\(\int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx\) [1013]

Optimal result

Integrand size = 35, antiderivative size = 90 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\frac {i (c-i c \tan (e+f x))^{5/2}}{7 f (a+i a \tan (e+f x))^{7/2}}+\frac {i (c-i c \tan (e+f x))^{5/2}}{35 a f (a+i a \tan (e+f x))^{5/2}} \] Output:

1/7*I*(c-I*c*tan(f*x+e))^(5/2)/f/(a+I*a*tan(f*x+e))^(7/2)+1/35*I*(c-I*c*ta 
n(f*x+e))^(5/2)/a/f/(a+I*a*tan(f*x+e))^(5/2)
 

Mathematica [A] (verified)

Time = 2.06 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.92 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\frac {c^2 (6+i \tan (e+f x)) (i+\tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}}{35 a^3 f (-i+\tan (e+f x))^3 \sqrt {a+i a \tan (e+f x)}} \] Input:

Integrate[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(7/2),x]
 

Output:

(c^2*(6 + I*Tan[e + f*x])*(I + Tan[e + f*x])^2*Sqrt[c - I*c*Tan[e + f*x]]) 
/(35*a^3*f*(-I + Tan[e + f*x])^3*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.

\(\displaystyle \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}}dx\)

\(\Big \downarrow \) 4006

\(\displaystyle \frac {a c \int \frac {(c-i c \tan (e+f x))^{3/2}}{(i \tan (e+f x) a+a)^{9/2}}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 55

\(\displaystyle \frac {a c \left (\frac {\int \frac {(c-i c \tan (e+f x))^{3/2}}{(i \tan (e+f x) a+a)^{7/2}}d\tan (e+f x)}{7 a}+\frac {i (c-i c \tan (e+f x))^{5/2}}{7 a c (a+i a \tan (e+f x))^{7/2}}\right )}{f}\)

\(\Big \downarrow \) 48

\(\displaystyle \frac {a c \left (\frac {i (c-i c \tan (e+f x))^{5/2}}{35 a^2 c (a+i a \tan (e+f x))^{5/2}}+\frac {i (c-i c \tan (e+f x))^{5/2}}{7 a c (a+i a \tan (e+f x))^{7/2}}\right )}{f}\)

Input:

Int[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(7/2),x]
 

Output:

(a*c*(((I/7)*(c - I*c*Tan[e + f*x])^(5/2))/(a*c*(a + I*a*Tan[e + f*x])^(7/ 
2)) + ((I/35)*(c - I*c*Tan[e + f*x])^(5/2))/(a^2*c*(a + I*a*Tan[e + f*x])^ 
(5/2))))/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 = 0.38 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.97

method	result	size
derivativedivides	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (5 i \tan \left (f x +e \right )-\tan \left (f x +e \right )^{2}-6\right )}{35 f \,a^{4} \left (-\tan \left (f x +e \right )+i\right )^{5}}\)	\(87\)
default	\(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (5 i \tan \left (f x +e \right )-\tan \left (f x +e \right )^{2}-6\right )}{35 f \,a^{4} \left (-\tan \left (f x +e \right )+i\right )^{5}}\)	\(87\)

Input:

int((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(7/2),x,method=_RETURNVERB 
OSE)
 

Output:

1/35/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*c^2/a^4*(1+t 
an(f*x+e)^2)*(5*I*tan(f*x+e)-tan(f*x+e)^2-6)/(-tan(f*x+e)+I)^5
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.94 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\frac {{\left (7 i \, c^{2} e^{\left (4 i \, f x + 4 i \, e\right )} + 12 i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + 5 i \, c^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-7 i \, f x - 7 i \, e\right )}}{70 \, a^{4} f} \] Input:

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm=" 
fricas")
 

Output:

1/70*(7*I*c^2*e^(4*I*f*x + 4*I*e) + 12*I*c^2*e^(2*I*f*x + 2*I*e) + 5*I*c^2 
)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(- 
7*I*f*x - 7*I*e)/(a^4*f)
 

Sympy [F]

\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\int \frac {\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {7}{2}}}\, dx \] Input:

integrate((c-I*c*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(7/2),x)
 

Output:

Integral((-I*c*(tan(e + f*x) + I))**(5/2)/(I*a*(tan(e + f*x) - I))**(7/2), 
 x)
 

Maxima [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.04 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\frac {{\left (5 i \, c^{2} \cos \left (7 \, f x + 7 \, e\right ) + 7 i \, c^{2} \cos \left (\frac {5}{7} \, \arctan \left (\sin \left (7 \, f x + 7 \, e\right ), \cos \left (7 \, f x + 7 \, e\right )\right )\right ) + 5 \, c^{2} \sin \left (7 \, f x + 7 \, e\right ) + 7 \, c^{2} \sin \left (\frac {5}{7} \, \arctan \left (\sin \left (7 \, f x + 7 \, e\right ), \cos \left (7 \, f x + 7 \, e\right )\right )\right )\right )} \sqrt {c}}{70 \, a^{\frac {7}{2}} f} \] Input:

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm=" 
maxima")
 

Output:

1/70*(5*I*c^2*cos(7*f*x + 7*e) + 7*I*c^2*cos(5/7*arctan2(sin(7*f*x + 7*e), 
 cos(7*f*x + 7*e))) + 5*c^2*sin(7*f*x + 7*e) + 7*c^2*sin(5/7*arctan2(sin(7 
*f*x + 7*e), cos(7*f*x + 7*e))))*sqrt(c)/(a^(7/2)*f)
 

Giac [F]

\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\int { \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {7}{2}}} \,d x } \] Input:

integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(7/2),x, algorithm=" 
giac")
 

Output:

integrate((-I*c*tan(f*x + e) + c)^(5/2)/(I*a*tan(f*x + e) + a)^(7/2), x)
 

Mupad [B] (verification not implemented)

Time = 3.53 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.79 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\frac {c^2\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (7\,\sin \left (4\,e+4\,f\,x\right )+12\,\sin \left (6\,e+6\,f\,x\right )+5\,\sin \left (8\,e+8\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )\,7{}\mathrm {i}+\cos \left (6\,e+6\,f\,x\right )\,12{}\mathrm {i}+\cos \left (8\,e+8\,f\,x\right )\,5{}\mathrm {i}\right )}{140\,a^4\,f} \] Input:

int((c - c*tan(e + f*x)*1i)^(5/2)/(a + a*tan(e + f*x)*1i)^(7/2),x)
 

Output:

(c^2*((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(4*e + 4*f*x)*7i + cos(6*e + 6*f*x)*12i + cos(8*e + 8 
*f*x)*5i + 7*sin(4*e + 4*f*x) + 12*sin(6*e + 6*f*x) + 5*sin(8*e + 8*f*x))) 
/(140*a^4*f)
 

Reduce [F]

\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{7/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, c^{2} \left (\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )+2 \left (\int -\frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{5} i +3 \tan \left (f x +e \right )^{4}-2 \tan \left (f x +e \right )^{3} i +2 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) \tan \left (f x +e \right )^{2} f i +2 \left (\int -\frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{3}}{\tan \left (f x +e \right )^{5} i +3 \tan \left (f x +e \right )^{4}-2 \tan \left (f x +e \right )^{3} i +2 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) f i +6 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{5} i +3 \tan \left (f x +e \right )^{4}-2 \tan \left (f x +e \right )^{3} i +2 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) \tan \left (f x +e \right )^{2} f i +6 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\tan \left (f x +e \right )^{5} i +3 \tan \left (f x +e \right )^{4}-2 \tan \left (f x +e \right )^{3} i +2 \tan \left (f x +e \right )^{2}-3 \tan \left (f x +e \right ) i -1}d x \right ) f i \right )}{a^{4} f \left (\tan \left (f x +e \right )^{2}+1\right )} \] Input:

int((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(7/2),x)
 

Output:

(sqrt(c)*sqrt(a)*c**2*(sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1 
)*tan(e + f*x) + 2*int(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i 
 + 1)*tan(e + f*x)**3)/(tan(e + f*x)**5*i + 3*tan(e + f*x)**4 - 2*tan(e + 
f*x)**3*i + 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*tan(e + f*x)**2*f 
*i + 2*int(( - sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e 
+ f*x)**3)/(tan(e + f*x)**5*i + 3*tan(e + f*x)**4 - 2*tan(e + f*x)**3*i + 
2*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*f*i + 6*int((sqrt(tan(e + f*x 
)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**5*i + 3* 
tan(e + f*x)**4 - 2*tan(e + f*x)**3*i + 2*tan(e + f*x)**2 - 3*tan(e + f*x) 
*i - 1),x)*tan(e + f*x)**2*f*i + 6*int((sqrt(tan(e + f*x)*i + 1)*sqrt( - t 
an(e + f*x)*i + 1)*tan(e + f*x))/(tan(e + f*x)**5*i + 3*tan(e + f*x)**4 - 
2*tan(e + f*x)**3*i + 2*tan(e + f*x)**2 - 3*tan(e + f*x)*i - 1),x)*f*i))/( 
a**4*f*(tan(e + f*x)**2 + 1))