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

Optimal. Leaf size=146 \[ -\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{2 e}+\frac {1}{2} x^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {b n \text {Li}_2(-e x)}{2 e^2}+\frac {b n \log (e x+1)}{4 e^2}-\frac {1}{4} b n x^2 \log (e x+1)-\frac {3 b n x}{4 e}+\frac {1}{4} b n x^2 \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {2395, 43, 2376, 2391} \[ -\frac {b n \text {PolyLog}(2,-e x)}{2 e^2}-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {1}{2} x^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )+\frac {x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (e x+1)}{4 e^2}-\frac {1}{4} b n x^2 \log (e x+1)-\frac {3 b n x}{4 e}+\frac {1}{4} b n x^2 \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 \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 )}{2 e}-\frac {1}{4} x^2 \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)}{2 e^2}+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-(b n) \int \left (\frac {1}{2 e}-\frac {x}{4}-\frac {\log (1+e x)}{2 e^2 x}+\frac {1}{2} x \log (1+e x)\right ) \, dx\\ &=-\frac {b n x}{2 e}+\frac {1}{8} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {1}{4} x^2 \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)}{2 e^2}+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {1}{2} (b n) \int x \log (1+e x) \, dx+\frac {(b n) \int \frac {\log (1+e x)}{x} \, dx}{2 e^2}\\ &=-\frac {b n x}{2 e}+\frac {1}{8} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b n x^2 \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 e^2}+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {b n \text {Li}_2(-e x)}{2 e^2}+\frac {1}{4} (b e n) \int \frac {x^2}{1+e x} \, dx\\ &=-\frac {b n x}{2 e}+\frac {1}{8} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b n x^2 \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 e^2}+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {b n \text {Li}_2(-e x)}{2 e^2}+\frac {1}{4} (b e n) \int \left (-\frac {1}{e^2}+\frac {x}{e}+\frac {1}{e^2 (1+e x)}\right ) \, dx\\ &=-\frac {3 b n x}{4 e}+\frac {1}{4} b n x^2+\frac {x \left (a+b \log \left (c x^n\right )\right )}{2 e}-\frac {1}{4} x^2 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log (1+e x)}{4 e^2}-\frac {1}{4} b n x^2 \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 e^2}+\frac {1}{2} x^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {b n \text {Li}_2(-e x)}{2 e^2}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 131, normalized size = 0.90 \[ \frac {-a e^2 x^2+2 a e^2 x^2 \log (e x+1)+2 a e x-2 a \log (e x+1)+b \left (2 \left (e^2 x^2-1\right ) \log (e x+1)+e x (2-e x)\right ) \log \left (c x^n\right )+b e^2 n x^2-b e^2 n x^2 \log (e x+1)-2 b n \text {Li}_2(-e x)-3 b e n x+b n \log (e x+1)}{4 e^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.62, size = 0, normalized size = 0.00 \[ {\rm integral}\left (b x \log \left (c x^{n}\right ) \log \left (e x + 1\right ) + a x \log \left (e x + 1\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 0.29, size = 725, normalized size = 4.97 \[ \frac {3 a}{4 e^{2}}-\frac {a \,x^{2}}{4}+\left (\frac {b \,x^{2} \ln \left (e x +1\right )}{2}-\frac {\left (e^{2} x^{2}-2 e x +2 \ln \left (e x +1\right )\right ) b}{4 e^{2}}\right ) \ln \left (x^{n}\right )-\frac {b n \dilog \left (e x +1\right )}{2 e^{2}}-\frac {b n}{e^{2}}-\frac {b \,x^{2} \ln \relax (c )}{4}+\frac {b n \,x^{2}}{4}+\frac {b x \ln \relax (c )}{2 e}-\frac {b \ln \relax (c ) \ln \left (e x +1\right )}{2 e^{2}}+\frac {3 b \ln \relax (c )}{4 e^{2}}-\frac {a \ln \left (e x +1\right )}{2 e^{2}}+\frac {a \,x^{2} \ln \left (e x +1\right )}{2}+\frac {b \,x^{2} \ln \relax (c ) \ln \left (e x +1\right )}{2}-\frac {3 i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{8 e^{2}}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{8}-\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e x +1\right )}{4}-\frac {i \pi b x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4 e}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e x +1\right )}{4 e^{2}}-\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x +1\right )}{4}-\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8}-\frac {i \pi b x \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 e}+\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x +1\right )}{4 e^{2}}+\frac {3 i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8 e^{2}}+\frac {3 i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{8 e^{2}}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +1\right )}{4 e^{2}}-\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +1\right )}{4 e^{2}}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +1\right )}{4}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x +1\right )}{4}+\frac {i \pi b \,x^{2} \mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{8}+\frac {i \pi b x \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 e}+\frac {i \pi b x \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 e}-\frac {3 i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{8 e^{2}}+\frac {a x}{2 e}-\frac {3 b n x}{4 e}+\frac {b n \ln \left (e x +1\right )}{4 e^{2}}-\frac {b n \,x^{2} \ln \left (e x +1\right )}{4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*a/e^2-1/4*a*x^2+(1/2*b*x^2*ln(e*x+1)-1/4*b*(e^2*x^2-2*e*x+2*ln(e*x+1))/e^2)*ln(x^n)-1/e^2*b*n-1/4*ln(c)*b*
x^2+1/4*b*n*x^2+1/2*b/e*x*ln(c)-1/2/e^2*ln(e*x+1)*b*ln(c)-1/2/e^2*b*n*dilog(e*x+1)+3/4/e^2*b*ln(c)-1/2*a/e^2*l
n(e*x+1)+1/2*x^2*ln(e*x+1)*a+1/2*b*ln(c)*ln(e*x+1)*x^2+1/4*I/e*x*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2+1/4*I*Pi*b*c
sgn(I*c*x^n)^2*csgn(I*c)*ln(e*x+1)*x^2+1/4*I/e^2*ln(e*x+1)*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/4*I/e*x*
Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-3/8*I/e^2*Pi*b*csgn(I*c*x^n)^3+1/8*I*Pi*b*x^2*csgn(I*c*x^n)^3+1/8*I*P
i*b*x^2*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+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*x*Pi*
b*csgn(I*c*x^n)^2*csgn(I*c)-1/4*I/e^2*ln(e*x+1)*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2+3/8*I/e^2*Pi*b*csgn(I*x^n)*cs
gn(I*c*x^n)^2-1/8*I*Pi*b*x^2*csgn(I*x^n)*csgn(I*c*x^n)^2-1/4*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*ln(e*x
+1)*x^2+1/4*I/e^2*ln(e*x+1)*Pi*b*csgn(I*c*x^n)^3-1/4*I/e*x*Pi*b*csgn(I*c*x^n)^3-1/4*I*Pi*b*csgn(I*c*x^n)^3*ln(
e*x+1)*x^2+3/8*I/e^2*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)-1/8*I*Pi*b*x^2*csgn(I*c*x^n)^2*csgn(I*c)-1/4*I/e^2*ln(e*x+
1)*Pi*b*csgn(I*c*x^n)^2*csgn(I*c)-3/8*I/e^2*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)+1/2*a/e*x-3/4*b/e*n*x+1/4
*b*n*ln(e*x+1)/e^2-1/4*b*n*x^2*ln(e*x+1)

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maxima [A]  time = 1.40, size = 178, normalized size = 1.22 \[ -\frac {{\left (\log \left (e x + 1\right ) \log \relax (x) + {\rm Li}_2\left (-e x\right )\right )} b n}{2 \, e^{2}} + \frac {{\left (b {\left (n - 2 \, \log \relax (c)\right )} - 2 \, a\right )} \log \left (e x + 1\right )}{4 \, e^{2}} - \frac {{\left (a e^{2} - {\left (e^{2} n - e^{2} \log \relax (c)\right )} b\right )} x^{2} + {\left ({\left (3 \, e n - 2 \, e \log \relax (c)\right )} b - 2 \, a e\right )} x - {\left ({\left (2 \, a e^{2} - {\left (e^{2} n - 2 \, e^{2} \log \relax (c)\right )} b\right )} x^{2} + 2 \, b n \log \relax (x)\right )} \log \left (e x + 1\right ) + {\left (b e^{2} x^{2} - 2 \, b e x - 2 \, {\left (b e^{2} x^{2} - b\right )} \log \left (e x + 1\right )\right )} \log \left (x^{n}\right )}{4 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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