Integrand size = 45, antiderivative size = 152 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {-i A-B}{3 f (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}+\frac {i A}{3 c f (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}+\frac {2 A \tan (e+f x)}{3 a c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
2/3*A*tan(f*x+e)/a/c/f/(a+I*a*tan(f*x+e))^(1/2)/(c-I*c*tan(f*x+e))^(1/2)+1 /3*I*A/c/f/(c-I*c*tan(f*x+e))^(1/2)/(a+I*a*tan(f*x+e))^(3/2)+1/3*(-I*A-B)/ f/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2)
Time = 5.92 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.52 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\cos ^2(e+f x) \left (-B+3 A \tan (e+f x)+2 A \tan ^3(e+f x)\right )}{3 a c f \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}} \]
Integrate[(A + B*Tan[e + f*x])/((a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[ e + f*x])^(3/2)),x]
(Cos[e + f*x]^2*(-B + 3*A*Tan[e + f*x] + 2*A*Tan[e + f*x]^3))/(3*a*c*f*Sqr t[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])
Time = 0.41 (sec) , antiderivative size = 160, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {3042, 4071, 87, 55, 41}
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.
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}dx\) |
| \(\Big \downarrow \) 4071 |
| \(\displaystyle \frac {a c \int \frac {A+B \tan (e+f x)}{(i \tan (e+f x) a+a)^{5/2} (c-i c \tan (e+f x))^{5/2}}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle \frac {a c \left (\frac {A \int \frac {1}{(i \tan (e+f x) a+a)^{5/2} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{c}-\frac {B+i A}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}\right )}{f}\) |
| \(\Big \downarrow \) 55 |
| \(\displaystyle \frac {a c \left (\frac {A \left (\frac {2 \int \frac {1}{(i \tan (e+f x) a+a)^{3/2} (c-i c \tan (e+f x))^{3/2}}d\tan (e+f x)}{3 a}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}\right )}{c}-\frac {B+i A}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}\right )}{f}\) |
| \(\Big \downarrow \) 41 |
| \(\displaystyle \frac {a c \left (\frac {A \left (\frac {2 \tan (e+f x)}{3 a^2 c \sqrt {a+i a \tan (e+f x)} \sqrt {c-i c \tan (e+f x)}}+\frac {i}{3 a c (a+i a \tan (e+f x))^{3/2} \sqrt {c-i c \tan (e+f x)}}\right )}{c}-\frac {B+i A}{3 a c (a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}}\right )}{f}\) |
(a*c*(-1/3*(I*A + B)/(a*c*(a + I*a*Tan[e + f*x])^(3/2)*(c - I*c*Tan[e + f* x])^(3/2)) + (A*((I/3)/(a*c*(a + I*a*Tan[e + f*x])^(3/2)*Sqrt[c - I*c*Tan[ e + f*x]]) + (2*Tan[e + f*x])/(3*a^2*c*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c - I*c*Tan[e + f*x]])))/c))/f
3.9.41.3.1 Defintions of rubi rules used
Int[1/(((a_) + (b_.)*(x_))^(3/2)*((c_) + (d_.)*(x_))^(3/2)), x_Symbol] :> S imp[x/(a*c*Sqrt[a + b*x]*Sqrt[c + d*x]), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[b*c + a*d, 0]
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_.)*(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 = 0.40 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.74
| method | result | size |
| derivativedivides | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 A \tan \left (f x +e \right )^{5}+5 A \tan \left (f x +e \right )^{3}-B \tan \left (f x +e \right )^{2}+3 A \tan \left (f x +e \right )-B \right )}{3 f \,a^{2} c^{2} \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}\) | \(113\) |
| default | \(-\frac {\sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (2 A \tan \left (f x +e \right )^{5}+5 A \tan \left (f x +e \right )^{3}-B \tan \left (f x +e \right )^{2}+3 A \tan \left (f x +e \right )-B \right )}{3 f \,a^{2} c^{2} \left (i-\tan \left (f x +e \right )\right )^{3} \left (i+\tan \left (f x +e \right )\right )^{3}}\) | \(113\) |
| parts | \(-\frac {A \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}\, \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \left (1+\tan \left (f x +e \right )^{2}\right ) \tan \left (f x +e \right ) \left (2 \tan \left (f x +e \right )^{2}+3\right )}{3 f \,a^{2} c^{2} \left (i+\tan \left (f x +e \right )\right )^{3} \left (i-\tan \left (f x +e \right )\right )^{3}}+\frac {B \left (1+\tan \left (f x +e \right )^{2}\right ) \sqrt {-c \left (i \tan \left (f x +e \right )-1\right )}\, \sqrt {a \left (1+i \tan \left (f x +e \right )\right )}}{3 f \,c^{2} a^{2} \left (i+\tan \left (f x +e \right )\right )^{3} \left (i-\tan \left (f x +e \right )\right )^{3}}\) | \(174\) |
int((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/2),x,m ethod=_RETURNVERBOSE)
-1/3/f*(a*(1+I*tan(f*x+e)))^(1/2)*(-c*(I*tan(f*x+e)-1))^(1/2)/a^2/c^2*(2*A *tan(f*x+e)^5+5*A*tan(f*x+e)^3-B*tan(f*x+e)^2+3*A*tan(f*x+e)-B)/(I-tan(f*x +e))^3/(I+tan(f*x+e))^3
Time = 0.24 (sec) , antiderivative size = 150, normalized size of antiderivative = 0.99 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left ({\left (-i \, A - B\right )} e^{\left (8 i \, f x + 8 i \, e\right )} - 2 \, {\left (5 i \, A + 2 \, B\right )} e^{\left (6 i \, f x + 6 i \, e\right )} + 8 \, B e^{\left (5 i \, f x + 5 i \, e\right )} - 6 \, B e^{\left (4 i \, f x + 4 i \, e\right )} + 8 \, B e^{\left (3 i \, f x + 3 i \, e\right )} - 2 \, {\left (-5 i \, A + 2 \, B\right )} e^{\left (2 i \, f x + 2 i \, e\right )} + i \, A - B\right )} \sqrt {\frac {a}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}} e^{\left (-3 i \, f x - 3 i \, e\right )}}{24 \, a^{2} c^{2} f} \]
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/ 2),x, algorithm="fricas")
1/24*((-I*A - B)*e^(8*I*f*x + 8*I*e) - 2*(5*I*A + 2*B)*e^(6*I*f*x + 6*I*e) + 8*B*e^(5*I*f*x + 5*I*e) - 6*B*e^(4*I*f*x + 4*I*e) + 8*B*e^(3*I*f*x + 3* I*e) - 2*(-5*I*A + 2*B)*e^(2*I*f*x + 2*I*e) + I*A - B)*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))*e^(-3*I*f*x - 3*I*e)/(a^2 *c^2*f)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\int \frac {A + B \tan {\left (e + f x \right )}}{\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {3}{2}} \left (- i c \left (\tan {\left (e + f x \right )} + i\right )\right )^{\frac {3}{2}}}\, dx \]
Integral((A + B*tan(e + f*x))/((I*a*(tan(e + f*x) - I))**(3/2)*(-I*c*(tan( e + f*x) + I))**(3/2)), x)
Time = 0.42 (sec) , antiderivative size = 200, normalized size of antiderivative = 1.32 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {{\left (3 \, {\left (3 i \, A - B\right )} \cos \left (2 \, f x + 2 \, e\right ) - 3 \, {\left (3 \, A + i \, B\right )} \sin \left (2 \, f x + 2 \, e\right ) - 2 \, B\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3 \, {\left (-3 i \, A - B\right )} \cos \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + {\left (3 \, {\left (3 \, A + i \, B\right )} \cos \left (2 \, f x + 2 \, e\right ) + 3 \, {\left (3 i \, A - B\right )} \sin \left (2 \, f x + 2 \, e\right ) + 2 \, A\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right ) + 3 \, {\left (3 \, A - i \, B\right )} \sin \left (\frac {1}{2} \, \arctan \left (\sin \left (2 \, f x + 2 \, e\right ), \cos \left (2 \, f x + 2 \, e\right )\right )\right )}{24 \, a^{\frac {3}{2}} c^{\frac {3}{2}} f} \]
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/ 2),x, algorithm="maxima")
1/24*((3*(3*I*A - B)*cos(2*f*x + 2*e) - 3*(3*A + I*B)*sin(2*f*x + 2*e) - 2 *B)*cos(3/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + 3*(-3*I*A - B)* cos(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))) + (3*(3*A + I*B)*cos( 2*f*x + 2*e) + 3*(3*I*A - B)*sin(2*f*x + 2*e) + 2*A)*sin(3/2*arctan2(sin(2 *f*x + 2*e), cos(2*f*x + 2*e))) + 3*(3*A - I*B)*sin(1/2*arctan2(sin(2*f*x + 2*e), cos(2*f*x + 2*e))))/(a^(3/2)*c^(3/2)*f)
\[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\int { \frac {B \tan \left (f x + e\right ) + A}{{\left (i \, a \tan \left (f x + e\right ) + a\right )}^{\frac {3}{2}} {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \]
integrate((A+B*tan(f*x+e))/(a+I*a*tan(f*x+e))^(3/2)/(c-I*c*tan(f*x+e))^(3/ 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)^(3/2)), x)
Time = 9.38 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.30 \[ \int \frac {A+B \tan (e+f x)}{(a+i a \tan (e+f x))^{3/2} (c-i c \tan (e+f x))^{3/2}} \, dx=\frac {\sqrt {\frac {a\,\left (\cos \left (2\,e+2\,f\,x\right )+1+\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (10\,A\,\sin \left (2\,e+2\,f\,x\right )-3\,B+A\,\cos \left (2\,e+2\,f\,x\right )\,8{}\mathrm {i}+A\,\cos \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}-4\,B\,\cos \left (2\,e+2\,f\,x\right )-B\,\cos \left (4\,e+4\,f\,x\right )-A\,9{}\mathrm {i}+A\,\sin \left (4\,e+4\,f\,x\right )+B\,\sin \left (2\,e+2\,f\,x\right )\,2{}\mathrm {i}+B\,\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )}{24\,a^2\,c\,f\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}} \]
(((a*(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)*8i - 3*B - A*9i + A*cos(4*e + 4*f*x)*1i - 4*B*c os(2*e + 2*f*x) - B*cos(4*e + 4*f*x) + 10*A*sin(2*e + 2*f*x) + A*sin(4*e + 4*f*x) + B*sin(2*e + 2*f*x)*2i + B*sin(4*e + 4*f*x)*1i))/(24*a^2*c*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 ))