Integrand size = 20, antiderivative size = 98 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^3} \, dx=-\frac {a B-(A c-a C) x}{4 a c \left (a+c x^2\right )^2}+\frac {(3 A c+a C) x}{8 a^2 c \left (a+c x^2\right )}+\frac {(3 A c+a C) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{3/2}} \]
1/4*(-a*B+(A*c-C*a)*x)/a/c/(c*x^2+a)^2+1/8*(3*A*c+C*a)*x/a^2/c/(c*x^2+a)+1 /8*(3*A*c+C*a)*arctan(x*c^(1/2)/a^(1/2))/a^(5/2)/c^(3/2)
Time = 0.05 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.92 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^3} \, dx=\frac {3 A c^2 x^3-a^2 (2 B+C x)+a c x \left (5 A+C x^2\right )}{8 a^2 c \left (a+c x^2\right )^2}+\frac {(3 A c+a C) \arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{3/2}} \]
(3*A*c^2*x^3 - a^2*(2*B + C*x) + a*c*x*(5*A + C*x^2))/(8*a^2*c*(a + c*x^2) ^2) + ((3*A*c + a*C)*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(3/2))
Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2345, 25, 27, 215, 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.
\(\displaystyle \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {\int -\frac {3 A c+a C}{c \left (c x^2+a\right )^2}dx}{4 a}-\frac {a B-x (A c-a C)}{4 a c \left (a+c x^2\right )^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {3 A c+a C}{c \left (c x^2+a\right )^2}dx}{4 a}-\frac {a B-x (A c-a C)}{4 a c \left (a+c x^2\right )^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(a C+3 A c) \int \frac {1}{\left (c x^2+a\right )^2}dx}{4 a c}-\frac {a B-x (A c-a C)}{4 a c \left (a+c x^2\right )^2}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {(a C+3 A c) \left (\frac {\int \frac {1}{c x^2+a}dx}{2 a}+\frac {x}{2 a \left (a+c x^2\right )}\right )}{4 a c}-\frac {a B-x (A c-a C)}{4 a c \left (a+c x^2\right )^2}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {(a C+3 A c) \left (\frac {\arctan \left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{2 a^{3/2} \sqrt {c}}+\frac {x}{2 a \left (a+c x^2\right )}\right )}{4 a c}-\frac {a B-x (A c-a C)}{4 a c \left (a+c x^2\right )^2}\) |
-1/4*(a*B - (A*c - a*C)*x)/(a*c*(a + c*x^2)^2) + ((3*A*c + a*C)*(x/(2*a*(a + c*x^2)) + ArcTan[(Sqrt[c]*x)/Sqrt[a]]/(2*a^(3/2)*Sqrt[c])))/(4*a*c)
3.1.60.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
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[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
Time = 0.55 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.85
method | result | size |
default | \(\frac {\frac {\left (3 A c +C a \right ) x^{3}}{8 a^{2}}+\frac {\left (5 A c -C a \right ) x}{8 a c}-\frac {B}{4 c}}{\left (c \,x^{2}+a \right )^{2}}+\frac {\left (3 A c +C a \right ) \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 a^{2} c \sqrt {a c}}\) | \(83\) |
risch | \(\frac {\frac {\left (3 A c +C a \right ) x^{3}}{8 a^{2}}+\frac {\left (5 A c -C a \right ) x}{8 a c}-\frac {B}{4 c}}{\left (c \,x^{2}+a \right )^{2}}-\frac {3 A \ln \left (c x +\sqrt {-a c}\right )}{16 \sqrt {-a c}\, a^{2}}-\frac {\ln \left (c x +\sqrt {-a c}\right ) C}{16 \sqrt {-a c}\, c a}+\frac {3 A \ln \left (-c x +\sqrt {-a c}\right )}{16 \sqrt {-a c}\, a^{2}}+\frac {\ln \left (-c x +\sqrt {-a c}\right ) C}{16 \sqrt {-a c}\, c a}\) | \(153\) |
(1/8*(3*A*c+C*a)/a^2*x^3+1/8*(5*A*c-C*a)/a/c*x-1/4*B/c)/(c*x^2+a)^2+1/8*(3 *A*c+C*a)/a^2/c/(a*c)^(1/2)*arctan(c*x/(a*c)^(1/2))
Time = 0.44 (sec) , antiderivative size = 314, normalized size of antiderivative = 3.20 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^3} \, dx=\left [-\frac {4 \, B a^{3} c - 2 \, {\left (C a^{2} c^{2} + 3 \, A a c^{3}\right )} x^{3} + {\left ({\left (C a c^{2} + 3 \, A c^{3}\right )} x^{4} + C a^{3} + 3 \, A a^{2} c + 2 \, {\left (C a^{2} c + 3 \, A a c^{2}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 2 \, {\left (C a^{3} c - 5 \, A a^{2} c^{2}\right )} x}{16 \, {\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}, -\frac {2 \, B a^{3} c - {\left (C a^{2} c^{2} + 3 \, A a c^{3}\right )} x^{3} - {\left ({\left (C a c^{2} + 3 \, A c^{3}\right )} x^{4} + C a^{3} + 3 \, A a^{2} c + 2 \, {\left (C a^{2} c + 3 \, A a c^{2}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + {\left (C a^{3} c - 5 \, A a^{2} c^{2}\right )} x}{8 \, {\left (a^{3} c^{4} x^{4} + 2 \, a^{4} c^{3} x^{2} + a^{5} c^{2}\right )}}\right ] \]
[-1/16*(4*B*a^3*c - 2*(C*a^2*c^2 + 3*A*a*c^3)*x^3 + ((C*a*c^2 + 3*A*c^3)*x ^4 + C*a^3 + 3*A*a^2*c + 2*(C*a^2*c + 3*A*a*c^2)*x^2)*sqrt(-a*c)*log((c*x^ 2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 2*(C*a^3*c - 5*A*a^2*c^2)*x)/(a^3*c ^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2), -1/8*(2*B*a^3*c - (C*a^2*c^2 + 3*A*a*c^ 3)*x^3 - ((C*a*c^2 + 3*A*c^3)*x^4 + C*a^3 + 3*A*a^2*c + 2*(C*a^2*c + 3*A*a *c^2)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + (C*a^3*c - 5*A*a^2*c^2)*x)/(a ^3*c^4*x^4 + 2*a^4*c^3*x^2 + a^5*c^2)]
Time = 0.59 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.59 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^3} \, dx=- \frac {\sqrt {- \frac {1}{a^{5} c^{3}}} \cdot \left (3 A c + C a\right ) \log {\left (- a^{3} c \sqrt {- \frac {1}{a^{5} c^{3}}} + x \right )}}{16} + \frac {\sqrt {- \frac {1}{a^{5} c^{3}}} \cdot \left (3 A c + C a\right ) \log {\left (a^{3} c \sqrt {- \frac {1}{a^{5} c^{3}}} + x \right )}}{16} + \frac {- 2 B a^{2} + x^{3} \cdot \left (3 A c^{2} + C a c\right ) + x \left (5 A a c - C a^{2}\right )}{8 a^{4} c + 16 a^{3} c^{2} x^{2} + 8 a^{2} c^{3} x^{4}} \]
-sqrt(-1/(a**5*c**3))*(3*A*c + C*a)*log(-a**3*c*sqrt(-1/(a**5*c**3)) + x)/ 16 + sqrt(-1/(a**5*c**3))*(3*A*c + C*a)*log(a**3*c*sqrt(-1/(a**5*c**3)) + x)/16 + (-2*B*a**2 + x**3*(3*A*c**2 + C*a*c) + x*(5*A*a*c - C*a**2))/(8*a* *4*c + 16*a**3*c**2*x**2 + 8*a**2*c**3*x**4)
Time = 0.28 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.00 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^3} \, dx=\frac {{\left (C a c + 3 \, A c^{2}\right )} x^{3} - 2 \, B a^{2} - {\left (C a^{2} - 5 \, A a c\right )} x}{8 \, {\left (a^{2} c^{3} x^{4} + 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )}} + \frac {{\left (C a + 3 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} \]
1/8*((C*a*c + 3*A*c^2)*x^3 - 2*B*a^2 - (C*a^2 - 5*A*a*c)*x)/(a^2*c^3*x^4 + 2*a^3*c^2*x^2 + a^4*c) + 1/8*(C*a + 3*A*c)*arctan(c*x/sqrt(a*c))/(sqrt(a* c)*a^2*c)
Time = 0.27 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.86 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^3} \, dx=\frac {{\left (C a + 3 \, A c\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \, \sqrt {a c} a^{2} c} + \frac {C a c x^{3} + 3 \, A c^{2} x^{3} - C a^{2} x + 5 \, A a c x - 2 \, B a^{2}}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} c} \]
1/8*(C*a + 3*A*c)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*a^2*c) + 1/8*(C*a*c*x^3 + 3*A*c^2*x^3 - C*a^2*x + 5*A*a*c*x - 2*B*a^2)/((c*x^2 + a)^2*a^2*c)
Time = 12.90 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.90 \[ \int \frac {A+B x+C x^2}{\left (a+c x^2\right )^3} \, dx=\frac {\frac {x^3\,\left (3\,A\,c+C\,a\right )}{8\,a^2}-\frac {B}{4\,c}+\frac {x\,\left (5\,A\,c-C\,a\right )}{8\,a\,c}}{a^2+2\,a\,c\,x^2+c^2\,x^4}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (3\,A\,c+C\,a\right )}{8\,a^{5/2}\,c^{3/2}} \]