Integrand size = 22, antiderivative size = 84 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right ) \, dx=\frac {1}{8} b e n r x^2-\frac {1}{8} e r x^2 \left (2 a-b n+2 b \log \left (c x^n\right )\right )-\frac {1}{4} b n x^2 \left (d+e \log \left (f x^r\right )\right )+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right ) \] Output:
1/8*b*e*n*r*x^2-1/8*e*r*x^2*(2*a-b*n+2*b*ln(c*x^n))-1/4*b*n*x^2*(d+e*ln(f* x^r))+1/2*x^2*(a+b*ln(c*x^n))*(d+e*ln(f*x^r))
Time = 0.08 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.81 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right ) \, dx=\frac {1}{4} x^2 \left (2 a d-b d n-a e r+b e n r+e (2 a-b n) \log \left (f x^r\right )+b \log \left (c x^n\right ) \left (2 d-e r+2 e \log \left (f x^r\right )\right )\right ) \] Input:
Integrate[x*(a + b*Log[c*x^n])*(d + e*Log[f*x^r]),x]
Output:
(x^2*(2*a*d - b*d*n - a*e*r + b*e*n*r + e*(2*a - b*n)*Log[f*x^r] + b*Log[c *x^n]*(2*d - e*r + 2*e*Log[f*x^r])))/4
Time = 0.28 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.04, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.136, Rules used = {2813, 27, 2741}
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.
\(\displaystyle \int x \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right ) \, dx\) |
\(\Big \downarrow \) 2813 |
\(\displaystyle -e r \int \frac {1}{4} x \left (2 a-b n+2 b \log \left (c x^n\right )\right )dx+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right )-\frac {1}{4} b n x^2 \left (d+e \log \left (f x^r\right )\right )\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {1}{4} e r \int x \left (2 a-b n+2 b \log \left (c x^n\right )\right )dx+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right )-\frac {1}{4} b n x^2 \left (d+e \log \left (f x^r\right )\right )\) |
\(\Big \downarrow \) 2741 |
\(\displaystyle \frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right )-\frac {1}{4} e r \left (\frac {1}{2} x^2 \left (2 a+2 b \log \left (c x^n\right )-b n\right )-\frac {1}{2} b n x^2\right )-\frac {1}{4} b n x^2 \left (d+e \log \left (f x^r\right )\right )\) |
Input:
Int[x*(a + b*Log[c*x^n])*(d + e*Log[f*x^r]),x]
Output:
-1/4*(e*r*(-1/2*(b*n*x^2) + (x^2*(2*a - b*n + 2*b*Log[c*x^n]))/2)) - (b*n* x^2*(d + e*Log[f*x^r]))/4 + (x^2*(a + b*Log[c*x^n])*(d + e*Log[f*x^r]))/2
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[(d*x)^(m + 1)*((a + b*Log[c*x^n])/(d*(m + 1))), x] - Simp[b*n*((d*x)^( m + 1)/(d*(m + 1)^2)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.) + Log[(f_.)*(x_)^(r_ .)]*(e_.))*((g_.)*(x_))^(m_.), x_Symbol] :> With[{u = IntHide[(g*x)^m*(a + b*Log[c*x^n])^p, x]}, Simp[(d + e*Log[f*x^r]) u, x] - Simp[e*r Int[Simp lifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, m, n, p, r}, x] && !(EqQ[p, 1] && EqQ[a, 0] && NeQ[d, 0])
Time = 1.96 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.57
method | result | size |
parallelrisch | \(-\frac {x^{2} \ln \left (f \,x^{r}\right ) b e \,n^{7}-2 x^{2} \ln \left (f \,x^{r}\right ) a e \,n^{6}-x^{2} b e \,n^{7} r +x^{2} a e \,n^{6} r -2 x^{2} \ln \left (c \,x^{n}\right ) b d \,n^{6}+x^{2} b d \,n^{7}-2 x^{2} a d \,n^{6}+x^{2} \ln \left (c \,x^{n}\right ) b e \,n^{6} r -2 x^{2} \ln \left (c \,x^{n}\right ) \ln \left (f \,x^{r}\right ) b e \,n^{6}}{4 n^{6}}\) | \(132\) |
risch | \(\text {Expression too large to display}\) | \(1640\) |
Input:
int(x*(a+b*ln(c*x^n))*(d+e*ln(f*x^r)),x,method=_RETURNVERBOSE)
Output:
-1/4*(x^2*ln(f*x^r)*b*e*n^7-2*x^2*ln(f*x^r)*a*e*n^6-x^2*b*e*n^7*r+x^2*a*e* n^6*r-2*x^2*ln(c*x^n)*b*d*n^6+x^2*b*d*n^7-2*x^2*a*d*n^6+x^2*ln(c*x^n)*b*e* n^6*r-2*x^2*ln(c*x^n)*ln(f*x^r)*b*e*n^6)/n^6
Time = 0.07 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.52 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right ) \, dx=\frac {1}{2} \, b e n r x^{2} \log \left (x\right )^{2} - \frac {1}{4} \, {\left (b e r - 2 \, b d\right )} x^{2} \log \left (c\right ) - \frac {1}{4} \, {\left (b d n - 2 \, a d - {\left (b e n - a e\right )} r\right )} x^{2} + \frac {1}{4} \, {\left (2 \, b e x^{2} \log \left (c\right ) - {\left (b e n - 2 \, a e\right )} x^{2}\right )} \log \left (f\right ) + \frac {1}{2} \, {\left (b e r x^{2} \log \left (c\right ) + b e n x^{2} \log \left (f\right ) + {\left (b d n - {\left (b e n - a e\right )} r\right )} x^{2}\right )} \log \left (x\right ) \] Input:
integrate(x*(a+b*log(c*x^n))*(d+e*log(f*x^r)),x, algorithm="fricas")
Output:
1/2*b*e*n*r*x^2*log(x)^2 - 1/4*(b*e*r - 2*b*d)*x^2*log(c) - 1/4*(b*d*n - 2 *a*d - (b*e*n - a*e)*r)*x^2 + 1/4*(2*b*e*x^2*log(c) - (b*e*n - 2*a*e)*x^2) *log(f) + 1/2*(b*e*r*x^2*log(c) + b*e*n*x^2*log(f) + (b*d*n - (b*e*n - a*e )*r)*x^2)*log(x)
Time = 0.83 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.50 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right ) \, dx=\frac {a d x^{2}}{2} - \frac {a e r x^{2}}{4} + \frac {a e x^{2} \log {\left (f x^{r} \right )}}{2} - \frac {b d n x^{2}}{4} + \frac {b d x^{2} \log {\left (c x^{n} \right )}}{2} + \frac {b e n r x^{2}}{4} - \frac {b e n x^{2} \log {\left (f x^{r} \right )}}{4} - \frac {b e r x^{2} \log {\left (c x^{n} \right )}}{4} + \frac {b e x^{2} \log {\left (c x^{n} \right )} \log {\left (f x^{r} \right )}}{2} \] Input:
integrate(x*(a+b*ln(c*x**n))*(d+e*ln(f*x**r)),x)
Output:
a*d*x**2/2 - a*e*r*x**2/4 + a*e*x**2*log(f*x**r)/2 - b*d*n*x**2/4 + b*d*x* *2*log(c*x**n)/2 + b*e*n*r*x**2/4 - b*e*n*x**2*log(f*x**r)/4 - b*e*r*x**2* log(c*x**n)/4 + b*e*x**2*log(c*x**n)*log(f*x**r)/2
Time = 0.04 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.21 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right ) \, dx=-\frac {1}{4} \, b d n x^{2} - \frac {1}{4} \, a e r x^{2} + \frac {1}{2} \, b d x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a e x^{2} \log \left (f x^{r}\right ) + \frac {1}{4} \, {\left ({\left (r - \log \left (f\right )\right )} x^{2} - x^{2} \log \left (x^{r}\right )\right )} b e n + \frac {1}{2} \, a d x^{2} - \frac {1}{4} \, {\left (r x^{2} - 2 \, x^{2} \log \left (f x^{r}\right )\right )} b e \log \left (c x^{n}\right ) \] Input:
integrate(x*(a+b*log(c*x^n))*(d+e*log(f*x^r)),x, algorithm="maxima")
Output:
-1/4*b*d*n*x^2 - 1/4*a*e*r*x^2 + 1/2*b*d*x^2*log(c*x^n) + 1/2*a*e*x^2*log( f*x^r) + 1/4*((r - log(f))*x^2 - x^2*log(x^r))*b*e*n + 1/2*a*d*x^2 - 1/4*( r*x^2 - 2*x^2*log(f*x^r))*b*e*log(c*x^n)
Time = 0.12 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.79 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right ) \, dx=\frac {1}{2} \, b e n r x^{2} \log \left (x\right )^{2} - \frac {1}{2} \, b e n r x^{2} \log \left (x\right ) + \frac {1}{2} \, b e r x^{2} \log \left (c\right ) \log \left (x\right ) + \frac {1}{2} \, b e n x^{2} \log \left (f\right ) \log \left (x\right ) + \frac {1}{4} \, b e n r x^{2} - \frac {1}{4} \, b e r x^{2} \log \left (c\right ) - \frac {1}{4} \, b e n x^{2} \log \left (f\right ) + \frac {1}{2} \, b e x^{2} \log \left (c\right ) \log \left (f\right ) + \frac {1}{2} \, b d n x^{2} \log \left (x\right ) + \frac {1}{2} \, a e r x^{2} \log \left (x\right ) - \frac {1}{4} \, b d n x^{2} - \frac {1}{4} \, a e r x^{2} + \frac {1}{2} \, b d x^{2} \log \left (c\right ) + \frac {1}{2} \, a e x^{2} \log \left (f\right ) + \frac {1}{2} \, a d x^{2} \] Input:
integrate(x*(a+b*log(c*x^n))*(d+e*log(f*x^r)),x, algorithm="giac")
Output:
1/2*b*e*n*r*x^2*log(x)^2 - 1/2*b*e*n*r*x^2*log(x) + 1/2*b*e*r*x^2*log(c)*l og(x) + 1/2*b*e*n*x^2*log(f)*log(x) + 1/4*b*e*n*r*x^2 - 1/4*b*e*r*x^2*log( c) - 1/4*b*e*n*x^2*log(f) + 1/2*b*e*x^2*log(c)*log(f) + 1/2*b*d*n*x^2*log( x) + 1/2*a*e*r*x^2*log(x) - 1/4*b*d*n*x^2 - 1/4*a*e*r*x^2 + 1/2*b*d*x^2*lo g(c) + 1/2*a*e*x^2*log(f) + 1/2*a*d*x^2
Time = 25.44 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.98 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right ) \, dx=\ln \left (f\,x^r\right )\,\left (\frac {a\,e\,x^2}{2}-\frac {b\,e\,n\,x^2}{4}+\frac {b\,e\,x^2\,\ln \left (c\,x^n\right )}{2}\right )+x^2\,\left (\frac {a\,d}{2}-\frac {b\,d\,n}{4}-\frac {a\,e\,r}{4}+\frac {b\,e\,n\,r}{4}\right )+\frac {b\,x^2\,\ln \left (c\,x^n\right )\,\left (2\,d-e\,r\right )}{4} \] Input:
int(x*(d + e*log(f*x^r))*(a + b*log(c*x^n)),x)
Output:
log(f*x^r)*((a*e*x^2)/2 - (b*e*n*x^2)/4 + (b*e*x^2*log(c*x^n))/2) + x^2*(( a*d)/2 - (b*d*n)/4 - (a*e*r)/4 + (b*e*n*r)/4) + (b*x^2*log(c*x^n)*(2*d - e *r))/4
Time = 0.16 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.99 \[ \int x \left (a+b \log \left (c x^n\right )\right ) \left (d+e \log \left (f x^r\right )\right ) \, dx=\frac {x^{2} \left (2 \,\mathrm {log}\left (x^{n} c \right ) \mathrm {log}\left (x^{r} f \right ) b e +2 \,\mathrm {log}\left (x^{n} c \right ) b d -\mathrm {log}\left (x^{n} c \right ) b e r +2 \,\mathrm {log}\left (x^{r} f \right ) a e -\mathrm {log}\left (x^{r} f \right ) b e n +2 a d -a e r -b d n +b e n r \right )}{4} \] Input:
int(x*(a+b*log(c*x^n))*(d+e*log(f*x^r)),x)
Output:
(x**2*(2*log(x**n*c)*log(x**r*f)*b*e + 2*log(x**n*c)*b*d - log(x**n*c)*b*e *r + 2*log(x**r*f)*a*e - log(x**r*f)*b*e*n + 2*a*d - a*e*r - b*d*n + b*e*n *r))/4