3.9 \(\int \frac{(a+b \log (c x^n)) \log (1+e x)}{x^4} \, dx\)

Optimal. Leaf size=195 \[ -\frac{1}{3} b e^3 n \text{PolyLog}(2,-e x)+\frac{1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{3} e^3 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac{e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}-\frac{e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}-\frac{\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{3 x^3}+\frac{4 b e^2 n}{9 x}-\frac{1}{6} b e^3 n \log ^2(x)+\frac{1}{9} b e^3 n \log (x)-\frac{1}{9} b e^3 n \log (e x+1)-\frac{5 b e n}{36 x^2}-\frac{b n \log (e x+1)}{9 x^3} \]

[Out]

(-5*b*e*n)/(36*x^2) + (4*b*e^2*n)/(9*x) + (b*e^3*n*Log[x])/9 - (b*e^3*n*Log[x]^2)/6 - (e*(a + b*Log[c*x^n]))/(
6*x^2) + (e^2*(a + b*Log[c*x^n]))/(3*x) + (e^3*Log[x]*(a + b*Log[c*x^n]))/3 - (b*e^3*n*Log[1 + e*x])/9 - (b*n*
Log[1 + e*x])/(9*x^3) - (e^3*(a + b*Log[c*x^n])*Log[1 + e*x])/3 - ((a + b*Log[c*x^n])*Log[1 + e*x])/(3*x^3) -
(b*e^3*n*PolyLog[2, -(e*x)])/3

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Rubi [A]  time = 0.106717, antiderivative size = 195, 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} \[ -\frac{1}{3} b e^3 n \text{PolyLog}(2,-e x)+\frac{1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{3} e^3 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac{e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}-\frac{e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}-\frac{\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{3 x^3}+\frac{4 b e^2 n}{9 x}-\frac{1}{6} b e^3 n \log ^2(x)+\frac{1}{9} b e^3 n \log (x)-\frac{1}{9} b e^3 n \log (e x+1)-\frac{5 b e n}{36 x^2}-\frac{b n \log (e x+1)}{9 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*b*e*n)/(36*x^2) + (4*b*e^2*n)/(9*x) + (b*e^3*n*Log[x])/9 - (b*e^3*n*Log[x]^2)/6 - (e*(a + b*Log[c*x^n]))/(
6*x^2) + (e^2*(a + b*Log[c*x^n]))/(3*x) + (e^3*Log[x]*(a + b*Log[c*x^n]))/3 - (b*e^3*n*Log[1 + e*x])/9 - (b*n*
Log[1 + e*x])/(9*x^3) - (e^3*(a + b*Log[c*x^n])*Log[1 + e*x])/3 - ((a + b*Log[c*x^n])*Log[1 + e*x])/(3*x^3) -
(b*e^3*n*PolyLog[2, -(e*x)])/3

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^4} \, dx &=-\frac{e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac{e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac{1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{3} e^3 \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)}{3 x^3}-(b n) \int \left (-\frac{e}{6 x^3}+\frac{e^2}{3 x^2}+\frac{e^3 \log (x)}{3 x}-\frac{\log (1+e x)}{3 x^4}-\frac{e^3 \log (1+e x)}{3 x}\right ) \, dx\\ &=-\frac{b e n}{12 x^2}+\frac{b e^2 n}{3 x}-\frac{e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac{e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac{1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{3} e^3 \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)}{3 x^3}+\frac{1}{3} (b n) \int \frac{\log (1+e x)}{x^4} \, dx-\frac{1}{3} \left (b e^3 n\right ) \int \frac{\log (x)}{x} \, dx+\frac{1}{3} \left (b e^3 n\right ) \int \frac{\log (1+e x)}{x} \, dx\\ &=-\frac{b e n}{12 x^2}+\frac{b e^2 n}{3 x}-\frac{1}{6} b e^3 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac{e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac{1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{b n \log (1+e x)}{9 x^3}-\frac{1}{3} e^3 \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)}{3 x^3}-\frac{1}{3} b e^3 n \text{Li}_2(-e x)+\frac{1}{9} (b e n) \int \frac{1}{x^3 (1+e x)} \, dx\\ &=-\frac{b e n}{12 x^2}+\frac{b e^2 n}{3 x}-\frac{1}{6} b e^3 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac{e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac{1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{b n \log (1+e x)}{9 x^3}-\frac{1}{3} e^3 \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)}{3 x^3}-\frac{1}{3} b e^3 n \text{Li}_2(-e x)+\frac{1}{9} (b e n) \int \left (\frac{1}{x^3}-\frac{e}{x^2}+\frac{e^2}{x}-\frac{e^3}{1+e x}\right ) \, dx\\ &=-\frac{5 b e n}{36 x^2}+\frac{4 b e^2 n}{9 x}+\frac{1}{9} b e^3 n \log (x)-\frac{1}{6} b e^3 n \log ^2(x)-\frac{e \left (a+b \log \left (c x^n\right )\right )}{6 x^2}+\frac{e^2 \left (a+b \log \left (c x^n\right )\right )}{3 x}+\frac{1}{3} e^3 \log (x) \left (a+b \log \left (c x^n\right )\right )-\frac{1}{9} b e^3 n \log (1+e x)-\frac{b n \log (1+e x)}{9 x^3}-\frac{1}{3} e^3 \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)}{3 x^3}-\frac{1}{3} b e^3 n \text{Li}_2(-e x)\\ \end{align*}

Mathematica [A]  time = 0.079695, size = 206, normalized size = 1.06 \[ -\frac{12 b e^3 n x^3 \text{PolyLog}(2,-e x)-4 e^3 x^3 \log (x) \left (3 a+3 b \log \left (c x^n\right )+b n\right )-12 a e^2 x^2+12 a e^3 x^3 \log (e x+1)+6 a e x+12 a \log (e x+1)+12 b e^3 x^3 \log (e x+1) \log \left (c x^n\right )-12 b e^2 x^2 \log \left (c x^n\right )+6 b e x \log \left (c x^n\right )+12 b \log (e x+1) \log \left (c x^n\right )-16 b e^2 n x^2+6 b e^3 n x^3 \log ^2(x)+4 b e^3 n x^3 \log (e x+1)+5 b e n x+4 b n \log (e x+1)}{36 x^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(6*a*e*x + 5*b*e*n*x - 12*a*e^2*x^2 - 16*b*e^2*n*x^2 + 6*b*e^3*n*x^3*Log[x]^2 + 6*b*e*x*Log[c*x^n] - 12*b*e^2
*x^2*Log[c*x^n] - 4*e^3*x^3*Log[x]*(3*a + b*n + 3*b*Log[c*x^n]) + 12*a*Log[1 + e*x] + 4*b*n*Log[1 + e*x] + 12*
a*e^3*x^3*Log[1 + e*x] + 4*b*e^3*n*x^3*Log[1 + e*x] + 12*b*Log[c*x^n]*Log[1 + e*x] + 12*b*e^3*x^3*Log[c*x^n]*L
og[1 + e*x] + 12*b*e^3*n*x^3*PolyLog[2, -(e*x)])/(36*x^3)

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Maple [C]  time = 0.103, size = 796, normalized size = 4.1 \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^4,x)

[Out]

(-1/3*b/x^3*ln(e*x+1)-1/6*b*e*(2*e^2*ln(e*x+1)*x^2-2*e^2*ln(x)*x^2-2*e*x+1)/x^2)*ln(x^n)-1/6*b*e^3*n*ln(x)^2-1
/9*b*e^3*n*ln(e*x+1)-1/9*b*n*ln(e*x+1)/x^3+4/9*b*e^2*n/x-5/36*b*e*n/x^2+1/9*n*e^3*b*ln(e*x)-1/3*b*ln(c)*ln(e*x
+1)/x^3+1/3*e^3*b*ln(c)*ln(e*x)+1/3*b*ln(c)*e^2/x-1/6*e*b*ln(c)/x^2-1/3*b*ln(c)*e^3*ln(e*x+1)-1/6*I*Pi*b*csgn(
I*x^n)*csgn(I*c*x^n)^2*ln(e*x+1)/x^3+1/6*I*e^3*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2*ln(e*x)+1/6*I*e^3*Pi*b*csgn(I*x^
n)*csgn(I*c*x^n)^2*ln(e*x)+1/6*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2*e^2/x-1/6*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2
*e^3*ln(e*x+1)-1/12*I*e*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2/x^2-1/6*I*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2*ln(e*x+1)/x^3-
1/12*I*e*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2/x^2+1/6*I*Pi*b*csgn(I*c)*csgn(I*c*x^n)^2*e^2/x-1/6*I*Pi*b*csgn(I*c)*
csgn(I*c*x^n)^2*e^3*ln(e*x+1)-1/6*I*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*e^2/x+1/6*I*Pi*b*csgn(I*c)*csgn(I
*x^n)*csgn(I*c*x^n)*ln(e*x+1)/x^3+1/6*I*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*e^3*ln(e*x+1)+1/12*I*e*Pi*b*c
sgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)/x^2-1/6*I*e^3*Pi*b*csgn(I*c)*csgn(I*x^n)*csgn(I*c*x^n)*ln(e*x)-1/3*ln(e*x+1
)/x^3*a+1/3*a*e^3*ln(e*x)-1/3*b*e^3*n*dilog(e*x+1)-1/6*a*e/x^2+1/3*a*e^2/x-1/3*a*e^3*ln(e*x+1)+1/12*I*e*Pi*b*c
sgn(I*c*x^n)^3/x^2-1/6*I*e^3*Pi*b*csgn(I*c*x^n)^3*ln(e*x)+1/6*I*Pi*b*csgn(I*c*x^n)^3*ln(e*x+1)/x^3-1/6*I*Pi*b*
csgn(I*c*x^n)^3*e^2/x+1/6*I*Pi*b*csgn(I*c*x^n)^3*e^3*ln(e*x+1)

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Maxima [A]  time = 1.30591, size = 313, normalized size = 1.61 \begin{align*} -\frac{1}{3} \,{\left (\log \left (e x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-e x\right )\right )} b e^{3} n - \frac{1}{9} \,{\left (3 \, a e^{3} +{\left (e^{3} n + 3 \, e^{3} \log \left (c\right )\right )} b\right )} \log \left (e x + 1\right ) - \frac{6 \, b e^{3} n x^{3} \log \left (x\right )^{2} - 4 \,{\left (3 \, a e^{3} +{\left (e^{3} n + 3 \, e^{3} \log \left (c\right )\right )} b\right )} x^{3} \log \left (x\right ) - 4 \,{\left (3 \, a e^{2} +{\left (4 \, e^{2} n + 3 \, e^{2} \log \left (c\right )\right )} b\right )} x^{2} +{\left ({\left (5 \, e n + 6 \, e \log \left (c\right )\right )} b + 6 \, a e\right )} x - 4 \,{\left (3 \, b e^{3} n x^{3} \log \left (x\right ) - b{\left (n + 3 \, \log \left (c\right )\right )} - 3 \, a\right )} \log \left (e x + 1\right ) - 6 \,{\left (2 \, b e^{3} x^{3} \log \left (x\right ) + 2 \, b e^{2} x^{2} - b e x - 2 \,{\left (b e^{3} x^{3} + b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{36 \, x^{3}} \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^4,x, algorithm="maxima")

[Out]

-1/3*(log(e*x + 1)*log(x) + dilog(-e*x))*b*e^3*n - 1/9*(3*a*e^3 + (e^3*n + 3*e^3*log(c))*b)*log(e*x + 1) - 1/3
6*(6*b*e^3*n*x^3*log(x)^2 - 4*(3*a*e^3 + (e^3*n + 3*e^3*log(c))*b)*x^3*log(x) - 4*(3*a*e^2 + (4*e^2*n + 3*e^2*
log(c))*b)*x^2 + ((5*e*n + 6*e*log(c))*b + 6*a*e)*x - 4*(3*b*e^3*n*x^3*log(x) - b*(n + 3*log(c)) - 3*a)*log(e*
x + 1) - 6*(2*b*e^3*x^3*log(x) + 2*b*e^2*x^2 - b*e*x - 2*(b*e^3*x^3 + b)*log(e*x + 1))*log(x^n))/x^3

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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^{4}}, 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^4,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n)*log(e*x + 1) + a*log(e*x + 1))/x^4, 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**4,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^{4}}\,{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^4,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*log(e*x + 1)/x^4, x)