3.763 \(\int \cot ^{\frac{7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)
Optimal. Leaf size=176 \[ \frac{4 a^2 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{(4+4 i) a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 i a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d} \]
[Out]
((-4 - 4*I)*a^(5/2)*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]])/d + (4*a^2*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d - (((2*I)/3)*a*Cot[c + d*x]^
(3/2)*(a + I*a*Tan[c + d*x])^(3/2))/d - (2*Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(5/2))/(5*d)
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Rubi [A] time = 0.352847, antiderivative size = 176, normalized size of antiderivative = 1.,
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, 3548, 3545, 3544, 205} \[ \frac{4 a^2 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{(4+4 i) a^{5/2} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{(1+i) \sqrt{a} \sqrt{\tan (c+d x)}}{\sqrt{a+i a \tan (c+d x)}}\right )}{d}-\frac{2 i a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(5/2),x]
[Out]
((-4 - 4*I)*a^(5/2)*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]])/d + (4*a^2*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d - (((2*I)/3)*a*Cot[c + d*x]^
(3/2)*(a + I*a*Tan[c + d*x])^(3/2))/d - (2*Cot[c + d*x]^(5/2)*(a + I*a*Tan[c + d*x])^(5/2))/(5*d)
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]
Rule 3548
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Si
mp[(d*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f*m*(c^2 + d^2)), x] + Dist[a/(a*c - b*d), Int[(a
+ b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*
d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && EqQ[m + n + 1, 0] && !LtQ[m, -1]
Rule 3545
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(a*b*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m - 1)*(a*c - b*d)), x] + Dist[(2*a^2)/(
a*c - b*d), 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] && GtQ[m, 1/2]
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 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]
Rubi steps
\begin{align*} \int \cot ^{\frac{7}{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{(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac{7}{2}}(c+d x)} \, dx\\ &=-\frac{2 \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}+\left (i \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac{5}{2}}(c+d x)} \, dx\\ &=-\frac{2 i a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}-\left (2 a \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{(a+i a \tan (c+d x))^{3/2}}{\tan ^{\frac{3}{2}}(c+d x)} \, dx\\ &=\frac{4 a^2 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}-\left (4 i a^2 \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\\ &=\frac{4 a^2 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}-\frac{\left (8 a^4 \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 )}{d}\\ &=-\frac{(4+4 i) a^{5/2} \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)}}{d}+\frac{4 a^2 \sqrt{\cot (c+d x)} \sqrt{a+i a \tan (c+d x)}}{d}-\frac{2 i a \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{2 \cot ^{\frac{5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d}\\ \end{align*}
Mathematica [A] time = 1.8098, size = 161, normalized size = 0.91 \[ \frac{16 a^2 e^{i (c+d x)} \cos ^2(c+d x) \sqrt{\cot (c+d x)} \left (e^{i (c+d x)} \left (-35 e^{2 i (c+d x)}+26 e^{4 i (c+d x)}+15\right )-15 \left (-1+e^{2 i (c+d x)}\right )^{5/2} \tanh ^{-1}\left (\frac{e^{i (c+d x)}}{\sqrt{-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{15 d \left (-1+e^{4 i (c+d x)}\right )^2} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^(5/2),x]
[Out]
(16*a^2*E^(I*(c + d*x))*(E^(I*(c + d*x))*(15 - 35*E^((2*I)*(c + d*x)) + 26*E^((4*I)*(c + d*x))) - 15*(-1 + E^(
(2*I)*(c + d*x)))^(5/2)*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)*(c + d*x))]])*Cos[c + d*x]^2*Sqrt[Cot[c + d
*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(15*d*(-1 + E^((4*I)*(c + d*x)))^2)
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Maple [B] time = 0.401, size = 1149, normalized size = 6.5 \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)^(7/2)*(a+I*a*tan(d*x+c))^(5/2),x)
[Out]
-1/15/d*2^(1/2)*a^2*(-11*I*sin(d*x+c)*cos(d*x+c)*2^(1/2)-60*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d
*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)*sin(d*x+c)-60*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)
/sin(d*x+c))^(1/2)*2^(1/2)+1)*sin(d*x+c)+30*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*ln(-(((cos(d*x+c)-1)/sin(d*x+c
))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos
(d*x+c)+sin(d*x+c)-1))*cos(d*x+c)^2*sin(d*x+c)-60*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*ar
ctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)-60*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)
*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)-1)-30*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x
+c)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*
x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1))-30*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*ln(-(((cos(d*x
+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c)+1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2
)*sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1))*sin(d*x+c)+60*I*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1
)/sin(d*x+c))^(1/2)*2^(1/2)-1)*cos(d*x+c)^2*sin(d*x+c)+52*I*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+60*I*((cos(d*x+c)-
1)/sin(d*x+c))^(1/2)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)+1)*cos(d*x+c)^2*sin(d*x+c)+52*2^(1/2)*co
s(d*x+c)^3-38*I*2^(1/2)*sin(d*x+c)+60*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(
d*x+c))^(1/2)*2^(1/2)+1)+60*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*arctan(((cos(d*x+c)-1)/sin(d*x+c))^(1
/2)*2^(1/2)-1)+30*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*sin(d*x+c)*ln(-(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*
sin(d*x+c)+cos(d*x+c)+sin(d*x+c)-1)/(((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-cos(d*x+c)-sin(d*x+c
)+1))-41*2^(1/2)*cos(d*x+c)^2-49*2^(1/2)*cos(d*x+c)+38*2^(1/2))*(cos(d*x+c)/sin(d*x+c))^(7/2)*(a*(I*sin(d*x+c)
+cos(d*x+c))/cos(d*x+c))^(1/2)*sin(d*x+c)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^3
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Maxima [B] time = 2.36995, size = 1702, normalized size = 9.67 \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)^(7/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
1/225*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(((900*I + 900)*a^2*cos(3*d*x +
3*c) - (930*I + 930)*a^2*cos(d*x + c) + (900*I - 900)*a^2*sin(3*d*x + 3*c) - (930*I - 930)*a^2*sin(d*x + c))*c
os(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (-(900*I - 900)*a^2*cos(3*d*x + 3*c) + (930*I - 930)
*a^2*cos(d*x + c) + (900*I + 900)*a^2*sin(3*d*x + 3*c) - (930*I + 930)*a^2*sin(d*x + c))*sin(3/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a) + ((-(900*I - 900)*a^2*cos(2*d*x + 2*c)^2 - (900*I - 900)*a^2*sin
(2*d*x + 2*c)^2 + (1800*I - 1800)*a^2*cos(2*d*x + 2*c) - (900*I - 900)*a^2)*arctan2(2*(cos(2*d*x + 2*c)^2 + si
n(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*
sin(d*x + c), 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2
*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 2*cos(d*x + c)) + (-(450*I + 450)*a^2*cos(2*d*x + 2*c)^2 - (450*I + 450)
*a^2*sin(2*d*x + 2*c)^2 + (900*I + 900)*a^2*cos(2*d*x + 2*c) - (450*I + 450)*a^2)*log(4*cos(d*x + c)^2 + 4*sin
(d*x + c)^2 + 4*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*arctan2(sin(2*
d*x + 2*c), cos(2*d*x + 2*c) - 1))^2 + sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1))^2) + 8*(cos(2*
d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(d*x + c)*cos(1/2*arctan2(sin(2*d*x + 2*
c), cos(2*d*x + 2*c) - 1)) + sin(d*x + c)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))))*(cos(2*d
*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + (((900*I + 900)*a^2*cos(5*d*x + 5*c
) - (750*I + 750)*a^2*cos(3*d*x + 3*c) + (210*I + 210)*a^2*cos(d*x + c) + (900*I - 900)*a^2*sin(5*d*x + 5*c) -
(750*I - 750)*a^2*sin(3*d*x + 3*c) + (210*I - 210)*a^2*sin(d*x + c))*cos(5/2*arctan2(sin(2*d*x + 2*c), cos(2*
d*x + 2*c) - 1)) + ((-(240*I + 240)*a^2*cos(d*x + c) - (240*I - 240)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (2
40*I + 240)*a^2*cos(d*x + c) + (-(240*I + 240)*a^2*cos(d*x + c) - (240*I - 240)*a^2*sin(d*x + c))*sin(2*d*x +
2*c)^2 - (240*I - 240)*a^2*sin(d*x + c) + ((480*I + 480)*a^2*cos(d*x + c) + (480*I - 480)*a^2*sin(d*x + c))*co
s(2*d*x + 2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + (-(900*I - 900)*a^2*cos(5*d*x + 5*c
) + (750*I - 750)*a^2*cos(3*d*x + 3*c) - (210*I - 210)*a^2*cos(d*x + c) + (900*I + 900)*a^2*sin(5*d*x + 5*c) -
(750*I + 750)*a^2*sin(3*d*x + 3*c) + (210*I + 210)*a^2*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*
d*x + 2*c) - 1)) + (((240*I - 240)*a^2*cos(d*x + c) - (240*I + 240)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 + (24
0*I - 240)*a^2*cos(d*x + c) + ((240*I - 240)*a^2*cos(d*x + c) - (240*I + 240)*a^2*sin(d*x + c))*sin(2*d*x + 2*
c)^2 - (240*I + 240)*a^2*sin(d*x + c) + (-(480*I - 480)*a^2*cos(d*x + c) + (480*I + 480)*a^2*sin(d*x + c))*cos
(2*d*x + 2*c))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)))*sqrt(a))/((cos(2*d*x + 2*c)^2 + sin(2
*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(5/4)*d)
________________________________________________________________________________________
Fricas [B] time = 1.33904, size = 1170, normalized size = 6.65 \begin{align*} \frac{8 \, \sqrt{2}{\left (26 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 35 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 15 \, a^{2}\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}} e^{\left (i \, d x + i \, c\right )} - 15 \, \sqrt{\frac{32 i \, a^{5}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2}\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}} e^{\left (i \, d x + i \, c\right )} + \sqrt{\frac{32 i \, a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\right ) + 15 \, \sqrt{\frac{32 i \, a^{5}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{2}\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}} e^{\left (i \, d x + i \, c\right )} - \sqrt{\frac{32 i \, a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{4 \, a^{2}}\right )}{30 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
1/30*(8*sqrt(2)*(26*a^2*e^(4*I*d*x + 4*I*c) - 35*a^2*e^(2*I*d*x + 2*I*c) + 15*a^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))*e^(I*d*x + I*c) - 15*sqrt(32*I*a^5/d^2)*(d*
e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) - a^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))*e^(I*d*x + I*c) + sqrt(3
2*I*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2) + 15*sqrt(32*I*a^5/d^2)*(d*e^(4*I*d*x + 4*I*c) -
2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) - a^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))*e^(I*d*x + I*c) - sqrt(32*I*a^5/d^2)*d*e^(2*I
*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/a^2))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)
________________________________________________________________________________________
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)**(7/2)*(a+I*a*tan(d*x+c))**(5/2),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{5}{2}} \cot \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^(7/2), x)