3.1.8 \(\int \frac {(a+b \log (c x^n)) \log (1+e x)}{x^3} \, dx\) [8]

Optimal. Leaf size=163 \[ -\frac {3 b e n}{4 x}-\frac {1}{4} b e^2 n \log (x)+\frac {1}{4} b e^2 n \log ^2(x)-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac {1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{4} b e^2 n \log (1+e x)-\frac {b n \log (1+e x)}{4 x^2}+\frac {1}{2} e^2 \left (a+b \log \left (c x^n\right )\right ) \log (1+e x)-\frac {\left (a+b \log \left (c x^n\right )\right ) \log (1+e x)}{2 x^2}+\frac {1}{2} b e^2 n \text {Li}_2(-e x) \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {2442, 46, 2423, 2338, 2438} \begin {gather*} \frac {1}{2} b e^2 n \text {PolyLog}(2,-e x)-\frac {1}{2} e^2 \log (x) \left (a+b \log \left (c x^n\right )\right )+\frac {1}{2} e^2 \log (e x+1) \left (a+b \log \left (c x^n\right )\right )-\frac {e \left (a+b \log \left (c x^n\right )\right )}{2 x}-\frac {\log (e x+1) \left (a+b \log \left (c x^n\right )\right )}{2 x^2}+\frac {1}{4} b e^2 n \log ^2(x)-\frac {1}{4} b e^2 n \log (x)+\frac {1}{4} b e^2 n \log (e x+1)-\frac {b n \log (e x+1)}{4 x^2}-\frac {3 b e n}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*Log[c*x^n])*Log[1 + e*x])/x^3,x]

[Out]

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

Rule 46

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] && Lt
Q[m + n + 2, 0])

Rule 2338

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 2423

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 2438

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 2442

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

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Mathematica [A]
time = 0.06, size = 215, normalized size = 1.32 \begin {gather*} -\frac {1}{4} b e^2 \log (x) \left (n+2 \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+\frac {b \left (-e n-2 e \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{4 x}-\frac {a \log (1+e x)}{2 x^2}+\frac {1}{4} b e^2 \left (n+2 \left (-n \log (x)+\log \left (c x^n\right )\right )\right ) \log (1+e x)-\frac {b \left (n+2 n \log (x)+2 \left (-n \log (x)+\log \left (c x^n\right )\right )\right ) \log (1+e x)}{4 x^2}+\frac {1}{2} a e \left (-\frac {1}{x}-e \log (x)+e \log (1+e x)\right )+\frac {1}{2} b e n \left (-\frac {1}{x}-\frac {\log (x)}{x}-\frac {1}{2} e \log ^2(x)+e^2 \left (\frac {\log (x) \log (1+e x)}{e}+\frac {\text {Li}_2(-e x)}{e}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Log[c*x^n])*Log[1 + e*x])/x^3,x]

[Out]

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

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 647, normalized size = 3.97

method	result	size
risch	\(-\frac {\ln \left (e x +1\right ) a}{2 x^{2}}-\frac {e^{2} b \ln \left (c \right ) \ln \left (e x \right )}{2}-\frac {b \ln \left (c \right ) \ln \left (e x +1\right )}{2 x^{2}}-\frac {e b \ln \left (c \right )}{2 x}+\frac {b \ln \left (c \right ) e^{2} \ln \left (e x +1\right )}{2}-\frac {a \,e^{2} \ln \left (e x \right )}{2}+\frac {a \,e^{2} \ln \left (e x +1\right )}{2}-\frac {a e}{2 x}+\frac {e^{2} b n \dilog \left (e x +1\right )}{2}+\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 x^{2}}+\frac {i e \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4 x}-\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 ) e^{2} \ln \left (e x +1\right )}{4}+\frac {i e^{2} \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 \right )}{4}-\frac {n \,e^{2} b \ln \left (e x \right )}{4}+\left (-\frac {b \ln \left (e x +1\right )}{2 x^{2}}-\frac {b e \left (e x \ln \left (x \right )-e \ln \left (e x +1\right ) x +1\right )}{2 x}\right ) \ln \left (x^{n}\right )-\frac {i e^{2} \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x \right )}{4}-\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 x^{2}}-\frac {i e \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 x}+\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{2} \ln \left (e x +1\right )}{4}-\frac {3 b e n}{4 x}-\frac {b n \ln \left (e x +1\right )}{4 x^{2}}-\frac {i e \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 x}-\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 x^{2}}+\frac {b \,e^{2} n \ln \left (x \right )^{2}}{4}+\frac {b \,e^{2} n \ln \left (e x +1\right )}{4}-\frac {i e^{2} \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e x \right )}{4}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} e^{2} \ln \left (e x +1\right )}{4}+\frac {i e^{2} \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x \right )}{4}+\frac {i e \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 x}+\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e x +1\right )}{4 x^{2}}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} e^{2} \ln \left (e x +1\right )}{4}\)	\(647\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))*ln(e*x+1)/x^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.34, size = 178, normalized size = 1.09 \begin {gather*} \frac {1}{2} \, {\left (\log \left (x e + 1\right ) \log \left (x\right ) + {\rm Li}_2\left (-x e\right )\right )} b n e^{2} + \frac {1}{4} \, {\left (b {\left (n + 2 \, \log \left (c\right )\right )} + 2 \, a\right )} e^{2} \log \left (x e + 1\right ) + \frac {b n x^{2} e^{2} \log \left (x\right )^{2} - {\left (b {\left (n + 2 \, \log \left (c\right )\right )} + 2 \, a\right )} x^{2} e^{2} \log \left (x\right ) - {\left (b {\left (3 \, n + 2 \, \log \left (c\right )\right )} + 2 \, a\right )} x e - {\left (2 \, b n x^{2} e^{2} \log \left (x\right ) + b {\left (n + 2 \, \log \left (c\right )\right )} + 2 \, a\right )} \log \left (x e + 1\right ) - 2 \, {\left (b x^{2} e^{2} \log \left (x\right ) + b x e - {\left (b x^{2} e^{2} - b\right )} \log \left (x e + 1\right )\right )} \log \left (x^{n}\right )}{4 \, x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(e*x+1)/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(e*x+1)/x^3,x, algorithm="fricas")

[Out]

integral((b*log(c*x^n)*log(x*e + 1) + a*log(x*e + 1))/x^3, x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))*ln(e*x+1)/x**3,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*log(c*x^n))*log(e*x+1)/x^3,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*log(x*e + 1)/x^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\ln \left (e\,x+1\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(e*x + 1)*(a + b*log(c*x^n)))/x^3,x)

[Out]

int((log(e*x + 1)*(a + b*log(c*x^n)))/x^3, x)

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