\(\int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx\) [222]

Optimal result

Integrand size = 28, antiderivative size = 125 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{3/2} d}+\frac {\sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}+\frac {7 \sqrt {\tan (c+d x)}}{6 a d \sqrt {a+i a \tan (c+d x)}} \] Output:

(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+1/3*tan(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^(3/2)+7/6*tan(d*x+c 
)^(1/2)/a/d/(a+I*a*tan(d*x+c))^(1/2)
 

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.10 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {-\frac {3 i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}}{a^2}+\frac {4 \tan (c+d x)}{(a+i a \tan (c+d x))^{3/2}}+\frac {14 \tan (c+d x)}{a \sqrt {a+i a \tan (c+d x)}}}{12 d \sqrt {\tan (c+d x)}} \] Input:

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

Output:

(((-3*I)*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]])/a^2 + (4*Tan[c + d*x])/(a + I*a*Tan[c 
+ d*x])^(3/2) + (14*Tan[c + d*x])/(a*Sqrt[a + I*a*Tan[c + d*x]]))/(12*d*Sq 
rt[Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4042, 27, 3042, 4079, 27, 3042, 4027, 218}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}}dx\)

\(\Big \downarrow \) 4042

\(\displaystyle \frac {\int \frac {5 a-2 i a \tan (c+d x)}{2 \sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{3 a^2}+\frac {\sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {5 a-2 i a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {\sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {5 a-2 i a \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}dx}{6 a^2}+\frac {\sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4079

\(\displaystyle \frac {\frac {\int \frac {3 a^2 \sqrt {i \tan (c+d x) a+a}}{2 \sqrt {\tan (c+d x)}}dx}{a^2}+\frac {7 a \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {3}{2} \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {7 a \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {3}{2} \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx+\frac {7 a \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {\frac {7 a \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i a^2 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{6 a^2}+\frac {\sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {\frac {\left (\frac {3}{2}-\frac {3 i}{2}\right ) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {7 a \sqrt {\tan (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}}{6 a^2}+\frac {\sqrt {\tan (c+d x)}}{3 d (a+i a \tan (c+d x))^{3/2}}\)

Input:

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

Output:

Sqrt[Tan[c + d*x]]/(3*d*(a + I*a*Tan[c + d*x])^(3/2)) + (((3/2 - (3*I)/2)* 
Sqrt[a]*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + 
d*x]]])/d + (7*a*Sqrt[Tan[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]]))/(6*a^ 
2)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 218

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

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4027

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-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] && N 
eQ[c^2 + d^2, 0]
 

rule 4042

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(a + b*Tan[e + f*x])^m*((c + d*Tan[e 
 + f*x])^(n + 1)/(2*f*m*(b*c - a*d))), x] + Simp[1/(2*a*m*(b*c - a*d))   In 
t[(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 4079

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + 
(f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim 
p[(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] + Simp[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[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] /; Free 
Q[{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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 463 vs. \(2 (98 ) = 196\).

Time = 1.62 (sec) , antiderivative size = 464, normalized size of antiderivative = 3.71

method	result	size
derivativedivides	\(\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (3 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}-9 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+28 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}+9 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-36 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}-3 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +64 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{24 d \,a^{2} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{3} \sqrt {-i a}}\)	\(464\)
default	\(\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (3 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}-9 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+28 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}+9 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-36 i \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}-3 \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a +64 \tan \left (d x +c \right ) \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{24 d \,a^{2} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \left (-\tan \left (d x +c \right )+i\right )^{3} \sqrt {-i a}}\)	\(464\)

Input:

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

Output:

1/24/d*tan(d*x+c)^(1/2)*(a*(1+I*tan(d*x+c)))^(1/2)*(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)^3-9*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)+28*I*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan( 
d*x+c)^2+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)^2-36*I*(a*tan 
(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)-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+64*tan(d*x+c)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/ 
2))/a^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. 320 vs. \(2 (91) = 182\).

Time = 0.10 (sec) , antiderivative size = 320, normalized size of antiderivative = 2.56 \[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {{\left (3 \, a^{2} d \sqrt {-\frac {i}{2 \, a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {1}{2} \, a^{2} d \sqrt {-\frac {i}{2 \, a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - 3 \, a^{2} d \sqrt {-\frac {i}{2 \, a^{3} d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (-\frac {1}{2} \, a^{2} d \sqrt {-\frac {i}{2 \, a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (8 \, e^{\left (4 i \, d x + 4 i \, c\right )} + 9 \, e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{12 \, a^{2} d} \] Input:

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

Output:

1/12*(3*a^2*d*sqrt(-1/2*I/(a^3*d^2))*e^(3*I*d*x + 3*I*c)*log(1/2*a^2*d*sqr 
t(-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*sqrt(-1/2*I/(a^3*d^2))*e^(3*I*d*x + 3*I* 
c)*log(-1/2*a^2*d*sqrt(-1/2*I/(a^3*d^2))*e^(I*d*x + I*c) + 1/4*sqrt(2)*sqr 
t(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)) + 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) 
)*(8*e^(4*I*d*x + 4*I*c) + 9*e^(2*I*d*x + 2*I*c) + 1))*e^(-3*I*d*x - 3*I*c 
)/(a^2*d)
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

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

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeDone
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx=\frac {4 \sqrt {a}\, \left (\left (\int -\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{4 \tan \left (d x +c \right )^{4}-4 \tan \left (d x +c \right )^{3} i +4 \tan \left (d x +c \right )^{2}-4 \tan \left (d x +c \right ) i}d x \right ) i +\int -\frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{4 \tan \left (d x +c \right )^{3}-4 \tan \left (d x +c \right )^{2} i +4 \tan \left (d x +c \right )-4 i}d x \right )}{a^{2}} \] Input:

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

Output:

(4*sqrt(a)*(int(( - sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(4*tan(c 
+ d*x)**4 - 4*tan(c + d*x)**3*i + 4*tan(c + d*x)**2 - 4*tan(c + d*x)*i),x) 
*i + int(( - sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(4*tan(c + d*x)* 
*3 - 4*tan(c + d*x)**2*i + 4*tan(c + d*x) - 4*i),x)))/a**2