Integrand size = 14, antiderivative size = 73 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=-\frac {a+b x^2}{4 b \log ^2\left (c \left (a+b x^2\right )\right )}-\frac {a+b x^2}{4 b \log \left (c \left (a+b x^2\right )\right )}+\frac {\operatorname {LogIntegral}\left (c \left (a+b x^2\right )\right )}{4 b c} \] Output:
-1/4*(b*x^2+a)/b/ln((b*x^2+a)*c)^2-1/4*(b*x^2+a)/b/ln((b*x^2+a)*c)+1/4*Li( (b*x^2+a)*c)/b/c
Time = 0.03 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.75 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=\frac {-\frac {\left (a+b x^2\right ) \left (1+\log \left (c \left (a+b x^2\right )\right )\right )}{\log ^2\left (c \left (a+b x^2\right )\right )}+\frac {\operatorname {LogIntegral}\left (c \left (a+b x^2\right )\right )}{c}}{4 b} \] Input:
Integrate[x/Log[c*(a + b*x^2)]^3,x]
Output:
(-(((a + b*x^2)*(1 + Log[c*(a + b*x^2)]))/Log[c*(a + b*x^2)]^2) + LogInteg ral[c*(a + b*x^2)]/c)/(4*b)
Time = 0.45 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {2904, 2836, 2734, 2734, 2735}
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.
\(\displaystyle \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx\) |
\(\Big \downarrow \) 2904 |
\(\displaystyle \frac {1}{2} \int \frac {1}{\log ^3\left (c \left (b x^2+a\right )\right )}dx^2\) |
\(\Big \downarrow \) 2836 |
\(\displaystyle \frac {\int \frac {1}{\log ^3\left (c \left (b x^2+a\right )\right )}d\left (b x^2+a\right )}{2 b}\) |
\(\Big \downarrow \) 2734 |
\(\displaystyle \frac {\frac {1}{2} \int \frac {1}{\log ^2\left (c \left (b x^2+a\right )\right )}d\left (b x^2+a\right )-\frac {a+b x^2}{2 \log ^2\left (c \left (a+b x^2\right )\right )}}{2 b}\) |
\(\Big \downarrow \) 2734 |
\(\displaystyle \frac {\frac {1}{2} \left (\int \frac {1}{\log \left (c \left (b x^2+a\right )\right )}d\left (b x^2+a\right )-\frac {a+b x^2}{\log \left (c \left (a+b x^2\right )\right )}\right )-\frac {a+b x^2}{2 \log ^2\left (c \left (a+b x^2\right )\right )}}{2 b}\) |
\(\Big \downarrow \) 2735 |
\(\displaystyle \frac {\frac {1}{2} \left (\frac {\operatorname {LogIntegral}\left (c \left (b x^2+a\right )\right )}{c}-\frac {a+b x^2}{\log \left (c \left (a+b x^2\right )\right )}\right )-\frac {a+b x^2}{2 \log ^2\left (c \left (a+b x^2\right )\right )}}{2 b}\) |
Input:
Int[x/Log[c*(a + b*x^2)]^3,x]
Output:
(-1/2*(a + b*x^2)/Log[c*(a + b*x^2)]^2 + (-((a + b*x^2)/Log[c*(a + b*x^2)] ) + LogIntegral[c*(a + b*x^2)]/c)/2)/(2*b)
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_), x_Symbol] :> Simp[x*((a + b *Log[c*x^n])^(p + 1)/(b*n*(p + 1))), x] - Simp[1/(b*n*(p + 1)) Int[(a + b *Log[c*x^n])^(p + 1), x], x] /; FreeQ[{a, b, c, n}, x] && LtQ[p, -1] && Int egerQ[2*p]
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[((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 = 0.74 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96
method | result | size |
derivativedivides | \(\frac {-\frac {c \left (b \,x^{2}+a \right )}{2 {\ln \left (c \left (b \,x^{2}+a \right )\right )}^{2}}-\frac {c \left (b \,x^{2}+a \right )}{2 \ln \left (c \left (b \,x^{2}+a \right )\right )}-\frac {\operatorname {expIntegral}_{1}\left (-\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2}}{2 b c}\) | \(70\) |
default | \(\frac {-\frac {c \left (b \,x^{2}+a \right )}{2 {\ln \left (c \left (b \,x^{2}+a \right )\right )}^{2}}-\frac {c \left (b \,x^{2}+a \right )}{2 \ln \left (c \left (b \,x^{2}+a \right )\right )}-\frac {\operatorname {expIntegral}_{1}\left (-\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{2}}{2 b c}\) | \(70\) |
risch | \(-\frac {\ln \left (c \left (b \,x^{2}+a \right )\right ) b \,x^{2}+b \,x^{2}+\ln \left (c \left (b \,x^{2}+a \right )\right ) a +a}{4 b {\ln \left (c \left (b \,x^{2}+a \right )\right )}^{2}}-\frac {\operatorname {expIntegral}_{1}\left (-\ln \left (c \left (b \,x^{2}+a \right )\right )\right )}{4 b c}\) | \(75\) |
Input:
int(x/ln(c*(b*x^2+a))^3,x,method=_RETURNVERBOSE)
Output:
1/2/b/c*(-1/2/ln(c*(b*x^2+a))^2*c*(b*x^2+a)-1/2/ln(c*(b*x^2+a))*c*(b*x^2+a )-1/2*Ei(1,-ln(c*(b*x^2+a))))
Time = 0.06 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.08 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=-\frac {b c x^{2} - \log \left (b c x^{2} + a c\right )^{2} \operatorname {log\_integral}\left (b c x^{2} + a c\right ) + a c + {\left (b c x^{2} + a c\right )} \log \left (b c x^{2} + a c\right )}{4 \, b c \log \left (b c x^{2} + a c\right )^{2}} \] Input:
integrate(x/log((b*x^2+a)*c)^3,x, algorithm="fricas")
Output:
-1/4*(b*c*x^2 - log(b*c*x^2 + a*c)^2*log_integral(b*c*x^2 + a*c) + a*c + ( b*c*x^2 + a*c)*log(b*c*x^2 + a*c))/(b*c*log(b*c*x^2 + a*c)^2)
Time = 0.75 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.96 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=\frac {\begin {cases} \frac {x^{2}}{2 \log {\left (a c \right )}} & \text {for}\: b = 0 \\0 & \text {for}\: c = 0 \\\frac {\operatorname {Ei}{\left (\log {\left (a c + b c x^{2} \right )} \right )}}{2 b c} & \text {otherwise} \end {cases}}{2} + \frac {- a - b x^{2} + \left (- a - b x^{2}\right ) \log {\left (c \left (a + b x^{2}\right ) \right )}}{4 b \log {\left (c \left (a + b x^{2}\right ) \right )}^{2}} \] Input:
integrate(x/ln((b*x**2+a)*c)**3,x)
Output:
Piecewise((x**2/(2*log(a*c)), Eq(b, 0)), (0, Eq(c, 0)), (Ei(log(a*c + b*c* x**2))/(2*b*c), True))/2 + (-a - b*x**2 + (-a - b*x**2)*log(c*(a + b*x**2) ))/(4*b*log(c*(a + b*x**2))**2)
\[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=\int { \frac {x}{\log \left ({\left (b x^{2} + a\right )} c\right )^{3}} \,d x } \] Input:
integrate(x/log((b*x^2+a)*c)^3,x, algorithm="maxima")
Output:
-1/4*(b*x^2*(log(c) + 1) + a*(log(c) + 1) + (b*x^2 + a)*log(b*x^2 + a))/(b *log(b*x^2 + a)^2 + 2*b*log(b*x^2 + a)*log(c) + b*log(c)^2) + integrate(1/ 2*x/(log(b*x^2 + a) + log(c)), x)
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=-\frac {\frac {b c x^{2} + a c}{\log \left (b c x^{2} + a c\right )} + \frac {b c x^{2} + a c}{\log \left (b c x^{2} + a c\right )^{2}} - {\rm Ei}\left (\log \left (b c x^{2} + a c\right )\right )}{4 \, b c} \] Input:
integrate(x/log((b*x^2+a)*c)^3,x, algorithm="giac")
Output:
-1/4*((b*c*x^2 + a*c)/log(b*c*x^2 + a*c) + (b*c*x^2 + a*c)/log(b*c*x^2 + a *c)^2 - Ei(log(b*c*x^2 + a*c)))/(b*c)
Time = 25.74 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.01 \[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=\frac {\mathrm {logint}\left (c\,\left (b\,x^2+a\right )\right )}{4\,b\,c}-\frac {\frac {a\,c}{4}+\ln \left (c\,\left (b\,x^2+a\right )\right )\,\left (\frac {b\,c\,x^2}{4}+\frac {a\,c}{4}\right )+\frac {b\,c\,x^2}{4}}{b\,c\,{\ln \left (c\,\left (b\,x^2+a\right )\right )}^2} \] Input:
int(x/log(c*(a + b*x^2))^3,x)
Output:
logint(c*(a + b*x^2))/(4*b*c) - ((a*c)/4 + log(c*(a + b*x^2))*((a*c)/4 + ( b*c*x^2)/4) + (b*c*x^2)/4)/(b*c*log(c*(a + b*x^2))^2)
\[ \int \frac {x}{\log ^3\left (c \left (a+b x^2\right )\right )} \, dx=\int \frac {x}{\mathrm {log}\left (b c \,x^{2}+a c \right )^{3}}d x \] Input:
int(x/log((b*x^2+a)*c)^3,x)
Output:
int(x/log(a*c + b*c*x**2)**3,x)