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]

(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)

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Rubi [A]  time = 0.0657028, antiderivative size = 82, normalized size of antiderivative = 1., 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 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]

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 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])

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*}

Mathematica [A]  time = 0.0690505, 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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Maple [C]  time = 0.102, size = 243, normalized size = 3. \begin{align*} -{\frac{b\ln \left ({x}^{n} \right ) }{4\, \left ( e{x}^{2}+d \right ) ^{2}e}}-{\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}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-i\pi \,b{d}^{2}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) -i\pi \,b{d}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{3}+i\pi \,b{d}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -4\,\ln \left ( x \right ) bden{x}^{2}+2\,\ln \left ( e{x}^{2}+d \right ) bden{x}^{2}-bden{x}^{2}-2\,\ln \left ( x \right ) b{d}^{2}n+\ln \left ( e{x}^{2}+d \right ) b{d}^{2}n+2\,\ln \left ( c \right ) b{d}^{2}-b{d}^{2}n+2\,a{d}^{2}}{8\,e{d}^{2} \left ( e{x}^{2}+d \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.18669, size = 147, normalized size = 1.79 \begin{align*} \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 )}} \end{align*}

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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Fricas [A]  time = 1.39759, size = 259, normalized size = 3.16 \begin{align*} \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 )}} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [A]  time = 1.27594, size = 184, normalized size = 2.24 \begin{align*} -\frac{b n x^{4} e^{2} \log \left (x^{2} e + d\right ) - 2 \, b n x^{4} e^{2} \log \left (x\right ) + 2 \, b d n x^{2} e \log \left (x^{2} e + d\right ) - 4 \, b d n x^{2} e \log \left (x\right ) - 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 \left (c\right ) + 2 \, a d^{2}}{8 \,{\left (d^{2} x^{4} e^{3} + 2 \, d^{3} x^{2} e^{2} + d^{4} e\right )}} \end{align*}

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)