Integrand size = 21, antiderivative size = 142 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\frac {b d^2 e n}{80 x^4}+\frac {b d e^2 n}{15 x^3}+\frac {3 b e^3 n}{20 x^2}+\frac {b e^4 n}{5 d x}-\frac {b n (d+e x)^5}{25 d^2 x^5}-\frac {b e^5 n \log (x)}{20 d^2}-\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {e (d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{20 d^2 x^4} \] Output:
1/80*b*d^2*e*n/x^4+1/15*b*d*e^2*n/x^3+3/20*b*e^3*n/x^2+1/5*b*e^4*n/d/x-1/2 5*b*n*(e*x+d)^5/d^2/x^5-1/20*b*e^5*n*ln(x)/d^2-1/5*(e*x+d)^4*(a+b*ln(c*x^n ))/d/x^5+1/20*e*(e*x+d)^4*(a+b*ln(c*x^n))/d^2/x^4
Time = 0.06 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.80 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {60 a \left (4 d^3+15 d^2 e x+20 d e^2 x^2+10 e^3 x^3\right )+b n \left (48 d^3+225 d^2 e x+400 d e^2 x^2+300 e^3 x^3\right )+60 b \left (4 d^3+15 d^2 e x+20 d e^2 x^2+10 e^3 x^3\right ) \log \left (c x^n\right )}{1200 x^5} \] Input:
Integrate[((d + e*x)^3*(a + b*Log[c*x^n]))/x^6,x]
Output:
-1/1200*(60*a*(4*d^3 + 15*d^2*e*x + 20*d*e^2*x^2 + 10*e^3*x^3) + b*n*(48*d ^3 + 225*d^2*e*x + 400*d*e^2*x^2 + 300*e^3*x^3) + 60*b*(4*d^3 + 15*d^2*e*x + 20*d*e^2*x^2 + 10*e^3*x^3)*Log[c*x^n])/x^5
Time = 0.34 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2772, 27, 87, 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.
\(\displaystyle \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx\) |
\(\Big \downarrow \) 2772 |
\(\displaystyle -b n \int -\frac {(4 d-e x) (d+e x)^4}{20 d^2 x^6}dx+\frac {e (d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{20 d^2 x^4}-\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b n \int \frac {(4 d-e x) (d+e x)^4}{x^6}dx}{20 d^2}+\frac {e (d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{20 d^2 x^4}-\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}\) |
\(\Big \downarrow \) 87 |
\(\displaystyle \frac {b n \left (-e \int \frac {(d+e x)^4}{x^5}dx-\frac {4 (d+e x)^5}{5 x^5}\right )}{20 d^2}+\frac {e (d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{20 d^2 x^4}-\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {b n \left (-e \int \left (\frac {d^4}{x^5}+\frac {4 e d^3}{x^4}+\frac {6 e^2 d^2}{x^3}+\frac {4 e^3 d}{x^2}+\frac {e^4}{x}\right )dx-\frac {4 (d+e x)^5}{5 x^5}\right )}{20 d^2}+\frac {e (d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{20 d^2 x^4}-\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {e (d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{20 d^2 x^4}-\frac {(d+e x)^4 \left (a+b \log \left (c x^n\right )\right )}{5 d x^5}+\frac {b n \left (-e \left (-\frac {d^4}{4 x^4}-\frac {4 d^3 e}{3 x^3}-\frac {3 d^2 e^2}{x^2}-\frac {4 d e^3}{x}+e^4 \log (x)\right )-\frac {4 (d+e x)^5}{5 x^5}\right )}{20 d^2}\) |
Input:
Int[((d + e*x)^3*(a + b*Log[c*x^n]))/x^6,x]
Output:
(b*n*((-4*(d + e*x)^5)/(5*x^5) - e*(-1/4*d^4/x^4 - (4*d^3*e)/(3*x^3) - (3* d^2*e^2)/x^2 - (4*d*e^3)/x + e^4*Log[x])))/(20*d^2) - ((d + e*x)^4*(a + b* Log[c*x^n]))/(5*d*x^5) + (e*(d + e*x)^4*(a + b*Log[c*x^n]))/(20*d^2*x^4)
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_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
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 = 0.53 (sec) , antiderivative size = 134, normalized size of antiderivative = 0.94
method | result | size |
parallelrisch | \(-\frac {600 b \ln \left (c \,x^{n}\right ) e^{3} x^{3}+300 b \,e^{3} n \,x^{3}+600 a \,e^{3} x^{3}+1200 b \ln \left (c \,x^{n}\right ) d \,e^{2} x^{2}+400 b d \,e^{2} n \,x^{2}+1200 a \,e^{2} x^{2} d +900 b \ln \left (c \,x^{n}\right ) d^{2} e x +225 b \,d^{2} e n x +900 a \,d^{2} e x +240 b \ln \left (c \,x^{n}\right ) d^{3}+48 b \,d^{3} n +240 a \,d^{3}}{1200 x^{5}}\) | \(134\) |
risch | \(-\frac {b \left (10 e^{3} x^{3}+20 d \,e^{2} x^{2}+15 d^{2} e x +4 d^{3}\right ) \ln \left (x^{n}\right )}{20 x^{5}}-\frac {300 i \pi b \,e^{3} x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+1200 a \,e^{2} x^{2} d +900 a \,d^{2} e x +240 a \,d^{3}-300 i \pi b \,e^{3} x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+120 i \pi b \,d^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+600 \ln \left (c \right ) b \,e^{3} x^{3}+1200 \ln \left (c \right ) b d \,e^{2} x^{2}+900 \ln \left (c \right ) b \,d^{2} e x -300 i \pi b \,e^{3} x^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+450 i \pi b \,d^{2} e x \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+450 i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right ) e x +600 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+600 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+240 d^{3} b \ln \left (c \right )-120 i \pi b \,d^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+600 a \,e^{3} x^{3}+48 b \,d^{3} n -120 i \pi b \,d^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+300 i \pi b \,e^{3} x^{3} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )-600 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-450 i \pi b \,d^{2} e x \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+120 i \pi b \,d^{3} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-450 i \pi b \,d^{2} e x \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-600 i \pi b d \,e^{2} x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+225 b \,d^{2} e n x +400 b d \,e^{2} n \,x^{2}+300 b \,e^{3} n \,x^{3}}{1200 x^{5}}\) | \(571\) |
Input:
int((e*x+d)^3*(a+b*ln(c*x^n))/x^6,x,method=_RETURNVERBOSE)
Output:
-1/1200/x^5*(600*b*ln(c*x^n)*e^3*x^3+300*b*e^3*n*x^3+600*a*e^3*x^3+1200*b* ln(c*x^n)*d*e^2*x^2+400*b*d*e^2*n*x^2+1200*a*e^2*x^2*d+900*b*ln(c*x^n)*d^2 *e*x+225*b*d^2*e*n*x+900*a*d^2*e*x+240*b*ln(c*x^n)*d^3+48*b*d^3*n+240*a*d^ 3)
Time = 0.07 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.09 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {48 \, b d^{3} n + 240 \, a d^{3} + 300 \, {\left (b e^{3} n + 2 \, a e^{3}\right )} x^{3} + 400 \, {\left (b d e^{2} n + 3 \, a d e^{2}\right )} x^{2} + 225 \, {\left (b d^{2} e n + 4 \, a d^{2} e\right )} x + 60 \, {\left (10 \, b e^{3} x^{3} + 20 \, b d e^{2} x^{2} + 15 \, b d^{2} e x + 4 \, b d^{3}\right )} \log \left (c\right ) + 60 \, {\left (10 \, b e^{3} n x^{3} + 20 \, b d e^{2} n x^{2} + 15 \, b d^{2} e n x + 4 \, b d^{3} n\right )} \log \left (x\right )}{1200 \, x^{5}} \] Input:
integrate((e*x+d)^3*(a+b*log(c*x^n))/x^6,x, algorithm="fricas")
Output:
-1/1200*(48*b*d^3*n + 240*a*d^3 + 300*(b*e^3*n + 2*a*e^3)*x^3 + 400*(b*d*e ^2*n + 3*a*d*e^2)*x^2 + 225*(b*d^2*e*n + 4*a*d^2*e)*x + 60*(10*b*e^3*x^3 + 20*b*d*e^2*x^2 + 15*b*d^2*e*x + 4*b*d^3)*log(c) + 60*(10*b*e^3*n*x^3 + 20 *b*d*e^2*n*x^2 + 15*b*d^2*e*n*x + 4*b*d^3*n)*log(x))/x^5
Time = 0.73 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.18 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=- \frac {a d^{3}}{5 x^{5}} - \frac {3 a d^{2} e}{4 x^{4}} - \frac {a d e^{2}}{x^{3}} - \frac {a e^{3}}{2 x^{2}} - \frac {b d^{3} n}{25 x^{5}} - \frac {b d^{3} \log {\left (c x^{n} \right )}}{5 x^{5}} - \frac {3 b d^{2} e n}{16 x^{4}} - \frac {3 b d^{2} e \log {\left (c x^{n} \right )}}{4 x^{4}} - \frac {b d e^{2} n}{3 x^{3}} - \frac {b d e^{2} \log {\left (c x^{n} \right )}}{x^{3}} - \frac {b e^{3} n}{4 x^{2}} - \frac {b e^{3} \log {\left (c x^{n} \right )}}{2 x^{2}} \] Input:
integrate((e*x+d)**3*(a+b*ln(c*x**n))/x**6,x)
Output:
-a*d**3/(5*x**5) - 3*a*d**2*e/(4*x**4) - a*d*e**2/x**3 - a*e**3/(2*x**2) - b*d**3*n/(25*x**5) - b*d**3*log(c*x**n)/(5*x**5) - 3*b*d**2*e*n/(16*x**4) - 3*b*d**2*e*log(c*x**n)/(4*x**4) - b*d*e**2*n/(3*x**3) - b*d*e**2*log(c* x**n)/x**3 - b*e**3*n/(4*x**2) - b*e**3*log(c*x**n)/(2*x**2)
Time = 0.04 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {b e^{3} n}{4 \, x^{2}} - \frac {b e^{3} \log \left (c x^{n}\right )}{2 \, x^{2}} - \frac {b d e^{2} n}{3 \, x^{3}} - \frac {a e^{3}}{2 \, x^{2}} - \frac {b d e^{2} \log \left (c x^{n}\right )}{x^{3}} - \frac {3 \, b d^{2} e n}{16 \, x^{4}} - \frac {a d e^{2}}{x^{3}} - \frac {3 \, b d^{2} e \log \left (c x^{n}\right )}{4 \, x^{4}} - \frac {b d^{3} n}{25 \, x^{5}} - \frac {3 \, a d^{2} e}{4 \, x^{4}} - \frac {b d^{3} \log \left (c x^{n}\right )}{5 \, x^{5}} - \frac {a d^{3}}{5 \, x^{5}} \] Input:
integrate((e*x+d)^3*(a+b*log(c*x^n))/x^6,x, algorithm="maxima")
Output:
-1/4*b*e^3*n/x^2 - 1/2*b*e^3*log(c*x^n)/x^2 - 1/3*b*d*e^2*n/x^3 - 1/2*a*e^ 3/x^2 - b*d*e^2*log(c*x^n)/x^3 - 3/16*b*d^2*e*n/x^4 - a*d*e^2/x^3 - 3/4*b* d^2*e*log(c*x^n)/x^4 - 1/25*b*d^3*n/x^5 - 3/4*a*d^2*e/x^4 - 1/5*b*d^3*log( c*x^n)/x^5 - 1/5*a*d^3/x^5
Time = 0.11 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.15 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {{\left (10 \, b e^{3} n x^{3} + 20 \, b d e^{2} n x^{2} + 15 \, b d^{2} e n x + 4 \, b d^{3} n\right )} \log \left (x\right )}{20 \, x^{5}} - \frac {300 \, b e^{3} n x^{3} + 600 \, b e^{3} x^{3} \log \left (c\right ) + 400 \, b d e^{2} n x^{2} + 600 \, a e^{3} x^{3} + 1200 \, b d e^{2} x^{2} \log \left (c\right ) + 225 \, b d^{2} e n x + 1200 \, a d e^{2} x^{2} + 900 \, b d^{2} e x \log \left (c\right ) + 48 \, b d^{3} n + 900 \, a d^{2} e x + 240 \, b d^{3} \log \left (c\right ) + 240 \, a d^{3}}{1200 \, x^{5}} \] Input:
integrate((e*x+d)^3*(a+b*log(c*x^n))/x^6,x, algorithm="giac")
Output:
-1/20*(10*b*e^3*n*x^3 + 20*b*d*e^2*n*x^2 + 15*b*d^2*e*n*x + 4*b*d^3*n)*log (x)/x^5 - 1/1200*(300*b*e^3*n*x^3 + 600*b*e^3*x^3*log(c) + 400*b*d*e^2*n*x ^2 + 600*a*e^3*x^3 + 1200*b*d*e^2*x^2*log(c) + 225*b*d^2*e*n*x + 1200*a*d* e^2*x^2 + 900*b*d^2*e*x*log(c) + 48*b*d^3*n + 900*a*d^2*e*x + 240*b*d^3*lo g(c) + 240*a*d^3)/x^5
Time = 27.35 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.85 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=-\frac {x^3\,\left (10\,a\,e^3+5\,b\,e^3\,n\right )+x\,\left (15\,a\,d^2\,e+\frac {15\,b\,d^2\,e\,n}{4}\right )+4\,a\,d^3+x^2\,\left (20\,a\,d\,e^2+\frac {20\,b\,d\,e^2\,n}{3}\right )+\frac {4\,b\,d^3\,n}{5}}{20\,x^5}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d^3}{5}+\frac {3\,b\,d^2\,e\,x}{4}+b\,d\,e^2\,x^2+\frac {b\,e^3\,x^3}{2}\right )}{x^5} \] Input:
int(((a + b*log(c*x^n))*(d + e*x)^3)/x^6,x)
Output:
- (x^3*(10*a*e^3 + 5*b*e^3*n) + x*(15*a*d^2*e + (15*b*d^2*e*n)/4) + 4*a*d^ 3 + x^2*(20*a*d*e^2 + (20*b*d*e^2*n)/3) + (4*b*d^3*n)/5)/(20*x^5) - (log(c *x^n)*((b*d^3)/5 + (b*e^3*x^3)/2 + (3*b*d^2*e*x)/4 + b*d*e^2*x^2))/x^5
Time = 0.16 (sec) , antiderivative size = 133, normalized size of antiderivative = 0.94 \[ \int \frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{x^6} \, dx=\frac {-240 \,\mathrm {log}\left (x^{n} c \right ) b \,d^{3}-900 \,\mathrm {log}\left (x^{n} c \right ) b \,d^{2} e x -1200 \,\mathrm {log}\left (x^{n} c \right ) b d \,e^{2} x^{2}-600 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{3} x^{3}-240 a \,d^{3}-900 a \,d^{2} e x -1200 a d \,e^{2} x^{2}-600 a \,e^{3} x^{3}-48 b \,d^{3} n -225 b \,d^{2} e n x -400 b d \,e^{2} n \,x^{2}-300 b \,e^{3} n \,x^{3}}{1200 x^{5}} \] Input:
int((e*x+d)^3*(a+b*log(c*x^n))/x^6,x)
Output:
( - 240*log(x**n*c)*b*d**3 - 900*log(x**n*c)*b*d**2*e*x - 1200*log(x**n*c) *b*d*e**2*x**2 - 600*log(x**n*c)*b*e**3*x**3 - 240*a*d**3 - 900*a*d**2*e*x - 1200*a*d*e**2*x**2 - 600*a*e**3*x**3 - 48*b*d**3*n - 225*b*d**2*e*n*x - 400*b*d*e**2*n*x**2 - 300*b*e**3*n*x**3)/(1200*x**5)