3.233 \(\int \frac {x (a+b \log (c x^n))}{(d+e x^2)^3} \, dx\)

Optimal. Leaf size=82 \[ -\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e}+\frac {b n \log (x)}{4 d^2 e}+\frac {b n}{8 d e \left (d+e x^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2338, 266, 44} \[ -\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e}+\frac {b n \log (x)}{4 d^2 e}+\frac {b n}{8 d e \left (d+e x^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(x*(a + b*Log[c*x^n]))/(d + e*x^2)^3,x]

[Out]

(b*n)/(8*d*e*(d + e*x^2)) + (b*n*Log[x])/(4*d^2*e) - (a + b*Log[c*x^n])/(4*e*(d + e*x^2)^2) - (b*n*Log[d + e*x
^2])/(8*d^2*e)

Rule 44

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] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2338

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*Log[c*x^n])^p)/(e*r*(q + 1)), x] - Dist[(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] || GtQ[f, 0]) && NeQ[r, n] && NeQ[q, -1]

Rubi steps

\begin {align*} \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^3} \, dx &=-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}+\frac {(b n) \int \frac {1}{x \left (d+e x^2\right )^2} \, dx}{4 e}\\ &=-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}+\frac {(b n) \operatorname {Subst}\left (\int \frac {1}{x (d+e x)^2} \, dx,x,x^2\right )}{8 e}\\ &=-\frac {a+b \log \left (c x^n\right )}{4 e \left (d+e x^2\right )^2}+\frac {(b n) \operatorname {Subst}\left (\int \left (\frac {1}{d^2 x}-\frac {e}{d (d+e x)^2}-\frac {e}{d^2 (d+e x)}\right ) \, dx,x,x^2\right )}{8 e}\\ &=\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}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 111, normalized size = 1.35 \[ \frac {-a-b \left (\log \left (c x^n\right )-n \log (x)\right )}{4 e \left (d+e x^2\right )^2}-\frac {b n \log \left (d+e x^2\right )}{8 d^2 e}+\frac {b n \log (x)}{4 d^2 e}+\frac {b n}{8 d e \left (d+e x^2\right )}-\frac {b n \log (x)}{4 e \left (d+e x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x*(a + b*Log[c*x^n]))/(d + e*x^2)^3,x]

[Out]

(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)

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fricas [A]  time = 0.67, size = 118, normalized size = 1.44 \[ \frac {b d e n x^{2} + b d^{2} n - 2 \, b d^{2} \log \relax (c) - 2 \, a d^{2} - {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2} + b d^{2} n\right )} \log \left (e x^{2} + d\right ) + 2 \, {\left (b e^{2} n x^{4} + 2 \, b d e n x^{2}\right )} \log \relax (x)}{8 \, {\left (d^{2} e^{3} x^{4} + 2 \, d^{3} e^{2} x^{2} + d^{4} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="fricas")

[Out]

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)

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giac [A]  time = 0.29, size = 136, normalized size = 1.66 \[ -\frac {b n x^{4} e^{2} \log \left (x^{2} e + d\right ) - 2 \, b n x^{4} e^{2} \log \relax (x) + 2 \, b d n x^{2} e \log \left (x^{2} e + d\right ) - 4 \, b d n x^{2} e \log \relax (x) - b d n x^{2} e + b d^{2} n \log \left (x^{2} e + d\right ) - b d^{2} n + 2 \, b d^{2} \log \relax (c) + 2 \, a d^{2}}{8 \, {\left (d^{2} x^{4} e^{3} + 2 \, d^{3} x^{2} e^{2} + d^{4} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="giac")

[Out]

-1/8*(b*n*x^4*e^2*log(x^2*e + d) - 2*b*n*x^4*e^2*log(x) + 2*b*d*n*x^2*e*log(x^2*e + d) - 4*b*d*n*x^2*e*log(x)
- b*d*n*x^2*e + b*d^2*n*log(x^2*e + d) - b*d^2*n + 2*b*d^2*log(c) + 2*a*d^2)/(d^2*x^4*e^3 + 2*d^3*x^2*e^2 + d^
4*e)

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maple [C]  time = 0.21, size = 243, normalized size = 2.96 \[ -\frac {b \ln \left (x^{n}\right )}{4 \left (e \,x^{2}+d \right )^{2} e}-\frac {-2 b \,e^{2} n \,x^{4} \ln \relax (x )+b \,e^{2} n \,x^{4} \ln \left (e \,x^{2}+d \right )-4 b d e n \,x^{2} \ln \relax (x )+2 b d e n \,x^{2} \ln \left (e \,x^{2}+d \right )-i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )+i \pi b \,d^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}+i \pi b \,d^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}-i \pi b \,d^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}-b d e n \,x^{2}-2 b \,d^{2} n \ln \relax (x )+b \,d^{2} n \ln \left (e \,x^{2}+d \right )-b \,d^{2} n +2 b \,d^{2} \ln \relax (c )+2 a \,d^{2}}{8 \left (e \,x^{2}+d \right )^{2} d^{2} e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(b*ln(c*x^n)+a)/(e*x^2+d)^3,x)

[Out]

-1/4*b/e/(e*x^2+d)^2*ln(x^n)-1/8*(-2*b*e^2*n*x^4*ln(x)+b*e^2*n*x^4*ln(e*x^2+d)+I*Pi*b*d^2*csgn(I*x^n)*csgn(I*c
*x^n)^2-I*Pi*b*d^2*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)-I*Pi*b*d^2*csgn(I*c*x^n)^3+I*Pi*b*d^2*csgn(I*c)*csgn(I*
c*x^n)^2-4*b*d*e*n*x^2*ln(x)+2*b*d*e*n*x^2*ln(e*x^2+d)-b*d*e*n*x^2-2*b*d^2*n*ln(x)+b*d^2*n*ln(e*x^2+d)+2*b*d^2
*ln(c)-b*d^2*n+2*a*d^2)/e/d^2/(e*x^2+d)^2

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maxima [A]  time = 0.52, size = 109, normalized size = 1.33 \[ \frac {1}{8} \, b n {\left (\frac {1}{d e^{2} x^{2} + d^{2} e} - \frac {\log \left (e x^{2} + d\right )}{d^{2} e} + \frac {\log \left (x^{2}\right )}{d^{2} e}\right )} - \frac {b \log \left (c x^{n}\right )}{4 \, {\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\right )}} - \frac {a}{4 \, {\left (e^{3} x^{4} + 2 \, d e^{2} x^{2} + d^{2} e\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*log(c*x^n))/(e*x^2+d)^3,x, algorithm="maxima")

[Out]

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)

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mupad [B]  time = 3.68, size = 109, normalized size = 1.33 \[ \frac {\frac {b\,n}{2}-a+\frac {b\,e\,n\,x^2}{2\,d}}{4\,d^2\,e+8\,d\,e^2\,x^2+4\,e^3\,x^4}-\frac {b\,\ln \left (c\,x^n\right )}{4\,e\,\left (d^2+2\,d\,e\,x^2+e^2\,x^4\right )}-\frac {b\,n\,\ln \left (e\,x^2+d\right )}{8\,d^2\,e}+\frac {b\,n\,\ln \relax (x)}{4\,d^2\,e} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*(a + b*log(c*x^n)))/(d + e*x^2)^3,x)

[Out]

((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)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(a+b*ln(c*x**n))/(e*x**2+d)**3,x)

[Out]

Timed out

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