\(\int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx\) [1009]

Optimal result

Integrand size = 35, antiderivative size = 154 \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {3 i \sqrt {a} c^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}-\frac {3 i c^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}-\frac {i c \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{3/2}}{2 f} \] Output:

-3*I*a^(1/2)*c^(5/2)*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-3/2*I*c^2*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e 
))^(1/2)/f-1/2*I*c*(a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(3/2)/f
 

Mathematica [A] (verified)

Time = 0.86 (sec) , antiderivative size = 126, normalized size of antiderivative = 0.82 \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {c^2 \sqrt {c-i c \tan (e+f x)} \left (6 i \sqrt {a} \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right )+\sqrt {1-i \tan (e+f x)} (4 i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}\right )}{2 f \sqrt {1-i \tan (e+f x)}} \] Input:

Integrate[Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2),x]
 

Output:

-1/2*(c^2*Sqrt[c - I*c*Tan[e + f*x]]*((6*I)*Sqrt[a]*ArcSin[Sqrt[a + I*a*Ta 
n[e + f*x]]/(Sqrt[2]*Sqrt[a])] + Sqrt[1 - I*Tan[e + f*x]]*(4*I + Tan[e + f 
*x])*Sqrt[a + I*a*Tan[e + f*x]]))/(f*Sqrt[1 - I*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 4006, 60, 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.

\(\displaystyle \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2}dx\)

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 45

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

\(\Big \downarrow \) 218

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

Input:

Int[Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(5/2),x]
 

Output:

(a*c*(((-1/2*I)*Sqrt[a + I*a*Tan[e + f*x]]*(c - I*c*Tan[e + f*x])^(3/2))/a 
 + (3*c*(((-2*I)*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt 
[a]*Sqrt[c - I*c*Tan[e + f*x]])])/Sqrt[a] - (I*Sqrt[a + I*a*Tan[e + f*x]]* 
Sqrt[c - I*c*Tan[e + f*x]])/a))/2))/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 = 0.33 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, c^{2} \left (4 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-3 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c +\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{2 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(153\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, c^{2} \left (4 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}-3 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c +\tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{2 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(153\)

Input:

int((a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(5/2),x,method=_RETURNVERB 
OSE)
 

Output:

-1/2/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*c^2*(4*I*(a* 
c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)-3*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)*(a*c*(1+tan(f*x+e)^ 
2))^(1/2)*(a*c)^(1/2))/(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. 361 vs. \(2 (112) = 224\).

Time = 0.12 (sec) , antiderivative size = 361, normalized size of antiderivative = 2.34 \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=\frac {3 \, \sqrt {\frac {a c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a c^{5}}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}}\right ) - 3 \, \sqrt {\frac {a c^{5}}{f^{2}}} {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a c^{5}}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + c^{2}}\right ) - 4 \, {\left (3 i \, c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + 5 i \, c^{2} e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{4 \, {\left (f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \] Input:

integrate((a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(5/2),x, algorithm=" 
fricas")
 

Output:

1/4*(3*sqrt(a*c^5/f^2)*(f*e^(2*I*f*x + 2*I*e) + f)*log(4*(2*(c^2*e^(3*I*f* 
x + 3*I*e) + c^2*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*c^5/f^2)*(I*f*e^(2*I*f*x + 2*I*e) - I 
*f))/(c^2*e^(2*I*f*x + 2*I*e) + c^2)) - 3*sqrt(a*c^5/f^2)*(f*e^(2*I*f*x + 
2*I*e) + f)*log(4*(2*(c^2*e^(3*I*f*x + 3*I*e) + c^2*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*c^ 
5/f^2)*(-I*f*e^(2*I*f*x + 2*I*e) + I*f))/(c^2*e^(2*I*f*x + 2*I*e) + c^2)) 
- 4*(3*I*c^2*e^(3*I*f*x + 3*I*e) + 5*I*c^2*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)))/(f*e^(2*I*f*x + 2*I 
*e) + f)
 

Sympy [F]

\[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=\int \sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}\, dx \] Input:

integrate((a+I*a*tan(f*x+e))**(1/2)*(c-I*c*tan(f*x+e))**(5/2),x)
 

Output:

Integral(sqrt(I*a*(tan(e + f*x) - I))*(-I*c*(tan(e + f*x) + I))**(5/2), 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. 445 vs. \(2 (112) = 224\).

Time = 0.26 (sec) , antiderivative size = 445, normalized size of antiderivative = 2.89 \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {{\left (12 \, c^{2} \cos \left (3 \, f x + 3 \, e\right ) + 20 \, c^{2} \cos \left (f x + e\right ) + 12 i \, c^{2} \sin \left (3 \, f x + 3 \, e\right ) + 20 i \, c^{2} \sin \left (f x + e\right ) + 6 \, {\left (c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 i \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + c^{2}\right )} \arctan \left (\cos \left (f x + e\right ), \sin \left (f x + e\right ) + 1\right ) + 6 \, {\left (c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + i \, c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 i \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + c^{2}\right )} \arctan \left (\cos \left (f x + e\right ), -\sin \left (f x + e\right ) + 1\right ) + 3 \, {\left (i \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) - c^{2} \sin \left (4 \, f x + 4 \, e\right ) - 2 \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) + i \, c^{2}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \sin \left (f x + e\right ) + 1\right ) + 3 \, {\left (-i \, c^{2} \cos \left (4 \, f x + 4 \, e\right ) - 2 i \, c^{2} \cos \left (2 \, f x + 2 \, e\right ) + c^{2} \sin \left (4 \, f x + 4 \, e\right ) + 2 \, c^{2} \sin \left (2 \, f x + 2 \, e\right ) - i \, c^{2}\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \sin \left (f x + e\right ) + 1\right )\right )} \sqrt {a} \sqrt {c}}{-4 \, f {\left (i \, \cos \left (4 \, f x + 4 \, e\right ) + 2 i \, \cos \left (2 \, f x + 2 \, e\right ) - \sin \left (4 \, f x + 4 \, e\right ) - 2 \, \sin \left (2 \, f x + 2 \, e\right ) + i\right )}} \] Input:

integrate((a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(5/2),x, algorithm=" 
maxima")
 

Output:

-(12*c^2*cos(3*f*x + 3*e) + 20*c^2*cos(f*x + e) + 12*I*c^2*sin(3*f*x + 3*e 
) + 20*I*c^2*sin(f*x + e) + 6*(c^2*cos(4*f*x + 4*e) + 2*c^2*cos(2*f*x + 2* 
e) + I*c^2*sin(4*f*x + 4*e) + 2*I*c^2*sin(2*f*x + 2*e) + c^2)*arctan2(cos( 
f*x + e), sin(f*x + e) + 1) + 6*(c^2*cos(4*f*x + 4*e) + 2*c^2*cos(2*f*x + 
2*e) + I*c^2*sin(4*f*x + 4*e) + 2*I*c^2*sin(2*f*x + 2*e) + c^2)*arctan2(co 
s(f*x + e), -sin(f*x + e) + 1) + 3*(I*c^2*cos(4*f*x + 4*e) + 2*I*c^2*cos(2 
*f*x + 2*e) - c^2*sin(4*f*x + 4*e) - 2*c^2*sin(2*f*x + 2*e) + I*c^2)*log(c 
os(f*x + e)^2 + sin(f*x + e)^2 + 2*sin(f*x + e) + 1) + 3*(-I*c^2*cos(4*f*x 
 + 4*e) - 2*I*c^2*cos(2*f*x + 2*e) + c^2*sin(4*f*x + 4*e) + 2*c^2*sin(2*f* 
x + 2*e) - I*c^2)*log(cos(f*x + e)^2 + sin(f*x + e)^2 - 2*sin(f*x + e) + 1 
))*sqrt(a)*sqrt(c)/(f*(-4*I*cos(4*f*x + 4*e) - 8*I*cos(2*f*x + 2*e) + 4*si 
n(4*f*x + 4*e) + 8*sin(2*f*x + 2*e) - 4*I))
 

Giac [F]

\[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=\int { \sqrt {i \, a \tan \left (f x + e\right ) + a} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} \,d x } \] Input:

integrate((a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(5/2),x, algorithm=" 
giac")
 

Output:

integrate(sqrt(I*a*tan(f*x + e) + a)*(-I*c*tan(f*x + e) + c)^(5/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=\int \sqrt {a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \] Input:

int((a + a*tan(e + f*x)*1i)^(1/2)*(c - c*tan(e + f*x)*1i)^(5/2),x)
 

Output:

int((a + a*tan(e + f*x)*1i)^(1/2)*(c - c*tan(e + f*x)*1i)^(5/2), x)
 

Reduce [F]

\[ \int \sqrt {a+i a \tan (e+f x)} (c-i c \tan (e+f x))^{5/2} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, c^{2} \left (-2 \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, i -\left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}d x \right ) f +\left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}d x \right ) f \right )}{f} \] Input:

int((a+I*a*tan(f*x+e))^(1/2)*(c-I*c*tan(f*x+e))^(5/2),x)
 

Output:

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