3.8.65 \(\int \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [765]

Optimal. Leaf size=179 \[ \frac {2 (-1)^{3/4} a^{5/2} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}+\frac {(4+4 i) a^{5/2} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{d}-\frac {2 a^2 \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \]

[Out]

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

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Rubi [A]
time = 0.32, antiderivative size = 179, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 9, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {4326, 3634, 3682, 3625, 211, 3680, 65, 223, 209} \begin {gather*} \frac {2 (-1)^{3/4} a^{5/2} \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {ArcTan}\left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{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 a^2 \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 65

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 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 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 3680

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 3682

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A*b + a*B)/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x]
, x] - Dist[B/b, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[e + f*x]), 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] && NeQ[A*b + a*B, 0]

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

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Mathematica [A]
time = 2.58, size = 176, normalized size = 0.98 \begin {gather*} -\frac {\sqrt {2} a^2 e^{-i (c+d x)} \left (\sqrt {2} e^{i (c+d x)}-2 \sqrt {2} \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 )+\sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {\sqrt {2} e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )\right ) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[2]*a^2*(Sqrt[2]*E^(I*(c + d*x)) - 2*Sqrt[2]*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqr
t[-1 + E^((2*I)*(c + d*x))]] + Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[-1 + E^((
2*I)*(c + d*x))]])*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^(I*(c + d*x))))

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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. 887 vs. \(2 (144 ) = 288\).
time = 49.08, size = 888, normalized size = 4.96

method	result	size
default	\(-\frac {\left (i \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+1\right ) \sqrt {2}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )-i \ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-1\right ) \sqrt {2}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )-2 i \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\right ) \sqrt {2}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )+8 i \sin \left (d x +c \right ) \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}+1\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+8 i \sin \left (d x +c \right ) \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}-1\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+4 i \sin \left (d x +c \right ) \ln \left (-\frac {\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )-\sin \left (d x +c \right )-\cos \left (d x +c \right )+1}{\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )+\sin \left (d x +c \right )+\cos \left (d x +c \right )-1}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+\ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+1\right ) \sqrt {2}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )-\ln \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-1\right ) \sqrt {2}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )+2 \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\right ) \sqrt {2}\, \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )-8 \sin \left (d x +c \right ) \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}+1\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-8 \sin \left (d x +c \right ) \arctan \left (\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}-1\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-4 \sin \left (d x +c \right ) \ln \left (-\frac {\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )+\sin \left (d x +c \right )+\cos \left (d x +c \right )-1}{\sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\, \sin \left (d x +c \right )-\sin \left (d x +c \right )-\cos \left (d x +c \right )+1}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+2 i \sin \left (d x +c \right ) \sqrt {2}+2 \cos \left (d x +c \right ) \sqrt {2}-2 \sqrt {2}\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}} \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sqrt {2}\, a^{2}}{2 d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \cos \left (d x +c \right )}\)	\(888\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*(I*ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-I*ln(
((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-2*I*arctan(((-1+co
s(d*x+c))/sin(d*x+c))^(1/2))*2^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+8*I*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)*sin(d*x+c)+8*I*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)*sin(d*x+c)+4*I*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)*sin(d*x+c)+ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+1)*2^(1/
2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)-ln(((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-1)*2^(1/2)*((-1+cos(d*x
+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+2*arctan(((-1+cos(d*x+c))/sin(d*x+c))^(1/2))*2^(1/2)*((-1+cos(d*x+c))/sin(d*
x+c))^(1/2)*sin(d*x+c)-8*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)-8*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)-4*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+co
s(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*
I*sin(d*x+c)*2^(1/2)+2*cos(d*x+c)*2^(1/2)-2*2^(1/2))*sin(d*x+c)*(cos(d*x+c)/sin(d*x+c))^(3/2)*(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)*2^(1/2)*a^2

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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. 570 vs. \(2 (137) = 274\).
time = 0.84, size = 570, normalized size = 3.18 \begin {gather*} -\frac {8 \, \sqrt {2} a^{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}} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {4 i \, a^{5}}{d^{2}}} d \log \left (-\frac {16 \, {\left (3 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} + \sqrt {2} \sqrt {\frac {4 i \, a^{5}}{d^{2}}} {\left (i \, d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, d 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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) + \sqrt {\frac {4 i \, a^{5}}{d^{2}}} d \log \left (-\frac {16 \, {\left (3 \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} + \sqrt {2} \sqrt {\frac {4 i \, a^{5}}{d^{2}}} {\left (-i \, d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, d 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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{a}\right ) - \sqrt {\frac {128 i \, a^{5}}{d^{2}}} d \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 ) + \sqrt {\frac {128 i \, a^{5}}{d^{2}}} d \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 )}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(8*sqrt(2)*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(4*I*a^5/d^2)*d*log(-16*(3*a^3*e^(2*I*d*x + 2*I*c) - a^3 + sqrt(2)*sqrt(4*I*a^5/d^2)
*(I*d*e^(3*I*d*x + 3*I*c) - I*d*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)))*e^(-2*I*d*x - 2*I*c)/a) + sqrt(4*I*a^5/d^2)*d*log(-16*(3*a^3*e^(2*I*d*x + 2*
I*c) - a^3 + sqrt(2)*sqrt(4*I*a^5/d^2)*(-I*d*e^(3*I*d*x + 3*I*c) + I*d*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)))*e^(-2*I*d*x - 2*I*c)/a) - sqrt(128*I*
a^5/d^2)*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)
+ sqrt(128*I*a^5/d^2)*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

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 8569 deep

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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)^(3/2)*(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)^(3/2), 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 )}^{3/2}\,{\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)^(3/2)*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

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

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