3.8.63 \(\int \cot ^{\frac {7}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [763]
Optimal. Leaf size=176 \[ -\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {4 a^2 \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {2 i a \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 \cot ^{\frac {5}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}{5 d} \]
[Out]
(-4-4*I)*a^(5/2)*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)/d+4*a^2*cot(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d-2/3*I*a*cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))^(3/2)/d-
2/5*cot(d*x+c)^(5/2)*(a+I*a*tan(d*x+c))^(5/2)/d
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Rubi [A]
time = 0.24, antiderivative size = 176, 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 = {4326, 3629,
3626, 3625, 211} \begin {gather*} -\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 {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 {gather*}
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 211
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]
Rule 3625
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 3626
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 3629
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-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 4326
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 \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 ) \text {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*}
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Mathematica [A]
time = 1.98, size = 161, normalized size = 0.91 \begin {gather*} \frac {16 a^2 e^{i (c+d x)} \left (e^{i (c+d x)} \left (15-35 e^{2 i (c+d x)}+26 e^{4 i (c+d x)}\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 ) \cos ^2(c+d x) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{15 d \left (-1+e^{4 i (c+d x)}\right )^2} \end {gather*}
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] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 1148 vs. \(2 (143 ) = 286\).
time = 47.78, size = 1149, normalized size = 6.53
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result |
size |
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\(\text {Expression too large to display}\) |
\(1149\) |
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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,method=_RETURNVERBOSE)
[Out]
-1/15/d*(-30*I*sin(d*x+c)*ln(-(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-sin(d*x+c)-cos(d*x+c)+1)/
(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))*((-1+cos(d*x+c))/sin(d*x+c))^
(1/2)-11*I*sin(d*x+c)*cos(d*x+c)*2^(1/2)-60*I*sin(d*x+c)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)*
((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+60*I*sin(d*x+c)*cos(d*x+c)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1
/2)-1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-60*sin(d*x+c)*cos(d*x+c)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)
*2^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-60*sin(d*x+c)*cos(d*x+c)^2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^
(1/2)*2^(1/2)-1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-30*sin(d*x+c)*cos(d*x+c)^2*ln(-(((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-si
n(d*x+c)-cos(d*x+c)+1))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+60*I*sin(d*x+c)*cos(d*x+c)^2*arctan(((-1+cos(d*x+c)
)/sin(d*x+c))^(1/2)*2^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+30*I*sin(d*x+c)*cos(d*x+c)^2*ln(-(((-1+cos(d
*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-sin(d*x+c)-cos(d*x+c)+1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/
2)*sin(d*x+c)+sin(d*x+c)+cos(d*x+c)-1))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-60*I*sin(d*x+c)*arctan(((-1+cos(d*x
+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+52*I*sin(d*x+c)*cos(d*x+c)^2*2^(1/2)+52*c
os(d*x+c)^3*2^(1/2)-38*I*sin(d*x+c)*2^(1/2)+60*sin(d*x+c)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)+1)
*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+60*sin(d*x+c)*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)-1)*((-1+co
s(d*x+c))/sin(d*x+c))^(1/2)+30*sin(d*x+c)*ln(-(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)+sin(d*x+c
)+cos(d*x+c)-1)/(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2)*sin(d*x+c)-sin(d*x+c)-cos(d*x+c)+1))*((-1+cos(d*x+
c))/sin(d*x+c))^(1/2)-41*2^(1/2)*cos(d*x+c)^2-49*cos(d*x+c)*2^(1/2)+38*2^(1/2))*sin(d*x+c)*(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)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^3*2^(1/2)*a^2
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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than
twice the leaf count of optimal. 1267 vs. \(2 (134) = 268\).
time = 0.63, size = 1267, normalized size = 7.20 \begin {gather*} \text {Too large to display} \end {gather*}
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]
-2/15*(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*((-(30*I + 30)*a^2*cos(3*d*x + 3
*c) + (31*I + 31)*a^2*cos(d*x + c) - (30*I - 30)*a^2*sin(3*d*x + 3*c) + (31*I - 31)*a^2*sin(d*x + c))*cos(3/2*
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + ((30*I - 30)*a^2*cos(3*d*x + 3*c) - (31*I - 31)*a^2*cos(d*x
+ c) - (30*I + 30)*a^2*sin(3*d*x + 3*c) + (31*I + 31)*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) + 15*(2*((I - 1)*a^2*cos(2*d*x + 2*c)^2 + (I - 1)*a^2*sin(2*d*x + 2*c)^2 - (2*I -
2)*a^2*cos(2*d*x + 2*c) + (I - 1)*a^2)*arctan2(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)*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)) + ((I + 1)*a^2*cos(2*d*x + 2*c)^2 + (I + 1)*a^2*sin(2*d*x + 2*c)^2 - (2*I + 2)*a^2*cos(2*d*x
+ 2*c) + (I + 1)*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(s
in(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) + ((-(30*I + 30)*a^2*cos(5*d*x + 5*c) + (25*I + 25)*a^2*cos(3*d*x + 3*c) - (7*I + 7)*a^2*cos(d*x
+ c) - (30*I - 30)*a^2*sin(5*d*x + 5*c) + (25*I - 25)*a^2*sin(3*d*x + 3*c) - (7*I - 7)*a^2*sin(d*x + c))*cos(
5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 8*(((I + 1)*a^2*cos(d*x + c) + (I - 1)*a^2*sin(d*x + c)
)*cos(2*d*x + 2*c)^2 + (I + 1)*a^2*cos(d*x + c) + ((I + 1)*a^2*cos(d*x + c) + (I - 1)*a^2*sin(d*x + c))*sin(2*
d*x + 2*c)^2 + (I - 1)*a^2*sin(d*x + c) + 2*(-(I + 1)*a^2*cos(d*x + c) - (I - 1)*a^2*sin(d*x + c))*cos(2*d*x +
2*c))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + ((30*I - 30)*a^2*cos(5*d*x + 5*c) - (25*I -
25)*a^2*cos(3*d*x + 3*c) + (7*I - 7)*a^2*cos(d*x + c) - (30*I + 30)*a^2*sin(5*d*x + 5*c) + (25*I + 25)*a^2*sin
(3*d*x + 3*c) - (7*I + 7)*a^2*sin(d*x + c))*sin(5/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) - 1)) + 8*((-(I
- 1)*a^2*cos(d*x + c) + (I + 1)*a^2*sin(d*x + c))*cos(2*d*x + 2*c)^2 - (I - 1)*a^2*cos(d*x + c) + (-(I - 1)*a
^2*cos(d*x + c) + (I + 1)*a^2*sin(d*x + c))*sin(2*d*x + 2*c)^2 + (I + 1)*a^2*sin(d*x + c) + 2*((I - 1)*a^2*cos
(d*x + c) - (I + 1)*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)
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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. 409 vs. \(2 (134) = 268\).
time = 0.80, size = 409, normalized size = 2.32 \begin {gather*} \frac {16 \, \sqrt {2} {\left (26 \, a^{2} e^{\left (5 i \, d x + 5 i \, c\right )} - 35 \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + 15 \, a^{2} e^{\left (i \, d x + i \, c\right )}\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}} - 15 \, \sqrt {\frac {128 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 (16 i \, a^{3} e^{\left (i \, d x + i \, c\right )} + \sqrt {2} \sqrt {\frac {128 i \, a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) + 15 \, \sqrt {\frac {128 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 (16 i \, a^{3} e^{\left (i \, d x + i \, c\right )} - \sqrt {2} \sqrt {\frac {128 i \, a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - 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}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right )}{60 \, {\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 {gather*}
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]
1/60*(16*sqrt(2)*(26*a^2*e^(5*I*d*x + 5*I*c) - 35*a^2*e^(3*I*d*x + 3*I*c) + 15*a^2*e^(I*d*x + I*c))*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)) - 15*sqrt(128*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*(16*I*a^3*e^(I*d*x + I*c) + sqrt(2)*sqrt(128*I*a^
5/d^2)*(d*e^(2*I*d*x + 2*I*c) - 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)))*e^(-I*d*x - I*c)/a^2) + 15*sqrt(128*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*(16*I*a^3*e^(I*d*x + I*c) - sqrt(2)*sqrt(128*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - 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)))*e^(-I*d*x - I*c)/a^2))
/(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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
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 = 0.00, 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(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)
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\mathrm {cot}\left (c+d\,x\right )}^{7/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^(5/2),x)
[Out]
int(cot(c + d*x)^(7/2)*(a + a*tan(c + d*x)*1i)^(5/2), x)
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