Integrand size = 21, antiderivative size = 75 \[ \int \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {b d^2 n}{9 x^3}-\frac {b d e n}{2 x^2}-\frac {b e^2 n}{x}+\frac {b e^3 n \log (x)}{3 d}-\frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{3 d x^3} \] Output:
-1/9*b*d^2*n/x^3-1/2*b*d*e*n/x^2-b*e^2*n/x+1/3*b*e^3*n*ln(x)/d-1/3*(e*x+d) ^3*(a+b*ln(c*x^n))/d/x^3
Time = 0.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01 \[ \int \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {6 a \left (d^2+3 d e x+3 e^2 x^2\right )+b n \left (2 d^2+9 d e x+18 e^2 x^2\right )+6 b \left (d^2+3 d e x+3 e^2 x^2\right ) \log \left (c x^n\right )}{18 x^3} \] Input:
Integrate[((d + e*x)^2*(a + b*Log[c*x^n]))/x^4,x]
Output:
-1/18*(6*a*(d^2 + 3*d*e*x + 3*e^2*x^2) + b*n*(2*d^2 + 9*d*e*x + 18*e^2*x^2 ) + 6*b*(d^2 + 3*d*e*x + 3*e^2*x^2)*Log[c*x^n])/x^3
Time = 0.29 (sec) , antiderivative size = 74, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2772, 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.
\(\displaystyle \int \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx\) |
\(\Big \downarrow \) 2772 |
\(\displaystyle -b n \int -\frac {(d+e x)^3}{3 d x^4}dx-\frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b n \int \frac {(d+e x)^3}{x^4}dx}{3 d}-\frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {b n \int \left (\frac {d^3}{x^4}+\frac {3 e d^2}{x^3}+\frac {3 e^2 d}{x^2}+\frac {e^3}{x}\right )dx}{3 d}-\frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b n \left (-\frac {d^3}{3 x^3}-\frac {3 d^2 e}{2 x^2}-\frac {3 d e^2}{x}+e^3 \log (x)\right )}{3 d}-\frac {(d+e x)^3 \left (a+b \log \left (c x^n\right )\right )}{3 d x^3}\) |
Input:
Int[((d + e*x)^2*(a + b*Log[c*x^n]))/x^4,x]
Output:
(b*n*(-1/3*d^3/x^3 - (3*d^2*e)/(2*x^2) - (3*d*e^2)/x + e^3*Log[x]))/(3*d) - ((d + e*x)^3*(a + b*Log[c*x^n]))/(3*d*x^3)
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_.))*(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 = 0.38 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.21
method | result | size |
parallelrisch | \(-\frac {18 b \ln \left (c \,x^{n}\right ) e^{2} x^{2}+18 b \,e^{2} n \,x^{2}+18 a \,e^{2} x^{2}+18 b \ln \left (c \,x^{n}\right ) e x d +9 b d e n x +18 a e x d +6 b \ln \left (c \,x^{n}\right ) d^{2}+2 b \,d^{2} n +6 a \,d^{2}}{18 x^{3}}\) | \(91\) |
risch | \(-\frac {b \left (3 e^{2} x^{2}+3 e x d +d^{2}\right ) \ln \left (x^{n}\right )}{3 x^{3}}-\frac {-3 i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-9 i \pi b \,e^{2} x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}-9 i \pi b d e x \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+9 i \pi b d e x \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+18 \ln \left (c \right ) b \,e^{2} x^{2}+18 b \,e^{2} n \,x^{2}+18 a \,e^{2} x^{2}+3 i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+9 i \pi b \,e^{2} x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}+9 i \pi b d e x \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-9 i \pi b \,e^{2} x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+18 \ln \left (c \right ) b d e x +9 b d e n x +18 a e x d -9 i \pi b d e x \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+3 i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-3 i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+9 i \pi b \,e^{2} x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+6 d^{2} b \ln \left (c \right )+2 b \,d^{2} n +6 a \,d^{2}}{18 x^{3}}\) | \(401\) |
Input:
int((e*x+d)^2*(a+b*ln(c*x^n))/x^4,x,method=_RETURNVERBOSE)
Output:
-1/18/x^3*(18*b*ln(c*x^n)*e^2*x^2+18*b*e^2*n*x^2+18*a*e^2*x^2+18*b*ln(c*x^ n)*e*x*d+9*b*d*e*n*x+18*a*e*x*d+6*b*ln(c*x^n)*d^2+2*b*d^2*n+6*a*d^2)
Time = 0.07 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.37 \[ \int \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {2 \, b d^{2} n + 6 \, a d^{2} + 18 \, {\left (b e^{2} n + a e^{2}\right )} x^{2} + 9 \, {\left (b d e n + 2 \, a d e\right )} x + 6 \, {\left (3 \, b e^{2} x^{2} + 3 \, b d e x + b d^{2}\right )} \log \left (c\right ) + 6 \, {\left (3 \, b e^{2} n x^{2} + 3 \, b d e n x + b d^{2} n\right )} \log \left (x\right )}{18 \, x^{3}} \] Input:
integrate((e*x+d)^2*(a+b*log(c*x^n))/x^4,x, algorithm="fricas")
Output:
-1/18*(2*b*d^2*n + 6*a*d^2 + 18*(b*e^2*n + a*e^2)*x^2 + 9*(b*d*e*n + 2*a*d *e)*x + 6*(3*b*e^2*x^2 + 3*b*d*e*x + b*d^2)*log(c) + 6*(3*b*e^2*n*x^2 + 3* b*d*e*n*x + b*d^2*n)*log(x))/x^3
Time = 0.37 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.39 \[ \int \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=- \frac {a d^{2}}{3 x^{3}} - \frac {a d e}{x^{2}} - \frac {a e^{2}}{x} - \frac {b d^{2} n}{9 x^{3}} - \frac {b d^{2} \log {\left (c x^{n} \right )}}{3 x^{3}} - \frac {b d e n}{2 x^{2}} - \frac {b d e \log {\left (c x^{n} \right )}}{x^{2}} - \frac {b e^{2} n}{x} - \frac {b e^{2} \log {\left (c x^{n} \right )}}{x} \] Input:
integrate((e*x+d)**2*(a+b*ln(c*x**n))/x**4,x)
Output:
-a*d**2/(3*x**3) - a*d*e/x**2 - a*e**2/x - b*d**2*n/(9*x**3) - b*d**2*log( c*x**n)/(3*x**3) - b*d*e*n/(2*x**2) - b*d*e*log(c*x**n)/x**2 - b*e**2*n/x - b*e**2*log(c*x**n)/x
Time = 0.04 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.33 \[ \int \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {b e^{2} n}{x} - \frac {b e^{2} \log \left (c x^{n}\right )}{x} - \frac {b d e n}{2 \, x^{2}} - \frac {a e^{2}}{x} - \frac {b d e \log \left (c x^{n}\right )}{x^{2}} - \frac {b d^{2} n}{9 \, x^{3}} - \frac {a d e}{x^{2}} - \frac {b d^{2} \log \left (c x^{n}\right )}{3 \, x^{3}} - \frac {a d^{2}}{3 \, x^{3}} \] Input:
integrate((e*x+d)^2*(a+b*log(c*x^n))/x^4,x, algorithm="maxima")
Output:
-b*e^2*n/x - b*e^2*log(c*x^n)/x - 1/2*b*d*e*n/x^2 - a*e^2/x - b*d*e*log(c* x^n)/x^2 - 1/9*b*d^2*n/x^3 - a*d*e/x^2 - 1/3*b*d^2*log(c*x^n)/x^3 - 1/3*a* d^2/x^3
Time = 0.11 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.47 \[ \int \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {{\left (3 \, b e^{2} n x^{2} + 3 \, b d e n x + b d^{2} n\right )} \log \left (x\right )}{3 \, x^{3}} - \frac {18 \, b e^{2} n x^{2} + 18 \, b e^{2} x^{2} \log \left (c\right ) + 9 \, b d e n x + 18 \, a e^{2} x^{2} + 18 \, b d e x \log \left (c\right ) + 2 \, b d^{2} n + 18 \, a d e x + 6 \, b d^{2} \log \left (c\right ) + 6 \, a d^{2}}{18 \, x^{3}} \] Input:
integrate((e*x+d)^2*(a+b*log(c*x^n))/x^4,x, algorithm="giac")
Output:
-1/3*(3*b*e^2*n*x^2 + 3*b*d*e*n*x + b*d^2*n)*log(x)/x^3 - 1/18*(18*b*e^2*n *x^2 + 18*b*e^2*x^2*log(c) + 9*b*d*e*n*x + 18*a*e^2*x^2 + 18*b*d*e*x*log(c ) + 2*b*d^2*n + 18*a*d*e*x + 6*b*d^2*log(c) + 6*a*d^2)/x^3
Time = 27.29 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.09 \[ \int \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=-\frac {x^2\,\left (3\,a\,e^2+3\,b\,e^2\,n\right )+a\,d^2+x\,\left (3\,a\,d\,e+\frac {3\,b\,d\,e\,n}{2}\right )+\frac {b\,d^2\,n}{3}}{3\,x^3}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,d^2}{3}+b\,d\,e\,x+b\,e^2\,x^2\right )}{x^3} \] Input:
int(((a + b*log(c*x^n))*(d + e*x)^2)/x^4,x)
Output:
- (x^2*(3*a*e^2 + 3*b*e^2*n) + a*d^2 + x*(3*a*d*e + (3*b*d*e*n)/2) + (b*d^ 2*n)/3)/(3*x^3) - (log(c*x^n)*((b*d^2)/3 + b*e^2*x^2 + b*d*e*x))/x^3
Time = 0.17 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.20 \[ \int \frac {(d+e x)^2 \left (a+b \log \left (c x^n\right )\right )}{x^4} \, dx=\frac {-6 \,\mathrm {log}\left (x^{n} c \right ) b \,d^{2}-18 \,\mathrm {log}\left (x^{n} c \right ) b d e x -18 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{2} x^{2}-6 a \,d^{2}-18 a d e x -18 a \,e^{2} x^{2}-2 b \,d^{2} n -9 b d e n x -18 b \,e^{2} n \,x^{2}}{18 x^{3}} \] Input:
int((e*x+d)^2*(a+b*log(c*x^n))/x^4,x)
Output:
( - 6*log(x**n*c)*b*d**2 - 18*log(x**n*c)*b*d*e*x - 18*log(x**n*c)*b*e**2* x**2 - 6*a*d**2 - 18*a*d*e*x - 18*a*e**2*x**2 - 2*b*d**2*n - 9*b*d*e*n*x - 18*b*e**2*n*x**2)/(18*x**3)