Integrand size = 45, antiderivative size = 104 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {(i A-B) (c-i c \tan (e+f x))^{3/2}}{5 f (a+i a \tan (e+f x))^{5/2}}+\frac {(i A+4 B) (c-i c \tan (e+f x))^{3/2}}{15 a f (a+i a \tan (e+f x))^{3/2}} \]
1/5*(I*A-B)*(c-I*c*tan(f*x+e))^(3/2)/f/(a+I*a*tan(f*x+e))^(5/2)+1/15*(I*A+ 4*B)*(c-I*c*tan(f*x+e))^(3/2)/a/f/(a+I*a*tan(f*x+e))^(3/2)
Time = 5.37 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=-\frac {i c (i+\tan (e+f x)) (-4 i A-B+(A-4 i B) \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{15 a^2 f (-i+\tan (e+f x))^2 \sqrt {a+i a \tan (e+f x)}} \]
Integrate[((A + B*Tan[e + f*x])*(c - I*c*Tan[e + f*x])^(3/2))/(a + I*a*Tan [e + f*x])^(5/2),x]
((-1/15*I)*c*(I + Tan[e + f*x])*((-4*I)*A - B + (A - (4*I)*B)*Tan[e + f*x] )*Sqrt[c - I*c*Tan[e + f*x]])/(a^2*f*(-I + Tan[e + f*x])^2*Sqrt[a + I*a*Ta n[e + f*x]])
Time = 0.38 (sec) , antiderivative size = 113, 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.
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(a+i a \tan (e+f x))^{5/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(c-i c \tan (e+f x))^{3/2} (A+B \tan (e+f x))}{(a+i a \tan (e+f x))^{5/2}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {(A+B \tan (e+f x)) \sqrt {c-i c \tan (e+f x)}}{(i \tan (e+f x) a+a)^{7/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {a c \left (\frac {(A-4 i B) \int \frac {\sqrt {c-i c \tan (e+f x)}}{(i \tan (e+f x) a+a)^{5/2}}d\tan (e+f x)}{5 a}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{5 a c (a+i a \tan (e+f x))^{5/2}}\right )}{f}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle \frac {a c \left (\frac {i (A-4 i B) (c-i c \tan (e+f x))^{3/2}}{15 a^2 c (a+i a \tan (e+f x))^{3/2}}+\frac {(-B+i A) (c-i c \tan (e+f x))^{3/2}}{5 a c (a+i a \tan (e+f x))^{5/2}}\right )}{f}\) |
(a*c*(((I*A - B)*(c - I*c*Tan[e + f*x])^(3/2))/(5*a*c*(a + I*a*Tan[e + f*x ])^(5/2)) + ((I/15)*(A - (4*I)*B)*(c - I*c*Tan[e + f*x])^(3/2))/(a^2*c*(a + I*a*Tan[e + f*x])^(3/2))))/f
3.9.46.3.1 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 = 0.41 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.88
| method | result | size |
| derivativedivides | \(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan \left (f x +e \right )^{2}\right ) \left (i A \tan \left (f x +e \right )-i B +4 B \tan \left (f x +e \right )+4 A \right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\) | \(92\) |
| default | \(\frac {i \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan \left (f x +e \right )^{2}\right ) \left (i A \tan \left (f x +e \right )-i B +4 B \tan \left (f x +e \right )+4 A \right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\) | \(92\) |
| parts | \(\frac {A \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan \left (f x +e \right )^{2}\right ) \left (4 i-\tan \left (f x +e \right )\right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}-\frac {i B \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, c \left (1+\tan \left (f x +e \right )^{2}\right ) \left (i-4 \tan \left (f x +e \right )\right )}{15 f \,a^{3} \left (i-\tan \left (f x +e \right )\right )^{4}}\) | \(153\) |
int((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/2),x,m ethod=_RETURNVERBOSE)
1/15*I/f*(-c*(I*tan(f*x+e)-1))^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)/a^3*c*(1+t an(f*x+e)^2)*(I*A*tan(f*x+e)-I*B+4*B*tan(f*x+e)+4*A)/(I-tan(f*x+e))^4
Time = 0.25 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.94 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=-\frac {{\left (5 \, {\left (-i \, A - B\right )} c e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, {\left (-4 i \, A - B\right )} c e^{\left (2 i \, f x + 2 i \, e\right )} + 3 \, {\left (-i \, A + B\right )} c\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-5 i \, f x - 5 i \, e\right )}}{30 \, a^{3} f} \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/ 2),x, algorithm="fricas")
-1/30*(5*(-I*A - B)*c*e^(4*I*f*x + 4*I*e) + 2*(-4*I*A - B)*c*e^(2*I*f*x + 2*I*e) + 3*(-I*A + B)*c)*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)
\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (e + f x \right )}\right )}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}\, dx \]
Integral((-I*c*(tan(e + f*x) + I))**(3/2)*(A + B*tan(e + f*x))/(I*a*(tan(e + f*x) - I))**(5/2), x)
Time = 0.43 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.47 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {30 \, {\left (5 \, {\left (A - i \, B\right )} c \cos \left (4 \, f x + 4 \, e\right ) + 2 \, {\left (4 \, A - i \, B\right )} c \cos \left (2 \, f x + 2 \, e\right ) - 5 \, {\left (-i \, A - B\right )} c \sin \left (4 \, f x + 4 \, e\right ) - 2 \, {\left (-4 i \, A - B\right )} c \sin \left (2 \, f x + 2 \, e\right ) + 3 \, {\left (A + i \, B\right )} c\right )} \sqrt {a} \sqrt {c}}{-900 \, {\left (i \, a^{3} \cos \left (7 \, f x + 7 \, e\right ) + i \, a^{3} \cos \left (5 \, f x + 5 \, e\right ) - a^{3} \sin \left (7 \, f x + 7 \, e\right ) - a^{3} \sin \left (5 \, f x + 5 \, e\right )\right )} f} \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/ 2),x, algorithm="maxima")
30*(5*(A - I*B)*c*cos(4*f*x + 4*e) + 2*(4*A - I*B)*c*cos(2*f*x + 2*e) - 5* (-I*A - B)*c*sin(4*f*x + 4*e) - 2*(-4*I*A - B)*c*sin(2*f*x + 2*e) + 3*(A + I*B)*c)*sqrt(a)*sqrt(c)/((-900*I*a^3*cos(7*f*x + 7*e) - 900*I*a^3*cos(5*f *x + 5*e) + 900*a^3*sin(7*f*x + 7*e) + 900*a^3*sin(5*f*x + 5*e))*f)
\[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\int { \frac {{\left (B \tan \left (f x + e\right ) + A\right )} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
integrate((A+B*tan(f*x+e))*(c-I*c*tan(f*x+e))^(3/2)/(a+I*a*tan(f*x+e))^(5/ 2),x, algorithm="giac")
integrate((B*tan(f*x + e) + A)*(-I*c*tan(f*x + e) + c)^(3/2)/(I*a*tan(f*x + e) + a)^(5/2), x)
Time = 10.10 (sec) , antiderivative size = 240, normalized size of antiderivative = 2.31 \[ \int \frac {(A+B \tan (e+f x)) (c-i c \tan (e+f x))^{3/2}}{(a+i a \tan (e+f x))^{5/2}} \, dx=\frac {c\,\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (A\,\cos \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,8{}\mathrm {i}+A\,\cos \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}+5\,B\,\cos \left (2\,e+2\,f\,x\right )+2\,B\,\cos \left (4\,e+4\,f\,x\right )-3\,B\,\cos \left (6\,e+6\,f\,x\right )+5\,A\,\sin \left (2\,e+2\,f\,x\right )+8\,A\,\sin \left (4\,e+4\,f\,x\right )+3\,A\,\sin \left (6\,e+6\,f\,x\right )-B\,\sin \left (2\,e+2\,f\,x\right )\,5{}\mathrm {i}-B\,\sin \left (4\,e+4\,f\,x\right )\,2{}\mathrm {i}+B\,\sin \left (6\,e+6\,f\,x\right )\,3{}\mathrm {i}\right )}{60\,a^3\,f} \]
(c*((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)*(A*cos(2*e + 2*f*x)*5i + A*cos(4*e + 4*f*x)*8i + A*cos(6*e + 6*f*x)*3i + 5*B*cos(2*e + 2*f*x) + 2*B*cos(4*e + 4*f*x) - 3*B*cos(6*e + 6*f*x) + 5*A*sin(2*e + 2*f*x) + 8*A*sin(4*e + 4*f*x) + 3*A*sin(6*e + 6*f*x ) - B*sin(2*e + 2*f*x)*5i - B*sin(4*e + 4*f*x)*2i + B*sin(6*e + 6*f*x)*3i) )/(60*a^3*f)