\(\int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt {\tan (c+d x)}} \, dx\) [196]

Optimal result

Integrand size = 28, antiderivative size = 104 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt {\tan (c+d x)}} \, dx=\frac {2 \sqrt [4]{-1} a^{3/2} \arctan \left (\frac {(-1)^{3/4} \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d}+\frac {(2-2 i) a^{3/2} \text {arctanh}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{d} \] Output:

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.49 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt {\tan (c+d x)}} \, dx=\frac {2 (i a \tan (c+d x))^{3/2} \left (\sqrt {a} \text {arcsinh}\left (\frac {\sqrt {i a \tan (c+d x)}}{\sqrt {a}}\right ) \sqrt {1+i \tan (c+d x)}-\sqrt {2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {i a \tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right ) \sqrt {a+i a \tan (c+d x)}\right )}{d \tan ^{\frac {3}{2}}(c+d x) \sqrt {a+i a \tan (c+d x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4038

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 218

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 65

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

\(\Big \downarrow \) 216

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

Input:

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

Output:

(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]*Sq 
rt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]])/d
 

Defintions of rubi rules used

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 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. 325 vs. \(2 (82 ) = 164\).

Time = 1.71 (sec) , antiderivative size = 326, normalized size of antiderivative = 3.13

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\tan \left (d x +c \right )}\, a^{2} \left (i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right )+4 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}-\sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right )-2 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right )\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(326\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {\tan \left (d x +c \right )}\, a^{2} \left (i \sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right )+4 i \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right ) \sqrt {-i a}-\sqrt {i a}\, \sqrt {2}\, \ln \left (-\frac {-2 \sqrt {2}\, \sqrt {-i a}\, \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}+i a -3 a \tan \left (d x +c \right )}{\tan \left (d x +c \right )+i}\right )-2 \sqrt {-i a}\, \ln \left (\frac {2 i a \tan \left (d x +c \right )+2 \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}+a}{2 \sqrt {i a}}\right )\right )}{2 d \sqrt {a \tan \left (d x +c \right ) \left (1+i \tan \left (d x +c \right )\right )}\, \sqrt {i a}\, \sqrt {-i a}}\)	\(326\)

Input:

int((a+I*a*tan(d*x+c))^(3/2)/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*(I*(I*a)^(1/2)*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))+4*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)-(I*a)^ 
(1/2)*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))-2*(-I*a)^(1/2)*ln(1/2*(2*I*a*ta 
n(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*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. 451 vs. \(2 (76) = 152\).

Time = 0.10 (sec) , antiderivative size = 451, normalized size of antiderivative = 4.34 \[ \int \frac {(a+i a \tan (c+d x))^{3/2}}{\sqrt {\tan (c+d x)}} \, dx=\frac {1}{2} \, \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) - \frac {1}{2} \, \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (2 \, \sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - \sqrt {-\frac {8 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, a}\right ) - \frac {1}{2} \, \sqrt {-\frac {4 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + \sqrt {-\frac {4 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) + \frac {1}{2} \, \sqrt {-\frac {4 i \, a^{3}}{d^{2}}} \log \left (\frac {{\left (\sqrt {2} {\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {-i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - \sqrt {-\frac {4 i \, a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) \] Input:

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

Output:

1/2*sqrt(-8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*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(-8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/ 
a) - 1/2*sqrt(-8*I*a^3/d^2)*log(1/2*(2*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a) 
*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(-8*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - 
I*c)/a) - 1/2*sqrt(-4*I*a^3/d^2)*log((sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)* 
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(-4*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I 
*c)/a) + 1/2*sqrt(-4*I*a^3/d^2)*log((sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*s 
qrt(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(-4*I*a^3/d^2)*d*e^(I*d*x + I*c))*e^(-I*d*x - I* 
c)/a)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-2)]

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

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

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

Output:

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

Reduce [F]

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

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

Output:

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