\(\int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2} \, dx\) [1007]

Optimal result

Integrand size = 35, antiderivative size = 168 \[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {3 i a^{5/2} c^{5/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{4 f}+\frac {3 a^2 c^2 \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{8 f}+\frac {a c \tan (e+f x) (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{4 f} \] Output:

-3/4*I*a^(5/2)*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+3/8*a^2*c^2*tan(f*x+e)*(a+I*a*tan(f*x+e))^(1/2)*( 
c-I*c*tan(f*x+e))^(1/2)/f+1/4*a*c*tan(f*x+e)*(a+I*a*tan(f*x+e))^(3/2)*(c-I 
*c*tan(f*x+e))^(3/2)/f
 

Mathematica [A] (verified)

Time = 1.79 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.90 \[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2} \, dx=\frac {-6 i a^{5/2} c^3 \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {1-i \tan (e+f x)} \sqrt {a+i a \tan (e+f x)}+a^3 c^3 \tan (e+f x) \left (5+7 \tan ^2(e+f x)+2 \tan ^4(e+f x)\right )}{8 f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \] Input:

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

Output:

((-6*I)*a^(5/2)*c^3*ArcSin[Sqrt[a + I*a*Tan[e + f*x]]/(Sqrt[2]*Sqrt[a])]*S 
qrt[1 - I*Tan[e + f*x]]*Sqrt[a + I*a*Tan[e + f*x]] + a^3*c^3*Tan[e + f*x]* 
(5 + 7*Tan[e + f*x]^2 + 2*Tan[e + f*x]^4))/(8*f*Sqrt[a + I*a*Tan[e + f*x]] 
*Sqrt[c - I*c*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 162, normalized size of antiderivative = 0.96, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 4006, 40, 40, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 40

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

\(\Big \downarrow \) 40

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

\(\Big \downarrow \) 45

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

\(\Big \downarrow \) 218

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

Input:

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

Output:

(a*c*((Tan[e + f*x]*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3 
/2))/4 + (3*a*c*((-I)*Sqrt[a]*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + 
 f*x]])/(Sqrt[a]*Sqrt[c - I*c*Tan[e + f*x]])] + (Tan[e + f*x]*Sqrt[a + I*a 
*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])/2))/4))/f
 

Defintions of rubi rules used

rule 40

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(m_), x_Symbol] :> Simp[x* 
(a + b*x)^m*((c + d*x)^m/(2*m + 1)), x] + Simp[2*a*c*(m/(2*m + 1))   Int[(a 
 + b*x)^(m - 1)*(c + d*x)^(m - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ 
b*c + a*d, 0] && IGtQ[m + 1/2, 0]
 

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 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 = 0.31 (sec) , antiderivative size = 164, normalized size of antiderivative = 0.98

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} c^{2} \left (2 \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3} \sqrt {a c}+3 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c +5 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{8 f \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}\)	\(164\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} c^{2} \left (2 \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \tan \left (f x +e \right )^{3} \sqrt {a c}+3 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c +5 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{8 f \sqrt {a c}\, \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}}\)	\(164\)

Input:

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

Output:

1/8/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^2*c^2*(2*(a 
*c*(1+tan(f*x+e)^2))^(1/2)*tan(f*x+e)^3*(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+5*tan(f*x+e)*(a* 
c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2))/(a*c)^(1/2)/(a*c*(1+tan(f*x+e)^2))^ 
(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. 507 vs. \(2 (128) = 256\).

Time = 0.12 (sec) , antiderivative size = 507, normalized size of antiderivative = 3.02 \[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2} \, dx=\frac {3 \, \sqrt {\frac {a^{5} c^{5}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{2} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} 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^{5} c^{5}}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{a^{2} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2}}\right ) - 3 \, \sqrt {\frac {a^{5} c^{5}}{f^{2}}} {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{2} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} 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^{5} c^{5}}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{a^{2} c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c^{2}}\right ) + 4 \, {\left (-3 i \, a^{2} c^{2} e^{\left (7 i \, f x + 7 i \, e\right )} - 11 i \, a^{2} c^{2} e^{\left (5 i \, f x + 5 i \, e\right )} + 11 i \, a^{2} c^{2} e^{\left (3 i \, f x + 3 i \, e\right )} + 3 i \, a^{2} 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}}}{16 \, {\left (f e^{\left (6 i \, f x + 6 i \, e\right )} + 3 \, f e^{\left (4 i \, f x + 4 i \, e\right )} + 3 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \] Input:

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

Output:

1/16*(3*sqrt(a^5*c^5/f^2)*(f*e^(6*I*f*x + 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) 
 + 3*f*e^(2*I*f*x + 2*I*e) + f)*log(4*(2*(a^2*c^2*e^(3*I*f*x + 3*I*e) + a^ 
2*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^5*c^5/f^2)*(I*f*e^(2*I*f*x + 2*I*e) - I*f))/(a^2 
*c^2*e^(2*I*f*x + 2*I*e) + a^2*c^2)) - 3*sqrt(a^5*c^5/f^2)*(f*e^(6*I*f*x + 
 6*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)*log(4*(2* 
(a^2*c^2*e^(3*I*f*x + 3*I*e) + a^2*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^5*c^5/f^2)*(-I* 
f*e^(2*I*f*x + 2*I*e) + I*f))/(a^2*c^2*e^(2*I*f*x + 2*I*e) + a^2*c^2)) + 4 
*(-3*I*a^2*c^2*e^(7*I*f*x + 7*I*e) - 11*I*a^2*c^2*e^(5*I*f*x + 5*I*e) + 11 
*I*a^2*c^2*e^(3*I*f*x + 3*I*e) + 3*I*a^2*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^(6*I*f*x + 6 
*I*e) + 3*f*e^(4*I*f*x + 4*I*e) + 3*f*e^(2*I*f*x + 2*I*e) + f)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

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. 1173 vs. \(2 (128) = 256\).

Time = 0.37 (sec) , antiderivative size = 1173, normalized size of antiderivative = 6.98 \[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

-(12*a^2*c^2*cos(7/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 44*a^2 
*c^2*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 44*a^2*c^2*cos 
(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 12*a^2*c^2*cos(1/2*arc 
tan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*I*a^2*c^2*sin(7/2*arctan2(s 
in(2*f*x + 2*e), cos(2*f*x + 2*e))) + 44*I*a^2*c^2*sin(5/2*arctan2(sin(2*f 
*x + 2*e), cos(2*f*x + 2*e))) - 44*I*a^2*c^2*sin(3/2*arctan2(sin(2*f*x + 2 
*e), cos(2*f*x + 2*e))) - 12*I*a^2*c^2*sin(1/2*arctan2(sin(2*f*x + 2*e), c 
os(2*f*x + 2*e))) + 6*(a^2*c^2*cos(8*f*x + 8*e) + 4*a^2*c^2*cos(6*f*x + 6* 
e) + 6*a^2*c^2*cos(4*f*x + 4*e) + 4*a^2*c^2*cos(2*f*x + 2*e) + I*a^2*c^2*s 
in(8*f*x + 8*e) + 4*I*a^2*c^2*sin(6*f*x + 6*e) + 6*I*a^2*c^2*sin(4*f*x + 4 
*e) + 4*I*a^2*c^2*sin(2*f*x + 2*e) + a^2*c^2)*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) + 6*(a^2*c^2*cos(8*f*x + 8*e) + 4*a^2*c^2*cos(6*f*x + 6* 
e) + 6*a^2*c^2*cos(4*f*x + 4*e) + 4*a^2*c^2*cos(2*f*x + 2*e) + I*a^2*c^2*s 
in(8*f*x + 8*e) + 4*I*a^2*c^2*sin(6*f*x + 6*e) + 6*I*a^2*c^2*sin(4*f*x + 4 
*e) + 4*I*a^2*c^2*sin(2*f*x + 2*e) + a^2*c^2)*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) + 3*(I*a^2*c^2*cos(8*f*x + 8*e) + 4*I*a^2*c^2*cos(6*f*x 
 + 6*e) + 6*I*a^2*c^2*cos(4*f*x + 4*e) + 4*I*a^2*c^2*cos(2*f*x + 2*e) - a^ 
2*c^2*sin(8*f*x + 8*e) - 4*a^2*c^2*sin(6*f*x + 6*e) - 6*a^2*c^2*sin(4*f...
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{5/2} \, dx=\sqrt {c}\, \sqrt {a}\, a^{2} c^{2} \left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{4}d x +2 \left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}d x \right )+\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}d x \right ) \] Input:

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

Output:

sqrt(c)*sqrt(a)*a**2*c**2*(int(sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f* 
x)*i + 1)*tan(e + f*x)**4,x) + 2*int(sqrt(tan(e + f*x)*i + 1)*sqrt( - tan( 
e + f*x)*i + 1)*tan(e + f*x)**2,x) + int(sqrt(tan(e + f*x)*i + 1)*sqrt( - 
tan(e + f*x)*i + 1),x))