\(\int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} \, dx\) [999]

Optimal result

Integrand size = 35, antiderivative size = 159 \[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} \, dx=-\frac {i a^{5/2} c^{3/2} \arctan \left (\frac {\sqrt {c} \sqrt {a+i a \tan (e+f x)}}{\sqrt {a} \sqrt {c-i c \tan (e+f x)}}\right )}{f}+\frac {a^2 c \tan (e+f x) \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}{2 f}+\frac {i a (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}{3 f} \] Output:

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

Mathematica [A] (verified)

Time = 1.53 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.86 \[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} \, dx=\frac {a^2 c^2 \left (-6 i \sqrt {a} \arcsin \left (\frac {\sqrt {a+i a \tan (e+f x)}}{\sqrt {2} \sqrt {a}}\right ) \sqrt {1-i \tan (e+f x)}+\sqrt {a+i a \tan (e+f x)} \left (2 i+5 \tan (e+f x)-i \tan ^2(e+f x)+2 \tan ^3(e+f x)\right )\right )}{6 f \sqrt {c-i c \tan (e+f x)}} \] Input:

Integrate[(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2),x]
 

Output:

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

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 157, normalized size of antiderivative = 0.99, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {3042, 4006, 59, 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.

\(\displaystyle \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2}dx\)

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 59

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

\(\Big \downarrow \) 40

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

\(\Big \downarrow \) 45

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

\(\Big \downarrow \) 218

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

Input:

Int[(a + I*a*Tan[e + f*x])^(5/2)*(c - I*c*Tan[e + f*x])^(3/2),x]
 

Output:

(a*c*(((I/3)*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f*x])^(3/2))/c 
+ a*((-I)*Sqrt[a]*Sqrt[c]*ArcTan[(Sqrt[c]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqr 
t[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)))/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 59

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[2*c*(n/(m + n + 1) 
)   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && 
EqQ[b*c + a*d, 0] && IGtQ[m + 1/2, 0] && IGtQ[n + 1/2, 0] && LtQ[m, n]
 

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 = 186, normalized size of antiderivative = 1.17

method	result	size
derivativedivides	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} c \left (2 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{2}+2 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+3 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c +3 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{6 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(186\)
default	\(\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a^{2} c \left (2 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\, \tan \left (f x +e \right )^{2}+2 i \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}+3 \ln \left (\frac {a c \tan \left (f x +e \right )+\sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}{\sqrt {a c}}\right ) a c +3 \tan \left (f x +e \right ) \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}\right )}{6 f \sqrt {a c \left (1+\tan \left (f x +e \right )^{2}\right )}\, \sqrt {a c}}\)	\(186\)

Input:

int((a+I*a*tan(f*x+e))^(5/2)*(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERB 
OSE)
 

Output:

1/6/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a^2*c*(2*I*(a 
*c*(1+tan(f*x+e)^2))^(1/2)*(a*c)^(1/2)*tan(f*x+e)^2+2*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+3*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. 432 vs. \(2 (119) = 238\).

Time = 0.10 (sec) , antiderivative size = 432, normalized size of antiderivative = 2.72 \[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} \, dx=\frac {3 \, \sqrt {\frac {a^{5} c^{3}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{2} c e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{5} c^{3}}{f^{2}}} {\left (i \, f e^{\left (2 i \, f x + 2 i \, e\right )} - i \, f\right )}\right )}}{a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c}\right ) - 3 \, \sqrt {\frac {a^{5} c^{3}}{f^{2}}} {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )} \log \left (\frac {4 \, {\left (2 \, {\left (a^{2} c e^{\left (3 i \, f x + 3 i \, e\right )} + a^{2} c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} - \sqrt {\frac {a^{5} c^{3}}{f^{2}}} {\left (-i \, f e^{\left (2 i \, f x + 2 i \, e\right )} + i \, f\right )}\right )}}{a^{2} c e^{\left (2 i \, f x + 2 i \, e\right )} + a^{2} c}\right ) - 4 \, {\left (3 i \, a^{2} c e^{\left (5 i \, f x + 5 i \, e\right )} - 8 i \, a^{2} c e^{\left (3 i \, f x + 3 i \, e\right )} - 3 i \, a^{2} c e^{\left (i \, f x + i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{12 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \] Input:

integrate((a+I*a*tan(f*x+e))^(5/2)*(c-I*c*tan(f*x+e))^(3/2),x, algorithm=" 
fricas")
 

Output:

1/12*(3*sqrt(a^5*c^3/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) 
 + f)*log(4*(2*(a^2*c*e^(3*I*f*x + 3*I*e) + a^2*c*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^ 
3/f^2)*(I*f*e^(2*I*f*x + 2*I*e) - I*f))/(a^2*c*e^(2*I*f*x + 2*I*e) + a^2*c 
)) - 3*sqrt(a^5*c^3/f^2)*(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) 
+ f)*log(4*(2*(a^2*c*e^(3*I*f*x + 3*I*e) + a^2*c*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^3 
/f^2)*(-I*f*e^(2*I*f*x + 2*I*e) + I*f))/(a^2*c*e^(2*I*f*x + 2*I*e) + a^2*c 
)) - 4*(3*I*a^2*c*e^(5*I*f*x + 5*I*e) - 8*I*a^2*c*e^(3*I*f*x + 3*I*e) - 3* 
I*a^2*c*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^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)
 

Sympy [F(-1)]

Timed out. \[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(f*x+e))**(5/2)*(c-I*c*tan(f*x+e))**(3/2),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. 887 vs. \(2 (119) = 238\).

Time = 0.27 (sec) , antiderivative size = 887, normalized size of antiderivative = 5.58 \[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} \, dx=\text {Too large to display} \] Input:

integrate((a+I*a*tan(f*x+e))^(5/2)*(c-I*c*tan(f*x+e))^(3/2),x, algorithm=" 
maxima")
 

Output:

-(12*a^2*c*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 32*a^2*c 
*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 12*a^2*c*cos(1/2*a 
rctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 12*I*a^2*c*sin(5/2*arctan2(s 
in(2*f*x + 2*e), cos(2*f*x + 2*e))) - 32*I*a^2*c*sin(3/2*arctan2(sin(2*f*x 
 + 2*e), cos(2*f*x + 2*e))) - 12*I*a^2*c*sin(1/2*arctan2(sin(2*f*x + 2*e), 
 cos(2*f*x + 2*e))) + 6*(a^2*c*cos(6*f*x + 6*e) + 3*a^2*c*cos(4*f*x + 4*e) 
 + 3*a^2*c*cos(2*f*x + 2*e) + I*a^2*c*sin(6*f*x + 6*e) + 3*I*a^2*c*sin(4*f 
*x + 4*e) + 3*I*a^2*c*sin(2*f*x + 2*e) + a^2*c)*arctan2(cos(1/2*arctan2(si 
n(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*cos(6*f*x + 6*e) + 3*a^2*c*cos(4*f*x + 4*e) 
 + 3*a^2*c*cos(2*f*x + 2*e) + I*a^2*c*sin(6*f*x + 6*e) + 3*I*a^2*c*sin(4*f 
*x + 4*e) + 3*I*a^2*c*sin(2*f*x + 2*e) + a^2*c)*arctan2(cos(1/2*arctan2(si 
n(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*cos(6*f*x + 6*e) + 3*I*a^2*c*cos(4*f*x + 
 4*e) + 3*I*a^2*c*cos(2*f*x + 2*e) - a^2*c*sin(6*f*x + 6*e) - 3*a^2*c*sin( 
4*f*x + 4*e) - 3*a^2*c*sin(2*f*x + 2*e) + I*a^2*c)*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), co 
s(2*f*x + 2*e)))^2 + 2*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e)) 
) + 1) + 3*(-I*a^2*c*cos(6*f*x + 6*e) - 3*I*a^2*c*cos(4*f*x + 4*e) - 3*I*a 
^2*c*cos(2*f*x + 2*e) + a^2*c*sin(6*f*x + 6*e) + 3*a^2*c*sin(4*f*x + 4*...
 

Giac [F]

\[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} \, dx=\int { {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}} \,d x } \] Input:

integrate((a+I*a*tan(f*x+e))^(5/2)*(c-I*c*tan(f*x+e))^(3/2),x, algorithm=" 
giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} \, dx=\int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (a+i a \tan (e+f x))^{5/2} (c-i c \tan (e+f x))^{3/2} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, a^{2} c \left (\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2} i +\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, i +3 \left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}d x \right ) f +3 \left (\int \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}d x \right ) f \right )}{3 f} \] Input:

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

Output:

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