Integrand size = 19, antiderivative size = 48 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{16} b d n x^4-\frac {1}{25} b e n x^5+\frac {1}{20} \left (5 d x^4+4 e x^5\right ) \left (a+b \log \left (c x^n\right )\right ) \] Output:
-1/16*b*d*n*x^4-1/25*b*e*n*x^5+1/20*(4*e*x^5+5*d*x^4)*(a+b*ln(c*x^n))
Time = 0.04 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{400} x^4 \left (20 a (5 d+4 e x)-b n (25 d+16 e x)+20 b (5 d+4 e x) \log \left (c x^n\right )\right ) \] Input:
Integrate[x^3*(d + e*x)*(a + b*Log[c*x^n]),x]
Output:
(x^4*(20*a*(5*d + 4*e*x) - b*n*(25*d + 16*e*x) + 20*b*(5*d + 4*e*x)*Log[c* x^n]))/400
Time = 0.23 (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.
\(\displaystyle \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2771 |
\(\displaystyle \frac {1}{20} \left (5 d x^4+4 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )-b n \int \frac {1}{20} x^3 (5 d+4 e x)dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{20} \left (5 d x^4+4 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{20} b n \int x^3 (5 d+4 e x)dx\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {1}{20} \left (5 d x^4+4 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{20} b n \int \left (4 e x^4+5 d x^3\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{20} \left (5 d x^4+4 e x^5\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{20} b n \left (\frac {5 d x^4}{4}+\frac {4 e x^5}{5}\right )\) |
Input:
Int[x^3*(d + e*x)*(a + b*Log[c*x^n]),x]
Output:
-1/20*(b*n*((5*d*x^4)/4 + (4*e*x^5)/5)) + ((5*d*x^4 + 4*e*x^5)*(a + b*Log[ c*x^n]))/20
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 = 0.30 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(\frac {x^{5} \ln \left (c \,x^{n}\right ) b e}{5}-\frac {b e n \,x^{5}}{25}+\frac {e a \,x^{5}}{5}+\frac {x^{4} \ln \left (c \,x^{n}\right ) b d}{4}-\frac {b d n \,x^{4}}{16}+\frac {d a \,x^{4}}{4}\) | \(58\) |
risch | \(\frac {b \,x^{4} \left (4 e x +5 d \right ) \ln \left (x^{n}\right )}{20}+\frac {i \pi b e \,x^{5} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{10}-\frac {i \pi b e \,x^{5} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{10}-\frac {i \pi b e \,x^{5} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{10}+\frac {i \pi b e \,x^{5} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{10}+\frac {\ln \left (c \right ) b e \,x^{5}}{5}-\frac {b e n \,x^{5}}{25}+\frac {e a \,x^{5}}{5}+\frac {i \pi b d \,x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b d \,x^{4} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{8}-\frac {i \pi b d \,x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{8}+\frac {i \pi b d \,x^{4} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{8}+\frac {\ln \left (c \right ) b d \,x^{4}}{4}-\frac {b d n \,x^{4}}{16}+\frac {d a \,x^{4}}{4}\) | \(264\) |
Input:
int(x^3*(e*x+d)*(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)
Output:
1/5*x^5*ln(c*x^n)*b*e-1/25*b*e*n*x^5+1/5*e*a*x^5+1/4*x^4*ln(c*x^n)*b*d-1/1 6*b*d*n*x^4+1/4*d*a*x^4
Time = 0.07 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.44 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{25} \, {\left (b e n - 5 \, a e\right )} x^{5} - \frac {1}{16} \, {\left (b d n - 4 \, a d\right )} x^{4} + \frac {1}{20} \, {\left (4 \, b e x^{5} + 5 \, b d x^{4}\right )} \log \left (c\right ) + \frac {1}{20} \, {\left (4 \, b e n x^{5} + 5 \, b d n x^{4}\right )} \log \left (x\right ) \] Input:
integrate(x^3*(e*x+d)*(a+b*log(c*x^n)),x, algorithm="fricas")
Output:
-1/25*(b*e*n - 5*a*e)*x^5 - 1/16*(b*d*n - 4*a*d)*x^4 + 1/20*(4*b*e*x^5 + 5 *b*d*x^4)*log(c) + 1/20*(4*b*e*n*x^5 + 5*b*d*n*x^4)*log(x)
Time = 0.38 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.38 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {a d x^{4}}{4} + \frac {a e x^{5}}{5} - \frac {b d n x^{4}}{16} + \frac {b d x^{4} \log {\left (c x^{n} \right )}}{4} - \frac {b e n x^{5}}{25} + \frac {b e x^{5} \log {\left (c x^{n} \right )}}{5} \] Input:
integrate(x**3*(e*x+d)*(a+b*ln(c*x**n)),x)
Output:
a*d*x**4/4 + a*e*x**5/5 - b*d*n*x**4/16 + b*d*x**4*log(c*x**n)/4 - b*e*n*x **5/25 + b*e*x**5*log(c*x**n)/5
Time = 0.03 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.19 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{25} \, b e n x^{5} + \frac {1}{5} \, b e x^{5} \log \left (c x^{n}\right ) - \frac {1}{16} \, b d n x^{4} + \frac {1}{5} \, a e x^{5} + \frac {1}{4} \, b d x^{4} \log \left (c x^{n}\right ) + \frac {1}{4} \, a d x^{4} \] Input:
integrate(x^3*(e*x+d)*(a+b*log(c*x^n)),x, algorithm="maxima")
Output:
-1/25*b*e*n*x^5 + 1/5*b*e*x^5*log(c*x^n) - 1/16*b*d*n*x^4 + 1/5*a*e*x^5 + 1/4*b*d*x^4*log(c*x^n) + 1/4*a*d*x^4
Time = 0.12 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.44 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{5} \, b e n x^{5} \log \left (x\right ) - \frac {1}{25} \, b e n x^{5} + \frac {1}{5} \, b e x^{5} \log \left (c\right ) + \frac {1}{4} \, b d n x^{4} \log \left (x\right ) - \frac {1}{16} \, b d n x^{4} + \frac {1}{5} \, a e x^{5} + \frac {1}{4} \, b d x^{4} \log \left (c\right ) + \frac {1}{4} \, a d x^{4} \] Input:
integrate(x^3*(e*x+d)*(a+b*log(c*x^n)),x, algorithm="giac")
Output:
1/5*b*e*n*x^5*log(x) - 1/25*b*e*n*x^5 + 1/5*b*e*x^5*log(c) + 1/4*b*d*n*x^4 *log(x) - 1/16*b*d*n*x^4 + 1/5*a*e*x^5 + 1/4*b*d*x^4*log(c) + 1/4*a*d*x^4
Time = 28.31 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,e\,x^5}{5}+\frac {b\,d\,x^4}{4}\right )+\frac {d\,x^4\,\left (4\,a-b\,n\right )}{16}+\frac {e\,x^5\,\left (5\,a-b\,n\right )}{25} \] Input:
int(x^3*(a + b*log(c*x^n))*(d + e*x),x)
Output:
log(c*x^n)*((b*d*x^4)/4 + (b*e*x^5)/5) + (d*x^4*(4*a - b*n))/16 + (e*x^5*( 5*a - b*n))/25
Time = 0.16 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.98 \[ \int x^3 (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x^{4} \left (100 \,\mathrm {log}\left (x^{n} c \right ) b d +80 \,\mathrm {log}\left (x^{n} c \right ) b e x +100 a d +80 a e x -25 b d n -16 b e n x \right )}{400} \] Input:
int(x^3*(e*x+d)*(a+b*log(c*x^n)),x)
Output:
(x**4*(100*log(x**n*c)*b*d + 80*log(x**n*c)*b*e*x + 100*a*d + 80*a*e*x - 2 5*b*d*n - 16*b*e*n*x))/400