\(\int (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)} \, dx\) [993]

Optimal result

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

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

Mathematica [A] (verified)

Time = 1.01 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.66 \[ \int (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {4 i a^{3/2} \sqrt {c} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {i a \sqrt {c-i c \tan (e+f x)} \left (2 \sqrt {a} \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )+\sqrt {1-i \tan (e+f x)} \sqrt {a+i a \tan (e+f x)}\right )}{f \sqrt {1-i \tan (e+f x)}} \] Input:

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

Output:

((-4*I)*a^(3/2)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[ 
a]*Sqrt[c - I*c*Tan[e + f*x]])])/f + (I*a*Sqrt[c - I*c*Tan[e + f*x]]*(2*Sq 
rt[a]*ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])] + Sqrt[1 - I*Ta 
n[e + f*x]]*Sqrt[a + I*a*Tan[e + f*x]]))/(f*Sqrt[1 - I*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.31 (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, 60, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 45

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

\(\Big \downarrow \) 218

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

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]])])/Sqrt[c] + (I*Sqrt[a + I*a*Tan[e + f*x]]*Sqr 
t[c - I*c*Tan[e + f*x]])/c))/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 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[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 [A] (verified)

Time = 0.34 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.15

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+\ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \right )}{f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(122\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+\ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c \right )}{f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(122\)

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:

1/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a*(I*(a*c*(1+ta 
n(f*x+e)^2))^(1/2)*(a*c)^(1/2)+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)/(a*c*(1+tan(f*x+e)^2))^(1/2)/(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. 289 vs. \(2 (78) = 156\).

Time = 0.09 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.73 \[ \int (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)} \, dx=-\frac {-4 i \, a \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (i \, f x + i \, e\right )} - \sqrt {\frac {a^{3} c}{f^{2}}} f \log \left (\frac {4 \, {\left (2 \, {\left (a e^{\left (3 i \, f x + 3 i \, e\right )} + a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{3} c}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right ) + \sqrt {\frac {a^{3} c}{f^{2}}} f \log \left (\frac {4 \, {\left (2 \, {\left (a e^{\left (3 i \, f x + 3 i \, e\right )} + a e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{3} c}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{a e^{\left (2 i \, f x + 2 i \, e\right )} + a}\right )}{2 \, f} \] 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*(-4*I*a*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) 
 + 1))*e^(I*f*x + I*e) - sqrt(a^3*c/f^2)*f*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)) - sqrt(a^3*c/f^2)*(I*f*e^(2*I*f*x + 2*I*e) - I*f))/(a*e^(2 
*I*f*x + 2*I*e) + a)) + sqrt(a^3*c/f^2)*f*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)) - sqrt(a^3*c/f^2)*(-I*f*e^(2*I*f*x + 2*I*e) + I*f))/(a*e^(2 
*I*f*x + 2*I*e) + a)))/f
 

Sympy [F]

\[ \int (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)} \, dx=\int \left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \sqrt {- i c \left (\tan {\left (e + f x \right )} + i\right )}\, dx \] 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. 447 vs. \(2 (78) = 156\).

Time = 0.27 (sec) , antiderivative size = 447, normalized size of antiderivative = 4.22 \[ \int (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)} \, dx =\text {Too large to display} \] Input:

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

Output:

-(2*(a*cos(2*f*x + 2*e) + I*a*sin(2*f*x + 2*e) + a)*arctan2(cos(1/2*arctan 
2(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*(a*cos(2*f*x + 2*e) + I*a*sin(2*f*x + 2*e) + a 
)*arctan2(cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))), -sin(1/2*a 
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 1) - 4*a*cos(1/2*arctan2(sin 
(2*f*x + 2*e), cos(2*f*x + 2*e))) - (-I*a*cos(2*f*x + 2*e) + a*sin(2*f*x + 
 2*e) - I*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) - (I*a*cos(2*f*x + 2*e) - a*sin 
(2*f*x + 2*e) + I*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*I*a*sin(1/2*arctan2 
(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))*sqrt(a)*sqrt(c)/(f*(-2*I*cos(2*f*x 
+ 2*e) + 2*sin(2*f*x + 2*e) - 2*I))
 

Giac [F]

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

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