Integrand size = 16, antiderivative size = 51 \[ \int \frac {1}{x^3 \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {e^{\frac {2 a}{b n}} \left (c x^n\right )^{2/n} \operatorname {ExpIntegralEi}\left (-\frac {2 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b n x^2} \] Output:
exp(2*a/b/n)*(c*x^n)^(2/n)*Ei((-2*a-2*b*ln(c*x^n))/b/n)/b/n/x^2
Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \frac {1}{x^3 \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {e^{\frac {2 a}{b n}} \left (c x^n\right )^{2/n} \operatorname {ExpIntegralEi}\left (-\frac {2 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b n x^2} \] Input:
Integrate[1/(x^3*(a + b*Log[c*x^n])),x]
Output:
(E^((2*a)/(b*n))*(c*x^n)^(2/n)*ExpIntegralEi[(-2*(a + b*Log[c*x^n]))/(b*n) ])/(b*n*x^2)
Time = 0.25 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {2747, 2609}
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.
\(\displaystyle \int \frac {1}{x^3 \left (a+b \log \left (c x^n\right )\right )} \, dx\) |
\(\Big \downarrow \) 2747 |
\(\displaystyle \frac {\left (c x^n\right )^{2/n} \int \frac {\left (c x^n\right )^{-2/n}}{a+b \log \left (c x^n\right )}d\log \left (c x^n\right )}{n x^2}\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {e^{\frac {2 a}{b n}} \left (c x^n\right )^{2/n} \operatorname {ExpIntegralEi}\left (-\frac {2 \left (a+b \log \left (c x^n\right )\right )}{b n}\right )}{b n x^2}\) |
Input:
Int[1/(x^3*(a + b*Log[c*x^n])),x]
Output:
(E^((2*a)/(b*n))*(c*x^n)^(2/n)*ExpIntegralEi[(-2*(a + b*Log[c*x^n]))/(b*n) ])/(b*n*x^2)
Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))/((c_.) + (d_.)*(x_)), x_Symbol] :> Si mp[(F^(g*(e - c*(f/d)))/d)*ExpIntegralEi[f*g*(c + d*x)*(Log[F]/d)], x] /; F reeQ[{F, c, d, e, f, g}, x] && !TrueQ[$UseGamma]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_)*((d_.)*(x_))^(m_.), x_Symbol ] :> Simp[(d*x)^(m + 1)/(d*n*(c*x^n)^((m + 1)/n)) Subst[Int[E^(((m + 1)/n )*x)*(a + b*x)^p, x], x, Log[c*x^n]], x] /; FreeQ[{a, b, c, d, m, n, p}, x]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.71 (sec) , antiderivative size = 240, normalized size of antiderivative = 4.71
method | result | size |
risch | \(-\frac {c^{\frac {2}{n}} \left (x^{n}\right )^{\frac {2}{n}} {\mathrm e}^{\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 a}{n b}} \operatorname {expIntegral}_{1}\left (2 \ln \left (x \right )+\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+2 b \ln \left (c \right )+2 b \left (\ln \left (x^{n}\right )-n \ln \left (x \right )\right )+2 a}{n b}\right )}{n b \,x^{2}}\) | \(240\) |
Input:
int(1/x^3/(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)
Output:
-1/n/b/x^2*c^(2/n)*(x^n)^(2/n)*exp((I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*P i*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn (I*c*x^n)^2*csgn(I*c)+2*a)/n/b)*Ei(1,2*ln(x)+(I*Pi*b*csgn(I*x^n)*csgn(I*c* x^n)^2-I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*Pi*b*csgn(I*c*x^n)^3+I *Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+2*b*ln(c)+2*b*(ln(x^n)-n*ln(x))+2*a)/n/b)
Time = 0.07 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.82 \[ \int \frac {1}{x^3 \left (a+b \log \left (c x^n\right )\right )} \, dx=\frac {e^{\left (\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )} \operatorname {log\_integral}\left (\frac {e^{\left (-\frac {2 \, {\left (b \log \left (c\right ) + a\right )}}{b n}\right )}}{x^{2}}\right )}{b n} \] Input:
integrate(1/x^3/(a+b*log(c*x^n)),x, algorithm="fricas")
Output:
e^(2*(b*log(c) + a)/(b*n))*log_integral(e^(-2*(b*log(c) + a)/(b*n))/x^2)/( b*n)
\[ \int \frac {1}{x^3 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{x^{3} \left (a + b \log {\left (c x^{n} \right )}\right )}\, dx \] Input:
integrate(1/x**3/(a+b*ln(c*x**n)),x)
Output:
Integral(1/(x**3*(a + b*log(c*x**n))), x)
\[ \int \frac {1}{x^3 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}} \,d x } \] Input:
integrate(1/x^3/(a+b*log(c*x^n)),x, algorithm="maxima")
Output:
integrate(1/((b*log(c*x^n) + a)*x^3), x)
\[ \int \frac {1}{x^3 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int { \frac {1}{{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}} \,d x } \] Input:
integrate(1/x^3/(a+b*log(c*x^n)),x, algorithm="giac")
Output:
integrate(1/((b*log(c*x^n) + a)*x^3), x)
Timed out. \[ \int \frac {1}{x^3 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )} \,d x \] Input:
int(1/(x^3*(a + b*log(c*x^n))),x)
Output:
int(1/(x^3*(a + b*log(c*x^n))), x)
\[ \int \frac {1}{x^3 \left (a+b \log \left (c x^n\right )\right )} \, dx=\int \frac {1}{\mathrm {log}\left (x^{n} c \right ) b \,x^{3}+a \,x^{3}}d x \] Input:
int(1/x^3/(a+b*log(c*x^n)),x)
Output:
int(1/(log(x**n*c)*b*x**3 + a*x**3),x)