\(\int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\tan (c+d x)}} \, dx\) [204]

Optimal result

Integrand size = 28, antiderivative size = 139 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\tan (c+d x)}} \, dx=\frac {5 \sqrt [4]{-1} a^{5/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(4-4 i) a^{5/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \] Output:

5*(-1)^(1/4)*a^(5/2)*arctan((-1)^(3/4)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan 
(d*x+c))^(1/2))/d+(4-4*I)*a^(5/2)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/( 
a+I*a*tan(d*x+c))^(1/2))/d-a^2*tan(d*x+c)^(1/2)*(a+I*a*tan(d*x+c))^(1/2)/d
 

Mathematica [A] (verified)

Time = 0.89 (sec) , antiderivative size = 216, normalized size of antiderivative = 1.55 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\tan (c+d x)}} \, dx=\frac {a^3 \sqrt {\tan (c+d x)} \left (-5 i \sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) (-i+\tan (c+d x))+\sqrt {1+i \tan (c+d x)} \left ((-1-i \tan (c+d x)) \sqrt {i a \tan (c+d x)}+4 \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {a+i a \tan (c+d x)}\right )\right )}{d \sqrt {1+i \tan (c+d x)} \sqrt {i a \tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

(a^3*Sqrt[Tan[c + d*x]]*((-5*I)*Sqrt[a]*ArcSinh[Sqrt[I*a*Tan[c + d*x]]/Sqr 
t[a]]*(-I + Tan[c + d*x]) + Sqrt[1 + I*Tan[c + d*x]]*((-1 - I*Tan[c + d*x] 
)*Sqrt[I*a*Tan[c + d*x]] + 4*Sqrt[2]*ArcTanh[(Sqrt[2]*Sqrt[I*a*Tan[c + d*x 
]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[a + I*a*Tan[c + d*x]])))/(d*Sqrt[1 + 
I*Tan[c + d*x]]*Sqrt[I*a*Tan[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])
 

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.04, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 4039, 27, 3042, 4084, 3042, 4027, 218, 4082, 65, 216}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4039

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{2} a \int \frac {\sqrt {i \tan (c+d x) a+a} (5 i \tan (c+d x) a+3 a)}{\sqrt {\tan (c+d x)}}dx-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} a \int \frac {\sqrt {i \tan (c+d x) a+a} (5 i \tan (c+d x) a+3 a)}{\sqrt {\tan (c+d x)}}dx-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 4084

\(\displaystyle \frac {1}{2} a \left (8 a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-5 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx\right )-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{2} a \left (8 a \int \frac {\sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx-5 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx\right )-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {1}{2} a \left (-\frac {16 i a^3 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}-5 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx\right )-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 218

\(\displaystyle \frac {1}{2} a \left (\frac {(8-8 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-5 \int \frac {(a-i a \tan (c+d x)) \sqrt {i \tan (c+d x) a+a}}{\sqrt {\tan (c+d x)}}dx\right )-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 4082

\(\displaystyle \frac {1}{2} a \left (\frac {(8-8 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {5 a^2 \int \frac {1}{\sqrt {\tan (c+d x)} \sqrt {i \tan (c+d x) a+a}}d\tan (c+d x)}{d}\right )-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 65

\(\displaystyle \frac {1}{2} a \left (\frac {(8-8 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {10 a^2 \int \frac {1}{1-\frac {i a \tan (c+d x)}{i \tan (c+d x) a+a}}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}\right )-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {1}{2} a \left (\frac {10 \sqrt [4]{-1} a^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(8-8 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}\right )-\frac {a^2 \sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}\)

Input:

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

Output:

(a*((10*(-1)^(1/4)*a^(3/2)*ArcTan[((-1)^(3/4)*Sqrt[a]*Sqrt[Tan[c + d*x]])/ 
Sqrt[a + I*a*Tan[c + d*x]]])/d + ((8 - 8*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt[ 
a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d))/2 - (a^2*Sqrt[Tan[ 
c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/d
 

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 65

Int[1/(Sqrt[(b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[2   Sub 
st[Int[1/(b - d*x^2), x], x, Sqrt[b*x]/Sqrt[c + d*x]], x] /; FreeQ[{b, c, d 
}, x] &&  !GtQ[c, 0]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

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 4039

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*(a + b*Tan[e + f*x])^(m - 2)*((c + 
 d*Tan[e + f*x])^(n + 1)/(d*f*(m + n - 1))), x] + Simp[a/(d*(m + n - 1)) 
Int[(a + b*Tan[e + f*x])^(m - 2)*(c + d*Tan[e + f*x])^n*Simp[b*c*(m - 2) + 
a*d*(m + 2*n) + (a*c*(m - 2) + b*d*(3*m + 2*n - 4))*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] && IntegerQ[2*m] && GtQ[m, 1] && NeQ[m + n - 1, 0] 
 && (IntegerQ[m] || IntegersQ[2*m, 2*n])
 

rule 4082

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[b*(B/f)   Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^n, x], x, Tan[e + f*x]], 
x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[ 
a^2 + b^2, 0] && EqQ[A*b + a*B, 0]
 

rule 4084

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*b + a*B)/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n, x], x] 
 - Simp[B/b   Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(a - b*Tan[ 
e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - 
a*d, 0] && EqQ[a^2 + b^2, 0] && NeQ[A*b + a*B, 0]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 362 vs. \(2 (112 ) = 224\).

Time = 1.72 (sec) , antiderivative size = 363, normalized size of antiderivative = 2.61

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\tan \left (d x +c \right )}\, a^{2} \left (-2 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-5 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}+2 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +8 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -2 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(363\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\tan \left (d x +c \right )}\, a^{2} \left (-2 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-5 \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}+2 i \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +8 i \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a -2 \sqrt {2}\, \ln \left (\frac {2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}-i a +3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a \right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(363\)

Input:

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

Output:

1/2/d*(a*(1+I*tan(d*x+c)))^(1/2)*tan(d*x+c)^(1/2)*a^2*(-2*(-I*a)^(1/2)*(I* 
a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)-5*ln(1/2*(2*I*a*tan(d*x+c)+ 
2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I*a)^(1/2))*a*(-I* 
a)^(1/2)+2*I*2^(1/2)*(I*a)^(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+8*I*ln(1/2*( 
2*I*a*tan(d*x+c)+2*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)*(I*a)^(1/2)+a)/(I 
*a)^(1/2))*(-I*a)^(1/2)*a-2*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))*(I*a)^(1/2) 
*a)/(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)/(I*a)^(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. 542 vs. \(2 (105) = 210\).

Time = 0.10 (sec) , antiderivative size = 542, normalized size of antiderivative = 3.90 \[ \int \frac {(a+i a \tan (c+d x))^{5/2}}{\sqrt {\tan (c+d x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

-1/2*(2*sqrt(2)*a^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^(I*d*x + I*c) + sqrt(-25*I*a^5/ 
d^2)*d*log(1/5*(5*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^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)) + 2*sqrt(-25*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/a^2) - sq 
rt(-25*I*a^5/d^2)*d*log(1/5*(5*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^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)) - 2*sqrt(-25*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I 
*c)/a^2) - sqrt(-32*I*a^5/d^2)*d*log(1/4*(4*sqrt(2)*(a^2*e^(2*I*d*x + 2*I* 
c) + a^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)) + sqrt(-32*I*a^5/d^2)*d*e^(I*d*x + I*c))*e^ 
(-I*d*x - I*c)/a^2) + sqrt(-32*I*a^5/d^2)*d*log(1/4*(4*sqrt(2)*(a^2*e^(2*I 
*d*x + 2*I*c) + a^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)) - sqrt(-32*I*a^5/d^2)*d*e^(I*d*x 
 + I*c))*e^(-I*d*x - I*c)/a^2))/d
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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