Integrand size = 35, antiderivative size = 90 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {i (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {i (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}} \]
-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)
Time = 2.67 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {a (1+i \tan (e+f x)) (4 i+\tan (e+f x)) \sqrt {a+i a \tan (e+f x)}}{15 c^2 f (i+\tan (e+f x))^2 \sqrt {c-i c \tan (e+f x)}} \]
(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]])
Time = 0.28 (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 |
\(\displaystyle \frac {a c \left (-\frac {i (a+i a \tan (e+f x))^{3/2}}{15 a c^2 (c-i c \tan (e+f x))^{3/2}}-\frac {i (a+i a \tan (e+f x))^{3/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
(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
3.11.37.3.1 Defintions of rubi rules used
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_.)*(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])
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 = 0.80 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.79
method | result | size |
derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (4 i+\tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}}\) | \(71\) |
default | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan ^{2}\left (f x +e \right )\right ) \left (4 i+\tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (\tan \left (f x +e \right )+i\right )^{4}}\) | \(71\) |
risch | \(-\frac {i a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (3 \,{\mathrm e}^{4 i \left (f x +e \right )}+5 \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{30 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) | \(77\) |
-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)*(4*I+tan(f*x+e))/(tan(f*x+e)+I)^4
Time = 0.25 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.88 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (-3 i \, a e^{\left (7 i \, f x + 7 i \, e\right )} - 8 i \, a e^{\left (5 i \, f x + 5 i \, e\right )} - 5 i \, a e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \]
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)
\[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}}}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]
Time = 0.39 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.24 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {{\left (-3 i \, a \cos \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) - 5 i \, a \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3 \, a \sin \left (\frac {5}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 5 \, a \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )\right )} \sqrt {a}}{30 \, c^{\frac {5}{2}} f} \]
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)
\[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
Time = 6.18 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.51 \[ \int \frac {(a+i a \tan (e+f x))^{3/2}}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {a\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (-5\,\sin \left (2\,e+2\,f\,x\right )-3\,\sin \left (4\,e+4\,f\,x\right )+\cos \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{30\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]