Integrand size = 35, antiderivative size = 43 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {i (a+i a \tan (e+f x))^{3/2}}{3 f (c-i c \tan (e+f x))^{3/2}} \] Output:
-1/3*I*(a+I*a*tan(f*x+e))^(3/2)/f/(c-I*c*tan(f*x+e))^(3/2)
Time = 0.79 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.60 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {a (1+i \tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}{3 c f (i+\tan (e+f x)) \sqrt {c-i c \tan (e+f x)}} \] Input:
Integrate[(a + I*a*Tan[e + f*x])^(3/2)/(c - I*c*Tan[e + f*x])^(3/2),x]
Output:
(a*(1 + I*Tan[e + f*x])*Sqrt[a + I*a*Tan[e + f*x]])/(3*c*f*(I + Tan[e + f* x])*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.28 (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.
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{3/2}}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a c \int \frac {\sqrt {i \tan (e+f x) a+a}}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 48 |
\(\displaystyle -\frac {i (a+i a \tan (e+f x))^{3/2}}{3 f (c-i c \tan (e+f x))^{3/2}}\) |
Input:
Int[(a + I*a*Tan[e + f*x])^(3/2)/(c - I*c*Tan[e + f*x])^(3/2),x]
Output:
((-1/3*I)*(a + I*a*Tan[e + f*x])^(3/2))/(f*(c - I*c*Tan[e + f*x])^(3/2))
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 = 0.96 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81
method | result | size |
orering | \(-\frac {i \left (a +i a \tan \left (f x +e \right )\right )^{\frac {3}{2}}}{3 f \left (c -i c \tan \left (f x +e \right )\right )^{\frac {3}{2}}}\) | \(35\) |
derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right )}{3 f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{3}}\) | \(62\) |
default | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right )}{3 f \,c^{2} \left (i+\tan \left (f x +e \right )\right )^{3}}\) | \(62\) |
risch | \(-\frac {i a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, {\mathrm e}^{2 i \left (f x +e \right )}}{3 c \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) | \(63\) |
Input:
int((a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2),x,method=_RETURNVERB OSE)
Output:
-1/3*I*(a+I*a*tan(f*x+e))^(3/2)/f/(c-I*c*tan(f*x+e))^(3/2)
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 67 vs. \(2 (31) = 62\).
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.56 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left (-i \, a e^{\left (5 i \, f x + 5 i \, e\right )} - i \, a e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{3 \, c^{2} f} \] Input:
integrate((a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm=" fricas")
Output:
1/3*(-I*a*e^(5*I*f*x + 5*I*e) - 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^2*f)
\[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}}}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}}}\, dx \] Input:
integrate((a+I*a*tan(f*x+e))**(3/2)/(c-I*c*tan(f*x+e))**(3/2),x)
Output:
Integral((I*a*(tan(e + f*x) - I))**(3/2)/(-I*c*(tan(e + f*x) + I))**(3/2), x)
Time = 0.23 (sec) , antiderivative size = 35, normalized size of antiderivative = 0.81 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left (-i \, a \cos \left (3 \, f x + 3 \, e\right ) + a \sin \left (3 \, f x + 3 \, e\right )\right )} \sqrt {a}}{3 \, c^{\frac {3}{2}} f} \] Input:
integrate((a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm=" maxima")
Output:
1/3*(-I*a*cos(3*f*x + 3*e) + a*sin(3*f*x + 3*e))*sqrt(a)/(c^(3/2)*f)
\[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:
integrate((a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2),x, algorithm=" giac")
Output:
integrate((I*a*tan(f*x + e) + a)^(3/2)/(-I*c*tan(f*x + e) + c)^(3/2), x)
Time = 0.77 (sec) , antiderivative size = 108, normalized size of antiderivative = 2.51 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=-\frac {\sqrt {2}\,a\,\left (\cos \left (2\,f\,x\right )+\sin \left (2\,f\,x\right )\,1{}\mathrm {i}\right )\,\left (\cos \left (2\,e\right )+\sin \left (2\,e\right )\,1{}\mathrm {i}\right )\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,1{}\mathrm {i}}{6\,c\,f\,\sqrt {\frac {c}{\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}}}} \] Input:
int((a + a*tan(e + f*x)*1i)^(3/2)/(c - c*tan(e + f*x)*1i)^(3/2),x)
Output:
-(2^(1/2)*a*(cos(2*f*x) + sin(2*f*x)*1i)*(cos(2*e) + sin(2*e)*1i)*((a*(cos (2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*1i )/(6*c*f*(c/(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))^(1/2))
\[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {c}\, \sqrt {a}\, a \left (-3 \sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, \tan \left (f x +e \right )+\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}\, i +4 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{3}+\tan \left (f x +e \right )^{2} i +\tan \left (f x +e \right )+i}d x \right ) \tan \left (f x +e \right )^{2} f i +4 \left (\int \frac {\sqrt {\tan \left (f x +e \right ) i +1}\, \sqrt {-\tan \left (f x +e \right ) i +1}}{\tan \left (f x +e \right )^{3}+\tan \left (f x +e \right )^{2} i +\tan \left (f x +e \right )+i}d x \right ) f i \right )}{c^{2} f \left (\tan \left (f x +e \right )^{2}+1\right )} \] Input:
int((a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2),x)
Output:
(sqrt(c)*sqrt(a)*a*( - 3*sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*tan(e + f*x) + sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1)*i + 4*int((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f* x)**3 + tan(e + f*x)**2*i + tan(e + f*x) + i),x)*tan(e + f*x)**2*f*i + 4*i nt((sqrt(tan(e + f*x)*i + 1)*sqrt( - tan(e + f*x)*i + 1))/(tan(e + f*x)**3 + tan(e + f*x)**2*i + tan(e + f*x) + i),x)*f*i))/(c**2*f*(tan(e + f*x)**2 + 1))