Integrand size = 45, antiderivative size = 102 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {(i A+B) (a+i a \tan (e+f x))^{3/2}}{5 f (c-i c \tan (e+f x))^{5/2}}-\frac {(i A-4 B) (a+i a \tan (e+f x))^{3/2}}{15 c f (c-i c \tan (e+f x))^{3/2}} \]
-1/5*(I*A+B)*(a+I*a*tan(f*x+e))^(3/2)/f/(c-I*c*tan(f*x+e))^(5/2)-1/15*(I*A -4*B)*(a+I*a*tan(f*x+e))^(3/2)/c/f/(c-I*c*tan(f*x+e))^(3/2)
Time = 7.02 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\frac {a^2 \sec ^2(e+f x) (\cos (2 (e+f x))+i \sin (2 (e+f x))) (4 i A-B+(A+4 i B) \tan (e+f x))}{15 c^2 f (i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
Integrate[((a + I*a*Tan[e + f*x])^(3/2)*(A + B*Tan[e + f*x]))/(c - I*c*Tan [e + f*x])^(5/2),x]
(a^2*Sec[e + f*x]^2*(Cos[2*(e + f*x)] + I*Sin[2*(e + f*x)])*((4*I)*A - B + (A + (4*I)*B)*Tan[e + f*x]))/(15*c^2*f*(I + Tan[e + f*x])^2*Sqrt[a + I*a* Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.37 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.089, Rules used = {3042, 4071, 87, 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.
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {\sqrt {i \tan (e+f x) a+a} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{7/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {a c \left (\frac {(A+4 i B) \int \frac {\sqrt {i \tan (e+f x) a+a}}{(c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{5 c}-\frac {(B+i A) (a+i a \tan (e+f x))^{3/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle \frac {a c \left (-\frac {i (A+4 i B) (a+i a \tan (e+f x))^{3/2}}{15 a c^2 (c-i c \tan (e+f x))^{3/2}}-\frac {(B+i A) (a+i a \tan (e+f x))^{3/2}}{5 a c (c-i c \tan (e+f x))^{5/2}}\right )}{f}\) |
(a*c*(-1/5*((I*A + B)*(a + I*a*Tan[e + f*x])^(3/2))/(a*c*(c - I*c*Tan[e + f*x])^(5/2)) - ((I/15)*(A + (4*I)*B)*(a + I*a*Tan[e + f*x])^(3/2))/(a*c^2* (c - I*c*Tan[e + f*x])^(3/2))))/f
3.9.1.3.1 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_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Si mp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*(c + d*x)^(n - 1)*(A + B*x), x], x , Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
Time = 0.34 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (-4 A +i A \tan \left (f x +e \right )-i B -4 B \tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\) | \(90\) |
| default | \(\frac {i \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (-4 A +i A \tan \left (f x +e \right )-i B -4 B \tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\) | \(90\) |
| risch | \(-\frac {a \sqrt {\frac {a \,{\mathrm e}^{2 i \left (f x +e \right )}}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, \left (3 i A \,{\mathrm e}^{4 i \left (f x +e \right )}+3 B \,{\mathrm e}^{4 i \left (f x +e \right )}+5 i A \,{\mathrm e}^{2 i \left (f x +e \right )}-5 B \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{30 c^{2} \sqrt {\frac {c}{{\mathrm e}^{2 i \left (f x +e \right )}+1}}\, f}\) | \(104\) |
| parts | \(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (4 i+\tan \left (f x +e \right )\right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}-\frac {i B \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, a \left (1+\tan \left (f x +e \right )^{2}\right ) \left (4 \tan \left (f x +e \right )+i\right )}{15 f \,c^{3} \left (i+\tan \left (f x +e \right )\right )^{4}}\) | \(147\) |
int((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/2),x,m ethod=_RETURNVERBOSE)
1/15*I/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)*a/c^3*(1+t an(f*x+e)^2)*(-4*A+I*A*tan(f*x+e)-I*B-4*B*tan(f*x+e))/(I+tan(f*x+e))^4
Time = 0.25 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.96 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {{\left (3 \, {\left (i \, A + B\right )} a e^{\left (7 i \, f x + 7 i \, e\right )} + 2 \, {\left (4 i \, A - B\right )} a e^{\left (5 i \, f x + 5 i \, e\right )} + 5 \, {\left (i \, A - B\right )} a e^{\left (3 i \, f x + 3 i \, e\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{30 \, c^{3} f} \]
integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/ 2),x, algorithm="fricas")
-1/30*(3*(I*A + B)*a*e^(7*I*f*x + 7*I*e) + 2*(4*I*A - B)*a*e^(5*I*f*x + 5* I*e) + 5*(I*A - B)*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)
\[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {5}{2}}}\, dx \]
Integral((I*a*(tan(e + f*x) - I))**(3/2)*(A + B*tan(e + f*x))/(-I*c*(tan(e + f*x) + I))**(5/2), x)
Time = 0.41 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.51 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {30 \, {\left (3 \, {\left (A - i \, B\right )} a \cos \left (7 \, f x + 7 \, e\right ) + 2 \, {\left (4 \, A + i \, B\right )} a \cos \left (5 \, f x + 5 \, e\right ) + 5 \, {\left (A + i \, B\right )} a \cos \left (3 \, f x + 3 \, e\right ) - 3 \, {\left (-i \, A - B\right )} a \sin \left (7 \, f x + 7 \, e\right ) - 2 \, {\left (-4 i \, A + B\right )} a \sin \left (5 \, f x + 5 \, e\right ) - 5 \, {\left (-i \, A + B\right )} a \sin \left (3 \, f x + 3 \, e\right )\right )} \sqrt {a} \sqrt {c}}{-900 \, {\left (i \, c^{3} \cos \left (2 \, f x + 2 \, e\right ) - c^{3} \sin \left (2 \, f x + 2 \, e\right ) + i \, c^{3}\right )} f} \]
integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/ 2),x, algorithm="maxima")
-30*(3*(A - I*B)*a*cos(7*f*x + 7*e) + 2*(4*A + I*B)*a*cos(5*f*x + 5*e) + 5 *(A + I*B)*a*cos(3*f*x + 3*e) - 3*(-I*A - B)*a*sin(7*f*x + 7*e) - 2*(-4*I* A + B)*a*sin(5*f*x + 5*e) - 5*(-I*A + B)*a*sin(3*f*x + 3*e))*sqrt(a)*sqrt( c)/((-900*I*c^3*cos(2*f*x + 2*e) + 900*c^3*sin(2*f*x + 2*e) - 900*I*c^3)*f )
\[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}}}{{\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}}} \,d x } \]
integrate((a+I*a*tan(f*x+e))^(3/2)*(A+B*tan(f*x+e))/(c-I*c*tan(f*x+e))^(5/ 2),x, algorithm="giac")
integrate((B*tan(f*x + e) + A)*(I*a*tan(f*x + e) + a)^(3/2)/(-I*c*tan(f*x + e) + c)^(5/2), x)
Time = 9.82 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.86 \[ \int \frac {(a+i a \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(c-i c \tan (e+f x))^{5/2}} \, dx=-\frac {a\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}-5\,B\,\cos \left (2\,e+2\,f\,x\right )+3\,B\,\cos \left (4\,e+4\,f\,x\right )-5\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,A\,\sin \left (4\,e+4\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,3{}\mathrm {i}\right )}{30\,c^2\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]
-(a*((a*(cos(2*e + 2*f*x) + sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(A*cos(2*e + 2*f*x)*5i + A*cos(4*e + 4*f*x)*3i - 5*B*cos(2*e + 2 *f*x) + 3*B*cos(4*e + 4*f*x) - 5*A*sin(2*e + 2*f*x) - 3*A*sin(4*e + 4*f*x) - B*sin(2*e + 2*f*x)*5i + B*sin(4*e + 4*f*x)*3i))/(30*c^2*f*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2))