Integrand size = 19, antiderivative size = 57 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {b d n}{9 x^3}-\frac {b e n}{4 x^2}-\frac {d \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x^2} \] Output:
-1/9*b*d*n/x^3-1/4*b*e*n/x^2-1/3*d*(a+b*ln(c*x^n))/x^3-1/2*e*(a+b*ln(c*x^n ))/x^2
Time = 0.03 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {6 a (2 d+3 e x)+b n (4 d+9 e x)+6 b (2 d+3 e x) \log \left (c x^n\right )}{36 x^3} \] Input:
Integrate[((d + e*x)*(a + b*Log[c*x^n]))/x^4,x]
Output:
-1/36*(6*a*(2*d + 3*e*x) + b*n*(4*d + 9*e*x) + 6*b*(2*d + 3*e*x)*Log[c*x^n ])/x^3
Time = 0.25 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.05, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.211, Rules used = {2772, 27, 53, 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.
\(\displaystyle \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 2772 |
\(\displaystyle -b n \int -\frac {2 d+3 e x}{6 x^4}dx-\frac {d \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {1}{6} b n \int \frac {2 d+3 e x}{x^4}dx-\frac {d \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {1}{6} b n \int \left (\frac {2 d}{x^4}+\frac {3 e}{x^3}\right )dx-\frac {d \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {d \left (a+b \log \left (c x^n\right )\right )}{3 x^3}-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {1}{6} b n \left (-\frac {2 d}{3 x^3}-\frac {3 e}{2 x^2}\right )\) |
Input:
Int[((d + e*x)*(a + b*Log[c*x^n]))/x^4,x]
Output:
(b*n*((-2*d)/(3*x^3) - (3*e)/(2*x^2)))/6 - (d*(a + b*Log[c*x^n]))/(3*x^3) - (e*(a + b*Log[c*x^n]))/(2*x^2)
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, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[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[(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 = 0.15 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.84
method | result | size |
parallelrisch | \(-\frac {18 b e x \ln \left (c \,x^{n}\right )+9 b e n x +18 a e x +12 \ln \left (c \,x^{n}\right ) b d +4 b d n +12 d a}{36 x^{3}}\) | \(48\) |
risch | \(-\frac {b \left (3 e x +2 d \right ) \ln \left (x^{n}\right )}{6 x^{3}}-\frac {9 i \pi b e x \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-9 i \pi b e x \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-9 i \pi b e x \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+9 i \pi b e x \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+18 \ln \left (c \right ) b e x +9 b e n x +18 a e x +6 i \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-6 i \pi b d \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-6 i \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+6 i \pi b d \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+12 \ln \left (c \right ) b d +4 b d n +12 d a}{36 x^{3}}\) | \(235\) |
Input:
int((e*x+d)*(a+b*ln(c*x^n))/x^4,x,method=_RETURNVERBOSE)
Output:
-1/36/x^3*(18*b*e*x*ln(c*x^n)+9*b*e*n*x+18*a*e*x+12*ln(c*x^n)*b*d+4*b*d*n+ 12*d*a)
Time = 0.07 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {4 \, b d n + 12 \, a d + 9 \, {\left (b e n + 2 \, a e\right )} x + 6 \, {\left (3 \, b e x + 2 \, b d\right )} \log \left (c\right ) + 6 \, {\left (3 \, b e n x + 2 \, b d n\right )} \log \left (x\right )}{36 \, x^{3}} \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))/x^4,x, algorithm="fricas")
Output:
-1/36*(4*b*d*n + 12*a*d + 9*(b*e*n + 2*a*e)*x + 6*(3*b*e*x + 2*b*d)*log(c) + 6*(3*b*e*n*x + 2*b*d*n)*log(x))/x^3
Time = 0.39 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=- \frac {a d}{3 x^{3}} - \frac {a e}{2 x^{2}} - \frac {b d n}{9 x^{3}} - \frac {b d \log {\left (c x^{n} \right )}}{3 x^{3}} - \frac {b e n}{4 x^{2}} - \frac {b e \log {\left (c x^{n} \right )}}{2 x^{2}} \] Input:
integrate((e*x+d)*(a+b*ln(c*x**n))/x**4,x)
Output:
-a*d/(3*x**3) - a*e/(2*x**2) - b*d*n/(9*x**3) - b*d*log(c*x**n)/(3*x**3) - b*e*n/(4*x**2) - b*e*log(c*x**n)/(2*x**2)
Time = 0.04 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.00 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {b e n}{4 \, x^{2}} - \frac {b e \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac {b d n}{9 \, x^{3}} - \frac {a e}{2 \, x^{2}} - \frac {b d \log \left (c x^{n}\right )}{3 \, x^{3}} - \frac {a d}{3 \, x^{3}} \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))/x^4,x, algorithm="maxima")
Output:
-1/4*b*e*n/x^2 - 1/2*b*e*log(c*x^n)/x^2 - 1/9*b*d*n/x^3 - 1/2*a*e/x^2 - 1/ 3*b*d*log(c*x^n)/x^3 - 1/3*a*d/x^3
Time = 0.11 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.04 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {{\left (3 \, b e n x + 2 \, b d n\right )} \log \left (x\right )}{6 \, x^{3}} - \frac {9 \, b e n x + 18 \, b e x \log \left (c\right ) + 4 \, b d n + 18 \, a e x + 12 \, b d \log \left (c\right ) + 12 \, a d}{36 \, x^{3}} \] Input:
integrate((e*x+d)*(a+b*log(c*x^n))/x^4,x, algorithm="giac")
Output:
-1/6*(3*b*e*n*x + 2*b*d*n)*log(x)/x^3 - 1/36*(9*b*e*n*x + 18*b*e*x*log(c) + 4*b*d*n + 18*a*e*x + 12*b*d*log(c) + 12*a*d)/x^3
Time = 28.62 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.86 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {2\,a\,d+x\,\left (3\,a\,e+\frac {3\,b\,e\,n}{2}\right )+\frac {2\,b\,d\,n}{3}}{6\,x^3}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d}{3}+\frac {b\,e\,x}{2}\right )}{x^3} \] Input:
int(((a + b*log(c*x^n))*(d + e*x))/x^4,x)
Output:
- (2*a*d + x*(3*a*e + (3*b*e*n)/2) + (2*b*d*n)/3)/(6*x^3) - (log(c*x^n)*(( b*d)/3 + (b*e*x)/2))/x^3
Time = 0.15 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.82 \[ \int \frac {(d+e x) \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=\frac {-12 \,\mathrm {log}\left (x^{n} c \right ) b d -18 \,\mathrm {log}\left (x^{n} c \right ) b e x -12 a d -18 a e x -4 b d n -9 b e n x}{36 x^{3}} \] Input:
int((e*x+d)*(a+b*log(c*x^n))/x^4,x)
Output:
( - 12*log(x**n*c)*b*d - 18*log(x**n*c)*b*e*x - 12*a*d - 18*a*e*x - 4*b*d* n - 9*b*e*n*x)/(36*x**3)