Integrand size = 35, antiderivative size = 43 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {i (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \] Output:
1/5*I*(c-I*c*tan(f*x+e))^(5/2)/f/(a+I*a*tan(f*x+e))^(5/2)
Time = 1.50 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {i (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}} \] Input:
Integrate[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(5/2),x]
Output:
((I/5)*(c - I*c*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^(5/2))
Time = 0.27 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {3042, 4006, 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.
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a c \int \frac {(c-i c \tan (e+f x))^{3/2}}{(i \tan (e+f x) a+a)^{7/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle \frac {i (c-i c \tan (e+f x))^{5/2}}{5 f (a+i a \tan (e+f x))^{5/2}}\) |
Input:
Int[(c - I*c*Tan[e + f*x])^(5/2)/(a + I*a*Tan[e + f*x])^(5/2),x]
Output:
((I/5)*(c - I*c*Tan[e + f*x])^(5/2))/(f*(a + I*a*Tan[e + f*x])^(5/2))
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]
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]
Time = 1.02 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81
method | result | size |
orering | \(\frac {i \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f \left (a +i a \tan \left (f x +e \right )\right )^{\frac {5}{2}}}\) | \(35\) |
derivativedivides | \(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (i+\tan \left (f x +e \right )\right )}{5 f \,a^{3} \left (-\tan \left (f x +e \right )+i\right )^{4}}\) | \(75\) |
default | \(\frac {\sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c^{2} \left (1+\tan \left (f x +e \right )^{2}\right ) \left (i+\tan \left (f x +e \right )\right )}{5 f \,a^{3} \left (-\tan \left (f x +e \right )+i\right )^{4}}\) | \(75\) |
Input:
int((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x,method=_RETURNVERB OSE)
Output:
1/5*I*(c-I*c*tan(f*x+e))^(5/2)/f/(a+I*a*tan(f*x+e))^(5/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 71 vs. \(2 (31) = 62\).
Time = 0.07 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.65 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {{\left (i \, c^{2} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, c^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{5 \, a^{3} f} \] Input:
integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm=" fricas")
Output:
1/5*(I*c^2*e^(2*I*f*x + 2*I*e) + 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^(-5*I*f*x - 5*I*e)/(a^3*f)
\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((c-I*c*tan(f*x+e))**(5/2)/(a+I*a*tan(f*x+e))**(5/2),x)
Output:
Integral((-I*c*(tan(e + f*x) + I))**(5/2)/(I*a*(tan(e + f*x) - I))**(5/2), x)
Time = 0.21 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.91 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {{\left (i \, c^{2} \cos \left (5 \, f x + 5 \, e\right ) + c^{2} \sin \left (5 \, f x + 5 \, e\right )\right )} \sqrt {c}}{5 \, a^{\frac {5}{2}} f} \] Input:
integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm=" maxima")
Output:
1/5*(I*c^2*cos(5*f*x + 5*e) + c^2*sin(5*f*x + 5*e))*sqrt(c)/(a^(5/2)*f)
\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x, algorithm=" giac")
Output:
integrate((-I*c*tan(f*x + e) + c)^(5/2)/(I*a*tan(f*x + e) + a)^(5/2), x)
Time = 4.61 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.51 \[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {c^2\,{\left (\mathrm {tan}\left (e+f\,x\right )+1{}\mathrm {i}\right )}^5\,\sqrt {a\,\left (1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}\,\sqrt {-c\,\left (-1+\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}}{5\,a^3\,f\,{\left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}^3} \] Input:
int((c - c*tan(e + f*x)*1i)^(5/2)/(a + a*tan(e + f*x)*1i)^(5/2),x)
Output:
(c^2*(tan(e + f*x) + 1i)^5*(a*(tan(e + f*x)*1i + 1))^(1/2)*(-c*(tan(e + f* x)*1i - 1))^(1/2))/(5*a^3*f*(tan(e + f*x)^2 + 1)^3)
\[ \int \frac {(c-i c \tan (e+f x))^{5/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {\sqrt {c}\, c^{2} \left (-\left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}}{\sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}-2 \sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {\tan \left (f x +e \right ) i +1}}d x \right )+\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}}{\sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}-2 \sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {\tan \left (f x +e \right ) i +1}}d x +2 \left (\int \frac {\sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )}{\sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )^{2}-2 \sqrt {\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right ) i -\sqrt {\tan \left (f x +e \right ) i +1}}d x \right ) i \right )}{\sqrt {a}\, a^{2}} \] Input:
int((c-I*c*tan(f*x+e))^(5/2)/(a+I*a*tan(f*x+e))^(5/2),x)
Output:
(sqrt(c)*c**2*( - int(sqrt( - tan(e + f*x)*i + 1)/(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) + int((sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x)**2)/(sq rt(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) + 2*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)