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

Optimal. Leaf size=83 \[ -\frac{b d n \text{PolyLog}\left (2,-\frac{e x^2}{d}\right )}{4 e^2}-\frac{d \log \left (\frac{e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac{b n x^2}{4 e} \]

[Out]

-(b*n*x^2)/(4*e) + (x^2*(a + b*Log[c*x^n]))/(2*e) - (d*(a + b*Log[c*x^n])*Log[1 + (e*x^2)/d])/(2*e^2) - (b*d*n
*PolyLog[2, -((e*x^2)/d)])/(4*e^2)

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Rubi [A]  time = 0.142518, antiderivative size = 83, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {266, 43, 2351, 2304, 2337, 2391} \[ -\frac{b d n \text{PolyLog}\left (2,-\frac{e x^2}{d}\right )}{4 e^2}-\frac{d \log \left (\frac{e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac{b n x^2}{4 e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*n*x^2)/(4*e) + (x^2*(a + b*Log[c*x^n]))/(2*e) - (d*(a + b*Log[c*x^n])*Log[1 + (e*x^2)/d])/(2*e^2) - (b*d*n
*PolyLog[2, -((e*x^2)/d)])/(4*e^2)

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])

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2337

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[(f^m*Log[1 + (e*x^r)/d]*(a + b*Log[c*x^n])^p)/(e*r), x] - Dist[(b*f^m*n*p)/(e*r), Int[(Log[1 + (e*x^r)/d]*(
a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] &
& (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \frac{x^3 \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx &=\int \left (\frac{x \left (a+b \log \left (c x^n\right )\right )}{e}-\frac{d x \left (a+b \log \left (c x^n\right )\right )}{e \left (d+e x^2\right )}\right ) \, dx\\ &=\frac{\int x \left (a+b \log \left (c x^n\right )\right ) \, dx}{e}-\frac{d \int \frac{x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx}{e}\\ &=-\frac{b n x^2}{4 e}+\frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac{d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{e x^2}{d}\right )}{2 e^2}+\frac{(b d n) \int \frac{\log \left (1+\frac{e x^2}{d}\right )}{x} \, dx}{2 e^2}\\ &=-\frac{b n x^2}{4 e}+\frac{x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac{d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac{e x^2}{d}\right )}{2 e^2}-\frac{b d n \text{Li}_2\left (-\frac{e x^2}{d}\right )}{4 e^2}\\ \end{align*}

Mathematica [A]  time = 0.0706448, size = 135, normalized size = 1.63 \[ -\frac{2 b d n \text{PolyLog}\left (2,\frac{\sqrt{e} x}{\sqrt{-d}}\right )+2 b d n \text{PolyLog}\left (2,\frac{d \sqrt{e} x}{(-d)^{3/2}}\right )+2 d \log \left (\frac{\sqrt{e} x}{\sqrt{-d}}+1\right ) \left (a+b \log \left (c x^n\right )\right )+2 d \log \left (\frac{d \sqrt{e} x}{(-d)^{3/2}}+1\right ) \left (a+b \log \left (c x^n\right )\right )-2 e x^2 \left (a+b \log \left (c x^n\right )\right )+b e n x^2}{4 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*e*n*x^2 - 2*e*x^2*(a + b*Log[c*x^n]) + 2*d*(a + b*Log[c*x^n])*Log[1 + (Sqrt[e]*x)/Sqrt[-d]] + 2*d*(a + b*L
og[c*x^n])*Log[1 + (d*Sqrt[e]*x)/(-d)^(3/2)] + 2*b*d*n*PolyLog[2, (Sqrt[e]*x)/Sqrt[-d]] + 2*b*d*n*PolyLog[2, (
d*Sqrt[e]*x)/(-d)^(3/2)])/(4*e^2)

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Maple [C]  time = 0.174, size = 460, normalized size = 5.5 \begin{align*} \text{result too large to display} \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),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{2} \, a{\left (\frac{x^{2}}{e} - \frac{d \log \left (e x^{2} + d\right )}{e^{2}}\right )} + b \int \frac{x^{3} \log \left (c\right ) + x^{3} \log \left (x^{n}\right )}{e x^{2} + d}\,{d x} \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),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b x^{3} \log \left (c x^{n}\right ) + a x^{3}}{e x^{2} + d}, x\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),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 36.34, size = 180, normalized size = 2.17 \begin{align*} - \frac{a d \left (\begin{cases} \frac{x^{2}}{d} & \text{for}\: e = 0 \\\frac{\log{\left (d + e x^{2} \right )}}{e} & \text{otherwise} \end{cases}\right )}{2 e} + \frac{a x^{2}}{2 e} + \frac{b d n \left (\begin{cases} \frac{x^{2}}{2 d} & \text{for}\: e = 0 \\\frac{\begin{cases} \log{\left (d \right )} \log{\left (x \right )} - \frac{\operatorname{Li}_{2}\left (\frac{e x^{2} e^{i \pi }}{d}\right )}{2} & \text{for}\: \left |{x}\right | < 1 \\- \log{\left (d \right )} \log{\left (\frac{1}{x} \right )} - \frac{\operatorname{Li}_{2}\left (\frac{e x^{2} e^{i \pi }}{d}\right )}{2} & \text{for}\: \frac{1}{\left |{x}\right |} < 1 \\-{G_{2, 2}^{2, 0}\left (\begin{matrix} & 1, 1 \\0, 0 & \end{matrix} \middle |{x} \right )} \log{\left (d \right )} +{G_{2, 2}^{0, 2}\left (\begin{matrix} 1, 1 & \\ & 0, 0 \end{matrix} \middle |{x} \right )} \log{\left (d \right )} - \frac{\operatorname{Li}_{2}\left (\frac{e x^{2} e^{i \pi }}{d}\right )}{2} & \text{otherwise} \end{cases}}{e} & \text{otherwise} \end{cases}\right )}{2 e} - \frac{b d \left (\begin{cases} \frac{x^{2}}{d} & \text{for}\: e = 0 \\\frac{\log{\left (d + e x^{2} \right )}}{e} & \text{otherwise} \end{cases}\right ) \log{\left (c x^{n} \right )}}{2 e} - \frac{b n x^{2}}{4 e} + \frac{b x^{2} \log{\left (c x^{n} \right )}}{2 e} \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),x)

[Out]

-a*d*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))/(2*e) + a*x**2/(2*e) + b*d*n*Piecewise((x**2/(2*
d), Eq(e, 0)), (Piecewise((log(d)*log(x) - polylog(2, e*x**2*exp_polar(I*pi)/d)/2, Abs(x) < 1), (-log(d)*log(1
/x) - polylog(2, e*x**2*exp_polar(I*pi)/d)/2, 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0), ()), x)*log(d) +
meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylog(2, e*x**2*exp_polar(I*pi)/d)/2, True))/e, True))/(2*e
) - b*d*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))*log(c*x**n)/(2*e) - b*n*x**2/(4*e) + b*x**2*l
og(c*x**n)/(2*e)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{e x^{2} + d}\,{d x} \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),x, algorithm="giac")

[Out]

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