3.8 \(\int \frac{(a+b \log (c x^n)) \log (1+e x)}{x^3} \, dx\)

Optimal. Leaf size=163 \[ \frac{1}{2} b e^2 n \text{PolyLog}(2,-e x)-\frac{1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac{1}{2} e^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )-\frac{e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac{\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac{1}{4} b e^2 n \log ^2(x)-\frac{1}{4} b e^2 n \log (x)+\frac{1}{4} b e^2 n \log (e x+1)-\frac{b n \log (e x+1)}{4 x^2}-\frac{3 b e n}{4 x} \]

[Out]

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

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Rubi [A]  time = 0.0912584, antiderivative size = 163, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {2395, 44, 2376, 2301, 2391} \[ \frac{1}{2} b e^2 n \text{PolyLog}(2,-e x)-\frac{1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac{1}{2} e^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )-\frac{e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac{\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac{1}{4} b e^2 n \log ^2(x)-\frac{1}{4} b e^2 n \log (x)+\frac{1}{4} b e^2 n \log (e x+1)-\frac{b n \log (e x+1)}{4 x^2}-\frac{3 b e n}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*Log[c*x^n])*Log[1 + e*x])/x^3,x]

[Out]

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

Rule 2395

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))*((f_.) + (g_.)*(x_))^(q_.), x_Symbol] :> Simp[((f + g
*x)^(q + 1)*(a + b*Log[c*(d + e*x)^n]))/(g*(q + 1)), x] - Dist[(b*e*n)/(g*(q + 1)), Int[(f + g*x)^(q + 1)/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n, q}, x] && NeQ[e*f - d*g, 0] && NeQ[q, -1]

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 2376

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((g_.)*(x_))^(q_.), x_Sym
bol] :> With[{u = IntHide[(g*x)^q*Log[d*(e + f*x^m)^r], x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[Dist
[1/x, u, x], x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q + 1)/m] || (RationalQ[m] &
& RationalQ[q])) && NeQ[q, -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 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{\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{x^3} \, dx &=-\frac{e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac{1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac{1}{2} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 x^2}-(b n) \int \left (-\frac{e}{2 x^2}-\frac{e^2 \log (x)}{2 x}-\frac{\log (1+e x)}{2 x^3}+\frac{e^2 \log (1+e x)}{2 x}\right ) \, dx\\ &=-\frac{b e n}{2 x}-\frac{e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac{1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac{1}{2} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 x^2}+\frac{1}{2} (b n) \int \frac{\log (1+e x)}{x^3} \, dx+\frac{1}{2} \left (b e^2 n\right ) \int \frac{\log (x)}{x} \, dx-\frac{1}{2} \left (b e^2 n\right ) \int \frac{\log (1+e x)}{x} \, dx\\ &=-\frac{b e n}{2 x}+\frac{1}{4} b e^2 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac{1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{b n \log (1+e x)}{4 x^2}+\frac{1}{2} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 x^2}+\frac{1}{2} b e^2 n \text{Li}_2(-e x)+\frac{1}{4} (b e n) \int \frac{1}{x^2 (1+e x)} \, dx\\ &=-\frac{b e n}{2 x}+\frac{1}{4} b e^2 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac{1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{b n \log (1+e x)}{4 x^2}+\frac{1}{2} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 x^2}+\frac{1}{2} b e^2 n \text{Li}_2(-e x)+\frac{1}{4} (b e n) \int \left (\frac{1}{x^2}-\frac{e}{x}+\frac{e^2}{1+e x}\right ) \, dx\\ &=-\frac{3 b e n}{4 x}-\frac{1}{4} b e^2 n \log (x)+\frac{1}{4} b e^2 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac{1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac{1}{4} b e^2 n \log (1+e x)-\frac{b n \log (1+e x)}{4 x^2}+\frac{1}{2} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac{\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 x^2}+\frac{1}{2} b e^2 n \text{Li}_2(-e x)\\ \end{align*}

Mathematica [A]  time = 0.0701962, size = 215, normalized size = 1.32 \[ \frac{1}{2} b e n \left (e^2 \left (\frac{\text{PolyLog}(2,-e x)}{e}+\frac{\log (x) \log (e x+1)}{e}\right )-\frac{1}{2} e \log ^2(x)-\frac{1}{x}-\frac{\log (x)}{x}\right )-\frac{a \log (e x+1)}{2 x^2}+\frac{1}{2} a e \left (-e \log (x)+e \log (e x+1)-\frac{1}{x}\right )-\frac{1}{4} b e^2 \log (x) \left (2 \left (\log \left (c x^n\right )-n \log (x)\right )+n\right )+\frac{1}{4} b e^2 \log (e x+1) \left (2 \left (\log \left (c x^n\right )-n \log (x)\right )+n\right )+\frac{b \left (-2 e \left (\log \left (c x^n\right )-n \log (x)\right )-e n\right )}{4 x}-\frac{b \log (e x+1) \left (2 \left (\log \left (c x^n\right )-n \log (x)\right )+2 n \log (x)+n\right )}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Log[c*x^n])*Log[1 + e*x])/x^3,x]

[Out]

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

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Maple [C]  time = 0.095, size = 647, normalized size = 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))*ln(e*x+1)/x^3,x)

[Out]

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

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Maxima [A]  time = 1.34184, size = 262, normalized size = 1.61 \begin{align*} \frac{1}{2} \,{\left (\log \left (e x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-e x\right )\right )} b e^{2} n + \frac{1}{4} \,{\left (2 \, a e^{2} +{\left (e^{2} n + 2 \, e^{2} \log \left (c\right )\right )} b\right )} \log \left (e x + 1\right ) + \frac{b e^{2} n x^{2} \log \left (x\right )^{2} -{\left (2 \, a e^{2} +{\left (e^{2} n + 2 \, e^{2} \log \left (c\right )\right )} b\right )} x^{2} \log \left (x\right ) -{\left ({\left (3 \, e n + 2 \, e \log \left (c\right )\right )} b + 2 \, a e\right )} x -{\left (2 \, b e^{2} n x^{2} \log \left (x\right ) + b{\left (n + 2 \, \log \left (c\right )\right )} + 2 \, a\right )} \log \left (e x + 1\right ) - 2 \,{\left (b e^{2} x^{2} \log \left (x\right ) + b e x -{\left (b e^{2} x^{2} - b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{4 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(e*x+1)/x^3,x, algorithm="maxima")

[Out]

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

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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 ) \log \left (e x + 1\right ) + a \log \left (e x + 1\right )}{x^{3}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(e*x+1)/x^3,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n)*log(e*x + 1) + a*log(e*x + 1))/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))*ln(e*x+1)/x**3,x)

[Out]

Timed out

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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 )} \log \left (e x + 1\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))*log(e*x+1)/x^3,x, algorithm="giac")

[Out]

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