrt[1 + E^((2*I)*(c
+ d*x))]*(5*I + 2*Cot[c + d*x])*Csc[c + d*x])*(Cos[(d*x)/2] + I*Sin[(d*x)/2]))/(4*Sqrt[2]*d*E^((I/2)*(2*c + 3*
d*x)))
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Maple [B] time = 0.351, size = 1142, normalized size = 6.2 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2),x)
[Out]
1/8/d*a*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-16*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*
2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3*2^(1/2)+16*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a
rctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)+16*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arct
an(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)*2^(1/2)+16*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2
)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)^3*2^(1/2)+11*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))
*cos(d*x+c)^3+11*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x
+c)-11*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3+16*2
^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2)*sin(d*x+c)/cos(d*x+c))+11*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)
+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))-14*I*cos(d*x+c)^3+10*I*cos(d*x+c)-16*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)*2^(1/2)
-11*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2-11*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))
*cos(d*x+c)-4*I*cos(d*x+c)^2-16*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))-16*I*2^(1/2)*cos(d*x+c)^2*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*
arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+14*cos(d*x+c)^2*sin(d*x+c)-11*(-2*cos(d*x+c)/(cos(d*x
+c)+1))^(1/2)*ln(((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-cos(d*x+c)+1)/sin(d*x+c))+11*I*(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+10*cos(d*x+c)*sin(d*x+c))/(I*sin(d*x+c)
+cos(d*x+c)-1)/sin(d*x+c)/(cos(d*x+c)+1)
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Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.47703, size = 1515, normalized size = 8.23 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
1/8*(2*sqrt(2)*(7*a*e^(4*I*d*x + 4*I*c) + 4*a*e^(2*I*d*x + 2*I*c) - 3*a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(
I*d*x + I*c) - 8*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/d^2)*log((sqrt(2)*sqrt
(a^3/d^2)*d*e^(2*I*d*x + 2*I*c) + sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d
*x + I*c))*e^(-2*I*d*x - 2*I*c)/a) + 8*sqrt(2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a^3/
d^2)*log(-(sqrt(2)*sqrt(a^3/d^2)*d*e^(2*I*d*x + 2*I*c) - sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*
x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-2*I*d*x - 2*I*c)/a) + 11*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c
) + d)*sqrt(a^3/d^2)*log((sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c
) + 2*sqrt(a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a) - 11*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*
x + 2*I*c) + d)*sqrt(a^3/d^2)*log((sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*
d*x + I*c) - 2*sqrt(a^3/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2
*I*d*x + 2*I*c) + d)
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**(3/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \cot \left (d x + c\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^(3/2)*cot(d*x + c)^3, x)