Integrand size = 19, antiderivative size = 48 \[ \int x^2 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{9} b d n x^3-\frac {1}{16} b e n x^4+\frac {1}{12} \left (4 d x^3+3 e x^4\right ) \left (a+b \log \left (c x^n\right )\right ) \] Output:
-1/9*b*d*n*x^3-1/16*b*e*n*x^4+1/12*(3*e*x^4+4*d*x^3)*(a+b*ln(c*x^n))
Time = 0.03 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.94 \[ \int x^2 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{144} x^3 \left (48 a d-16 b d n+36 a e x-9 b e n x+12 b (4 d+3 e x) \log \left (c x^n\right )\right ) \] Input:
Integrate[x^2*(d + e*x)*(a + b*Log[c*x^n]),x]
Output:
(x^3*(48*a*d - 16*b*d*n + 36*a*e*x - 9*b*e*n*x + 12*b*(4*d + 3*e*x)*Log[c* x^n]))/144
Time = 0.24 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, 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.
\(\displaystyle \int x^2 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2771 |
\(\displaystyle \frac {1}{12} \left (4 d x^3+3 e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-b n \int \frac {1}{12} x^2 (4 d+3 e x)dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{12} \left (4 d x^3+3 e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{12} b n \int x^2 (4 d+3 e x)dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{12} \left (4 d x^3+3 e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{12} b n \int \left (3 e x^3+4 d x^2\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{12} \left (4 d x^3+3 e x^4\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{12} b n \left (\frac {4 d x^3}{3}+\frac {3 e x^4}{4}\right )\) |
Input:
Int[x^2*(d + e*x)*(a + b*Log[c*x^n]),x]
Output:
-1/12*(b*n*((4*d*x^3)/3 + (3*e*x^4)/4)) + ((4*d*x^3 + 3*e*x^4)*(a + b*Log[ c*x^n]))/12
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 = 0.24 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(\frac {x^{4} b e \ln \left (c \,x^{n}\right )}{4}-\frac {b e n \,x^{4}}{16}+\frac {x^{4} a e}{4}+\frac {x^{3} \ln \left (c \,x^{n}\right ) b d}{3}-\frac {b d n \,x^{3}}{9}+\frac {x^{3} d a}{3}\) | \(58\) |
risch | \(\frac {b \,x^{3} \left (3 e x +4 d \right ) \ln \left (x^{n}\right )}{12}+\frac {i \pi b e \,x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b e \,x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{8}-\frac {i \pi b e \,x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{8}+\frac {i \pi b e \,x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{8}+\frac {\ln \left (c \right ) b e \,x^{4}}{4}-\frac {b e n \,x^{4}}{16}+\frac {x^{4} a e}{4}+\frac {i \pi b d \,x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{6}-\frac {i \pi b d \,x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{6}-\frac {i \pi b d \,x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{6}+\frac {i \pi b d \,x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{6}+\frac {\ln \left (c \right ) b d \,x^{3}}{3}-\frac {b d n \,x^{3}}{9}+\frac {x^{3} d a}{3}\) | \(264\) |
Input:
int(x^2*(e*x+d)*(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)
Output:
1/4*x^4*b*e*ln(c*x^n)-1/16*b*e*n*x^4+1/4*x^4*a*e+1/3*x^3*ln(c*x^n)*b*d-1/9 *b*d*n*x^3+1/3*x^3*d*a
Time = 0.09 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.44 \[ \int x^2 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} \, {\left (b e n - 4 \, a e\right )} x^{4} - \frac {1}{9} \, {\left (b d n - 3 \, a d\right )} x^{3} + \frac {1}{12} \, {\left (3 \, b e x^{4} + 4 \, b d x^{3}\right )} \log \left (c\right ) + \frac {1}{12} \, {\left (3 \, b e n x^{4} + 4 \, b d n x^{3}\right )} \log \left (x\right ) \] Input:
integrate(x^2*(e*x+d)*(a+b*log(c*x^n)),x, algorithm="fricas")
Output:
-1/16*(b*e*n - 4*a*e)*x^4 - 1/9*(b*d*n - 3*a*d)*x^3 + 1/12*(3*b*e*x^4 + 4* b*d*x^3)*log(c) + 1/12*(3*b*e*n*x^4 + 4*b*d*n*x^3)*log(x)
Time = 0.35 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.38 \[ \int x^2 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d x^{3}}{3} + \frac {a e x^{4}}{4} - \frac {b d n x^{3}}{9} + \frac {b d x^{3} \log {\left (c x^{n} \right )}}{3} - \frac {b e n x^{4}}{16} + \frac {b e x^{4} \log {\left (c x^{n} \right )}}{4} \] Input:
integrate(x**2*(e*x+d)*(a+b*ln(c*x**n)),x)
Output:
a*d*x**3/3 + a*e*x**4/4 - b*d*n*x**3/9 + b*d*x**3*log(c*x**n)/3 - b*e*n*x* *4/16 + b*e*x**4*log(c*x**n)/4
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int x^2 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} \, b e n x^{4} + \frac {1}{4} \, b e x^{4} \log \left (c x^{n}\right ) - \frac {1}{9} \, b d n x^{3} + \frac {1}{4} \, a e x^{4} + \frac {1}{3} \, b d x^{3} \log \left (c x^{n}\right ) + \frac {1}{3} \, a d x^{3} \] Input:
integrate(x^2*(e*x+d)*(a+b*log(c*x^n)),x, algorithm="maxima")
Output:
-1/16*b*e*n*x^4 + 1/4*b*e*x^4*log(c*x^n) - 1/9*b*d*n*x^3 + 1/4*a*e*x^4 + 1 /3*b*d*x^3*log(c*x^n) + 1/3*a*d*x^3
Time = 0.11 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.44 \[ \int x^2 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{4} \, b e n x^{4} \log \left (x\right ) - \frac {1}{16} \, b e n x^{4} + \frac {1}{4} \, b e x^{4} \log \left (c\right ) + \frac {1}{3} \, b d n x^{3} \log \left (x\right ) - \frac {1}{9} \, b d n x^{3} + \frac {1}{4} \, a e x^{4} + \frac {1}{3} \, b d x^{3} \log \left (c\right ) + \frac {1}{3} \, a d x^{3} \] Input:
integrate(x^2*(e*x+d)*(a+b*log(c*x^n)),x, algorithm="giac")
Output:
1/4*b*e*n*x^4*log(x) - 1/16*b*e*n*x^4 + 1/4*b*e*x^4*log(c) + 1/3*b*d*n*x^3 *log(x) - 1/9*b*d*n*x^3 + 1/4*a*e*x^4 + 1/3*b*d*x^3*log(c) + 1/3*a*d*x^3
Time = 28.07 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int x^2 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,e\,x^4}{4}+\frac {b\,d\,x^3}{3}\right )+\frac {d\,x^3\,\left (3\,a-b\,n\right )}{9}+\frac {e\,x^4\,\left (4\,a-b\,n\right )}{16} \] Input:
int(x^2*(a + b*log(c*x^n))*(d + e*x),x)
Output:
log(c*x^n)*((b*d*x^3)/3 + (b*e*x^4)/4) + (d*x^3*(3*a - b*n))/9 + (e*x^4*(4 *a - b*n))/16
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98 \[ \int x^2 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{3} \left (48 \,\mathrm {log}\left (x^{n} c \right ) b d +36 \,\mathrm {log}\left (x^{n} c \right ) b e x +48 a d +36 a e x -16 b d n -9 b e n x \right )}{144} \] Input:
int(x^2*(e*x+d)*(a+b*log(c*x^n)),x)
Output:
(x**3*(48*log(x**n*c)*b*d + 36*log(x**n*c)*b*e*x + 48*a*d + 36*a*e*x - 16* b*d*n - 9*b*e*n*x))/144