Integrand size = 23, antiderivative size = 68 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {b n}{8 e^2 \left (d+e x^2\right )}+\frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d e^2} \] Output:
-1/8*b*n/e^2/(e*x^2+d)+1/4*x^4*(a+b*ln(c*x^n))/d/(e*x^2+d)^2-1/8*b*n*ln(e* x^2+d)/d/e^2
Time = 0.14 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.90 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {2 a d^2+b d^2 n+4 a d e x^2+b d e n x^2-2 b n \left (d+e x^2\right )^2 \log (x)+2 b d \left (d+2 e x^2\right ) \log \left (c x^n\right )+b d^2 n \log \left (d+e x^2\right )+2 b d e n x^2 \log \left (d+e x^2\right )+b e^2 n x^4 \log \left (d+e x^2\right )}{8 d e^2 \left (d+e x^2\right )^2} \] Input:
Integrate[(x^3*(a + b*Log[c*x^n]))/(d + e*x^2)^3,x]
Output:
-1/8*(2*a*d^2 + b*d^2*n + 4*a*d*e*x^2 + b*d*e*n*x^2 - 2*b*n*(d + e*x^2)^2* Log[x] + 2*b*d*(d + 2*e*x^2)*Log[c*x^n] + b*d^2*n*Log[d + e*x^2] + 2*b*d*e *n*x^2*Log[d + e*x^2] + b*e^2*n*x^4*Log[d + e*x^2])/(d*e^2*(d + e*x^2)^2)
Time = 0.29 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2773, 243, 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.
\(\displaystyle \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 2773 |
\(\displaystyle \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}-\frac {b n \int \frac {x^3}{\left (e x^2+d\right )^2}dx}{4 d}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}-\frac {b n \int \frac {x^2}{\left (e x^2+d\right )^2}dx^2}{8 d}\) |
\(\Big \downarrow \) 49 |
\(\displaystyle \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}-\frac {b n \int \left (\frac {1}{e \left (e x^2+d\right )}-\frac {d}{e \left (e x^2+d\right )^2}\right )dx^2}{8 d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}-\frac {b n \left (\frac {d}{e^2 \left (d+e x^2\right )}+\frac {\log \left (d+e x^2\right )}{e^2}\right )}{8 d}\) |
Input:
Int[(x^3*(a + b*Log[c*x^n]))/(d + e*x^2)^3,x]
Output:
(x^4*(a + b*Log[c*x^n]))/(4*d*(d + e*x^2)^2) - (b*n*(d/(e^2*(d + e*x^2)) + Log[d + e*x^2]/e^2))/(8*d)
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[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* (x_)^(r_.))^(q_), x_Symbol] :> Simp[(f*x)^(m + 1)*(d + e*x^r)^(q + 1)*((a + b*Log[c*x^n])/(d*f*(m + 1))), x] - Simp[b*(n/(d*(m + 1))) Int[(f*x)^m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && Eq Q[m + r*(q + 1) + 1, 0] && NeQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(129\) vs. \(2(62)=124\).
Time = 1.08 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.91
method | result | size |
parallelrisch | \(\frac {-\ln \left (e \,x^{2}+d \right ) x^{4} b \,e^{2} n^{2}+2 x^{4} \ln \left (c \,x^{n}\right ) b \,e^{2} n -2 \ln \left (e \,x^{2}+d \right ) x^{2} b d e \,n^{2}-x^{2} b d e \,n^{2}-\ln \left (e \,x^{2}+d \right ) b \,d^{2} n^{2}-4 x^{2} a d e n -b \,d^{2} n^{2}-2 a \,d^{2} n}{8 e^{2} d n \left (e \,x^{2}+d \right )^{2}}\) | \(130\) |
risch | \(-\frac {b \left (2 e \,x^{2}+d \right ) \ln \left (x^{n}\right )}{4 \left (e \,x^{2}+d \right )^{2} e^{2}}-\frac {-i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )+2 i \pi b d e \,x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )-i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+2 i \pi b d e \,x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-2 \ln \left (x \right ) b \,e^{2} n \,x^{4}+\ln \left (e \,x^{2}+d \right ) b \,e^{2} n \,x^{4}-2 i \pi b d e \,x^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )+i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-2 i \pi b d e \,x^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-4 \ln \left (x \right ) b d e n \,x^{2}+2 \ln \left (e \,x^{2}+d \right ) b d e n \,x^{2}+4 \ln \left (c \right ) b d e \,x^{2}+b d e n \,x^{2}-2 \ln \left (x \right ) b \,d^{2} n +\ln \left (e \,x^{2}+d \right ) b \,d^{2} n +4 a e \,x^{2} d +2 d^{2} b \ln \left (c \right )+b \,d^{2} n +2 a \,d^{2}}{8 e^{2} d \left (e \,x^{2}+d \right )^{2}}\) | \(369\) |
Input:
int(x^3*(a+b*ln(c*x^n))/(e*x^2+d)^3,x,method=_RETURNVERBOSE)
Output:
1/8*(-ln(e*x^2+d)*x^4*b*e^2*n^2+2*x^4*ln(c*x^n)*b*e^2*n-2*ln(e*x^2+d)*x^2* b*d*e*n^2-x^2*b*d*e*n^2-ln(e*x^2+d)*b*d^2*n^2-4*x^2*a*d*e*n-b*d^2*n^2-2*a* d^2*n)/e^2/d/n/(e*x^2+d)^2
Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (62) = 124\).
Time = 0.09 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.85 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {2 \, b e^{2} n x^{4} \log \left (x\right ) - b d^{2} n - 2 \, a d^{2} - {\left (b d e n + 4 \, a d e\right )} x^{2} - {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (e x^{2} + d\right ) - 2 \, {\left (2 \, b d e x^{2} + b d^{2}\right )} \log \left (c\right )}{8 \, {\left (d e^{4} x^{4} + 2 \, d^{2} e^{3} x^{2} + d^{3} e^{2}\right )}} \] Input:
integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="fricas")
Output:
1/8*(2*b*e^2*n*x^4*log(x) - b*d^2*n - 2*a*d^2 - (b*d*e*n + 4*a*d*e)*x^2 - (b*e^2*n*x^4 + 2*b*d*e*n*x^2 + b*d^2*n)*log(e*x^2 + d) - 2*(2*b*d*e*x^2 + b*d^2)*log(c))/(d*e^4*x^4 + 2*d^2*e^3*x^2 + d^3*e^2)
Timed out. \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate(x**3*(a+b*ln(c*x**n))/(e*x**2+d)**3,x)
Output:
Timed out
Leaf count of result is larger than twice the leaf count of optimal. 128 vs. \(2 (62) = 124\).
Time = 0.04 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.88 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {1}{8} \, b n {\left (\frac {1}{e^{3} x^{2} + d e^{2}} + \frac {\log \left (e x^{2} + d\right )}{d e^{2}} - \frac {\log \left (x^{2}\right )}{d e^{2}}\right )} - \frac {{\left (2 \, e x^{2} + d\right )} b \log \left (c x^{n}\right )}{4 \, {\left (e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}\right )}} - \frac {{\left (2 \, e x^{2} + d\right )} a}{4 \, {\left (e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}\right )}} \] Input:
integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="maxima")
Output:
-1/8*b*n*(1/(e^3*x^2 + d*e^2) + log(e*x^2 + d)/(d*e^2) - log(x^2)/(d*e^2)) - 1/4*(2*e*x^2 + d)*b*log(c*x^n)/(e^4*x^4 + 2*d*e^3*x^2 + d^2*e^2) - 1/4* (2*e*x^2 + d)*a/(e^4*x^4 + 2*d*e^3*x^2 + d^2*e^2)
Leaf count of result is larger than twice the leaf count of optimal. 140 vs. \(2 (62) = 124\).
Time = 0.12 (sec) , antiderivative size = 140, normalized size of antiderivative = 2.06 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {{\left (2 \, b e n x^{2} + b d n\right )} \log \left (x\right )}{4 \, {\left (e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}\right )}} - \frac {b e n x^{2} + 4 \, b e x^{2} \log \left (c\right ) + 4 \, a e x^{2} + b d n + 2 \, b d \log \left (c\right ) + 2 \, a d}{8 \, {\left (e^{4} x^{4} + 2 \, d e^{3} x^{2} + d^{2} e^{2}\right )}} - \frac {b n \log \left (e x^{2} + d\right )}{8 \, d e^{2}} + \frac {b n \log \left (x\right )}{4 \, d e^{2}} \] Input:
integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="giac")
Output:
-1/4*(2*b*e*n*x^2 + b*d*n)*log(x)/(e^4*x^4 + 2*d*e^3*x^2 + d^2*e^2) - 1/8* (b*e*n*x^2 + 4*b*e*x^2*log(c) + 4*a*e*x^2 + b*d*n + 2*b*d*log(c) + 2*a*d)/ (e^4*x^4 + 2*d*e^3*x^2 + d^2*e^2) - 1/8*b*n*log(e*x^2 + d)/(d*e^2) + 1/4*b *n*log(x)/(d*e^2)
Time = 26.65 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.90 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b\,n\,\ln \left (x\right )}{4\,d\,e^2}-\frac {\ln \left (c\,x^n\right )\,\left (\frac {b\,x^2}{2\,e}+\frac {b\,d}{4\,e^2}\right )}{d^2+2\,d\,e\,x^2+e^2\,x^4}-\frac {b\,n\,\ln \left (e\,x^2+d\right )}{8\,d\,e^2}-\frac {\left (2\,a\,e+\frac {b\,e\,n}{2}\right )\,x^2+a\,d+\frac {b\,d\,n}{2}}{4\,d^2\,e^2+8\,d\,e^3\,x^2+4\,e^4\,x^4} \] Input:
int((x^3*(a + b*log(c*x^n)))/(d + e*x^2)^3,x)
Output:
(b*n*log(x))/(4*d*e^2) - (log(c*x^n)*((b*x^2)/(2*e) + (b*d)/(4*e^2)))/(d^2 + e^2*x^4 + 2*d*e*x^2) - (b*n*log(d + e*x^2))/(8*d*e^2) - (a*d + x^2*(2*a *e + (b*e*n)/2) + (b*d*n)/2)/(4*d^2*e^2 + 4*e^4*x^4 + 8*d*e^3*x^2)
Time = 0.18 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.75 \[ \int \frac {x^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {-2 \,\mathrm {log}\left (e \,x^{2}+d \right ) b \,d^{2} n -4 \,\mathrm {log}\left (e \,x^{2}+d \right ) b d e n \,x^{2}-2 \,\mathrm {log}\left (e \,x^{2}+d \right ) b \,e^{2} n \,x^{4}+4 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{2} x^{4}+4 a \,e^{2} x^{4}-b \,d^{2} n +b \,e^{2} n \,x^{4}}{16 d \,e^{2} \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:
int(x^3*(a+b*log(c*x^n))/(e*x^2+d)^3,x)
Output:
( - 2*log(d + e*x**2)*b*d**2*n - 4*log(d + e*x**2)*b*d*e*n*x**2 - 2*log(d + e*x**2)*b*e**2*n*x**4 + 4*log(x**n*c)*b*e**2*x**4 + 4*a*e**2*x**4 - b*d* *2*n + b*e**2*n*x**4)/(16*d*e**2*(d**2 + 2*d*e*x**2 + e**2*x**4))