3.3 \(\int x^2 (a+b \log (c x^n)) \log (1+e x) \, dx\)

Optimal. Leaf size=178 \[ \frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {1}{3} x^3 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \text {Li}_2(-e x)}{3 e^3}-\frac {b n \log (e x+1)}{9 e^3}+\frac {4 b n x}{9 e^2}-\frac {1}{9} b n x^3 \log (e x+1)-\frac {5 b n x^2}{36 e}+\frac {2}{27} b n x^3 \]

[Out]

4/9*b*n*x/e^2-5/36*b*n*x^2/e+2/27*b*n*x^3-1/3*x*(a+b*ln(c*x^n))/e^2+1/6*x^2*(a+b*ln(c*x^n))/e-1/9*x^3*(a+b*ln(
c*x^n))-1/9*b*n*ln(e*x+1)/e^3-1/9*b*n*x^3*ln(e*x+1)+1/3*(a+b*ln(c*x^n))*ln(e*x+1)/e^3+1/3*x^3*(a+b*ln(c*x^n))*
ln(e*x+1)+1/3*b*n*polylog(2,-e*x)/e^3

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Rubi [A]  time = 0.10, antiderivative size = 178, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2395, 43, 2376, 2391} \[ \frac {b n \text {PolyLog}(2,-e x)}{3 e^3}-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{3 e^3}+\frac {1}{3} x^3 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {4 b n x}{9 e^2}-\frac {b n \log (e x+1)}{9 e^3}-\frac {5 b n x^2}{36 e}-\frac {1}{9} b n x^3 \log (e x+1)+\frac {2}{27} b n x^3 \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*b*n*x)/(9*e^2) - (5*b*n*x^2)/(36*e) + (2*b*n*x^3)/27 - (x*(a + b*Log[c*x^n]))/(3*e^2) + (x^2*(a + b*Log[c*x
^n]))/(6*e) - (x^3*(a + b*Log[c*x^n]))/9 - (b*n*Log[1 + e*x])/(9*e^3) - (b*n*x^3*Log[1 + e*x])/9 + ((a + b*Log
[c*x^n])*Log[1 + e*x])/(3*e^3) + (x^3*(a + b*Log[c*x^n])*Log[1 + e*x])/3 + (b*n*PolyLog[2, -(e*x)])/(3*e^3)

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 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 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]

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]

Rubi steps

\begin {align*} \int x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x) \, dx &=-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 e^3}+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-(b n) \int \left (-\frac {1}{3 e^2}+\frac {x}{6 e}-\frac {x^2}{9}+\frac {\log (1+e x)}{3 e^3 x}+\frac {1}{3} x^2 \log (1+e x)\right ) \, dx\\ &=\frac {b n x}{3 e^2}-\frac {b n x^2}{12 e}+\frac {1}{27} b n x^3-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 e^3}+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {1}{3} (b n) \int x^2 \log (1+e x) \, dx-\frac {(b n) \int \frac {\log (1+e x)}{x} \, dx}{3 e^3}\\ &=\frac {b n x}{3 e^2}-\frac {b n x^2}{12 e}+\frac {1}{27} b n x^3-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b n x^3 \log (1+e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 e^3}+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)+\frac {b n \text {Li}_2(-e x)}{3 e^3}+\frac {1}{9} (b e n) \int \frac {x^3}{1+e x} \, dx\\ &=\frac {b n x}{3 e^2}-\frac {b n x^2}{12 e}+\frac {1}{27} b n x^3-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{9} b n x^3 \log (1+e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 e^3}+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)+\frac {b n \text {Li}_2(-e x)}{3 e^3}+\frac {1}{9} (b e n) \int \left (\frac {1}{e^3}-\frac {x}{e^2}+\frac {x^2}{e}-\frac {1}{e^3 (1+e x)}\right ) \, dx\\ &=\frac {4 b n x}{9 e^2}-\frac {5 b n x^2}{36 e}+\frac {2}{27} b n x^3-\frac {x \left (a+b \log \left (c x^n\right )\right )}{3 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{6 e}-\frac {1}{9} x^3 \left (a+b \log \left (c x^n\right )\right )-\frac {b n \log (1+e x)}{9 e^3}-\frac {1}{9} b n x^3 \log (1+e x)+\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{3 e^3}+\frac {1}{3} x^3 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)+\frac {b n \text {Li}_2(-e x)}{3 e^3}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 161, normalized size = 0.90 \[ \frac {-12 a e^3 x^3+36 a e^3 x^3 \log (e x+1)+18 a e^2 x^2-36 a e x+36 a \log (e x+1)+6 b \left (6 \left (e^3 x^3+1\right ) \log (e x+1)+e x \left (-2 e^2 x^2+3 e x-6\right )\right ) \log \left (c x^n\right )+8 b e^3 n x^3-12 b e^3 n x^3 \log (e x+1)-15 b e^2 n x^2+36 b n \text {Li}_2(-e x)+48 b e n x-12 b n \log (e x+1)}{108 e^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-36*a*e*x + 48*b*e*n*x + 18*a*e^2*x^2 - 15*b*e^2*n*x^2 - 12*a*e^3*x^3 + 8*b*e^3*n*x^3 + 36*a*Log[1 + e*x] - 1
2*b*n*Log[1 + e*x] + 36*a*e^3*x^3*Log[1 + e*x] - 12*b*e^3*n*x^3*Log[1 + e*x] + 6*b*Log[c*x^n]*(e*x*(-6 + 3*e*x
- 2*e^2*x^2) + 6*(1 + e^3*x^3)*Log[1 + e*x]) + 36*b*n*PolyLog[2, -(e*x)])/(108*e^3)

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fricas [F]  time = 0.70, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x^{2} \log \left (c x^{n}\right ) \log \left (e x + 1\right ) + a x^{2} \log \left (e x + 1\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))*log(e*x+1),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \log \left (c x^{n}\right ) + a\right )} x^{2} \log \left (e x + 1\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))*log(e*x+1),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.29, size = 870, normalized size = 4.89 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(b*ln(c*x^n)+a)*ln(e*x+1),x)

[Out]

-11/18*a/e^3-1/9*a*x^3+(1/3*b*x^3*ln(e*x+1)+1/18*b*(-2*e^3*x^3+3*e^2*x^2-6*e*x+6*ln(e*x+1))/e^3)*ln(x^n)-1/9*l
n(c)*b*x^3+1/6*a/e*x^2+2/27*b*n*x^3+1/3*b*ln(c)*ln(e*x+1)*x^3+71/108/e^3*b*n+1/6*b/e*x^2*ln(c)+1/3/e^3*ln(e*x+
1)*b*ln(c)+1/3/e^3*b*n*dilog(e*x+1)+1/3*x^3*ln(e*x+1)*a-1/3*b/e^2*x*ln(c)-11/18/e^3*b*ln(c)+1/3*a/e^3*ln(e*x+1
)+1/18*I*Pi*b*x^3*csgn(I*c*x^n)^3+11/36*I/e^3*Pi*b*csgn(I*c*x^n)^3+4/9*b/e^2*n*x-5/36*b/e*n*x^2-1/3*a/e^2*x-1/
6*I/e^2*x*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/6*I/e^2*x*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+1/6*I/e^3*ln(e*x+1)*Pi*b
*csgn(I*x^n)*csgn(I*c*x^n)^2+1/6*I/e^3*ln(e*x+1)*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+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^2*x*Pi*b*csgn(I*c*x^n)^3-1/6*I/e^3*ln(e*x+1)*Pi*b*csgn(I*c*x^n)^3-1/18*I*Pi*b*x
^3*csgn(I*x^n)*csgn(I*c*x^n)^2-1/18*I*Pi*b*x^3*csgn(I*c*x^n)^2*csgn(I*c)-1/12*I/e*x^2*Pi*b*csgn(I*c*x^n)^3-1/6
*I*Pi*b*csgn(I*c*x^n)^3*ln(e*x+1)*x^3+1/6*I/e^2*x*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/6*I/e^3*ln(e*x+1)
*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/6*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*ln(e*x+1)*x^3-1/12*I/
e*x^2*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-11/36*I/e^3*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-11/36*I/e^3*Pi*b*c
sgn(I*c*x^n)^2*csgn(I*c)+1/6*I*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)*ln(e*x+1)*x^3+1/18*I*Pi*b*x^3*csgn(I*x^n)*csgn(I
*c*x^n)*csgn(I*c)+1/12*I/e*x^2*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/12*I/e*x^2*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)+11
/36*I/e^3*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/9*b*n*ln(e*x+1)/e^3-1/9*b*n*x^3*ln(e*x+1)

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maxima [A]  time = 1.35, size = 220, normalized size = 1.24 \[ \frac {{\left (\log \left (e x + 1\right ) \log \relax (x) + {\rm Li}_2\left (-e x\right )\right )} b n}{3 \, e^{3}} - \frac {{\left (b {\left (n - 3 \, \log \relax (c)\right )} - 3 \, a\right )} \log \left (e x + 1\right )}{9 \, e^{3}} - \frac {4 \, {\left (3 \, a e^{3} - {\left (2 \, e^{3} n - 3 \, e^{3} \log \relax (c)\right )} b\right )} x^{3} - 3 \, {\left (6 \, a e^{2} - {\left (5 \, e^{2} n - 6 \, e^{2} \log \relax (c)\right )} b\right )} x^{2} - 12 \, {\left ({\left (4 \, e n - 3 \, e \log \relax (c)\right )} b - 3 \, a e\right )} x - 12 \, {\left ({\left (3 \, a e^{3} - {\left (e^{3} n - 3 \, e^{3} \log \relax (c)\right )} b\right )} x^{3} - 3 \, b n \log \relax (x)\right )} \log \left (e x + 1\right ) + 6 \, {\left (2 \, b e^{3} x^{3} - 3 \, b e^{2} x^{2} + 6 \, b e x - 6 \, {\left (b e^{3} x^{3} + b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{108 \, e^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*log(c*x^n))*log(e*x+1),x, algorithm="maxima")

[Out]

1/3*(log(e*x + 1)*log(x) + dilog(-e*x))*b*n/e^3 - 1/9*(b*(n - 3*log(c)) - 3*a)*log(e*x + 1)/e^3 - 1/108*(4*(3*
a*e^3 - (2*e^3*n - 3*e^3*log(c))*b)*x^3 - 3*(6*a*e^2 - (5*e^2*n - 6*e^2*log(c))*b)*x^2 - 12*((4*e*n - 3*e*log(
c))*b - 3*a*e)*x - 12*((3*a*e^3 - (e^3*n - 3*e^3*log(c))*b)*x^3 - 3*b*n*log(x))*log(e*x + 1) + 6*(2*b*e^3*x^3
- 3*b*e^2*x^2 + 6*b*e*x - 6*(b*e^3*x^3 + b)*log(e*x + 1))*log(x^n))/e^3

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\ln \left (e\,x+1\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*ln(c*x**n))*ln(e*x+1),x)

[Out]

Timed out

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