\(\int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx\) [215]

Optimal result

Integrand size = 28, antiderivative size = 120 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d}+\frac {1}{d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {3 \sqrt {a+i a \tan (c+d x)}}{a d \sqrt {\tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {-4 a-6 i a \tan (c+d x)+\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{2 a d \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

(-4*a - (6*I)*a*Tan[c + d*x] + 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]])/(2*a*d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.02, 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, 4081, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

1/(d*Sqrt[Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]]) + (((1 + I)*a^(3/2)*Ar 
cTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d 
- (6*a*Sqrt[a + I*a*Tan[c + d*x]])/(d*Sqrt[Tan[c + d*x]]))/(2*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 4081

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*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] - Simp[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] /; Fr 
eeQ[{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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 394 vs. \(2 (97 ) = 194\).

Time = 1.78 (sec) , antiderivative size = 395, normalized size of antiderivative = 3.29

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (2 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-\sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}-20 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+\sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+12 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}-8 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{4 d a \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}}\)	\(395\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \left (2 i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-\sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{3}-20 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )+\sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )+12 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \tan \left (d x +c \right )^{2}-8 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right )}{4 d a \sqrt {\tan \left (d x +c \right )}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\, \left (-\tan \left (d x +c \right )+i\right )^{2}}\)	\(395\)

Input:

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

Output:

-1/4/d*(a*(1+I*tan(d*x+c)))^(1/2)*(2*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)^2-2^(1/2)*ln(-(-2*2^(1/2)*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*t 
an(d*x+c)))^(1/2)+I*a-3*a*tan(d*x+c))/(tan(d*x+c)+I))*a*tan(d*x+c)^3-20*I* 
(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2)*tan(d*x+c)+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)+12*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^( 
1/2)*(-I*a)^(1/2)*tan(d*x+c)^2-8*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I 
*a)^(1/2))/a/tan(d*x+c)^(1/2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(-I*a) 
^(1/2)/(-tan(d*x+c)+I)^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. 355 vs. \(2 (90) = 180\).

Time = 0.11 (sec) , antiderivative size = 355, normalized size of antiderivative = 2.96 \[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=-\frac {2 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (5 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 4 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )} + {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {2 i}{a d^{2}}} \log \left (\frac {1}{4} i \, a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right ) - {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {2 i}{a d^{2}}} \log \left (-\frac {1}{4} i \, a d \sqrt {\frac {2 i}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} + \frac {1}{4} \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )}\right )}{4 \, {\left (a d e^{\left (3 i \, d x + 3 i \, c\right )} - a d e^{\left (i \, d x + i \, c\right )}\right )}} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \, dx=\frac {2 \sqrt {a}\, \left (-8 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right )^{2}-4 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}\, \tan \left (d x +c \right ) i -5 \sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}+2 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{3}+\tan \left (d x +c \right )}d x \right ) \tan \left (d x +c \right )^{3} d i +2 \left (\int \frac {\sqrt {\tan \left (d x +c \right )}\, \sqrt {\tan \left (d x +c \right ) i +1}}{\tan \left (d x +c \right )^{3}+\tan \left (d x +c \right )}d x \right ) \tan \left (d x +c \right ) d i \right )}{5 \tan \left (d x +c \right ) a d \left (\tan \left (d x +c \right )^{2}+1\right )} \] Input:

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

Output:

(2*sqrt(a)*( - 8*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)* 
*2 - 4*sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x)*i - 5*sqrt 
(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1) + 2*int((sqrt(tan(c + d*x))*sqrt(t 
an(c + d*x)*i + 1))/(tan(c + d*x)**3 + tan(c + d*x)),x)*tan(c + d*x)**3*d* 
i + 2*int((sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1))/(tan(c + d*x)**3 + 
 tan(c + d*x)),x)*tan(c + d*x)*d*i))/(5*tan(c + d*x)*a*d*(tan(c + d*x)**2 
+ 1))