Integrand size = 16, antiderivative size = 71 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx=\frac {\operatorname {ExpIntegralEi}\left (2 \log \left (c \left (a+b x^2\right )\right )\right )}{b^2 c^2}-\frac {x^2 \left (a+b x^2\right )}{2 b \log \left (c \left (a+b x^2\right )\right )}-\frac {a \operatorname {LogIntegral}\left (c \left (a+b x^2\right )\right )}{2 b^2 c} \]
Ei(2*ln(c*(b*x^2+a)))/b^2/c^2-1/2*a*Li(c*(b*x^2+a))/b^2/c-1/2*x^2*(b*x^2+a )/b/ln(c*(b*x^2+a))
\[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx=\int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx \]
Time = 0.39 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.30, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {2904, 2847, 2836, 2735, 2846, 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.
\(\displaystyle \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{2} \int \frac {x^2}{\log ^2\left (c \left (b x^2+a\right )\right )}dx^2\) |
\(\Big \downarrow \) 2847 |
\(\displaystyle \frac {1}{2} \left (\frac {a \int \frac {1}{\log \left (c \left (b x^2+a\right )\right )}dx^2}{b}+2 \int \frac {x^2}{\log \left (c \left (b x^2+a\right )\right )}dx^2-\frac {x^2 \left (a+b x^2\right )}{b \log \left (c \left (a+b x^2\right )\right )}\right )\) |
\(\Big \downarrow \) 2836 |
\(\displaystyle \frac {1}{2} \left (\frac {a \int \frac {1}{\log \left (c \left (b x^2+a\right )\right )}d\left (b x^2+a\right )}{b^2}+2 \int \frac {x^2}{\log \left (c \left (b x^2+a\right )\right )}dx^2-\frac {x^2 \left (a+b x^2\right )}{b \log \left (c \left (a+b x^2\right )\right )}\right )\) |
\(\Big \downarrow \) 2735 |
\(\displaystyle \frac {1}{2} \left (2 \int \frac {x^2}{\log \left (c \left (b x^2+a\right )\right )}dx^2+\frac {a \operatorname {LogIntegral}\left (c \left (b x^2+a\right )\right )}{b^2 c}-\frac {x^2 \left (a+b x^2\right )}{b \log \left (c \left (a+b x^2\right )\right )}\right )\) |
\(\Big \downarrow \) 2846 |
\(\displaystyle \frac {1}{2} \left (2 \int \left (\frac {b x^2+a}{b \log \left (c \left (b x^2+a\right )\right )}-\frac {a}{b \log \left (c \left (b x^2+a\right )\right )}\right )dx^2+\frac {a \operatorname {LogIntegral}\left (c \left (b x^2+a\right )\right )}{b^2 c}-\frac {x^2 \left (a+b x^2\right )}{b \log \left (c \left (a+b x^2\right )\right )}\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (2 \left (\frac {\operatorname {ExpIntegralEi}\left (2 \log \left (c \left (b x^2+a\right )\right )\right )}{b^2 c^2}-\frac {a \operatorname {LogIntegral}\left (c \left (b x^2+a\right )\right )}{b^2 c}\right )+\frac {a \operatorname {LogIntegral}\left (c \left (b x^2+a\right )\right )}{b^2 c}-\frac {x^2 \left (a+b x^2\right )}{b \log \left (c \left (a+b x^2\right )\right )}\right )\) |
(-((x^2*(a + b*x^2))/(b*Log[c*(a + b*x^2)])) + (a*LogIntegral[c*(a + b*x^2 )])/(b^2*c) + 2*(ExpIntegralEi[2*Log[c*(a + b*x^2)]]/(b^2*c^2) - (a*LogInt egral[c*(a + b*x^2)])/(b^2*c)))/2
3.2.25.3.1 Defintions of rubi rules used
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] : > Simp[1/e Subst[Int[(a + b*Log[c*x^n])^p, x], x, d + e*x], x] /; FreeQ[{ a, b, c, d, e, n, p}, x]
Int[((f_.) + (g_.)*(x_))^(q_.)/((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.) ]*(b_.)), x_Symbol] :> Int[ExpandIntegrand[(f + g*x)^q/(a + b*Log[c*(d + e* x)^n]), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] & & IGtQ[q, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_)*((f_.) + (g_. )*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)*(f + g*x)^q*((a + b*Log[c*(d + e *x)^n])^(p + 1)/(b*e*n*(p + 1))), x] + (-Simp[(q + 1)/(b*n*(p + 1)) Int[( f + g*x)^q*(a + b*Log[c*(d + e*x)^n])^(p + 1), x], x] + Simp[q*((e*f - d*g) /(b*e*n*(p + 1))) Int[(f + g*x)^(q - 1)*(a + b*Log[c*(d + e*x)^n])^(p + 1 ), x], x]) /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0] && Lt Q[p, -1] && GtQ[q, 0]
Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))^(q_.)*(x_)^(m _.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*L og[c*(d + e*x)^p])^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && IntegerQ[Simplify[(m + 1)/n]] && (GtQ[(m + 1)/n, 0] || IGtQ[q, 0]) & & !(EqQ[q, 1] && ILtQ[n, 0] && IGtQ[m, 0])
Time = 2.01 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.04
method | result | size |
risch | \(-\frac {x^{2} \left (b \,x^{2}+a \right )}{2 b \ln \left (c \left (b \,x^{2}+a \right )\right )}+\frac {a \,\operatorname {Ei}_{1}\left (-\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2 c \,b^{2}}-\frac {\operatorname {Ei}_{1}\left (-2 \ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{c^{2} b^{2}}\) | \(74\) |
default | \(\frac {-\frac {c^{2} \left (b \,x^{2}+a \right )^{2}}{\ln \left (c \left (b \,x^{2}+a \right )\right )}-2 \,\operatorname {Ei}_{1}\left (-2 \ln \left (c \left (b \,x^{2}+a \right )\right )\right )-c a \left (-\frac {c \left (b \,x^{2}+a \right )}{\ln \left (c \left (b \,x^{2}+a \right )\right )}-\operatorname {Ei}_{1}\left (-\ln \left (c \left (b \,x^{2}+a \right )\right )\right )\right )}{2 c^{2} b^{2}}\) | \(95\) |
-1/2*x^2*(b*x^2+a)/b/ln(c*(b*x^2+a))+1/2/c/b^2*a*Ei(1,-ln(c*(b*x^2+a)))-1/ c^2/b^2*Ei(1,-2*ln(c*(b*x^2+a)))
Time = 0.28 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.39 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx=-\frac {b^{2} c^{2} x^{4} + a b c^{2} x^{2} + {\left (a c \operatorname {log\_integral}\left (b c x^{2} + a c\right ) - 2 \, \operatorname {log\_integral}\left (b^{2} c^{2} x^{4} + 2 \, a b c^{2} x^{2} + a^{2} c^{2}\right )\right )} \log \left (b c x^{2} + a c\right )}{2 \, b^{2} c^{2} \log \left (b c x^{2} + a c\right )} \]
-1/2*(b^2*c^2*x^4 + a*b*c^2*x^2 + (a*c*log_integral(b*c*x^2 + a*c) - 2*log _integral(b^2*c^2*x^4 + 2*a*b*c^2*x^2 + a^2*c^2))*log(b*c*x^2 + a*c))/(b^2 *c^2*log(b*c*x^2 + a*c))
\[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx=\frac {- a x^{2} - b x^{4}}{2 b \log {\left (c \left (a + b x^{2}\right ) \right )}} + \frac {\int \frac {a x}{\log {\left (a c + b c x^{2} \right )}}\, dx + \int \frac {2 b x^{3}}{\log {\left (a c + b c x^{2} \right )}}\, dx}{b} \]
(-a*x**2 - b*x**4)/(2*b*log(c*(a + b*x**2))) + (Integral(a*x/log(a*c + b*c *x**2), x) + Integral(2*b*x**3/log(a*c + b*c*x**2), x))/b
\[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx=\int { \frac {x^{3}}{\log \left ({\left (b x^{2} + a\right )} c\right )^{2}} \,d x } \]
-1/2*(b*x^4 + a*x^2)/(b*log(b*x^2 + a) + b*log(c)) + integrate((2*b*x^3 + a*x)/(b*log(b*x^2 + a) + b*log(c)), x)
Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.41 \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx=\frac {a {\left (\frac {b c x^{2} + a c}{\log \left (b c x^{2} + a c\right )} - {\rm Ei}\left (\log \left (b c x^{2} + a c\right )\right )\right )}}{2 \, b^{2} c} - \frac {\frac {{\left (b c x^{2} + a c\right )}^{2}}{\log \left (b c x^{2} + a c\right )} - 2 \, {\rm Ei}\left (2 \, \log \left (b c x^{2} + a c\right )\right )}{2 \, b^{2} c^{2}} \]
1/2*a*((b*c*x^2 + a*c)/log(b*c*x^2 + a*c) - Ei(log(b*c*x^2 + a*c)))/(b^2*c ) - 1/2*((b*c*x^2 + a*c)^2/log(b*c*x^2 + a*c) - 2*Ei(2*log(b*c*x^2 + a*c)) )/(b^2*c^2)
Timed out. \[ \int \frac {x^3}{\log ^2\left (c \left (a+b x^2\right )\right )} \, dx=\int \frac {x^3}{{\ln \left (c\,\left (b\,x^2+a\right )\right )}^2} \,d x \]