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

Optimal. Leaf size=68 \[ \frac{x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}-\frac{b n}{8 e^2 \left (d+e x^2\right )}-\frac{b n \log \left (d+e x^2\right )}{8 d e^2} \]

[Out]

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

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Rubi [A]  time = 0.079546, antiderivative size = 68, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2335, 266, 43} \[ \frac{x^4 \left (a+b \log \left (c x^n\right )\right )}{4 d \left (d+e x^2\right )^2}-\frac{b n}{8 e^2 \left (d+e x^2\right )}-\frac{b n \log \left (d+e x^2\right )}{8 d e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2335

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] - Dist[(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] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -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 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.144152, size = 129, normalized size = 1.9 \[ -\frac{2 a d^2+4 a d e x^2+2 b d \left (d+2 e x^2\right ) \log \left (c x^n\right )+b d^2 n \log \left (d+e x^2\right )+b d^2 n+b e^2 n x^4 \log \left (d+e x^2\right )+b d e n x^2+2 b d e n x^2 \log \left (d+e x^2\right )-2 b n \log (x) \left (d+e x^2\right )^2}{8 d e^2 \left (d+e x^2\right )^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(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])/(8*d*e^2*(d + e*x^2)^2)

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Maple [C]  time = 0.117, size = 369, normalized size = 5.4 \begin{align*} -{\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{-2\,i\pi \,bde{x}^{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 ) ^{3}+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} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -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 \,bde{x}^{2} \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}{\it csgn} \left ( ic \right ) -2\,i\pi \,bde{x}^{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}{\it csgn} \left ( i{x}^{n} \right ){\it csgn} \left ( ic{x}^{n} \right ){\it csgn} \left ( ic \right ) +2\,i\pi \,bde{x}^{2}{\it csgn} \left ( i{x}^{n} \right ) \left ({\it csgn} \left ( ic{x}^{n} \right ) \right ) ^{2}-4\,\ln \left ( x \right ) bden{x}^{2}+2\,\ln \left ( e{x}^{2}+d \right ) bden{x}^{2}+4\,\ln \left ( c \right ) bde{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+4\,ade{x}^{2}+2\,\ln \left ( c \right ) b{d}^{2}+b{d}^{2}n+2\,a{d}^{2}}{8\,d{e}^{2} \left ( e{x}^{2}+d \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B]  time = 1.19274, size = 173, normalized size = 2.54 \begin{align*} -\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-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)

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Fricas [B]  time = 1.34009, size = 273, normalized size = 4.01 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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)

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

[Out]

Timed out

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Giac [B]  time = 1.33218, size = 189, normalized size = 2.78 \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 ) + b d n x^{2} e + 4 \, b d x^{2} e \log \left (c\right ) + 4 \, a d 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 x^{4} e^{4} + 2 \, d^{2} x^{2} e^{3} + d^{3} e^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(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) + b*d*n*x^2*e + 4*b*d*x
^2*e*log(c) + 4*a*d*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*x^4*e^4 + 2*d^2*x^
2*e^3 + d^3*e^2)