Integrand size = 22, antiderivative size = 142 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=-\frac {(a B-A c x) (d+e x)^2}{4 a c \left (a+c x^2\right )^2}-\frac {2 a e (2 A c d+a B e)-c \left (3 A c d^2+2 a B d e-a A e^2\right ) x}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\left (3 A c d^2+2 a B d e+a A e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{3/2}} \]
-1/4*(-A*c*x+B*a)*(e*x+d)^2/a/c/(c*x^2+a)^2+1/8*(-2*a*e*(2*A*c*d+B*a*e)+c* (-A*a*e^2+3*A*c*d^2+2*B*a*d*e)*x)/a^2/c^2/(c*x^2+a)+1/8*(A*a*e^2+3*A*c*d^2 +2*B*a*d*e)*arctan(x*c^(1/2)/a^(1/2))/a^(5/2)/c^(3/2)
Time = 0.07 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.11 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=\frac {\frac {\sqrt {a} \left (-4 a^2 B e^2+3 A c^2 d^2 x+a c e (2 B d+A e) x\right )}{a+c x^2}+\frac {2 a^{3/2} \left (a^2 B e^2+A c^2 d^2 x-a c (A e (2 d+e x)+B d (d+2 e x))\right )}{\left (a+c x^2\right )^2}+\sqrt {c} \left (3 A c d^2+2 a B d e+a A e^2\right ) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^2} \]
((Sqrt[a]*(-4*a^2*B*e^2 + 3*A*c^2*d^2*x + a*c*e*(2*B*d + A*e)*x))/(a + c*x ^2) + (2*a^(3/2)*(a^2*B*e^2 + A*c^2*d^2*x - a*c*(A*e*(2*d + e*x) + B*d*(d + 2*e*x))))/(a + c*x^2)^2 + Sqrt[c]*(3*A*c*d^2 + 2*a*B*d*e + a*A*e^2)*ArcT an[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^2)
Time = 0.30 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.11, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {685, 675, 218}
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.
\(\displaystyle \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 685 |
\(\displaystyle \frac {\int \frac {(d+e x) (3 A c d+2 a B e+A c e x)}{\left (c x^2+a\right )^2}dx}{4 a c}-\frac {(d+e x)^2 (a B-A c x)}{4 a c \left (a+c x^2\right )^2}\) |
\(\Big \downarrow \) 675 |
\(\displaystyle \frac {\frac {\left (a A e^2+2 a B d e+3 A c d^2\right ) \int \frac {1}{c x^2+a}dx}{2 a}+\frac {x \left (-a A e^2+2 a B d e+3 A c d^2\right )}{2 a \left (a+c x^2\right )}-\frac {e (a B e+2 A c d)}{c \left (a+c x^2\right )}}{4 a c}-\frac {(d+e x)^2 (a B-A c x)}{4 a c \left (a+c x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (a A e^2+2 a B d e+3 A c d^2\right )}{2 a^{3/2} \sqrt {c}}+\frac {x \left (-a A e^2+2 a B d e+3 A c d^2\right )}{2 a \left (a+c x^2\right )}-\frac {e (a B e+2 A c d)}{c \left (a+c x^2\right )}}{4 a c}-\frac {(d+e x)^2 (a B-A c x)}{4 a c \left (a+c x^2\right )^2}\) |
-1/4*((a*B - A*c*x)*(d + e*x)^2)/(a*c*(a + c*x^2)^2) + (-((e*(2*A*c*d + a* B*e))/(c*(a + c*x^2))) + ((3*A*c*d^2 + 2*a*B*d*e - a*A*e^2)*x)/(2*a*(a + c *x^2)) + ((3*A*c*d^2 + 2*a*B*d*e + a*A*e^2)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/( 2*a^(3/2)*Sqrt[c]))/(4*a*c)
3.14.49.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((d_) + (e_.)*(x_))*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_S ymbol] :> Simp[a*(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + (- Simp[(c*d*f - a*e*g)*x*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Simp[(a* e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)) Int[(a + c*x^2)^(p + 1), x], x]) / ; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] && !(IntegerQ[p] && NiceSqrtQ [(-a)*c])
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p _), x_Symbol] :> Simp[(d + e*x)^m*(a + c*x^2)^(p + 1)*((a*g - c*f*x)/(2*a*c *(p + 1))), x] - Simp[1/(2*a*c*(p + 1)) Int[(d + e*x)^(m - 1)*(a + c*x^2) ^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
Time = 0.30 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.04
method | result | size |
default | \(\frac {\frac {\left (A a \,e^{2}+3 A c \,d^{2}+2 B a d e \right ) x^{3}}{8 a^{2}}-\frac {B \,e^{2} x^{2}}{2 c}-\frac {\left (A a \,e^{2}-5 A c \,d^{2}+2 B a d e \right ) x}{8 a c}-\frac {2 A c d e +B a \,e^{2}+B c \,d^{2}}{4 c^{2}}}{\left (c \,x^{2}+a \right )^{2}}+\frac {\left (A a \,e^{2}+3 A c \,d^{2}+2 B a d e \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 a^{2} c \sqrt {a c}}\) | \(148\) |
risch | \(\frac {\frac {\left (A a \,e^{2}+3 A c \,d^{2}+2 B a d e \right ) x^{3}}{8 a^{2}}-\frac {B \,e^{2} x^{2}}{2 c}-\frac {\left (A a \,e^{2}-5 A c \,d^{2}+2 B a d e \right ) x}{8 a c}-\frac {2 A c d e +B a \,e^{2}+B c \,d^{2}}{4 c^{2}}}{\left (c \,x^{2}+a \right )^{2}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) A \,e^{2}}{16 \sqrt {-a c}\, c a}-\frac {3 \ln \left (c x +\sqrt {-a c}\right ) A \,d^{2}}{16 \sqrt {-a c}\, a^{2}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) B d e}{8 \sqrt {-a c}\, c a}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) A \,e^{2}}{16 \sqrt {-a c}\, c a}+\frac {3 \ln \left (-c x +\sqrt {-a c}\right ) A \,d^{2}}{16 \sqrt {-a c}\, a^{2}}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) B d e}{8 \sqrt {-a c}\, c a}\) | \(275\) |
(1/8*(A*a*e^2+3*A*c*d^2+2*B*a*d*e)/a^2*x^3-1/2*B*e^2*x^2/c-1/8*(A*a*e^2-5* A*c*d^2+2*B*a*d*e)/a/c*x-1/4*(2*A*c*d*e+B*a*e^2+B*c*d^2)/c^2)/(c*x^2+a)^2+ 1/8*(A*a*e^2+3*A*c*d^2+2*B*a*d*e)/a^2/c/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2) )
Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (128) = 256\).
Time = 0.77 (sec) , antiderivative size = 537, normalized size of antiderivative = 3.78 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=\left [-\frac {8 \, B a^{3} c e^{2} x^{2} + 4 \, B a^{3} c d^{2} + 8 \, A a^{3} c d e + 4 \, B a^{4} e^{2} - 2 \, {\left (3 \, A a c^{3} d^{2} + 2 \, B a^{2} c^{2} d e + A a^{2} c^{2} e^{2}\right )} x^{3} + {\left (3 \, A a^{2} c d^{2} + 2 \, B a^{3} d e + A a^{3} e^{2} + {\left (3 \, A c^{3} d^{2} + 2 \, B a c^{2} d e + A a c^{2} e^{2}\right )} x^{4} + 2 \, {\left (3 \, A a c^{2} d^{2} + 2 \, B a^{2} c d e + A a^{2} c e^{2}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (5 \, A a^{2} c^{2} d^{2} - 2 \, B a^{3} c d e - A a^{3} c e^{2}\right )} x}{16 \, {\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}, -\frac {4 \, B a^{3} c e^{2} x^{2} + 2 \, B a^{3} c d^{2} + 4 \, A a^{3} c d e + 2 \, B a^{4} e^{2} - {\left (3 \, A a c^{3} d^{2} + 2 \, B a^{2} c^{2} d e + A a^{2} c^{2} e^{2}\right )} x^{3} - {\left (3 \, A a^{2} c d^{2} + 2 \, B a^{3} d e + A a^{3} e^{2} + {\left (3 \, A c^{3} d^{2} + 2 \, B a c^{2} d e + A a c^{2} e^{2}\right )} x^{4} + 2 \, {\left (3 \, A a c^{2} d^{2} + 2 \, B a^{2} c d e + A a^{2} c e^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (5 \, A a^{2} c^{2} d^{2} - 2 \, B a^{3} c d e - A a^{3} c e^{2}\right )} x}{8 \, {\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}\right ] \]
[-1/16*(8*B*a^3*c*e^2*x^2 + 4*B*a^3*c*d^2 + 8*A*a^3*c*d*e + 4*B*a^4*e^2 - 2*(3*A*a*c^3*d^2 + 2*B*a^2*c^2*d*e + A*a^2*c^2*e^2)*x^3 + (3*A*a^2*c*d^2 + 2*B*a^3*d*e + A*a^3*e^2 + (3*A*c^3*d^2 + 2*B*a*c^2*d*e + A*a*c^2*e^2)*x^4 + 2*(3*A*a*c^2*d^2 + 2*B*a^2*c*d*e + A*a^2*c*e^2)*x^2)*sqrt(-a*c)*log((c* x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 2*(5*A*a^2*c^2*d^2 - 2*B*a^3*c*d* e - A*a^3*c*e^2)*x)/(a^3*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2), -1/8*(4*B*a^3 *c*e^2*x^2 + 2*B*a^3*c*d^2 + 4*A*a^3*c*d*e + 2*B*a^4*e^2 - (3*A*a*c^3*d^2 + 2*B*a^2*c^2*d*e + A*a^2*c^2*e^2)*x^3 - (3*A*a^2*c*d^2 + 2*B*a^3*d*e + A* a^3*e^2 + (3*A*c^3*d^2 + 2*B*a*c^2*d*e + A*a*c^2*e^2)*x^4 + 2*(3*A*a*c^2*d ^2 + 2*B*a^2*c*d*e + A*a^2*c*e^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - ( 5*A*a^2*c^2*d^2 - 2*B*a^3*c*d*e - A*a^3*c*e^2)*x)/(a^3*c^4*x^4 + 2*a^4*c^3 *x^2 + a^5*c^2)]
Time = 3.04 (sec) , antiderivative size = 274, normalized size of antiderivative = 1.93 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{a^{5} c^{3}}} \left (A a e^{2} + 3 A c d^{2} + 2 B a d e\right ) \log {\left (- a^{3} c \sqrt {- \frac {1}{a^{5} c^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{5} c^{3}}} \left (A a e^{2} + 3 A c d^{2} + 2 B a d e\right ) \log {\left (a^{3} c \sqrt {- \frac {1}{a^{5} c^{3}}} + x \right )}}{16} + \frac {- 4 A a^{2} c d e - 2 B a^{3} e^{2} - 2 B a^{2} c d^{2} - 4 B a^{2} c e^{2} x^{2} + x^{3} \left (A a c^{2} e^{2} + 3 A c^{3} d^{2} + 2 B a c^{2} d e\right ) + x \left (- A a^{2} c e^{2} + 5 A a c^{2} d^{2} - 2 B a^{2} c d e\right )}{8 a^{4} c^{2} + 16 a^{3} c^{3} x^{2} + 8 a^{2} c^{4} x^{4}} \]
-sqrt(-1/(a**5*c**3))*(A*a*e**2 + 3*A*c*d**2 + 2*B*a*d*e)*log(-a**3*c*sqrt (-1/(a**5*c**3)) + x)/16 + sqrt(-1/(a**5*c**3))*(A*a*e**2 + 3*A*c*d**2 + 2 *B*a*d*e)*log(a**3*c*sqrt(-1/(a**5*c**3)) + x)/16 + (-4*A*a**2*c*d*e - 2*B *a**3*e**2 - 2*B*a**2*c*d**2 - 4*B*a**2*c*e**2*x**2 + x**3*(A*a*c**2*e**2 + 3*A*c**3*d**2 + 2*B*a*c**2*d*e) + x*(-A*a**2*c*e**2 + 5*A*a*c**2*d**2 - 2*B*a**2*c*d*e))/(8*a**4*c**2 + 16*a**3*c**3*x**2 + 8*a**2*c**4*x**4)
Time = 0.29 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.30 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=-\frac {4 \, B a^{2} c e^{2} x^{2} + 2 \, B a^{2} c d^{2} + 4 \, A a^{2} c d e + 2 \, B a^{3} e^{2} - {\left (3 \, A c^{3} d^{2} + 2 \, B a c^{2} d e + A a c^{2} e^{2}\right )} x^{3} - {\left (5 \, A a c^{2} d^{2} - 2 \, B a^{2} c d e - A a^{2} c e^{2}\right )} x}{8 \, {\left (a^{2} c^{4} x^{4} + 2 \, a^{3} c^{3} x^{2} + a^{4} c^{2}\right )}} + \frac {{\left (3 \, A c d^{2} + 2 \, B a d e + A a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} \]
-1/8*(4*B*a^2*c*e^2*x^2 + 2*B*a^2*c*d^2 + 4*A*a^2*c*d*e + 2*B*a^3*e^2 - (3 *A*c^3*d^2 + 2*B*a*c^2*d*e + A*a*c^2*e^2)*x^3 - (5*A*a*c^2*d^2 - 2*B*a^2*c *d*e - A*a^2*c*e^2)*x)/(a^2*c^4*x^4 + 2*a^3*c^3*x^2 + a^4*c^2) + 1/8*(3*A* c*d^2 + 2*B*a*d*e + A*a*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c)
Time = 0.26 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.20 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=\frac {{\left (3 \, A c d^{2} + 2 \, B a d e + A a e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} + \frac {3 \, A c^{3} d^{2} x^{3} + 2 \, B a c^{2} d e x^{3} + A a c^{2} e^{2} x^{3} - 4 \, B a^{2} c e^{2} x^{2} + 5 \, A a c^{2} d^{2} x - 2 \, B a^{2} c d e x - A a^{2} c e^{2} x - 2 \, B a^{2} c d^{2} - 4 \, A a^{2} c d e - 2 \, B a^{3} e^{2}}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c^{2}} \]
1/8*(3*A*c*d^2 + 2*B*a*d*e + A*a*e^2)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2 *c) + 1/8*(3*A*c^3*d^2*x^3 + 2*B*a*c^2*d*e*x^3 + A*a*c^2*e^2*x^3 - 4*B*a^2 *c*e^2*x^2 + 5*A*a*c^2*d^2*x - 2*B*a^2*c*d*e*x - A*a^2*c*e^2*x - 2*B*a^2*c *d^2 - 4*A*a^2*c*d*e - 2*B*a^3*e^2)/((c*x^2 + a)^2*a^2*c^2)
Time = 10.88 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) (d+e x)^2}{\left (a+c x^2\right )^3} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (3\,A\,c\,d^2+2\,B\,a\,d\,e+A\,a\,e^2\right )}{8\,a^{5/2}\,c^{3/2}}-\frac {\frac {B\,c\,d^2+2\,A\,c\,d\,e+B\,a\,e^2}{4\,c^2}-\frac {x^3\,\left (3\,A\,c\,d^2+2\,B\,a\,d\,e+A\,a\,e^2\right )}{8\,a^2}+\frac {x\,\left (-5\,A\,c\,d^2+2\,B\,a\,d\,e+A\,a\,e^2\right )}{8\,a\,c}+\frac {B\,e^2\,x^2}{2\,c}}{a^2+2\,a\,c\,x^2+c^2\,x^4} \]