\(\int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [187]

Optimal result

Integrand size = 28, antiderivative size = 135 \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {\sqrt [4]{-1} \sqrt {a} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}-\frac {(1-i) \sqrt {a} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.21 \[ \int \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {\frac {i \sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {i a \tan (c+d x)}}{\sqrt {\tan (c+d x)}}-\frac {(-1)^{3/4} a \text {arcsinh}\left (\sqrt [4]{-1} \sqrt {\tan (c+d x)}\right ) \sqrt {1+i \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}+\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.77 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.06, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.393, Rules used = {3042, 4043, 27, 3042, 4038, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4043

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4038

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

\(\displaystyle \frac {\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}}{d}-\frac {i \int \frac {\sqrt {i \tan (c+d x) a+a} (\tan (c+d x) a+i a)}{\sqrt {\tan (c+d x)}}dx-\frac {4 i a^3 \int \frac {1}{-\frac {2 \tan (c+d x) a^2}{i \tan (c+d x) a+a}-i a}d\frac {\sqrt {\tan (c+d x)}}{\sqrt {i \tan (c+d x) a+a}}}{d}}{2 a}\)

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

-1/2*((2*(-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 + ((2 - 2*I)*a^(3/2)*ArcTanh[((1 + I)*Sqrt 
[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d)/a + (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 4038

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(3/2)/Sqrt[(c_.) + (d_.)*tan[(e_ 
.) + (f_.)*(x_)]], x_Symbol] :> Simp[2*a   Int[Sqrt[a + b*Tan[e + f*x]]/Sqr 
t[c + d*Tan[e + f*x]], x], x] + Simp[b/a   Int[(b + a*Tan[e + f*x])*(Sqrt[a 
 + b*Tan[e + f*x]]/Sqrt[c + d*Tan[e + f*x]]), 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 4043

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*(m + n - 1))), x] - Simp[1/(a*(m + n - 1))   Int[(a + b 
*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n - 2)*Simp[d*(b*c*m + a*d*(-1 + n)) 
 - a*c^2*(m + n - 1) + d*(b*d*m - a*c*(m + 2*n - 2))*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] && GtQ[n, 1] && NeQ[m + n - 1, 0] && (IntegerQ[n] 
 || 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]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 228 vs. \(2 (108 ) = 216\).

Time = 1.69 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.70

method	result	size
derivativedivides	\(\frac {\left (i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +2 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+\ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\tan \left (d x +c \right )}}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(229\)
default	\(\frac {\left (i \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right ) \sqrt {i a}\, a +2 \sqrt {-i a}\, \sqrt {i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+\ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) a \sqrt {-i a}\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\tan \left (d x +c \right )}}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(229\)

Input:

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

Output:

1/2/d*(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))*(I*a)^(1/2)*a+2*(-I*a)^(1/2) 
*(I*a)^(1/2)*(a*tan(d*x+c)*(1+I*tan(d*x+c)))^(1/2)+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))*(a*(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)/(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. 479 vs. \(2 (101) = 202\).

Time = 0.11 (sec) , antiderivative size = 479, normalized size of antiderivative = 3.55 \[ \int \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}} e^{\left (i \, d x + i \, c\right )} + d \sqrt {-\frac {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 )} + 2 \, d \sqrt {-\frac {i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - d \sqrt {-\frac {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 )} - 2 \, d \sqrt {-\frac {i \, a}{d^{2}}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - d \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 ) + d \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 )}{2 \, d} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

sqrt(a)*int(sqrt(tan(c + d*x))*sqrt(tan(c + d*x)*i + 1)*tan(c + d*x),x)