\(\int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx\) [191]

Optimal result

Integrand size = 28, antiderivative size = 120 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {(1-i) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {2 \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)}-\frac {2 i \sqrt {a+i a \tan (c+d x)}}{3 d \sqrt {\tan (c+d x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.94 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {\frac {3 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) (i a \tan (c+d x))^{3/2}}{a}-2 i (-i+\tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{3 d \tan ^{\frac {3}{2}}(c+d x)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.66 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.06, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.321, Rules used = {3042, 4044, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4044

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

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

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 4044

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[d*(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*(c^2 + d^2)*(n + 
1))   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1)*Simp[b*d*m - 
a*c*(n + 1) + a*d*(m + n + 1)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d 
, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 
0] && LtQ[n, -1] && (IntegerQ[n] || 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. 195 vs. \(2 (96 ) = 192\).

Time = 1.69 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.63

method	result	size
derivativedivides	\(\frac {\left (3 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-4 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{6 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}}\)	\(196\)
default	\(\frac {\left (3 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) a \tan \left (d x +c \right )^{2}-4 i \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \tan \left (d x +c \right )-4 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}{6 d \tan \left (d x +c \right )^{\frac {3}{2}} \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {-i a}}\)	\(196\)

Input:

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

Output:

1/6/d/tan(d*x+c)^(3/2)*(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 
)^2-4*I*(-I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)-4*(a 
*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(-I*a)^(1/2))*(a*(1+I*tan(d*x+c)))^(1/ 
2)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(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. 355 vs. \(2 (88) = 176\).

Time = 0.09 (sec) , antiderivative size = 355, normalized size of antiderivative = 2.96 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\frac {8 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (5 i \, d x + 5 i \, c\right )} + e^{\left (3 i \, d x + 3 i \, c\right )}\right )} - 3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} + d \sqrt {-\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + 3 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {2 i \, a}{d^{2}}} \log \left ({\left (\sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} - d \sqrt {-\frac {2 i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right )}{6 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:

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

Output:

1/6*(8*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^(5*I*d*x + 5*I*c) + e^(3*I*d*x + 3* 
I*c)) - 3*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-2*I* 
a/d^2)*log((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) + d*sqr 
t(-2*I*a/d^2)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + 3*(d*e^(4*I*d*x + 4*I*c 
) - 2*d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-2*I*a/d^2)*log((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) - d*sqrt(-2*I*a/d^2)*e^(I*d*x + I*c))* 
e^(-I*d*x - I*c)))/(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 974 vs. \(2 (88) = 176\).

Time = 0.26 (sec) , antiderivative size = 974, normalized size of antiderivative = 8.12 \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\text {Too large to display} \] Input:

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

Output:

1/6*(2*sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 
 1)*(((3*I + 3)*cos(3*d*x + 3*c) + (I + 1)*cos(d*x + c) + (3*I - 3)*sin(3* 
d*x + 3*c) + (I - 1)*sin(d*x + c))*cos(3/2*arctan2(sin(2*d*x + 2*c), -cos( 
2*d*x + 2*c) + 1)) + ((3*I - 3)*cos(3*d*x + 3*c) + (I - 1)*cos(d*x + c) - 
(3*I + 3)*sin(3*d*x + 3*c) - (I + 1)*sin(d*x + c))*sin(3/2*arctan2(sin(2*d 
*x + 2*c), -cos(2*d*x + 2*c) + 1)))*sqrt(a) + 3*(2*(-(I + 1)*cos(2*d*x + 2 
*c)^2 - (I + 1)*sin(2*d*x + 2*c)^2 + (2*I + 2)*cos(2*d*x + 2*c) - I - 1)*a 
rctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^ 
(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - cos(d*x 
+ c), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^( 
1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)) - sin(d*x + 
 c)) + (-(I - 1)*cos(2*d*x + 2*c)^2 - (I - 1)*sin(2*d*x + 2*c)^2 + (2*I - 
2)*cos(2*d*x + 2*c) - I + 1)*log(cos(d*x + c)^2 + sin(d*x + c)^2 + sqrt(co 
s(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)*(cos(1/2*a 
rctan2(sin(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2 + sin(1/2*arctan2(sin(2 
*d*x + 2*c), -cos(2*d*x + 2*c) + 1))^2) - 2*(cos(2*d*x + 2*c)^2 + sin(2*d* 
x + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*(cos(1/2*arctan2(sin(2*d*x + 2* 
c), -cos(2*d*x + 2*c) + 1))*sin(d*x + c) + cos(d*x + c)*sin(1/2*arctan2(si 
n(2*d*x + 2*c), -cos(2*d*x + 2*c) + 1)))))*(cos(2*d*x + 2*c)^2 + sin(2*d*x 
 + 2*c)^2 - 2*cos(2*d*x + 2*c) + 1)^(1/4)*sqrt(a) + 2*(((-(I + 1)*cos(d...
 

Giac [F(-2)]

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

integrate((a+I*a*tan(d*x+c))^(1/2)/tan(d*x+c)^(5/2),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. \[ \int \frac {\sqrt {a+i a \tan (c+d x)}}{\tan ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{{\mathrm {tan}\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

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