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

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

[Out]

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

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Rubi [A]  time = 0.176401, antiderivative size = 109, normalized size of antiderivative = 1.31, number of steps used = 6, number of rules used = 7, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.304, Rules used = {266, 44, 2351, 2304, 2301, 2337, 2391} \[ \frac{b e n \text{PolyLog}\left (2,-\frac{e x^2}{d}\right )}{4 d^2}-\frac{e \left (a+b \log \left (c x^n\right )\right )^2}{2 b d^2 n}+\frac{e \log \left (\frac{e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 d^2}-\frac{a+b \log \left (c x^n\right )}{2 d x^2}-\frac{b n}{4 d x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(b*n)/(4*d*x^2) - (a + b*Log[c*x^n])/(2*d*x^2) - (e*(a + b*Log[c*x^n])^2)/(2*b*d^2*n) + (e*(a + b*Log[c*x^n])
*Log[1 + (e*x^2)/d])/(2*d^2) + (b*e*n*PolyLog[2, -((e*x^2)/d)])/(4*d^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 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 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 2301

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*x^n])^2/(2*b*n), x] /; FreeQ[{a
, b, c, n}, x]

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

Mathematica [A]  time = 0.108763, size = 157, normalized size = 1.89 \[ \frac{2 b e n \text{PolyLog}\left (2,\frac{\sqrt{e} x}{\sqrt{-d}}\right )+2 b e n \text{PolyLog}\left (2,\frac{d \sqrt{e} x}{(-d)^{3/2}}\right )+2 e \log \left (\frac{\sqrt{e} x}{\sqrt{-d}}+1\right ) \left (a+b \log \left (c x^n\right )\right )+2 e \log \left (\frac{d \sqrt{e} x}{(-d)^{3/2}}+1\right ) \left (a+b \log \left (c x^n\right )\right )-\frac{2 d \left (a+b \log \left (c x^n\right )\right )}{x^2}-\frac{2 e \left (a+b \log \left (c x^n\right )\right )^2}{b n}-\frac{b d n}{x^2}}{4 d^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.146, size = 611, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*b*n*e/d^2*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n*e/d^2*ln(x)*ln(e*x^2+d)+1/2*b*n*e/d^2*ln(x)*ln
((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/4*I*b*Pi*csgn(I*c*x^n)^3/d/x^2-1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*e/d^2
*ln(x)-a*e/d^2*ln(x)-1/4*I*b*Pi*csgn(I*c*x^n)^3*e/d^2*ln(e*x^2+d)-b*ln(c)*e/d^2*ln(x)-1/2*b*ln(c)/d/x^2+1/4*I*
b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/d/x^2+1/4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*e/d^2*ln(e*x^2+d)+1/4*I*b*
Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e/d^2*ln(e*x^2+d)-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*e/d^2*ln(x)-1/2*a/d/x^
2-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/d/x^2-b*ln(x^n)*e/d^2*ln(x)+1/2*b*ln(x^n)*e/d^2*ln(e*x^2+d)+1/2*b*n*e
/d^2*ln(x)^2+1/2*b*ln(c)*e/d^2*ln(e*x^2+d)-1/2*b*ln(x^n)/d/x^2+1/2*I*b*Pi*csgn(I*c*x^n)^3*e/d^2*ln(x)-1/4*I*b*
Pi*csgn(I*c*x^n)^2*csgn(I*c)/d/x^2+1/2*b*n*e/d^2*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n*e/d^2*dilog((
e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*a*e/d^2*ln(e*x^2+d)-1/4*b*n/d/x^2+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csg
n(I*c)*e/d^2*ln(x)-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*e/d^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{e \log \left (e x^{2} + d\right )}{d^{2}} - \frac{2 \, e \log \left (x\right )}{d^{2}} - \frac{1}{d x^{2}}\right )} + b \int \frac{\log \left (c\right ) + \log \left (x^{n}\right )}{e x^{5} + d x^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*(e*log(e*x^2 + d)/d^2 - 2*e*log(x)/d^2 - 1/(d*x^2)) + b*integrate((log(c) + log(x^n))/(e*x^5 + d*x^3), x
)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{b \log \left (c x^{n}\right ) + a}{e x^{5} + d x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*log(c*x^n) + a)/(e*x^5 + d*x^3), x)

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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