Integrand size = 19, antiderivative size = 55 \[ \int x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \, dx=-\frac {1}{3} b p x^3 \left (d x^n\right )^{-3/n} \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (d x^n\right )}{n}\right )+\frac {1}{3} x^3 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \] Output:
-1/3*b*p*x^3*Ei(3*ln(d*x^n)/n)/((d*x^n)^(3/n))+1/3*x^3*(a+b*ln(c*ln(d*x^n) ^p))
Time = 0.09 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \, dx=\frac {1}{3} x^3 \left (a-b p \left (d x^n\right )^{-3/n} \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (d x^n\right )}{n}\right )+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \] Input:
Integrate[x^2*(a + b*Log[c*Log[d*x^n]^p]),x]
Output:
(x^3*(a - (b*p*ExpIntegralEi[(3*Log[d*x^n])/n])/(d*x^n)^(3/n) + b*Log[c*Lo g[d*x^n]^p]))/3
Time = 0.28 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3002, 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.
\(\displaystyle \int x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \, dx\) |
\(\Big \downarrow \) 3002 |
\(\displaystyle \frac {1}{3} x^3 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right )-\frac {1}{3} b n p \int \frac {x^2}{\log \left (d x^n\right )}dx\) |
\(\Big \downarrow \) 2747 |
\(\displaystyle \frac {1}{3} x^3 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right )-\frac {1}{3} b p x^3 \left (d x^n\right )^{-3/n} \int \frac {\left (d x^n\right )^{3/n}}{\log \left (d x^n\right )}d\log \left (d x^n\right )\) |
\(\Big \downarrow \) 2609 |
\(\displaystyle \frac {1}{3} x^3 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right )-\frac {1}{3} b p x^3 \left (d x^n\right )^{-3/n} \operatorname {ExpIntegralEi}\left (\frac {3 \log \left (d x^n\right )}{n}\right )\) |
Input:
Int[x^2*(a + b*Log[c*Log[d*x^n]^p]),x]
Output:
-1/3*(b*p*x^3*ExpIntegralEi[(3*Log[d*x^n])/n])/(d*x^n)^(3/n) + (x^3*(a + b *Log[c*Log[d*x^n]^p]))/3
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]
Int[((a_.) + Log[Log[(d_.)*(x_)^(n_.)]^(p_.)*(c_.)]*(b_.))*((e_.)*(x_))^(m_ .), x_Symbol] :> Simp[(e*x)^(m + 1)*((a + b*Log[c*Log[d*x^n]^p])/(e*(m + 1) )), x] - Simp[b*n*(p/(m + 1)) Int[(e*x)^m/Log[d*x^n], x], x] /; FreeQ[{a, b, c, d, e, m, n, p}, x] && NeQ[m, -1]
\[\int x^{2} \left (a +b \ln \left (c \ln \left (d \,x^{n}\right )^{p}\right )\right )d x\]
Input:
int(x^2*(a+b*ln(c*ln(d*x^n)^p)),x)
Output:
int(x^2*(a+b*ln(c*ln(d*x^n)^p)),x)
Time = 0.07 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.27 \[ \int x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \, dx=\frac {b d^{\frac {3}{n}} p x^{3} \log \left (n \log \left (x\right ) + \log \left (d\right )\right ) - b p \operatorname {log\_integral}\left (d^{\frac {3}{n}} x^{3}\right ) + {\left (b x^{3} \log \left (c\right ) + a x^{3}\right )} d^{\frac {3}{n}}}{3 \, d^{\frac {3}{n}}} \] Input:
integrate(x^2*(a+b*log(c*log(d*x^n)^p)),x, algorithm="fricas")
Output:
1/3*(b*d^(3/n)*p*x^3*log(n*log(x) + log(d)) - b*p*log_integral(d^(3/n)*x^3 ) + (b*x^3*log(c) + a*x^3)*d^(3/n))/d^(3/n)
\[ \int x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \, dx=\int x^{2} \left (a + b \log {\left (c \log {\left (d x^{n} \right )}^{p} \right )}\right )\, dx \] Input:
integrate(x**2*(a+b*ln(c*ln(d*x**n)**p)),x)
Output:
Integral(x**2*(a + b*log(c*log(d*x**n)**p)), x)
\[ \int x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \, dx=\int { {\left (b \log \left (c \log \left (d x^{n}\right )^{p}\right ) + a\right )} x^{2} \,d x } \] Input:
integrate(x^2*(a+b*log(c*log(d*x^n)^p)),x, algorithm="maxima")
Output:
1/3*a*x^3 + 1/3*(x^3*log(c) + x^3*log((log(d) + log(x^n))^p) - 3*n*p*integ rate(1/3*x^2/(log(d) + log(x^n)), x))*b
Time = 0.12 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02 \[ \int x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \, dx=\frac {1}{3} \, b p x^{3} \log \left (n \log \left (x\right ) + \log \left (d\right )\right ) + \frac {1}{3} \, b x^{3} \log \left (c\right ) + \frac {1}{3} \, a x^{3} - \frac {b p {\rm Ei}\left (\frac {3 \, \log \left (d\right )}{n} + 3 \, \log \left (x\right )\right )}{3 \, d^{\frac {3}{n}}} \] Input:
integrate(x^2*(a+b*log(c*log(d*x^n)^p)),x, algorithm="giac")
Output:
1/3*b*p*x^3*log(n*log(x) + log(d)) + 1/3*b*x^3*log(c) + 1/3*a*x^3 - 1/3*b* p*Ei(3*log(d)/n + 3*log(x))/d^(3/n)
Timed out. \[ \int x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \, dx=\int x^2\,\left (a+b\,\ln \left (c\,{\ln \left (d\,x^n\right )}^p\right )\right ) \,d x \] Input:
int(x^2*(a + b*log(c*log(d*x^n)^p)),x)
Output:
int(x^2*(a + b*log(c*log(d*x^n)^p)), x)
\[ \int x^2 \left (a+b \log \left (c \log ^p\left (d x^n\right )\right )\right ) \, dx=-\frac {\left (\int \frac {x^{2}}{\mathrm {log}\left (x^{n} d \right )}d x \right ) b n p}{3}+\frac {\mathrm {log}\left (\mathrm {log}\left (x^{n} d \right )^{p} c \right ) b \,x^{3}}{3}+\frac {a \,x^{3}}{3} \] Input:
int(x^2*(a+b*log(c*log(d*x^n)^p)),x)
Output:
( - int(x**2/log(x**n*d),x)*b*n*p + log(log(x**n*d)**p*c)*b*x**3 + a*x**3) /3