\(\int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\) [231]

Optimal result

Integrand size = 28, antiderivative size = 199 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {1}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {317 \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}} \]

[Out]

(1/8+1/8*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(5/2)/d+151/60/a^2/d/tan(d*x+c)
^(1/2)/(a+I*a*tan(d*x+c))^(1/2)-317/60*(a+I*a*tan(d*x+c))^(1/2)/a^3/d/tan(d*x+c)^(1/2)+1/5/d/tan(d*x+c)^(1/2)/
(a+I*a*tan(d*x+c))^(5/2)+17/30/a/d/tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.68 (sec) , antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3640, 3677, 3679, 12, 3625, 211} \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {317 \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {151}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {17}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {1}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}} \]

[In]

Int[1/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((1/8 + I/8)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(5/2)*d) + 1/(5*d*Sq
rt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + 17/(30*a*d*Sqrt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) +
151/(60*a^2*d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (317*Sqrt[a + I*a*Tan[c + d*x]])/(60*a^3*d*Sqr
t[Tan[c + d*x]])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3640

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[a*(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*Simp[b*c*m - a*d*(2*m + n + 1) + b*d*(m + n + 1)*Tan
[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2
+ d^2, 0] && LtQ[m, 0] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3677

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 3679

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]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {\int \frac {\frac {11 a}{2}-3 i a \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2} \\ & = \frac {1}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\frac {83 a^2}{4}-17 i a^2 \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4} \\ & = \frac {1}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\int \frac {\sqrt {a+i a \tan (c+d x)} \left (\frac {317 a^3}{8}-\frac {151}{4} i a^3 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{15 a^6} \\ & = \frac {1}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {317 \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {2 \int \frac {15 i a^4 \sqrt {a+i a \tan (c+d x)}}{16 \sqrt {\tan (c+d x)}} \, dx}{15 a^7} \\ & = \frac {1}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {317 \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {i \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{8 a^3} \\ & = \frac {1}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {317 \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}}+\frac {\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 )}{4 a d} \\ & = \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {1}{5 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {17}{30 a d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {151}{60 a^2 d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {317 \sqrt {a+i a \tan (c+d x)}}{60 a^3 d \sqrt {\tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.77 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.74 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {15 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}+\frac {2 \sqrt {a+i a \tan (c+d x)} \left (-120 i+615 \tan (c+d x)+800 i \tan ^2(c+d x)-317 \tan ^3(c+d x)\right )}{(-i+\tan (c+d x))^3}}{120 a^3 d \sqrt {\tan (c+d x)}} \]

[In]

Integrate[1/(Tan[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

(15*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]] + (2*S
qrt[a + I*a*Tan[c + d*x]]*(-120*I + 615*Tan[c + d*x] + (800*I)*Tan[c + d*x]^2 - 317*Tan[c + d*x]^3))/(-I + Tan
[c + d*x])^3)/(120*a^3*d*Sqrt[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. 619 vs. \(2 (158 ) = 316\).

Time = 0.99 (sec) , antiderivative size = 620, normalized size of antiderivative = 3.12

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (60 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 {2}\, a \left (\tan ^{4}\left (d x +c \right )\right )-15 \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 {2}\, a \left (\tan ^{5}\left (d x +c \right )\right )+1268 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (\tan ^{4}\left (d x +c \right )\right )-60 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 {2}\, a \left (\tan ^{2}\left (d x +c \right )\right )+90 \sqrt {2}\, \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 ) a \left (\tan ^{3}\left (d x +c \right )\right )-5660 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )-4468 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (\tan ^{3}\left (d x +c \right )\right )-15 \sqrt {2}\, \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 ) a \tan \left (d x +c \right )+2940 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+480 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{240 d \,a^{3} \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 )}\, \left (-\tan \left (d x +c \right )+i\right )^{4} \sqrt {-i a}}\)	\(620\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (60 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 {2}\, a \left (\tan ^{4}\left (d x +c \right )\right )-15 \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 {2}\, a \left (\tan ^{5}\left (d x +c \right )\right )+1268 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (\tan ^{4}\left (d x +c \right )\right )-60 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 {2}\, a \left (\tan ^{2}\left (d x +c \right )\right )+90 \sqrt {2}\, \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 ) a \left (\tan ^{3}\left (d x +c \right )\right )-5660 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )-4468 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (\tan ^{3}\left (d x +c \right )\right )-15 \sqrt {2}\, \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 ) a \tan \left (d x +c \right )+2940 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+480 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{240 d \,a^{3} \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 )}\, \left (-\tan \left (d x +c \right )+i\right )^{4} \sqrt {-i a}}\)	\(620\)

[In]

int(1/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/240/d*(a*(1+I*tan(d*x+c)))^(1/2)/a^3*(60*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))*2^(1/2)*a*tan(d*x+c)^4-15*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))*2^(1/2)*a*tan(d*x+c)^5+1268*(a*tan(d*x+c)*(1+I*tan(d
*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)^4-60*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))*2^(1/2)*a*tan(d*x+c)^2+90*2^(1/2)*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))*a*tan(d*x+c)^3-5660*(-I*a)^(1/2)*(a*tan(d*x+c
)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2-4468*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^3-
15*2^(1/2)*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))*a*tan(d*x+c)+2940*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)+480*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)*(-I*a)^(1/2))/tan(d*x+c)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-tan(d*x+c)+I)^4/(-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. 393 vs. \(2 (149) = 298\).

Time = 0.27 (sec) , antiderivative size = 393, normalized size of antiderivative = 1.97 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-463 i \, e^{\left (8 i \, d x + 8 i \, c\right )} - 269 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 220 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 29 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} - 30 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {i}{8 \, a^{5} d^{2}}} \log \left (i \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) + 30 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )} \sqrt {\frac {i}{8 \, a^{5} d^{2}}} \log \left (-i \, a^{3} d \sqrt {\frac {i}{8 \, a^{5} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )}{120 \, {\left (a^{3} d e^{\left (7 i \, d x + 7 i \, c\right )} - a^{3} d e^{\left (5 i \, d x + 5 i \, c\right )}\right )}} \]

[In]

integrate(1/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/120*(sqrt(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))*
(-463*I*e^(8*I*d*x + 8*I*c) - 269*I*e^(6*I*d*x + 6*I*c) + 220*I*e^(4*I*d*x + 4*I*c) + 29*I*e^(2*I*d*x + 2*I*c)
+ 3*I) - 30*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))*sqrt(1/8*I/(a^5*d^2))*log(I*a^3*d*sqrt(1/
8*I/(a^5*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(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^(2*I*d*x + 2*I*c) + 1)) + 30*(a^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x +
5*I*c))*sqrt(1/8*I/(a^5*d^2))*log(-I*a^3*d*sqrt(1/8*I/(a^5*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(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^(2*I*d*x + 2*I*c) + 1)))/(a
^3*d*e^(7*I*d*x + 7*I*c) - a^3*d*e^(5*I*d*x + 5*I*c))

Sympy [F(-1)]

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

[In]

integrate(1/tan(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

Maxima [F(-2)]

Exception generated. \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Exception raised: RuntimeError} \]

[In]

integrate(1/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

Giac [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1319 vs. \(2 (149) = 298\).

Time = 19.08 (sec) , antiderivative size = 1319, normalized size of antiderivative = 6.63 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

integrate(1/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/8*(-I*a*sqrt(abs(a)) + abs(a)^(3/2))*arctan(1/16*sqrt(2)*((sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a +
1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2
- 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^2 + 12*I))/(a^4*d) - sqrt(-2*(I*a*tan(d*x + c) + a)*
a + 2*a^2)*sqrt(I*a*tan(d*x + c) + a)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*
a + a^2)/a^2) + 1)*abs(a)/(a^5*d*tan(d*x + c)) + 1/15*sqrt(2)*(105*I*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*a
bs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c
) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^8*a*sqrt(abs(a)) - 105*(sqrt(2)*sqrt(I*a*ta
n(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/s
qrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^8*abs(a)^(3/2) + 2400*
(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*
a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^6*
a*sqrt(abs(a)) + 2400*I*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*
tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/
a^2) + 1)*abs(a)/a^2)^6*abs(a)^(3/2) - 21920*I*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3
/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(
d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^4*a*sqrt(abs(a)) + 21920*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*
abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x +
c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^4*abs(a)^(3/2) - 56320*(sqrt(2)*sqrt(I*a*t
an(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/
sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^2*a*sqrt(abs(a)) - 56
320*I*(sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*abs(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*
a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a
^2)^2*abs(a)^(3/2) + 50432*I*a*sqrt(abs(a)) - 50432*abs(a)^(3/2))/(((sqrt(2)*sqrt(I*a*tan(d*x + c) + a)*(-I*ab
s(a)/a + 1)*abs(a)^(3/2)/a^2 - sqrt(-2*(I*a*tan(d*x + c) + a)*a + 2*a^2)*(tan(d*x + c)/sqrt(((I*a*tan(d*x + c)
+ a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2)/a^2) + 1)*abs(a)/a^2)^2 - 4*I)^5*a^4*d)

Mupad [F(-1)]

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

[In]

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

[Out]

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