\(\int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx\) [1037]

Optimal result

Integrand size = 35, antiderivative size = 90 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {i (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}} \] Output:

-1/5*I*(a+I*a*tan(f*x+e))^(3/2)/f/(c-I*c*tan(f*x+e))^(5/2)-1/15*I*(a+I*a*t 
an(f*x+e))^(3/2)/c/f/(c-I*c*tan(f*x+e))^(3/2)
 

Mathematica [A] (verified)

Time = 1.37 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {a (1+i \tan (e+f x)) (4 i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}{15 c^2 f (i+\tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.10, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.114, Rules used = {3042, 4006, 55, 48}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a c \int \frac {\sqrt {i \tan (e+f x) a+a}}{(c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {\int \frac {\sqrt {i \tan (e+f x) a+a}}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{5 c}-\frac {i (a+i a \tan (e+f x))^{3/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {a c \left (-\frac {i (a+i a \tan (e+f x))^{3/2}}{15 a c^2 (c-i c \tan (e+f x))^{3/2}}-\frac {i (a+i a \tan (e+f x))^{3/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\)

Input:

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

Output:

(a*c*(((-1/5*I)*(a + I*a*Tan[e + f*x])^(3/2))/(a*c*(c - I*c*Tan[e + f*x])^ 
(5/2)) - ((I/15)*(a + I*a*Tan[e + f*x])^(3/2))/(a*c^2*(c - I*c*Tan[e + f*x 
])^(3/2))))/f
 

Defintions of rubi rules used

rule 48

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

rule 55

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] - Simp[d*(S 
implify[m + n + 2]/((b*c - a*d)*(m + 1)))   Int[(a + b*x)^Simplify[m + 1]*( 
c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && ILtQ[Simplify[m + n + 
 2], 0] && NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[ 
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (SumSimplerQ[m, 1] ||  !SumSimp 
lerQ[n, 1])
 

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 = 0.35 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.79

method	result	size
derivativedivides	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (\tan \left (f x +e \right )+4 i\right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(71\)
default	\(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (\tan \left (f x +e \right )+4 i\right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\)	\(71\)
risch	\(-\frac {i a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (3 \,{\mathrm e}^{4 i \left (f x +e \right )}+5 \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{30 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\)	\(77\)

Input:

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

Output:

-1/15/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a/c^3*(1+ta 
n(f*x+e)^2)*(tan(f*x+e)+4*I)/(I+tan(f*x+e))^4
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (-3 i \, a e^{\left (7 i \, f x + 7 i \, e\right )} - 8 i \, a e^{\left (5 i \, f x + 5 i \, e\right )} - 5 i \, a e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \] Input:

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

Output:

1/30*(-3*I*a*e^(7*I*f*x + 7*I*e) - 8*I*a*e^(5*I*f*x + 5*I*e) - 5*I*a*e^(3* 
I*f*x + 3*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^3*f)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.24 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (-3 i \, a \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 5 i \, a \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3 \, a \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 5 \, a \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{30 \, c^{\frac {5}{2}} f} \] Input:

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

Output:

1/30*(-3*I*a*cos(5/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) - 5*I*a* 
cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 3*a*sin(5/2*arctan2 
(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 5*a*sin(3/2*arctan2(sin(2*f*x + 2* 
e), cos(2*f*x + 2*e))))*sqrt(a)/(c^(5/2)*f)
 

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 2.55 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.51 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {a\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (-5\,\sin \left (2\,e+2\,f\,x\right )-3\,\sin \left (4\,e+4\,f\,x\right )+\cos \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{30\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \] Input:

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

Output:

-(a*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 
1))^(1/2)*(cos(2*e + 2*f*x)*5i + cos(4*e + 4*f*x)*3i - 5*sin(2*e + 2*f*x) 
- 3*sin(4*e + 4*f*x)))/(30*c^2*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)* 
1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))
 

Reduce [F]

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

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

Output:

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