\(\int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx\) [1018]

Optimal result

Integrand size = 35, antiderivative size = 106 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 i a^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.75 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=\frac {2 i a^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{\sqrt {c} f}-\frac {2 i a \sqrt {a+i a \tan (e+f x)}}{f \sqrt {c-i c \tan (e+f x)}} \] Input:

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

Output:

((2*I)*a^(3/2)*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]*Sqrt[c 
 - I*c*Tan[e + f*x]])])/(Sqrt[c]*f) - ((2*I)*a*Sqrt[a + I*a*Tan[e + f*x]]) 
/(f*Sqrt[c - I*c*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.02, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4006, 57, 45, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 57

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

\(\Big \downarrow \) 45

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

\(\Big \downarrow \) 218

\(\displaystyle \frac {a c \left (\frac {2 i \sqrt {a} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{c^{3/2}}-\frac {2 i \sqrt {a+i a \tan (e+f x)}}{c \sqrt {c-i c \tan (e+f x)}}\right )}{f}\)

Input:

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

Output:

(a*c*(((2*I)*Sqrt[a]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[a]* 
Sqrt[c - I*c*Tan[e + f*x]])])/c^(3/2) - ((2*I)*Sqrt[a + I*a*Tan[e + f*x]]) 
/(c*Sqrt[c - I*c*Tan[e + f*x]])))/f
 

Defintions of rubi rules used

rule 45

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] && EqQ[b*c + a*d, 0] &&  !GtQ[c, 0]
 

rule 57

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + 1))), x] - Simp[d*(n/(b*(m + 1))) 
Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] & 
& GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m 
+ n + 2, 0] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c 
, d, m, n, 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 4006

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (84 ) = 168\).

Time = 0.33 (sec) , antiderivative size = 267, normalized size of antiderivative = 2.52

method	result	size
derivativedivides	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}-i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -2 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-2 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+2 \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{f c \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{2} \sqrt {a c}}\)	\(267\)
default	\(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )^{2}-i \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c -2 i \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )-2 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \tan \left (f x +e \right )+2 \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{f c \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \left (i+\tan \left (f x +e \right )\right )^{2} \sqrt {a c}}\)	\(267\)

Input:

int((a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(1/2),x,method=_RETURNVERB 
OSE)
 

Output:

I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a/c*(I*ln((a*c* 
tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a*c*tan( 
f*x+e)^2-I*ln((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a 
*c)^(1/2))*a*c-2*I*(a*c)^(1/2)*(a*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)-2*l 
n((a*c*tan(f*x+e)+(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2))*a 
*c*tan(f*x+e)+2*(a*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c*(1+tan(f*x+ 
e)^2))^(1/2)/(I+tan(f*x+e))^2/(a*c)^(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. 321 vs. \(2 (78) = 156\).

Time = 0.11 (sec) , antiderivative size = 321, normalized size of antiderivative = 3.03 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {c f \sqrt {\frac {a^{3}}{c f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a e^{\left (3 i \, f x + 3 i \, e\right )} + a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, c f\right )} \sqrt {\frac {a^{3}}{c f^{2}}}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right ) - c f \sqrt {\frac {a^{3}}{c f^{2}}} \log \left (\frac {4 \, {\left (2 \, {\left (a e^{\left (3 i \, f x + 3 i \, e\right )} + a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - {\left (-i \, c f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c f\right )} \sqrt {\frac {a^{3}}{c f^{2}}}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right ) + 4 \, {\left (i \, a e^{\left (3 i \, f x + 3 i \, e\right )} + i \, a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{2 \, c f} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((I*a*(tan(e + f*x) - I))**(3/2)/sqrt(-I*c*(tan(e + f*x) + I)), 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. 319 vs. \(2 (78) = 156\).

Time = 0.24 (sec) , antiderivative size = 319, normalized size of antiderivative = 3.01 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{\sqrt {c-i c \tan (e+f x)}} \, dx=-\frac {{\left (-2 i \, a \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - 2 i \, a \arctan \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ), -\sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) + 4 i \, a \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + a \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - a \log \left (\cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} + \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )^{2} - 2 \, \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 1\right ) - 4 \, a \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{2 \, \sqrt {c} f} \] Input:

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

Output:

-1/2*(-2*I*a*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), 
 sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 2*I*a*arctan2 
(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*arctan2(si 
n(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) + 4*I*a*cos(1/2*arctan2(sin(2*f*x 
+ 2*e), cos(2*f*x + 2*e))) + a*log(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2 
*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + 
 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - a*log(cos(1 
/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)))^2 + sin(1/2*arctan2(sin(2* 
f*x + 2*e), cos(2*f*x + 2*e)))^2 - 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos 
(2*f*x + 2*e))) + 1) - 4*a*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2 
*e))))*sqrt(a)/(sqrt(c)*f)
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(c)*sqrt(a)*a*(sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*t 
an(e + f*x) - 2*sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*i - i 
nt((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/ 
(tan(e + f*x)**2 + 1),x)*tan(e + f*x)**2*f - int((sqrt(tan(e + f*x)*i + 1) 
*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(tan(e + f*x)**2 + 1),x)*f)) 
/(c*f*(tan(e + f*x)**2 + 1))