Integrand size = 16, antiderivative size = 55 \[ \int (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-b d n x-\frac {1}{4} b e n x^2-\frac {b d^2 n \log (x)}{2 e}+\frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{2 e} \] Output:
-b*d*n*x-1/4*b*e*n*x^2-1/2*b*d^2*n*ln(x)/e+1/2*(e*x+d)^2*(a+b*ln(c*x^n))/e
Time = 0.01 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00 \[ \int (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=a d x-b d n x+\frac {1}{2} a e x^2-\frac {1}{4} b e n x^2+b d x \log \left (c x^n\right )+\frac {1}{2} b e x^2 \log \left (c x^n\right ) \] Input:
Integrate[(d + e*x)*(a + b*Log[c*x^n]),x]
Output:
a*d*x - b*d*n*x + (a*e*x^2)/2 - (b*e*n*x^2)/4 + b*d*x*Log[c*x^n] + (b*e*x^ 2*Log[c*x^n])/2
Time = 0.22 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2750, 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.
\(\displaystyle \int (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx\) |
\(\Big \downarrow \) 2750 |
\(\displaystyle \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-b n \int \frac {(d+e x)^2}{2 e x}dx\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {b n \int \frac {(d+e x)^2}{x}dx}{2 e}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {b n \int \left (\frac {d^2}{x}+2 e d+e^2 x\right )dx}{2 e}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {b n \left (d^2 \log (x)+2 d e x+\frac {e^2 x^2}{2}\right )}{2 e}\) |
Input:
Int[(d + e*x)*(a + b*Log[c*x^n]),x]
Output:
-1/2*(b*n*(2*d*e*x + (e^2*x^2)/2 + d^2*Log[x]))/e + ((d + e*x)^2*(a + b*Lo g[c*x^n]))/(2*e)
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_.))*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^r)^q, x]}, Simp[(a + b*Log[c*x^n]) u, x] - Simp[b*n Int[SimplifyIntegrand[u/x, x], x], x]] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0]
Time = 0.23 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.91
method | result | size |
parallelrisch | \(\frac {b e \,x^{2} \ln \left (c \,x^{n}\right )}{2}-\frac {b e n \,x^{2}}{4}+\frac {a e \,x^{2}}{2}+x \ln \left (c \,x^{n}\right ) b d -b d n x +x d a\) | \(50\) |
default | \(x d a +\frac {a e \,x^{2}}{2}+b d \left (x \ln \left (c \,x^{n}\right )-x n \right )+\frac {b e \,x^{2} \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )}{2}-\frac {b e n \,x^{2}}{4}\) | \(52\) |
parts | \(a \left (\frac {1}{2} e \,x^{2}+d x \right )+b \left (x \ln \left (c \,x^{n}\right ) d -x d n +\frac {x^{2} e \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )}{2}-\frac {e n \,x^{2}}{4}\right )\) | \(52\) |
norman | \(\left (-\frac {1}{4} b e n +\frac {1}{2} e a \right ) x^{2}+\left (-b d n +d a \right ) x +b d x \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )+\frac {b e \,x^{2} \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )}{2}\) | \(54\) |
risch | \(\frac {b x \left (e x +2 d \right ) \ln \left (x^{n}\right )}{2}+\frac {i \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} b e \,x^{2}}{4}-\frac {i \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right ) b e \,x^{2}}{4}-\frac {i \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3} b e \,x^{2}}{4}+\frac {i \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right ) b e \,x^{2}}{4}+\frac {i \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} b d x}{2}-\frac {i \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right ) b d x}{2}-\frac {i \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3} b d x}{2}+\frac {i \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right ) b d x}{2}+\frac {\ln \left (c \right ) b e \,x^{2}}{2}-\frac {b e n \,x^{2}}{4}+x \ln \left (c \right ) b d +\frac {a e \,x^{2}}{2}-b d n x +x d a\) | \(245\) |
Input:
int((e*x+d)*(a+b*ln(c*x^n)),x,method=_RETURNVERBOSE)
Output:
1/2*b*e*x^2*ln(c*x^n)-1/4*b*e*n*x^2+1/2*a*e*x^2+x*ln(c*x^n)*b*d-b*d*n*x+x* d*a
Time = 0.08 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.11 \[ \int (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{4} \, {\left (b e n - 2 \, a e\right )} x^{2} - {\left (b d n - a d\right )} x + \frac {1}{2} \, {\left (b e x^{2} + 2 \, b d x\right )} \log \left (c\right ) + \frac {1}{2} \, {\left (b e n x^{2} + 2 \, b d n x\right )} \log \left (x\right ) \] Input:
integrate((e*x+d)*(a+b*log(c*x^n)),x, algorithm="fricas")
Output:
-1/4*(b*e*n - 2*a*e)*x^2 - (b*d*n - a*d)*x + 1/2*(b*e*x^2 + 2*b*d*x)*log(c ) + 1/2*(b*e*n*x^2 + 2*b*d*n*x)*log(x)
Time = 0.15 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.02 \[ \int (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=a d x + \frac {a e x^{2}}{2} - b d n x + b d x \log {\left (c x^{n} \right )} - \frac {b e n x^{2}}{4} + \frac {b e x^{2} \log {\left (c x^{n} \right )}}{2} \] Input:
integrate((e*x+d)*(a+b*ln(c*x**n)),x)
Output:
a*d*x + a*e*x**2/2 - b*d*n*x + b*d*x*log(c*x**n) - b*e*n*x**2/4 + b*e*x**2 *log(c*x**n)/2
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.89 \[ \int (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=-\frac {1}{4} \, b e n x^{2} + \frac {1}{2} \, b e x^{2} \log \left (c x^{n}\right ) - b d n x + \frac {1}{2} \, a e x^{2} + b d x \log \left (c x^{n}\right ) + a d x \] Input:
integrate((e*x+d)*(a+b*log(c*x^n)),x, algorithm="maxima")
Output:
-1/4*b*e*n*x^2 + 1/2*b*e*x^2*log(c*x^n) - b*d*n*x + 1/2*a*e*x^2 + b*d*x*lo g(c*x^n) + a*d*x
Time = 0.12 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.05 \[ \int (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {1}{2} \, b e n x^{2} \log \left (x\right ) - \frac {1}{4} \, b e n x^{2} + \frac {1}{2} \, b e x^{2} \log \left (c\right ) + b d n x \log \left (x\right ) - b d n x + \frac {1}{2} \, a e x^{2} + b d x \log \left (c\right ) + a d x \] Input:
integrate((e*x+d)*(a+b*log(c*x^n)),x, algorithm="giac")
Output:
1/2*b*e*n*x^2*log(x) - 1/4*b*e*n*x^2 + 1/2*b*e*x^2*log(c) + b*d*n*x*log(x) - b*d*n*x + 1/2*a*e*x^2 + b*d*x*log(c) + a*d*x
Time = 28.30 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.78 \[ \int (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\ln \left (c\,x^n\right )\,\left (\frac {b\,e\,x^2}{2}+b\,d\,x\right )+d\,x\,\left (a-b\,n\right )+\frac {e\,x^2\,\left (2\,a-b\,n\right )}{4} \] Input:
int((a + b*log(c*x^n))*(d + e*x),x)
Output:
log(c*x^n)*(b*d*x + (b*e*x^2)/2) + d*x*(a - b*n) + (e*x^2*(2*a - b*n))/4
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.82 \[ \int (d+e x) \left (a+b \log \left (c x^n\right )\right ) \, dx=\frac {x \left (4 \,\mathrm {log}\left (x^{n} c \right ) b d +2 \,\mathrm {log}\left (x^{n} c \right ) b e x +4 a d +2 a e x -4 b d n -b e n x \right )}{4} \] Input:
int((e*x+d)*(a+b*log(c*x^n)),x)
Output:
(x*(4*log(x**n*c)*b*d + 2*log(x**n*c)*b*e*x + 4*a*d + 2*a*e*x - 4*b*d*n - b*e*n*x))/4