\(\int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx\) [205]

Optimal result

Integrand size = 28, antiderivative size = 139 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 (-1)^{3/4} a^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(4+4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \]

[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))/d+(4+4*I)*a^(5/2)*ar
ctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d-2*a^2*(a+I*a*tan(d*x+c))^(1/2)/d/tan(d*x+c)^(
1/2)

Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3634, 3682, 3625, 211, 3680, 65, 223, 209} \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {2 (-1)^{3/4} a^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(4+4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \]

[In]

Int[(a + I*a*Tan[c + d*x])^(5/2)/Tan[c + d*x]^(3/2),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]]])/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]]])/d - (2*a^2*Sqrt[a + I*a*T
an[c + d*x]])/(d*Sqrt[Tan[c + d*x]])

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-2 \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 {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-(i a) \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\right ) \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx \\ & = -\frac {2 a^2 \sqrt {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {\left (i a^3\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\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} \text {arctanh}\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 {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {\left (2 i a^3\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} \text {arctanh}\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 {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}}-\frac {\left (2 i a^3\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} \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} \text {arctanh}\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 {a+i a \tan (c+d x)}}{d \sqrt {\tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.17 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.91 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\frac {a^2 \left (3 \sqrt [4]{-1} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) (1+i \tan (c+d x)) \sqrt {\tan (c+d x)}-5 i \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {i a \tan (c+d x)} (-i+\tan (c+d x))+2 \sqrt {1+i \tan (c+d x)} \left (-a-i a \tan (c+d x)+2 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}\right )\right )}{d \sqrt {1+i \tan (c+d x)} \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \]

[In]

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

[Out]

(a^2*(3*(-1)^(1/4)*a*ArcSinh[(-1)^(1/4)*Sqrt[Tan[c + d*x]]]*(1 + I*Tan[c + d*x])*Sqrt[Tan[c + d*x]] - (5*I)*Sq
rt[a]*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqrt[a]]*Sqrt[I*a*Tan[c + d*x]]*(-I + Tan[c + d*x]) + 2*Sqrt[1 + I*Tan[c
+ d*x]]*(-a - I*a*Tan[c + d*x] + 2*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]
]*Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])))/(d*Sqrt[1 + I*Tan[c + d*x]]*Sqrt[Tan[c + d*x]]*Sqrt[a +
I*a*Tan[c + d*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 385 vs. \(2 (112 ) = 224\).

Time = 1.03 (sec) , antiderivative size = 386, normalized size of antiderivative = 2.78

method	result	size
derivativedivides	\(\frac {\left (-i \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, a \tan \left (d x +c \right )-\ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, a \tan \left (d x +c \right )-i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\, \tan \left (d x +c \right )-4 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )-2 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2}}{d \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(386\)
default	\(\frac {\left (-i \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, a \tan \left (d x +c \right )-\ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, \sqrt {2}\, a \tan \left (d x +c \right )-i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\, \tan \left (d x +c \right )-4 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \tan \left (d x +c \right )-2 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{2}}{d \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(386\)

[In]

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

[Out]

1/d*(-I*ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d*x+c))/(tan(d*x+c)+I))*(
I*a)^(1/2)*2^(1/2)*a*tan(d*x+c)-ln((2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-I*a+3*a*tan(d
*x+c))/(tan(d*x+c)+I))*(I*a)^(1/2)*2^(1/2)*a*tan(d*x+c)-I*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(a*tan(d*x+c
)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d*x+c)-4*(-I*a)^(1/2)*ln(1/2*(2*I*a*tan(d*x+c)+2*(
a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*tan(d*x+c)-2*(-I*a)^(1/2)*(I*a)^(1/2)*(a*ta
n(d*x+c)*(1+I*tan(d*x+c)))^(1/2))*(a*(1+I*tan(d*x+c)))^(1/2)*a^2/tan(d*x+c)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c
)))^(1/2)/(I*a)^(1/2)/(-I*a)^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 626 vs. \(2 (105) = 210\).

Time = 0.26 (sec) , antiderivative size = 626, normalized size of antiderivative = 4.50 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=-\frac {4 \, \sqrt {2} {\left (i \, a^{2} e^{\left (3 i \, d x + 3 i \, c\right )} + i \, 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}} + \sqrt {\frac {32 i \, a^{5}}{d^{2}}} {\left (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}} + i \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) - \sqrt {\frac {32 i \, a^{5}}{d^{2}}} {\left (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}} - i \, \sqrt {\frac {32 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a^{2}}\right ) - \sqrt {\frac {4 i \, a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\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}} + i \, \sqrt {\frac {4 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right ) + \sqrt {\frac {4 i \, a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \log \left (\frac {{\left (\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}} - i \, \sqrt {\frac {4 i \, a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2}}\right )}{2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )}} \]

[In]

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

[Out]

-1/2*(4*sqrt(2)*(I*a^2*e^(3*I*d*x + 3*I*c) + I*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)) + sqrt(32*I*a^5/d^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)) + I*sqrt(32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2) - sqrt(32*I*a^5/d^
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)) - I*sqrt(32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e
^(-I*d*x - I*c)/a^2) - sqrt(4*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((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)) + I*sqrt(4*
I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2) + sqrt(4*I*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*log((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)) - I*sqrt(4*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2))/(d*e^(2*I*d*x + 2*I*c) - d
)

Sympy [F]

\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}}}{\tan ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx \]

[In]

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

[Out]

Integral((I*a*(tan(c + d*x) - I))**(5/2)/tan(c + d*x)**(3/2), x)

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\tan \left (d x + c\right )^{\frac {3}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Non regular value [0] was discarded and replaced randomly by 0=[89]Warning, replacing 89 by -23, a substitu
tion variab

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\tan ^{\frac {3}{2}}(c+d x)} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\mathrm {tan}\left (c+d\,x\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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