Integrand size = 27, antiderivative size = 457 \[ \int \frac {1}{x^3 \sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=-\frac {\sqrt {a+c x^2}}{2 a d x^2}+\frac {e \sqrt {a+c x^2}}{a d^2 x}+\frac {f \left (2 e^3-4 d e f-\left (e^2-d f\right ) \left (e-\sqrt {e^2-4 d f}\right )\right ) \text {arctanh}\left (\frac {2 a f-c \left (e-\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d^3 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f-e \sqrt {e^2-4 d f}\right )}}-\frac {f \left (2 e^3-4 d e f-\left (e^2-d f\right ) \left (e+\sqrt {e^2-4 d f}\right )\right ) \text {arctanh}\left (\frac {2 a f-c \left (e+\sqrt {e^2-4 d f}\right ) x}{\sqrt {2} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )} \sqrt {a+c x^2}}\right )}{\sqrt {2} d^3 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e^2-2 d f+e \sqrt {e^2-4 d f}\right )}}+\frac {c \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d}-\frac {\left (e^2-d f\right ) \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^3} \]
1/2*c*arctanh((c*x^2+a)^(1/2)/a^(1/2))/a^(3/2)/d-(-d*f+e^2)*arctanh((c*x^2 +a)^(1/2)/a^(1/2))/d^3/a^(1/2)-1/2*(c*x^2+a)^(1/2)/a/d/x^2+e*(c*x^2+a)^(1/ 2)/a/d^2/x+1/2*f*arctanh(1/2*(2*a*f-c*x*(e-(-4*d*f+e^2)^(1/2)))*2^(1/2)/(c *x^2+a)^(1/2)/(2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2))*(2*e^3-4 *d*e*f-(-d*f+e^2)*(e-(-4*d*f+e^2)^(1/2)))/d^3*2^(1/2)/(-4*d*f+e^2)^(1/2)/( 2*a*f^2+c*(e^2-2*d*f-e*(-4*d*f+e^2)^(1/2)))^(1/2)-1/2*f*arctanh(1/2*(2*a*f -c*x*(e+(-4*d*f+e^2)^(1/2)))*2^(1/2)/(c*x^2+a)^(1/2)/(2*a*f^2+c*(e^2-2*d*f +e*(-4*d*f+e^2)^(1/2)))^(1/2))*(2*e^3-4*d*e*f-(-d*f+e^2)*(e+(-4*d*f+e^2)^( 1/2)))/d^3*2^(1/2)/(-4*d*f+e^2)^(1/2)/(2*a*f^2+c*(e^2-2*d*f+e*(-4*d*f+e^2) ^(1/2)))^(1/2)
Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
Time = 0.61 (sec) , antiderivative size = 422, normalized size of antiderivative = 0.92 \[ \int \frac {1}{x^3 \sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\frac {\frac {d (-d+2 e x) \sqrt {a+c x^2}}{a x^2}-\frac {2 c d^2 \text {arctanh}\left (\frac {\sqrt {c} x-\sqrt {a+c x^2}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {4 \left (e^2-d f\right ) \text {arctanh}\left (\frac {-\sqrt {c} x+\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a}}+2 \text {RootSum}\left [a^2 f+2 a \sqrt {c} e \text {$\#$1}+4 c d \text {$\#$1}^2-2 a f \text {$\#$1}^2-2 \sqrt {c} e \text {$\#$1}^3+f \text {$\#$1}^4\&,\frac {a e^2 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )-a d f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right )+2 \sqrt {c} e^3 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-4 \sqrt {c} d e f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}-e^2 f \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2+d f^2 \log \left (-\sqrt {c} x+\sqrt {a+c x^2}-\text {$\#$1}\right ) \text {$\#$1}^2}{a \sqrt {c} e+4 c d \text {$\#$1}-2 a f \text {$\#$1}-3 \sqrt {c} e \text {$\#$1}^2+2 f \text {$\#$1}^3}\&\right ]}{2 d^3} \]
((d*(-d + 2*e*x)*Sqrt[a + c*x^2])/(a*x^2) - (2*c*d^2*ArcTanh[(Sqrt[c]*x - Sqrt[a + c*x^2])/Sqrt[a]])/a^(3/2) - (4*(e^2 - d*f)*ArcTanh[(-(Sqrt[c]*x) + Sqrt[a + c*x^2])/Sqrt[a]])/Sqrt[a] + 2*RootSum[a^2*f + 2*a*Sqrt[c]*e*#1 + 4*c*d*#1^2 - 2*a*f*#1^2 - 2*Sqrt[c]*e*#1^3 + f*#1^4 & , (a*e^2*f*Log[-(S qrt[c]*x) + Sqrt[a + c*x^2] - #1] - a*d*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c* x^2] - #1] + 2*Sqrt[c]*e^3*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - 4 *Sqrt[c]*d*e*f*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1 - e^2*f*Log[-(S qrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2 + d*f^2*Log[-(Sqrt[c]*x) + Sqrt[a + c*x^2] - #1]*#1^2)/(a*Sqrt[c]*e + 4*c*d*#1 - 2*a*f*#1 - 3*Sqrt[c]*e*#1^2 + 2*f*#1^3) & ])/(2*d^3)
Time = 1.59 (sec) , antiderivative size = 457, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {7279, 2009}
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.
| \(\displaystyle \int \frac {1}{x^3 \sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx\) |
| \(\Big \downarrow \) 7279 |
| \(\displaystyle \int \left (\frac {e^2-d f}{d^3 x \sqrt {a+c x^2}}+\frac {-f x \left (e^2-d f\right )-e \left (e^2-2 d f\right )}{d^3 \sqrt {a+c x^2} \left (d+e x+f x^2\right )}-\frac {e}{d^2 x^2 \sqrt {a+c x^2}}+\frac {1}{d x^3 \sqrt {a+c x^2}}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {c \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{2 a^{3/2} d}-\frac {\left (e^2-d f\right ) \text {arctanh}\left (\frac {\sqrt {a+c x^2}}{\sqrt {a}}\right )}{\sqrt {a} d^3}+\frac {f \left (-\left (e^2-d f\right ) \left (e-\sqrt {e^2-4 d f}\right )-4 d e f+2 e^3\right ) \text {arctanh}\left (\frac {2 a f-c x \left (e-\sqrt {e^2-4 d f}\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d^3 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (-e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}-\frac {f \left (-\left (e^2-d f\right ) \left (\sqrt {e^2-4 d f}+e\right )-4 d e f+2 e^3\right ) \text {arctanh}\left (\frac {2 a f-c x \left (\sqrt {e^2-4 d f}+e\right )}{\sqrt {2} \sqrt {a+c x^2} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}\right )}{\sqrt {2} d^3 \sqrt {e^2-4 d f} \sqrt {2 a f^2+c \left (e \sqrt {e^2-4 d f}-2 d f+e^2\right )}}+\frac {e \sqrt {a+c x^2}}{a d^2 x}-\frac {\sqrt {a+c x^2}}{2 a d x^2}\) |
-1/2*Sqrt[a + c*x^2]/(a*d*x^2) + (e*Sqrt[a + c*x^2])/(a*d^2*x) + (f*(2*e^3 - 4*d*e*f - (e^2 - d*f)*(e - Sqrt[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c*(e - Sqrt[e^2 - 4*d*f])*x)/(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c*x^2])])/(Sqrt[2]*d^3*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f - e*Sqrt[e^2 - 4*d*f])]) - (f*(2*e^3 - 4*d*e*f - (e^2 - d *f)*(e + Sqrt[e^2 - 4*d*f]))*ArcTanh[(2*a*f - c*(e + Sqrt[e^2 - 4*d*f])*x) /(Sqrt[2]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e*Sqrt[e^2 - 4*d*f])]*Sqrt[a + c *x^2])])/(Sqrt[2]*d^3*Sqrt[e^2 - 4*d*f]*Sqrt[2*a*f^2 + c*(e^2 - 2*d*f + e* Sqrt[e^2 - 4*d*f])]) + (c*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(2*a^(3/2)*d) - ((e^2 - d*f)*ArcTanh[Sqrt[a + c*x^2]/Sqrt[a]])/(Sqrt[a]*d^3)
3.1.70.3.1 Defintions of rubi rules used
Int[(u_)/((a_.) + (b_.)*(x_)^(n_.) + (c_.)*(x_)^(n2_.)), x_Symbol] :> With[ {v = RationalFunctionExpand[u/(a + b*x^n + c*x^(2*n)), x]}, Int[v, x] /; Su mQ[v]] /; FreeQ[{a, b, c}, x] && EqQ[n2, 2*n] && IGtQ[n, 0]
Time = 0.85 (sec) , antiderivative size = 777, normalized size of antiderivative = 1.70
| method | result | size |
| risch | \(-\frac {\sqrt {c \,x^{2}+a}\, \left (-2 e x +d \right )}{2 a \,d^{2} x^{2}}-\frac {\frac {4 f \left (2 a d f -2 e^{2} a +c \,d^{2}\right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \left (e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {a}}-\frac {2 f a \left (e \sqrt {-4 d f +e^{2}}+2 d f -e^{2}\right ) \sqrt {2}\, \ln \left (\frac {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 {\left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2} c -\frac {4 c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\sqrt {-4 d f +e^{2}}\, \left (e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}+\frac {2 f a \left (e^{2}-2 d f +e \sqrt {-4 d f +e^{2}}\right ) \sqrt {2}\, \ln \left (\frac {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 {\left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2} c -\frac {4 c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\sqrt {-4 d f +e^{2}}\, \left (-e +\sqrt {-4 d f +e^{2}}\right ) \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}}{2 a \,d^{2}}\) | \(777\) |
| default | \(-\frac {4 f \left (-\frac {\sqrt {c \,x^{2}+a}}{2 a \,x^{2}}+\frac {c \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{2 a^{\frac {3}{2}}}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right ) \left (e +\sqrt {-4 d f +e^{2}}\right )}+\frac {16 f^{2} e \sqrt {c \,x^{2}+a}}{\left (-e +\sqrt {-4 d f +e^{2}}\right )^{2} \left (e +\sqrt {-4 d f +e^{2}}\right )^{2} a x}-\frac {64 f^{3} \left (d f -e^{2}\right ) \ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {c \,x^{2}+a}}{x}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right )^{3} \left (e +\sqrt {-4 d f +e^{2}}\right )^{3} \sqrt {a}}-\frac {8 f^{3} \sqrt {2}\, \ln \left (\frac {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 {\left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2} c -\frac {4 c \left (e +\sqrt {-4 d f +e^{2}}\right ) \left (x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x +\frac {e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (e +\sqrt {-4 d f +e^{2}}\right )^{3} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}-\frac {8 f^{3} \sqrt {2}\, \ln \left (\frac {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}-\frac {c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {\sqrt {2}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}\, \sqrt {4 {\left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}^{2} c -\frac {4 c \left (e -\sqrt {-4 d f +e^{2}}\right ) \left (x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}\right )}{f}+\frac {-2 \sqrt {-4 d f +e^{2}}\, c e +4 a \,f^{2}-4 c d f +2 c \,e^{2}}{f^{2}}}}{2}}{x -\frac {-e +\sqrt {-4 d f +e^{2}}}{2 f}}\right )}{\left (-e +\sqrt {-4 d f +e^{2}}\right )^{3} \sqrt {-4 d f +e^{2}}\, \sqrt {\frac {-\sqrt {-4 d f +e^{2}}\, c e +2 a \,f^{2}-2 c d f +c \,e^{2}}{f^{2}}}}\) | \(827\) |
-1/2*(c*x^2+a)^(1/2)*(-2*e*x+d)/a/d^2/x^2-1/2/a/d^2*(4*f*(2*a*d*f-2*a*e^2+ c*d^2)/(-e+(-4*d*f+e^2)^(1/2))/(e+(-4*d*f+e^2)^(1/2))/a^(1/2)*ln((2*a+2*a^ (1/2)*(c*x^2+a)^(1/2))/x)-2*f*a*(e*(-4*d*f+e^2)^(1/2)+2*d*f-e^2)/(-4*d*f+e ^2)^(1/2)/(e+(-4*d*f+e^2)^(1/2))*2^(1/2)/(((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2- 2*c*d*f+c*e^2)/f^2)^(1/2)*ln((((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^ 2)/f^2-c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f)+1/2*2^( 1/2)*(((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*(4*(x+1/2* (e+(-4*d*f+e^2)^(1/2))/f)^2*c-4*c*(e+(-4*d*f+e^2)^(1/2))/f*(x+1/2*(e+(-4*d *f+e^2)^(1/2))/f)+2*((-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1 /2))/(x+1/2*(e+(-4*d*f+e^2)^(1/2))/f))+2*f*a*(e^2-2*d*f+e*(-4*d*f+e^2)^(1/ 2))/(-4*d*f+e^2)^(1/2)/(-e+(-4*d*f+e^2)^(1/2))*2^(1/2)/((-(-4*d*f+e^2)^(1/ 2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2)^(1/2)*ln(((-(-4*d*f+e^2)^(1/2)*c*e+2*a* f^2-2*c*d*f+c*e^2)/f^2-c*(e-(-4*d*f+e^2)^(1/2))/f*(x-1/2/f*(-e+(-4*d*f+e^2 )^(1/2)))+1/2*2^(1/2)*((-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2*c*d*f+c*e^2)/f^2 )^(1/2)*(4*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))^2*c-4*c*(e-(-4*d*f+e^2)^(1/2) )/f*(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))+2*(-(-4*d*f+e^2)^(1/2)*c*e+2*a*f^2-2 *c*d*f+c*e^2)/f^2)^(1/2))/(x-1/2/f*(-e+(-4*d*f+e^2)^(1/2)))))
Timed out. \[ \int \frac {1}{x^3 \sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \]
\[ \int \frac {1}{x^3 \sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {1}{x^{3} \sqrt {a + c x^{2}} \left (d + e x + f x^{2}\right )}\, dx \]
\[ \int \frac {1}{x^3 \sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\int { \frac {1}{\sqrt {c x^{2} + a} {\left (f x^{2} + e x + d\right )} x^{3}} \,d x } \]
Timed out. \[ \int \frac {1}{x^3 \sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {1}{x^3 \sqrt {a+c x^2} \left (d+e x+f x^2\right )} \, dx=\int \frac {1}{x^3\,\sqrt {c\,x^2+a}\,\left (f\,x^2+e\,x+d\right )} \,d x \]