\(\int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\) [1004]

Optimal result

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

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

Mathematica [A] (verified)

Time = 1.13 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {c (1-i \tan (e+f x)) (-4 i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{15 a^2 f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}} \] Input:

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

Output:

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

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4006

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

\(\Big \downarrow \) 55

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

\(\Big \downarrow \) 48

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

Input:

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

Output:

(a*c*(((I/5)*(c - I*c*Tan[e + f*x])^(3/2))/(a*c*(a + I*a*Tan[e + f*x])^(5/ 
2)) + ((I/15)*(c - I*c*Tan[e + f*x])^(3/2))/(a^2*c*(a + I*a*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 = 0.36 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.83

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan \left (f x +e \right )^{2}\right ) \left (4 i-\tan \left (f x +e \right )\right )}{15 f \,a^{3} \left (-\tan \left (f x +e \right )+i\right )^{4}}\)	\(75\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan \left (f x +e \right )^{2}\right ) \left (4 i-\tan \left (f x +e \right )\right )}{15 f \,a^{3} \left (-\tan \left (f x +e \right )+i\right )^{4}}\)	\(75\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.07 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {{\left (5 i \, c e^{\left (4 i \, f x + 4 i \, e\right )} + 8 i \, c e^{\left (2 i \, f x + 2 i \, e\right )} + 3 i \, c\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{30 \, a^{3} f} \] Input:

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

Output:

1/30*(5*I*c*e^(4*I*f*x + 4*I*e) + 8*I*c*e^(2*I*f*x + 2*I*e) + 3*I*c)*sqrt( 
a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-5*I*f*x 
 - 5*I*e)/(a^3*f)
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.96 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {{\left (3 i \, c \cos \left (5 \, f x + 5 \, e\right ) + 5 i \, c \cos \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right ) + 3 \, c \sin \left (5 \, f x + 5 \, e\right ) + 5 \, c \sin \left (\frac {3}{5} \, \arctan \left (\sin \left (5 \, f x + 5 \, e\right ), \cos \left (5 \, f x + 5 \, e\right )\right )\right )\right )} \sqrt {c}}{30 \, a^{\frac {5}{2}} f} \] Input:

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 3.10 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.77 \[ \int \frac {(c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {c\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (5\,\sin \left (2\,e+2\,f\,x\right )+8\,\sin \left (4\,e+4\,f\,x\right )+3\,\sin \left (6\,e+6\,f\,x\right )+\cos \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,8{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{60\,a^3\,f} \] Input:

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

Output:

(c*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1 
))^(1/2)*((c*(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)*8i + cos(6*e + 6*f* 
x)*3i + 5*sin(2*e + 2*f*x) + 8*sin(4*e + 4*f*x) + 3*sin(6*e + 6*f*x)))/(60 
*a^3*f)
 

Reduce [F]

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

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

Output:

(sqrt(c)*c*( - 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(tan(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(a)*a**2)