\(\int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx\) [223]

Optimal result

Integrand size = 28, antiderivative size = 162 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {1}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {25 \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}} \]

[Out]

(1/4+1/4*I)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/a^(3/2)/d+11/6/a/d/tan(d*x+c)^(1/
2)/(a+I*a*tan(d*x+c))^(1/2)-25/6*(a+I*a*tan(d*x+c))^(1/2)/a^2/d/tan(d*x+c)^(1/2)+1/3/d/tan(d*x+c)^(1/2)/(a+I*a
*tan(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.214, Rules used = {3640, 3677, 3679, 12, 3625, 211} \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac {25 \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {11}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {1}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]

[In]

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

[Out]

((1/4 + I/4)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/(a^(3/2)*d) + 1/(3*d*Sq
rt[Tan[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + 11/(6*a*d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) - (2
5*Sqrt[a + I*a*Tan[c + d*x]])/(6*a^2*d*Sqrt[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}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {\int \frac {\frac {7 a}{2}-2 i a \tan (c+d x)}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx}{3 a^2} \\ & = \frac {1}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11}{6 a 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 {25 a^2}{4}-\frac {11}{2} i a^2 \tan (c+d x)\right )}{\tan ^{\frac {3}{2}}(c+d x)} \, dx}{3 a^4} \\ & = \frac {1}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {25 \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {2 \int \frac {3 i a^3 \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{3 a^5} \\ & = \frac {1}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {25 \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}}+\frac {i \int \frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}} \, dx}{4 a^2} \\ & = \frac {1}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {25 \sqrt {a+i a \tan (c+d x)}}{6 a^2 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 )}{2 d} \\ & = \frac {\left (\frac {1}{4}+\frac {i}{4}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {1}{3 d \sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {11}{6 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {25 \sqrt {a+i a \tan (c+d x)}}{6 a^2 d \sqrt {\tan (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.16 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.83 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}+\frac {2 \sqrt {a+i a \tan (c+d x)} \left (12+39 i \tan (c+d x)-25 \tan ^2(c+d x)\right )}{(-i+\tan (c+d x))^2}}{12 a^2 d \sqrt {\tan (c+d x)}} \]

[In]

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

[Out]

(3*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*Sq
rt[a + I*a*Tan[c + d*x]]*(12 + (39*I)*Tan[c + d*x] - 25*Tan[c + d*x]^2))/(-I + Tan[c + d*x])^2)/(12*a^2*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. 504 vs. \(2 (128 ) = 256\).

Time = 1.01 (sec) , antiderivative size = 505, normalized size of antiderivative = 3.12

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (9 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{3}\left (d x +c \right )\right )-3 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )+100 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )-3 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+9 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )-204 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}-256 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+48 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{24 d \,a^{2} \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{3} \sqrt {-i a}}\)	\(505\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (9 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{3}\left (d x +c \right )\right )-3 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{4}\left (d x +c \right )\right )+100 \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{3}\left (d x +c \right )\right )-3 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+9 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \left (\tan ^{2}\left (d x +c \right )\right )-204 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}-256 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (\tan ^{2}\left (d x +c \right )\right )+48 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{24 d \,a^{2} \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{3} \sqrt {-i a}}\)	\(505\)

[In]

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

[Out]

1/24/d*(a*(1+I*tan(d*x+c)))^(1/2)/a^2*(9*I*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)^3-3*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)^4+100*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*t
an(d*x+c)))^(1/2)*tan(d*x+c)^3-3*I*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)+9*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)^2-204*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^
(1/2)*(-I*a)^(1/2)-256*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^2+48*I*(-I*a)^(1/2)*(a*
tan(d*x+c)*(1+I*tan(d*x+c)))^(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)^3/(
-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. 382 vs. \(2 (120) = 240\).

Time = 0.28 (sec) , antiderivative size = 382, normalized size of antiderivative = 2.36 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-38 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 25 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 14 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )} - 3 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {i}{2 \, a^{3} d^{2}}} \log \left (\frac {1}{2} i \, a^{2} d \sqrt {\frac {i}{2 \, a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) + 3 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )} \sqrt {\frac {i}{2 \, a^{3} d^{2}}} \log \left (-\frac {1}{2} i \, a^{2} d \sqrt {\frac {i}{2 \, a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )}{12 \, {\left (a^{2} d e^{\left (5 i \, d x + 5 i \, c\right )} - a^{2} d e^{\left (3 i \, d x + 3 i \, c\right )}\right )}} \]

[In]

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

[Out]

1/12*(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))*(
-38*I*e^(6*I*d*x + 6*I*c) - 25*I*e^(4*I*d*x + 4*I*c) + 14*I*e^(2*I*d*x + 2*I*c) + I) - 3*(a^2*d*e^(5*I*d*x + 5
*I*c) - a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/2*I/(a^3*d^2))*log(1/2*I*a^2*d*sqrt(1/2*I/(a^3*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)) + 3*(a^2*d*e^(5*I*d*x + 5*I*c) - a^2*d*e^(3*I*d*x + 3*I*c))*sqrt(1/2*I/(a^3*d^2))*l
og(-1/2*I*a^2*d*sqrt(1/2*I/(a^3*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^2*d*e^(5*I*d*x + 5*I*c) -
a^2*d*e^(3*I*d*x + 3*I*c))

Sympy [F]

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

[In]

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

[Out]

Integral(1/((I*a*(tan(c + d*x) - I))**(3/2)*tan(c + d*x)**(3/2)), x)

Maxima [F(-2)]

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

[In]

integrate(1/tan(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(3/2),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. 849 vs. \(2 (120) = 240\).

Time = 6.74 (sec) , antiderivative size = 849, normalized size of antiderivative = 5.24 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/4*(-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^3*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^4*d*tan(d*x + c)) + 2/3*sqrt(2)*(9*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)) - 9*(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) + 120*(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*a*sqrt
(abs(a)) + 120*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) - 208*I*a*sqrt(abs(a)) + 208*abs(a)^(3/2))/(((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 - 4*I)^3*a^3*d)

Mupad [F(-1)]

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

[In]

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

[Out]

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