Integrand size = 19, antiderivative size = 48 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {b d n}{x}-\frac {d \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{2 b n} \] Output:
-b*d*n/x-d*(a+b*ln(c*x^n))/x+1/2*e*(a+b*ln(c*x^n))^2/b/n
Time = 0.03 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=-\frac {b d n}{x}-\frac {d \left (a+b \log \left (c x^n\right )\right )}{x}+\frac {e \left (a+b \log \left (c x^n\right )\right )^2}{2 b n} \] Input:
Integrate[((d + e*x)*(a + b*Log[c*x^n]))/x^2,x]
Output:
-((b*d*n)/x) - (d*(a + b*Log[c*x^n]))/x + (e*(a + b*Log[c*x^n])^2)/(2*b*n)
Time = 0.28 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2772, 25, 2010, 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.
\(\displaystyle \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 2772 |
\(\displaystyle -b n \int -\frac {d-e x \log (x)}{x^2}dx-\frac {d \left (a+b \log \left (c x^n\right )\right )}{x}+e \log (x) \left (a+b \log \left (c x^n\right )\right )\) |
\(\Big \downarrow \) 25 |
\(\displaystyle b n \int \frac {d-e x \log (x)}{x^2}dx-\frac {d \left (a+b \log \left (c x^n\right )\right )}{x}+e \log (x) \left (a+b \log \left (c x^n\right )\right )\) |
\(\Big \downarrow \) 2010 |
\(\displaystyle b n \int \left (\frac {d}{x^2}-\frac {e \log (x)}{x}\right )dx-\frac {d \left (a+b \log \left (c x^n\right )\right )}{x}+e \log (x) \left (a+b \log \left (c x^n\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d \left (a+b \log \left (c x^n\right )\right )}{x}+e \log (x) \left (a+b \log \left (c x^n\right )\right )+b n \left (-\frac {d}{x}-\frac {1}{2} e \log ^2(x)\right )\) |
Input:
Int[((d + e*x)*(a + b*Log[c*x^n]))/x^2,x]
Output:
b*n*(-(d/x) - (e*Log[x]^2)/2) - (d*(a + b*Log[c*x^n]))/x + e*Log[x]*(a + b *Log[c*x^n])
Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x] , x] /; FreeQ[{c, m}, x] && SumQ[u] && !LinearQ[u, x] && !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]
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[(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] && IntegerQ[m] && !(EqQ[q , 1] && EqQ[m, -1])
Time = 0.19 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10
method | result | size |
parallelrisch | \(\frac {2 \ln \left (x \right ) x a e n +b e \ln \left (c \,x^{n}\right )^{2} x -2 \ln \left (c \,x^{n}\right ) b d n -2 b d \,n^{2}-2 a d n}{2 x n}\) | \(53\) |
risch | \(-\frac {b \left (-e x \ln \left (x \right )+d \right ) \ln \left (x^{n}\right )}{x}-\frac {-i \ln \left (x \right ) \pi b e \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2} x +i \ln \left (x \right ) \pi b e \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right ) x +i \ln \left (x \right ) \pi b e \operatorname {csgn}\left (i c \,x^{n}\right )^{3} x -i \ln \left (x \right ) \pi b e \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right ) x +i \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-i \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+b e n \ln \left (x \right )^{2} x -2 \ln \left (x \right ) \ln \left (c \right ) b e x -2 \ln \left (x \right ) e a x +2 \ln \left (c \right ) b d +2 b d n +2 d a}{2 x}\) | \(250\) |
Input:
int((e*x+d)*(a+b*ln(c*x^n))/x^2,x,method=_RETURNVERBOSE)
Output:
1/2/x*(2*ln(x)*x*a*e*n+b*e*ln(c*x^n)^2*x-2*ln(c*x^n)*b*d*n-2*b*d*n^2-2*a*d *n)/n
Time = 0.07 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {b e n x \log \left (x\right )^{2} - 2 \, b d n - 2 \, b d \log \left (c\right ) - 2 \, a d + 2 \, {\left (b e x \log \left (c\right ) - b d n + a e x\right )} \log \left (x\right )}{2 \, x} \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))/x^2,x, algorithm="fricas")
Output:
1/2*(b*e*n*x*log(x)^2 - 2*b*d*n - 2*b*d*log(c) - 2*a*d + 2*(b*e*x*log(c) - b*d*n + a*e*x)*log(x))/x
Time = 1.80 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.10 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=- \frac {a d}{x} + a e \log {\left (x \right )} + b d \left (- \frac {n}{x} - \frac {\log {\left (c x^{n} \right )}}{x}\right ) - b e \left (\begin {cases} - \log {\left (c \right )} \log {\left (x \right )} & \text {for}\: n = 0 \\- \frac {\log {\left (c x^{n} \right )}^{2}}{2 n} & \text {otherwise} \end {cases}\right ) \] Input:
integrate((e*x+d)*(a+b*ln(c*x**n))/x**2,x)
Output:
-a*d/x + a*e*log(x) + b*d*(-n/x - log(c*x**n)/x) - b*e*Piecewise((-log(c)* log(x), Eq(n, 0)), (-log(c*x**n)**2/(2*n), True))
Time = 0.03 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.02 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {b e \log \left (c x^{n}\right )^{2}}{2 \, n} + a e \log \left (x\right ) - \frac {b d n}{x} - \frac {b d \log \left (c x^{n}\right )}{x} - \frac {a d}{x} \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))/x^2,x, algorithm="maxima")
Output:
1/2*b*e*log(c*x^n)^2/n + a*e*log(x) - b*d*n/x - b*d*log(c*x^n)/x - a*d/x
Time = 0.11 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {1}{2} \, b e n \log \left (x\right )^{2} - b d n {\left (\frac {\log \left (x\right )}{x} + \frac {1}{x}\right )} + b e \log \left (c\right ) \log \left ({\left | x \right |}\right ) + a e \log \left ({\left | x \right |}\right ) - \frac {b d \log \left (c\right )}{x} - \frac {a d}{x} \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))/x^2,x, algorithm="giac")
Output:
1/2*b*e*n*log(x)^2 - b*d*n*(log(x)/x + 1/x) + b*e*log(c)*log(abs(x)) + a*e *log(abs(x)) - b*d*log(c)/x - a*d/x
Time = 29.33 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.23 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\ln \left (x\right )\,\left (a\,e+b\,e\,n\right )-\frac {a\,d+b\,d\,n}{x}-\frac {\ln \left (c\,x^n\right )\,\left (b\,d+b\,e\,x\right )}{x}+\frac {b\,e\,{\ln \left (c\,x^n\right )}^2}{2\,n} \] Input:
int(((a + b*log(c*x^n))*(d + e*x))/x^2,x)
Output:
log(x)*(a*e + b*e*n) - (a*d + b*d*n)/x - (log(c*x^n)*(b*d + b*e*x))/x + (b *e*log(c*x^n)^2)/(2*n)
Time = 0.15 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^2} \, dx=\frac {\mathrm {log}\left (x^{n} c \right )^{2} b e x -2 \,\mathrm {log}\left (x^{n} c \right ) b d n +2 \,\mathrm {log}\left (x \right ) a e n x -2 a d n -2 b d \,n^{2}}{2 n x} \] Input:
int((e*x+d)*(a+b*log(c*x^n))/x^2,x)
Output:
(log(x**n*c)**2*b*e*x - 2*log(x**n*c)*b*d*n + 2*log(x)*a*e*n*x - 2*a*d*n - 2*b*d*n**2)/(2*n*x)