Integrand size = 29, antiderivative size = 218 \[ \int \frac {1}{x^3 \left (d+e x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx=-\frac {\sqrt {a+b x^2+c x^4}}{2 a d x^2}+\frac {b \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{4 a^{3/2} d}+\frac {e \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 \sqrt {a} d^2}+\frac {e^2 \text {arctanh}\left (\frac {b d-2 a e+(2 c d-b e) x^2}{2 \sqrt {c d^2-b d e+a e^2} \sqrt {a+b x^2+c x^4}}\right )}{2 d^2 \sqrt {c d^2-b d e+a e^2}} \]
1/4*b*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2+a)^(1/2))/a^(3/2)/d+1/2 *e*arctanh(1/2*(b*x^2+2*a)/a^(1/2)/(c*x^4+b*x^2+a)^(1/2))/d^2/a^(1/2)+1/2* e^2*arctanh(1/2*(b*d-2*a*e+(-b*e+2*c*d)*x^2)/(a*e^2-b*d*e+c*d^2)^(1/2)/(c* x^4+b*x^2+a)^(1/2))/d^2/(a*e^2-b*d*e+c*d^2)^(1/2)-1/2*(c*x^4+b*x^2+a)^(1/2 )/a/d/x^2
Time = 1.01 (sec) , antiderivative size = 209, normalized size of antiderivative = 0.96 \[ \int \frac {1}{x^3 \left (d+e x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx=\frac {-\frac {2 d \sqrt {a+b x^2+c x^4}}{a x^2}+\frac {4 e^2 \sqrt {-c d^2+b d e-a e^2} \arctan \left (\frac {\sqrt {-c d^2+b d e-a e^2} x^2}{\sqrt {a} \left (d+e x^2\right )-d \sqrt {a+b x^2+c x^4}}\right )}{c d^2+e (-b d+a e)}+\frac {(b d+2 a e) \log \left (x^2\right )}{a^{3/2}}-\frac {(b d+2 a e) \log \left (a d^2 \left (2 a+b x^2-2 \sqrt {a} \sqrt {a+b x^2+c x^4}\right )\right )}{a^{3/2}}}{4 d^2} \]
((-2*d*Sqrt[a + b*x^2 + c*x^4])/(a*x^2) + (4*e^2*Sqrt[-(c*d^2) + b*d*e - a *e^2]*ArcTan[(Sqrt[-(c*d^2) + b*d*e - a*e^2]*x^2)/(Sqrt[a]*(d + e*x^2) - d *Sqrt[a + b*x^2 + c*x^4])])/(c*d^2 + e*(-(b*d) + a*e)) + ((b*d + 2*a*e)*Lo g[x^2])/a^(3/2) - ((b*d + 2*a*e)*Log[a*d^2*(2*a + b*x^2 - 2*Sqrt[a]*Sqrt[a + b*x^2 + c*x^4])])/a^(3/2))/(4*d^2)
Time = 0.42 (sec) , antiderivative size = 214, normalized size of antiderivative = 0.98, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {1578, 1289, 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.
| \(\displaystyle \int \frac {1}{x^3 \left (d+e x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx\) |
| \(\Big \downarrow \) 1578 |
| \(\displaystyle \frac {1}{2} \int \frac {1}{x^4 \left (e x^2+d\right ) \sqrt {c x^4+b x^2+a}}dx^2\) |
| \(\Big \downarrow \) 1289 |
| \(\displaystyle \frac {1}{2} \int \left (\frac {e^2}{d^2 \left (e x^2+d\right ) \sqrt {c x^4+b x^2+a}}-\frac {e}{d^2 x^2 \sqrt {c x^4+b x^2+a}}+\frac {1}{d x^4 \sqrt {c x^4+b x^2+a}}\right )dx^2\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {1}{2} \left (\frac {b \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{2 a^{3/2} d}+\frac {e^2 \text {arctanh}\left (\frac {-2 a e+x^2 (2 c d-b e)+b d}{2 \sqrt {a+b x^2+c x^4} \sqrt {a e^2-b d e+c d^2}}\right )}{d^2 \sqrt {a e^2-b d e+c d^2}}+\frac {e \text {arctanh}\left (\frac {2 a+b x^2}{2 \sqrt {a} \sqrt {a+b x^2+c x^4}}\right )}{\sqrt {a} d^2}-\frac {\sqrt {a+b x^2+c x^4}}{a d x^2}\right )\) |
(-(Sqrt[a + b*x^2 + c*x^4]/(a*d*x^2)) + (b*ArcTanh[(2*a + b*x^2)/(2*Sqrt[a ]*Sqrt[a + b*x^2 + c*x^4])])/(2*a^(3/2)*d) + (e*ArcTanh[(2*a + b*x^2)/(2*S qrt[a]*Sqrt[a + b*x^2 + c*x^4])])/(Sqrt[a]*d^2) + (e^2*ArcTanh[(b*d - 2*a* e + (2*c*d - b*e)*x^2)/(2*Sqrt[c*d^2 - b*d*e + a*e^2]*Sqrt[a + b*x^2 + c*x ^4])])/(d^2*Sqrt[c*d^2 - b*d*e + a*e^2]))/2
3.4.36.3.1 Defintions of rubi rules used
Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.)*((a_.) + (b_.)*(x _) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && ( IntegerQ[p] || (ILtQ[m, 0] && ILtQ[n, 0]))
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + (c_.)*(x_ )^4)^(p_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(d + e*x)^q*(a + b*x + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && Int egerQ[(m - 1)/2]
Time = 0.46 (sec) , antiderivative size = 206, normalized size of antiderivative = 0.94
| method | result | size |
| pseudoelliptic | \(\frac {-\ln \left (\frac {2 \sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, e +\left (b \,x^{2}+2 a \right ) e -d \left (2 c \,x^{2}+b \right )}{e \,x^{2}+d}\right ) a^{\frac {3}{2}} e \,x^{2}+\sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \left (x^{2} \left (a e +\frac {b d}{2}\right ) \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )-\sqrt {c \,x^{4}+b \,x^{2}+a}\, \sqrt {a}\, d \right )}{2 a^{\frac {3}{2}} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, d^{2} x^{2}}\) | \(206\) |
| risch | \(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{2 a d \,x^{2}}-\frac {-\frac {\left (2 a e +b d \right ) \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 d \sqrt {a}}+\frac {a e \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x^{2}+\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x^{2}+\frac {d}{e}}\right )}{d \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}}{2 d a}\) | \(250\) |
| default | \(\frac {-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{2 a \,x^{2}}+\frac {b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 a^{\frac {3}{2}}}}{d}+\frac {e \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 d^{2} \sqrt {a}}-\frac {e \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x^{2}+\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x^{2}+\frac {d}{e}}\right )}{2 d^{2} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\) | \(275\) |
| elliptic | \(-\frac {\sqrt {c \,x^{4}+b \,x^{2}+a}}{2 a d \,x^{2}}+\frac {b \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{4 d \,a^{\frac {3}{2}}}+\frac {e \ln \left (\frac {2 a +b \,x^{2}+2 \sqrt {a}\, \sqrt {c \,x^{4}+b \,x^{2}+a}}{x^{2}}\right )}{2 d^{2} \sqrt {a}}-\frac {e \ln \left (\frac {\frac {2 a \,e^{2}-2 b d e +2 c \,d^{2}}{e^{2}}+\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+2 \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}\, \sqrt {c \left (x^{2}+\frac {d}{e}\right )^{2}+\frac {\left (b e -2 c d \right ) \left (x^{2}+\frac {d}{e}\right )}{e}+\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}{x^{2}+\frac {d}{e}}\right )}{2 d^{2} \sqrt {\frac {a \,e^{2}-b d e +c \,d^{2}}{e^{2}}}}\) | \(276\) |
1/2/a^(3/2)*(-ln((2*(c*x^4+b*x^2+a)^(1/2)*((a*e^2-b*d*e+c*d^2)/e^2)^(1/2)* e+(b*x^2+2*a)*e-d*(2*c*x^2+b))/(e*x^2+d))*a^(3/2)*e*x^2+((a*e^2-b*d*e+c*d^ 2)/e^2)^(1/2)*(x^2*(a*e+1/2*b*d)*ln((2*a+b*x^2+2*a^(1/2)*(c*x^4+b*x^2+a)^( 1/2))/x^2)-(c*x^4+b*x^2+a)^(1/2)*a^(1/2)*d))/((a*e^2-b*d*e+c*d^2)/e^2)^(1/ 2)/d^2/x^2
Time = 0.68 (sec) , antiderivative size = 1414, normalized size of antiderivative = 6.49 \[ \int \frac {1}{x^3 \left (d+e x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx=\text {Too large to display} \]
[1/8*(2*sqrt(c*d^2 - b*d*e + a*e^2)*a^2*e^2*x^2*log(-((8*c^2*d^2 - 8*b*c*d *e + (b^2 + 4*a*c)*e^2)*x^4 - 8*a*b*d*e + 8*a^2*e^2 + (b^2 + 4*a*c)*d^2 + 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x^2 + 4*sqrt(c*x^4 + b*x^2 + a)*sqrt(c*d^2 - b*d*e + a*e^2)*((2*c*d - b*e)*x^2 + b*d - 2*a*e))/(e^2* x^4 + 2*d*e*x^2 + d^2)) + (b*c*d^3 - a*b*d*e^2 + 2*a^2*e^3 - (b^2 - 2*a*c) *d^2*e)*sqrt(a)*x^2*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 + 4*sqrt(c*x^4 + b *x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^2)/x^4) - 4*(a*c*d^3 - a*b*d^2*e + a ^2*d*e^2)*sqrt(c*x^4 + b*x^2 + a))/((a^2*c*d^4 - a^2*b*d^3*e + a^3*d^2*e^2 )*x^2), 1/8*(4*sqrt(-c*d^2 + b*d*e - a*e^2)*a^2*e^2*x^2*arctan(-1/2*sqrt(c *x^4 + b*x^2 + a)*sqrt(-c*d^2 + b*d*e - a*e^2)*((2*c*d - b*e)*x^2 + b*d - 2*a*e)/((c^2*d^2 - b*c*d*e + a*c*e^2)*x^4 + a*c*d^2 - a*b*d*e + a^2*e^2 + (b*c*d^2 - b^2*d*e + a*b*e^2)*x^2)) + (b*c*d^3 - a*b*d*e^2 + 2*a^2*e^3 - ( b^2 - 2*a*c)*d^2*e)*sqrt(a)*x^2*log(-((b^2 + 4*a*c)*x^4 + 8*a*b*x^2 + 4*sq rt(c*x^4 + b*x^2 + a)*(b*x^2 + 2*a)*sqrt(a) + 8*a^2)/x^4) - 4*(a*c*d^3 - a *b*d^2*e + a^2*d*e^2)*sqrt(c*x^4 + b*x^2 + a))/((a^2*c*d^4 - a^2*b*d^3*e + a^3*d^2*e^2)*x^2), 1/4*(sqrt(c*d^2 - b*d*e + a*e^2)*a^2*e^2*x^2*log(-((8* c^2*d^2 - 8*b*c*d*e + (b^2 + 4*a*c)*e^2)*x^4 - 8*a*b*d*e + 8*a^2*e^2 + (b^ 2 + 4*a*c)*d^2 + 2*(4*b*c*d^2 + 4*a*b*e^2 - (3*b^2 + 4*a*c)*d*e)*x^2 + 4*s qrt(c*x^4 + b*x^2 + a)*sqrt(c*d^2 - b*d*e + a*e^2)*((2*c*d - b*e)*x^2 + b* d - 2*a*e))/(e^2*x^4 + 2*d*e*x^2 + d^2)) - (b*c*d^3 - a*b*d*e^2 + 2*a^2...
\[ \int \frac {1}{x^3 \left (d+e x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {1}{x^{3} \left (d + e x^{2}\right ) \sqrt {a + b x^{2} + c x^{4}}}\, dx \]
\[ \int \frac {1}{x^3 \left (d+e x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx=\int { \frac {1}{\sqrt {c x^{4} + b x^{2} + a} {\left (e x^{2} + d\right )} x^{3}} \,d x } \]
Time = 0.30 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.95 \[ \int \frac {1}{x^3 \left (d+e x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx=\frac {e^{2} \arctan \left (-\frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} + b d e - a e^{2}}}\right )}{\sqrt {-c d^{2} + b d e - a e^{2}} d^{2}} - \frac {{\left (b d + 2 \, a e\right )} \arctan \left (-\frac {\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}}{\sqrt {-a}}\right )}{2 \, \sqrt {-a} a d^{2}} + \frac {{\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )} b + 2 \, a \sqrt {c}}{2 \, {\left ({\left (\sqrt {c} x^{2} - \sqrt {c x^{4} + b x^{2} + a}\right )}^{2} - a\right )} a d} \]
e^2*arctan(-((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*e + sqrt(c)*d)/sqrt(- c*d^2 + b*d*e - a*e^2))/(sqrt(-c*d^2 + b*d*e - a*e^2)*d^2) - 1/2*(b*d + 2* a*e)*arctan(-(sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a *d^2) + 1/2*((sqrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))*b + 2*a*sqrt(c))/(((s qrt(c)*x^2 - sqrt(c*x^4 + b*x^2 + a))^2 - a)*a*d)
Timed out. \[ \int \frac {1}{x^3 \left (d+e x^2\right ) \sqrt {a+b x^2+c x^4}} \, dx=\int \frac {1}{x^3\,\left (e\,x^2+d\right )\,\sqrt {c\,x^4+b\,x^2+a}} \,d x \]