Integrand size = 19, antiderivative size = 44 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=a e x-b e n x+b e x \log \left (c x^n\right )+\frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 b n} \] Output:
a*e*x-b*e*n*x+b*e*x*ln(c*x^n)+1/2*d*(a+b*ln(c*x^n))^2/b/n
Time = 0.01 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.98 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=a e x-b e n x+a d \log (x)+b e x \log \left (c x^n\right )+\frac {b d \log ^2\left (c x^n\right )}{2 n} \] Input:
Integrate[((d + e*x)*(a + b*Log[c*x^n]))/x,x]
Output:
a*e*x - b*e*n*x + a*d*Log[x] + b*e*x*Log[c*x^n] + (b*d*Log[c*x^n]^2)/(2*n)
Time = 0.26 (sec) , antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2788, 2009, 2738}
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.
\(\displaystyle \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx\) |
\(\Big \downarrow \) 2788 |
\(\displaystyle d \int \frac {a+b \log \left (c x^n\right )}{x}dx+e \int \left (a+b \log \left (c x^n\right )\right )dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle d \int \frac {a+b \log \left (c x^n\right )}{x}dx+e \left (a x+b x \log \left (c x^n\right )-b n x\right )\) |
\(\Big \downarrow \) 2738 |
\(\displaystyle \frac {d \left (a+b \log \left (c x^n\right )\right )^2}{2 b n}+e \left (a x+b x \log \left (c x^n\right )-b n x\right )\) |
Input:
Int[((d + e*x)*(a + b*Log[c*x^n]))/x,x]
Output:
(d*(a + b*Log[c*x^n])^2)/(2*b*n) + e*(a*x - b*n*x + b*x*Log[c*x^n])
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Lo g[c*x^n])^2/(2*b*n), x] /; FreeQ[{a, b, c, n}, x]
Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_) + (e_.)*(x_))^(q_.)) /(x_), x_Symbol] :> Simp[d Int[(d + e*x)^(q - 1)*((a + b*Log[c*x^n])^p/x) , x], x] + Simp[e Int[(d + e*x)^(q - 1)*(a + b*Log[c*x^n])^p, x], x] /; F reeQ[{a, b, c, d, e, n}, x] && IGtQ[p, 0] && GtQ[q, 0] && IntegerQ[2*q]
Time = 0.23 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.05
method | result | size |
default | \(\ln \left (x \right ) a d +a e x +b e x \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )+\frac {b d \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{2}}{2 n}-b e n x\) | \(46\) |
parts | \(\ln \left (x \right ) a d +a e x +b e x \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )+\frac {b d \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{2}}{2 n}-b e n x\) | \(46\) |
parallelrisch | \(\frac {2 x \ln \left (c \,x^{n}\right ) b e n -2 x b e \,n^{2}+2 \ln \left (x \right ) a d n +2 x a e n +b d \ln \left (c \,x^{n}\right )^{2}}{2 n}\) | \(51\) |
norman | \(\left (-b e n +e a \right ) x +\frac {d a \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )}{n}+b e x \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )+\frac {b d \ln \left (c \,{\mathrm e}^{n \ln \left (x \right )}\right )^{2}}{2 n}\) | \(56\) |
risch | \(\left (b e x +b d \ln \left (x \right )\right ) \ln \left (x^{n}\right )-\frac {b d n \ln \left (x \right )^{2}}{2}+\frac {i \pi b e x \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i \pi b e x \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi b e x \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i \pi b e x \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right ) b e x -b e n x +a e x +\frac {i \ln \left (x \right ) \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i \ln \left (x \right ) \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \ln \left (x \right ) \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i \ln \left (x \right ) \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (x \right ) \ln \left (c \right ) b d +\ln \left (x \right ) a d\) | \(238\) |
Input:
int((e*x+d)*(a+b*ln(c*x^n))/x,x,method=_RETURNVERBOSE)
Output:
ln(x)*a*d+a*e*x+b*e*x*ln(c*exp(n*ln(x)))+1/2*b*d/n*ln(c*exp(n*ln(x)))^2-b* e*n*x
Time = 0.07 (sec) , antiderivative size = 45, normalized size of antiderivative = 1.02 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {1}{2} \, b d n \log \left (x\right )^{2} + b e x \log \left (c\right ) - {\left (b e n - a e\right )} x + {\left (b e n x + b d \log \left (c\right ) + a d\right )} \log \left (x\right ) \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))/x,x, algorithm="fricas")
Output:
1/2*b*d*n*log(x)^2 + b*e*x*log(c) - (b*e*n - a*e)*x + (b*e*n*x + b*d*log(c ) + a*d)*log(x)
Time = 0.19 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.48 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\begin {cases} \frac {a d \log {\left (c x^{n} \right )}}{n} + a e x + \frac {b d \log {\left (c x^{n} \right )}^{2}}{2 n} - b e n x + b e x \log {\left (c x^{n} \right )} & \text {for}\: n \neq 0 \\\left (a + b \log {\left (c \right )}\right ) \left (d \log {\left (x \right )} + e x\right ) & \text {otherwise} \end {cases} \] Input:
integrate((e*x+d)*(a+b*ln(c*x**n))/x,x)
Output:
Piecewise((a*d*log(c*x**n)/n + a*e*x + b*d*log(c*x**n)**2/(2*n) - b*e*n*x + b*e*x*log(c*x**n), Ne(n, 0)), ((a + b*log(c))*(d*log(x) + e*x), True))
Time = 0.04 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.93 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=-b e n x + b e x \log \left (c x^{n}\right ) + a e x + \frac {b d \log \left (c x^{n}\right )^{2}}{2 \, n} + a d \log \left (x\right ) \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))/x,x, algorithm="maxima")
Output:
-b*e*n*x + b*e*x*log(c*x^n) + a*e*x + 1/2*b*d*log(c*x^n)^2/n + a*d*log(x)
Time = 0.11 (sec) , antiderivative size = 47, normalized size of antiderivative = 1.07 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=b e n x \log \left (x\right ) + \frac {1}{2} \, b d n \log \left (x\right )^{2} - {\left (b e n - b e \log \left (c\right ) - a e\right )} x + {\left (b d \log \left (c\right ) + a d\right )} \log \left (x\right ) \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))/x,x, algorithm="giac")
Output:
b*e*n*x*log(x) + 1/2*b*d*n*log(x)^2 - (b*e*n - b*e*log(c) - a*e)*x + (b*d* log(c) + a*d)*log(x)
Time = 28.58 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.91 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=a\,d\,\ln \left (x\right )+e\,x\,\left (a-b\,n\right )+b\,e\,x\,\ln \left (c\,x^n\right )+\frac {b\,d\,{\ln \left (c\,x^n\right )}^2}{2\,n} \] Input:
int(((a + b*log(c*x^n))*(d + e*x))/x,x)
Output:
a*d*log(x) + e*x*(a - b*n) + b*e*x*log(c*x^n) + (b*d*log(c*x^n)^2)/(2*n)
Time = 0.15 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.14 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x} \, dx=\frac {\mathrm {log}\left (x^{n} c \right )^{2} b d +2 \,\mathrm {log}\left (x^{n} c \right ) b e n x +2 \,\mathrm {log}\left (x \right ) a d n +2 a e n x -2 b e \,n^{2} x}{2 n} \] Input:
int((e*x+d)*(a+b*log(c*x^n))/x,x)
Output:
(log(x**n*c)**2*b*d + 2*log(x**n*c)*b*e*n*x + 2*log(x)*a*d*n + 2*a*e*n*x - 2*b*e*n**2*x)/(2*n)