\(\int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx\) [1156]

Optimal result

Integrand size = 32, antiderivative size = 151 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=-\frac {2 (-1)^{3/4} a^{3/2} \text {arctanh}\left (\frac {(-1)^{3/4} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {d} f}-\frac {2 i \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {c-i d} f} \] Output:

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

Mathematica [A] (verified)

Time = 1.17 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.34 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx=\frac {2 a \left (-\frac {\sqrt {2} \sqrt {-a (c-i d)} \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{i c+d}+\frac {\sqrt [4]{-1} \sqrt {a} \sqrt {c+i d} \arcsin \left (\frac {\sqrt [4]{-1} \sqrt {d} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c+i d}}\right ) \sqrt {\frac {c+d \tan (e+f x)}{c+i d}}}{\sqrt {d} \sqrt {c+d \tan (e+f x)}}\right )}{f} \] Input:

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

Output:

(2*a*(-((Sqrt[2]*Sqrt[-(a*(c - I*d))]*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[a 
+ I*a*Tan[e + f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])])/(I*c + d)) + ( 
(-1)^(1/4)*Sqrt[a]*Sqrt[c + I*d]*ArcSin[((-1)^(1/4)*Sqrt[d]*Sqrt[a + I*a*T 
an[e + f*x]])/(Sqrt[a]*Sqrt[c + I*d])]*Sqrt[(c + d*Tan[e + f*x])/(c + I*d) 
])/(Sqrt[d]*Sqrt[c + d*Tan[e + f*x]])))/f
 

Rubi [A] (verified)

Time = 0.75 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 4038, 3042, 4027, 221, 4082, 66, 221}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4038

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 221

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

\(\Big \downarrow \) 4082

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 66

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

rule 221

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x 
/Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[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. 982 vs. \(2 (114 ) = 228\).

Time = 0.63 (sec) , antiderivative size = 983, normalized size of antiderivative = 6.51

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {c +d \tan \left (f x +e \right )}\, a^{2} \left (i \ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c -i \ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d +\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c^{2}+\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d^{2}+i \sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) c +i \sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) d +\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c +\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d -\sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) c +\sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) d \right ) \sqrt {2}}{2 f \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \left (c^{2}+d^{2}\right ) \sqrt {i a d}\, \sqrt {-a \left (i d -c \right )}}\)	\(983\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {c +d \tan \left (f x +e \right )}\, a^{2} \left (i \ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c -i \ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d +\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c^{2}+\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d^{2}+i \sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) c +i \sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) d +\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, c +\ln \left (\frac {2 i a d \tan \left (f x +e \right )+i a c +2 \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {i a d}+a d}{2 \sqrt {i a d}}\right ) \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, d -\sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) c +\sqrt {i a d}\, \ln \left (\frac {3 a c +i a \tan \left (f x +e \right ) c -i a d +3 a d \tan \left (f x +e \right )+2 \sqrt {2}\, \sqrt {-a \left (i d -c \right )}\, \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}}{i+\tan \left (f x +e \right )}\right ) d \right ) \sqrt {2}}{2 f \sqrt {a \left (c +d \tan \left (f x +e \right )\right ) \left (1+i \tan \left (f x +e \right )\right )}\, \left (c^{2}+d^{2}\right ) \sqrt {i a d}\, \sqrt {-a \left (i d -c \right )}}\)	\(983\)

Input:

int((a+I*a*tan(f*x+e))^(3/2)/(c+d*tan(f*x+e))^(1/2),x,method=_RETURNVERBOS 
E)
 

Output:

1/2/f*(a*(1+I*tan(f*x+e)))^(1/2)*(c+d*tan(f*x+e))^(1/2)*a^2*(I*ln(1/2*(2*I 
*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d 
)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*c-I*ln(1/2*(2*I*a*d 
*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1 
/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*d+ln(1/2*(2*I*a*d*tan(f 
*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a* 
d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*c^2+ln(1/2*(2*I*a*d*tan(f*x+e 
)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/( 
I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*d^2+I*(I*a*d)^(1/2)*ln((3*a*c+I*a 
*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a*(c+d* 
tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*c+I*(I*a*d)^(1/2)*ln( 
(3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2 
)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*d+ln(1/2*(2 
*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a 
*d)^(1/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*c+ln(1/2*(2*I*a*d 
*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1 
/2)+a*d)/(I*a*d)^(1/2))*2^(1/2)*(-a*(I*d-c))^(1/2)*d-(I*a*d)^(1/2)*ln((3*a 
*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c))^(1/2)*(a 
*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan(f*x+e)))*c+(I*a*d)^(1/2) 
*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/2)*(-a*(I*d-c...
 

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. 533 vs. \(2 (109) = 218\).

Time = 0.09 (sec) , antiderivative size = 533, normalized size of antiderivative = 3.53 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume((d^2-2*c*d-c^2)>0)', see `assume 
?` for mor
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeError: Bad Argument TypeError: Bad Argument TypeRecursive ass 
umption s
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c+d \tan (e+f x)}} \, dx =\text {Too large to display} \] Input:

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

Output:

(sqrt(a)*a*( - 2*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c*i + 2 
*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*d + 2*int((sqrt(tan(e + 
 f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)*c**2* 
d + tan(e + f*x)*d**3 + c**3 + c*d**2),x)*c**3*d*f*i - 2*int((sqrt(tan(e + 
 f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)*c**2* 
d + tan(e + f*x)*d**3 + c**3 + c*d**2),x)*c**2*d**2*f + 2*int((sqrt(tan(e 
+ f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)*c**2 
*d + tan(e + f*x)*d**3 + c**3 + c*d**2),x)*c*d**3*f*i - 2*int((sqrt(tan(e 
+ f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x)**2)/(tan(e + f*x)*c**2 
*d + tan(e + f*x)*d**3 + c**3 + c*d**2),x)*d**4*f + int((sqrt(tan(e + f*x) 
*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)*c**2*d + tan( 
e + f*x)*d**3 + c**3 + c*d**2),x)*c**4*f*i - 2*int((sqrt(tan(e + f*x)*i + 
1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)*c**2*d + tan(e + f 
*x)*d**3 + c**3 + c*d**2),x)*c**3*d*f - 2*int((sqrt(tan(e + f*x)*i + 1)*sq 
rt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)*c**2*d + tan(e + f*x)*d 
**3 + c**3 + c*d**2),x)*c*d**3*f - int((sqrt(tan(e + f*x)*i + 1)*sqrt(tan( 
e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)*c**2*d + tan(e + f*x)*d**3 + c 
**3 + c*d**2),x)*d**4*f*i + int((sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x 
)*d + c)*tan(e + f*x))/(tan(e + f*x)*d + c),x)*c**2*f*i + int((sqrt(tan(e 
+ f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(tan(e + f*x)*d + ...