3.7 \(\int \frac{(a+b \log (c x^n)) \log (1+e x)}{x^2} \, dx\)

Optimal. Leaf size=107 \[ -b e n \text{PolyLog}(2,-e x)+e \log (x) \left (a+b \log \left (c x^n\right )\right )-e \log (e x+1) \left (a+b \log \left (c x^n\right )\right )-\frac{\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{x}-\frac{1}{2} b e n \log ^2(x)+b e n \log (x)-b e n \log (e x+1)-\frac{b n \log (e x+1)}{x} \]

[Out]

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

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Rubi [A]  time = 0.070474, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.35, Rules used = {2395, 36, 29, 31, 2376, 2301, 2391} \[ -b e n \text{PolyLog}(2,-e x)+e \log (x) \left (a+b \log \left (c x^n\right )\right )-e \log (e x+1) \left (a+b \log \left (c x^n\right )\right )-\frac{\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{x}-\frac{1}{2} b e n \log ^2(x)+b e n \log (x)-b e n \log (e x+1)-\frac{b n \log (e x+1)}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

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

Mathematica [A]  time = 0.0533662, size = 69, normalized size = 0.64 \[ -b e n \text{PolyLog}(2,-e x)+e \log (x) \left (a+b \log \left (c x^n\right )+b n\right )-\frac{(e x+1) \log (e x+1) \left (a+b \log \left (c x^n\right )+b n\right )}{x}-\frac{1}{2} b e n \log ^2(x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.078, size = 481, normalized size = 4.5 \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^2,x)

[Out]

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

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Maxima [A]  time = 1.35103, size = 173, normalized size = 1.62 \begin{align*} -{\left (\log \left (e x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-e x\right )\right )} b e n -{\left ({\left (e n + e \log \left (c\right )\right )} b + a e\right )} \log \left (e x + 1\right ) +{\left ({\left (e n + e \log \left (c\right )\right )} b + a e\right )} \log \left (x\right ) - \frac{b e n x \log \left (x\right )^{2} - 2 \,{\left (b e n x \log \left (x\right ) - b{\left (n + \log \left (c\right )\right )} - a\right )} \log \left (e x + 1\right ) - 2 \,{\left (b e x \log \left (x\right ) -{\left (b e x + b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{2 \, 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^2,x, algorithm="maxima")

[Out]

-(log(e*x + 1)*log(x) + dilog(-e*x))*b*e*n - ((e*n + e*log(c))*b + a*e)*log(e*x + 1) + ((e*n + e*log(c))*b + a
*e)*log(x) - 1/2*(b*e*n*x*log(x)^2 - 2*(b*e*n*x*log(x) - b*(n + log(c)) - a)*log(e*x + 1) - 2*(b*e*x*log(x) -
(b*e*x + b)*log(e*x + 1))*log(x^n))/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 ) \log \left (e x + 1\right ) + a \log \left (e x + 1\right )}{x^{2}}, 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^2,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n)*log(e*x + 1) + a*log(e*x + 1))/x^2, 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**2,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^{2}}\,{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^2,x, algorithm="giac")

[Out]

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