Integrand size = 21, antiderivative size = 82 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b n}{8 d e \left (d+e x^2\right )}+\frac {b n \log (x)}{4 d^2 e}-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e} \] Output:
1/8*b*n/d/e/(e*x^2+d)+1/4*b*n*ln(x)/d^2/e-1/4*(a+b*ln(c*x^n))/e/(e*x^2+d)^ 2-1/8*b*n*ln(e*x^2+d)/d^2/e
Time = 0.10 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.35 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b n}{8 d e \left (d+e x^2\right )}+\frac {b n \log (x)}{4 d^2 e}-\frac {b n \log (x)}{4 e \left (d+e x^2\right )^2}+\frac {-a-b \left (-n \log (x)+\log \left (c x^n\right )\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e} \] Input:
Integrate[(x*(a + b*Log[c*x^n]))/(d + e*x^2)^3,x]
Output:
(b*n)/(8*d*e*(d + e*x^2)) + (b*n*Log[x])/(4*d^2*e) - (b*n*Log[x])/(4*e*(d + e*x^2)^2) + (-a - b*(-(n*Log[x]) + Log[c*x^n]))/(4*e*(d + e*x^2)^2) - (b *n*Log[d + e*x^2])/(8*d^2*e)
Time = 0.28 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.87, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2776, 243, 54, 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.
\(\displaystyle \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx\) |
\(\Big \downarrow \) 2776 |
\(\displaystyle \frac {b n \int \frac {1}{x \left (e x^2+d\right )^2}dx}{4 e}-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {b n \int \frac {1}{x^2 \left (e x^2+d\right )^2}dx^2}{8 e}-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 54 |
\(\displaystyle \frac {b n \int \left (-\frac {e}{d^2 \left (e x^2+d\right )}-\frac {e}{d \left (e x^2+d\right )^2}+\frac {1}{d^2 x^2}\right )dx^2}{8 e}-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {b n \left (-\frac {\log \left (d+e x^2\right )}{d^2}+\frac {\log \left (x^2\right )}{d^2}+\frac {1}{d \left (d+e x^2\right )}\right )}{8 e}-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}\) |
Input:
Int[(x*(a + b*Log[c*x^n]))/(d + e*x^2)^3,x]
Output:
-1/4*(a + b*Log[c*x^n])/(e*(d + e*x^2)^2) + (b*n*(1/(d*(d + e*x^2)) + Log[ x^2]/d^2 - Log[d + e*x^2]/d^2))/(8*e)
Int[((a_) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[E xpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && ILtQ[m, 0] && IntegerQ[n] && !(IGtQ[n, 0] && LtQ[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_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_))^(q_.), x_Symbol] :> Simp[f^m*(d + e*x^r)^(q + 1)*((a + b*L og[c*x^n])^p/(e*r*(q + 1))), x] - Simp[b*f^m*n*(p/(e*r*(q + 1))) Int[(d + e*x^r)^(q + 1)*((a + b*Log[c*x^n])^(p - 1)/x), x], x] /; FreeQ[{a, b, c, d , e, f, m, n, q, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] && (IntegerQ[m] || G tQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]
Time = 1.05 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.74
method | result | size |
parallelrisch | \(\frac {-\ln \left (e \,x^{2}+d \right ) x^{4} b \,e^{3} n^{2}+2 x^{4} \ln \left (c \,x^{n}\right ) b \,e^{3} n -2 \ln \left (e \,x^{2}+d \right ) x^{2} b d \,e^{2} n^{2}+4 x^{2} \ln \left (c \,x^{n}\right ) b d \,e^{2} n +x^{2} b d \,e^{2} n^{2}-\ln \left (e \,x^{2}+d \right ) b \,d^{2} e \,n^{2}+b \,d^{2} e \,n^{2}-2 a \,d^{2} e n}{8 e^{2} d^{2} n \left (e \,x^{2}+d \right )^{2}}\) | \(143\) |
risch | \(-\frac {b \ln \left (x^{n}\right )}{4 e \left (e \,x^{2}+d \right )^{2}}-\frac {-2 \ln \left (x \right ) b \,e^{2} n \,x^{4}+\ln \left (e \,x^{2}+d \right ) b \,e^{2} n \,x^{4}+i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,d^{2} \operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )-i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{3}+i \pi b \,d^{2} \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )-4 \ln \left (x \right ) b d e n \,x^{2}+2 \ln \left (e \,x^{2}+d \right ) b d e n \,x^{2}-b d e n \,x^{2}-2 \ln \left (x \right ) b \,d^{2} n +\ln \left (e \,x^{2}+d \right ) b \,d^{2} n +2 d^{2} b \ln \left (c \right )-b \,d^{2} n +2 a \,d^{2}}{8 d^{2} e \left (e \,x^{2}+d \right )^{2}}\) | \(243\) |
Input:
int(x*(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^3*n^2+2*x^4*ln(c*x^n)*b*e^3*n-2*ln(e*x^2+d)*x^2* b*d*e^2*n^2+4*x^2*ln(c*x^n)*b*d*e^2*n+x^2*b*d*e^2*n^2-ln(e*x^2+d)*b*d^2*e* n^2+b*d^2*e*n^2-2*a*d^2*e*n)/e^2/d^2/n/(e*x^2+d)^2
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 1.44 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {b d e n x^{2} + b d^{2} n - 2 \, b d^{2} \log \left (c\right ) - 2 \, a d^{2} - {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (e x^{2} + d\right ) + 2 \, {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2}\right )} \log \left (x\right )}{8 \, {\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}} \] Input:
integrate(x*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="fricas")
Output:
1/8*(b*d*e*n*x^2 + b*d^2*n - 2*b*d^2*log(c) - 2*a*d^2 - (b*e^2*n*x^4 + 2*b *d*e*n*x^2 + b*d^2*n)*log(e*x^2 + d) + 2*(b*e^2*n*x^4 + 2*b*d*e*n*x^2)*log (x))/(d^2*e^3*x^4 + 2*d^3*e^2*x^2 + d^4*e)
Timed out. \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\text {Timed out} \] Input:
integrate(x*(a+b*ln(c*x**n))/(e*x**2+d)**3,x)
Output:
Timed out
Time = 0.04 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.33 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {1}{8} \, b n {\left (\frac {1}{d e^{2} x^{2} + d^{2} e} - \frac {\log \left (e x^{2} + d\right )}{d^{2} e} + \frac {\log \left (x^{2}\right )}{d^{2} e}\right )} - \frac {b \log \left (c x^{n}\right )}{4 \, {\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\right )}} - \frac {a}{4 \, {\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\right )}} \] Input:
integrate(x*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="maxima")
Output:
1/8*b*n*(1/(d*e^2*x^2 + d^2*e) - log(e*x^2 + d)/(d^2*e) + log(x^2)/(d^2*e) ) - 1/4*b*log(c*x^n)/(e^3*x^4 + 2*d*e^2*x^2 + d^2*e) - 1/4*a/(e^3*x^4 + 2* d*e^2*x^2 + d^2*e)
Time = 0.12 (sec) , antiderivative size = 116, normalized size of antiderivative = 1.41 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=-\frac {b d n \log \left (x\right )}{4 \, {\left (d e^{3} x^{4} + 2 \, d^{2} e^{2} x^{2} + d^{3} e\right )}} + \frac {b e n x^{2} + b d n - 2 \, b d \log \left (c\right ) - 2 \, a d}{8 \, {\left (d e^{3} x^{4} + 2 \, d^{2} e^{2} x^{2} + d^{3} e\right )}} - \frac {b n \log \left (e x^{2} + d\right )}{8 \, d^{2} e} + \frac {b n \log \left (x\right )}{4 \, d^{2} e} \] Input:
integrate(x*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="giac")
Output:
-1/4*b*d*n*log(x)/(d*e^3*x^4 + 2*d^2*e^2*x^2 + d^3*e) + 1/8*(b*e*n*x^2 + b *d*n - 2*b*d*log(c) - 2*a*d)/(d*e^3*x^4 + 2*d^2*e^2*x^2 + d^3*e) - 1/8*b*n *log(e*x^2 + d)/(d^2*e) + 1/4*b*n*log(x)/(d^2*e)
Time = 26.36 (sec) , antiderivative size = 109, normalized size of antiderivative = 1.33 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {\frac {b\,n}{2}-a+\frac {b\,e\,n\,x^2}{2\,d}}{4\,d^2\,e+8\,d\,e^2\,x^2+4\,e^3\,x^4}-\frac {b\,\ln \left (c\,x^n\right )}{4\,e\,\left (d^2+2\,d\,e\,x^2+e^2\,x^4\right )}-\frac {b\,n\,\ln \left (e\,x^2+d\right )}{8\,d^2\,e}+\frac {b\,n\,\ln \left (x\right )}{4\,d^2\,e} \] Input:
int((x*(a + b*log(c*x^n)))/(d + e*x^2)^3,x)
Output:
((b*n)/2 - a + (b*e*n*x^2)/(2*d))/(4*d^2*e + 4*e^3*x^4 + 8*d*e^2*x^2) - (b *log(c*x^n))/(4*e*(d^2 + e^2*x^4 + 2*d*e*x^2)) - (b*n*log(d + e*x^2))/(8*d ^2*e) + (b*n*log(x))/(4*d^2*e)
Time = 0.16 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.59 \[ \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx=\frac {-2 \,\mathrm {log}\left (e \,x^{2}+d \right ) b \,d^{2} n -4 \,\mathrm {log}\left (e \,x^{2}+d \right ) b d e n \,x^{2}-2 \,\mathrm {log}\left (e \,x^{2}+d \right ) b \,e^{2} n \,x^{4}+8 \,\mathrm {log}\left (x^{n} c \right ) b d e \,x^{2}+4 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{2} x^{4}-4 a \,d^{2}+b \,d^{2} n -b \,e^{2} n \,x^{4}}{16 d^{2} e \left (e^{2} x^{4}+2 d e \,x^{2}+d^{2}\right )} \] Input:
int(x*(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 + 8*log(x**n*c)*b*d*e*x**2 + 4*log(x**n*c)*b*e**2* x**4 - 4*a*d**2 + b*d**2*n - b*e**2*n*x**4)/(16*d**2*e*(d**2 + 2*d*e*x**2 + e**2*x**4))