Integrand size = 17, antiderivative size = 48 \[ \int x (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{4} b d n x^2-\frac {1}{9} b e n x^3+\frac {1}{6} \left (3 d x^2+2 e x^3\right ) \left (a+b \log \left (c x^n\right )\right ) \] Output:
-1/4*b*d*n*x^2-1/9*b*e*n*x^3+1/6*(2*e*x^3+3*d*x^2)*(a+b*ln(c*x^n))
Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int x (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{36} x^2 \left (6 a (3 d+2 e x)-b n (9 d+4 e x)+6 b (3 d+2 e x) \log \left (c x^n\right )\right ) \] Input:
Integrate[x*(d + e*x)*(a + b*Log[c*x^n]),x]
Output:
(x^2*(6*a*(3*d + 2*e*x) - b*n*(9*d + 4*e*x) + 6*b*(3*d + 2*e*x)*Log[c*x^n] ))/36
Time = 0.24 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2771, 27, 49, 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.
\(\displaystyle \int x (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2771 |
\(\displaystyle \frac {1}{6} \left (3 d x^2+2 e x^3\right ) \left (a+b \log \left (c x^n\right )\right )-b n \int \frac {1}{6} x (3 d+2 e x)dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} \left (3 d x^2+2 e x^3\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{6} b n \int x (3 d+2 e x)dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{6} \left (3 d x^2+2 e x^3\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{6} b n \int \left (2 e x^2+3 d x\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{6} \left (3 d x^2+2 e x^3\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{6} b n \left (\frac {3 d x^2}{2}+\frac {2 e x^3}{3}\right )\) |
Input:
Int[x*(d + e*x)*(a + b*Log[c*x^n]),x]
Output:
-1/6*(b*n*((3*d*x^2)/2 + (2*e*x^3)/3)) + ((3*d*x^2 + 2*e*x^3)*(a + b*Log[c *x^n]))/6
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 0] && IGtQ[m + n + 2, 0]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_ .))^(q_.), x_Symbol] :> With[{u = IntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Simp[b*n Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IGtQ[m, 0]
Time = 0.21 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(\frac {x^{3} b e \ln \left (c \,x^{n}\right )}{3}-\frac {b e n \,x^{3}}{9}+\frac {x^{3} a e}{3}+\frac {x^{2} \ln \left (c \,x^{n}\right ) b d}{2}-\frac {b d n \,x^{2}}{4}+\frac {d a \,x^{2}}{2}\) | \(58\) |
risch | \(\frac {b \,x^{2} \left (2 e x +3 d \right ) \ln \left (x^{n}\right )}{6}+\frac {i \pi b e \,x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{6}-\frac {i \pi b e \,x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{6}-\frac {i \pi b e \,x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{6}+\frac {i \pi b e \,x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{6}+\frac {\ln \left (c \right ) b e \,x^{3}}{3}-\frac {b e n \,x^{3}}{9}+\frac {x^{3} a e}{3}+\frac {i \pi b d \,x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{4}-\frac {i \pi b d \,x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{4}-\frac {i \pi b d \,x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{4}+\frac {i \pi b d \,x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{4}+\frac {\ln \left (c \right ) b d \,x^{2}}{2}-\frac {b d n \,x^{2}}{4}+\frac {d a \,x^{2}}{2}\) | \(264\) |
Input:
int(x*(e*x+d)*(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)
Output:
1/3*x^3*b*e*ln(c*x^n)-1/9*b*e*n*x^3+1/3*x^3*a*e+1/2*x^2*ln(c*x^n)*b*d-1/4* b*d*n*x^2+1/2*d*a*x^2
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.44 \[ \int x (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{9} \, {\left (b e n - 3 \, a e\right )} x^{3} - \frac {1}{4} \, {\left (b d n - 2 \, a d\right )} x^{2} + \frac {1}{6} \, {\left (2 \, b e x^{3} + 3 \, b d x^{2}\right )} \log \left (c\right ) + \frac {1}{6} \, {\left (2 \, b e n x^{3} + 3 \, b d n x^{2}\right )} \log \left (x\right ) \] Input:
integrate(x*(e*x+d)*(a+b*log(c*x^n)),x, algorithm="fricas")
Output:
-1/9*(b*e*n - 3*a*e)*x^3 - 1/4*(b*d*n - 2*a*d)*x^2 + 1/6*(2*b*e*x^3 + 3*b* d*x^2)*log(c) + 1/6*(2*b*e*n*x^3 + 3*b*d*n*x^2)*log(x)
Time = 0.20 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.38 \[ \int x (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d x^{2}}{2} + \frac {a e x^{3}}{3} - \frac {b d n x^{2}}{4} + \frac {b d x^{2} \log {\left (c x^{n} \right )}}{2} - \frac {b e n x^{3}}{9} + \frac {b e x^{3} \log {\left (c x^{n} \right )}}{3} \] Input:
integrate(x*(e*x+d)*(a+b*ln(c*x**n)),x)
Output:
a*d*x**2/2 + a*e*x**3/3 - b*d*n*x**2/4 + b*d*x**2*log(c*x**n)/2 - b*e*n*x* *3/9 + b*e*x**3*log(c*x**n)/3
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int x (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{9} \, b e n x^{3} + \frac {1}{3} \, b e x^{3} \log \left (c x^{n}\right ) - \frac {1}{4} \, b d n x^{2} + \frac {1}{3} \, a e x^{3} + \frac {1}{2} \, b d x^{2} \log \left (c x^{n}\right ) + \frac {1}{2} \, a d x^{2} \] Input:
integrate(x*(e*x+d)*(a+b*log(c*x^n)),x, algorithm="maxima")
Output:
-1/9*b*e*n*x^3 + 1/3*b*e*x^3*log(c*x^n) - 1/4*b*d*n*x^2 + 1/3*a*e*x^3 + 1/ 2*b*d*x^2*log(c*x^n) + 1/2*a*d*x^2
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.44 \[ \int x (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{3} \, b e n x^{3} \log \left (x\right ) - \frac {1}{9} \, b e n x^{3} + \frac {1}{3} \, b e x^{3} \log \left (c\right ) + \frac {1}{2} \, b d n x^{2} \log \left (x\right ) - \frac {1}{4} \, b d n x^{2} + \frac {1}{3} \, a e x^{3} + \frac {1}{2} \, b d x^{2} \log \left (c\right ) + \frac {1}{2} \, a d x^{2} \] Input:
integrate(x*(e*x+d)*(a+b*log(c*x^n)),x, algorithm="giac")
Output:
1/3*b*e*n*x^3*log(x) - 1/9*b*e*n*x^3 + 1/3*b*e*x^3*log(c) + 1/2*b*d*n*x^2* log(x) - 1/4*b*d*n*x^2 + 1/3*a*e*x^3 + 1/2*b*d*x^2*log(c) + 1/2*a*d*x^2
Time = 27.62 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int x (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,e\,x^3}{3}+\frac {b\,d\,x^2}{2}\right )+\frac {d\,x^2\,\left (2\,a-b\,n\right )}{4}+\frac {e\,x^3\,\left (3\,a-b\,n\right )}{9} \] Input:
int(x*(a + b*log(c*x^n))*(d + e*x),x)
Output:
log(c*x^n)*((b*d*x^2)/2 + (b*e*x^3)/3) + (d*x^2*(2*a - b*n))/4 + (e*x^3*(3 *a - b*n))/9
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98 \[ \int x (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{2} \left (18 \,\mathrm {log}\left (x^{n} c \right ) b d +12 \,\mathrm {log}\left (x^{n} c \right ) b e x +18 a d +12 a e x -9 b d n -4 b e n x \right )}{36} \] Input:
int(x*(e*x+d)*(a+b*log(c*x^n)),x)
Output:
(x**2*(18*log(x**n*c)*b*d + 12*log(x**n*c)*b*e*x + 18*a*d + 12*a*e*x - 9*b *d*n - 4*b*e*n*x))/36