3.5 \(\int (a+b \log (c x^n)) \log (1+e x) \, dx\)

Optimal. Leaf size=74 \[ \frac{b n \text{PolyLog}(2,-e x)}{e}+\frac{(e x+1) \log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{e}-x \left (a+b \log \left (c x^n\right )\right )-\frac{b n (e x+1) \log (e x+1)}{e}+2 b n x \]

[Out]

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

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Rubi [A]  time = 0.0886038, antiderivative size = 74, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.412, Rules used = {2389, 2295, 2370, 2411, 43, 2351, 2315} \[ \frac{b n \text{PolyLog}(2,-e x)}{e}+\frac{(e x+1) \log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{e}-x \left (a+b \log \left (c x^n\right )\right )-\frac{b n (e x+1) \log (e x+1)}{e}+2 b n x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2389

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.), x_Symbol] :> Dist[1/e, Subst[Int[(a + b*Log[c*
x^n])^p, x], x, d + e*x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2370

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

Rule 2411

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_.) + (g_.)*(x_))^(q_.)*((h_.) + (i_.)*(x_))
^(r_.), x_Symbol] :> Dist[1/e, Subst[Int[((g*x)/e)^q*((e*h - d*i)/e + (i*x)/e)^r*(a + b*Log[c*x^n])^p, x], x,
d + e*x], x] /; FreeQ[{a, b, c, d, e, f, g, h, i, n, p, q, r}, x] && EqQ[e*f - d*g, 0] && (IGtQ[p, 0] || IGtQ[
r, 0]) && IntegerQ[2*r]

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 2315

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[PolyLog[2, 1 - c*x]/e, x] /; FreeQ[{c, d, e}, x] &
& EqQ[e + c*d, 0]

Rubi steps

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

Mathematica [A]  time = 0.0311021, size = 90, normalized size = 1.22 \[ \frac{b n \text{PolyLog}(2,-e x)-a e x+a e x \log (e x+1)+a \log (e x+1)+b ((e x+1) \log (e x+1)-e x) \log \left (c x^n\right )+2 b e n x-b e n x \log (e x+1)-b n \log (e x+1)}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.064, size = 557, normalized size = 7.5 \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)

[Out]

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

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Maxima [A]  time = 1.31946, size = 170, normalized size = 2.3 \begin{align*} \frac{{\left (\log \left (e x + 1\right ) \log \left (x\right ) +{\rm Li}_2\left (-e x\right )\right )} b n}{e} - \frac{{\left (b{\left (n - \log \left (c\right )\right )} - a\right )} \log \left (e x + 1\right )}{e} + \frac{{\left ({\left (2 \, e n - e \log \left (c\right )\right )} b - a e\right )} x -{\left (b n \log \left (x\right ) +{\left ({\left (e n - e \log \left (c\right )\right )} b - a e\right )} x\right )} \log \left (e x + 1\right ) -{\left (b e x -{\left (b e x + b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{e} \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, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (b \log \left (c x^{n}\right ) \log \left (e x + 1\right ) + a \log \left (e x + 1\right ), 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, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \log \left (c x^{n}\right ) + a\right )} \log \left (e x + 1\right )\,{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, algorithm="giac")

[Out]

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