3.45.59 \(\int \frac {(15 e^3-30 x) \log (-\frac {5}{e^3-2 x})+(30 x+(-15 e^3+30 x) \log (-\frac {5}{e^3-2 x})) \log (x)}{4 e^3 x^2-8 x^3} \, dx\)

Optimal. Leaf size=23 \[ \frac {15 \log (x) \log \left (\frac {5}{-e^3+2 x}\right )}{4 x} \]

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Rubi [C]  time = 1.14, antiderivative size = 101, normalized size of antiderivative = 4.39, number of steps used = 26, number of rules used = 16, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.232, Rules used = {1593, 6688, 12, 14, 2395, 36, 31, 29, 6742, 2344, 2301, 2316, 2315, 2376, 2392, 2391} \begin {gather*} \frac {15 \text {Li}_2\left (\frac {2 x}{e^3}\right )}{2 e^3}+\frac {15 \text {Li}_2\left (1-\frac {2 x}{e^3}\right )}{2 e^3}+\frac {15 \log (x) \log \left (e^3-2 x\right )}{2 e^3}-\frac {15 (3-\log (2)) \log \left (e^3-2 x\right )}{2 e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}-\frac {45 \log (x)}{2 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((15*E^3 - 30*x)*Log[-5/(E^3 - 2*x)] + (30*x + (-15*E^3 + 30*x)*Log[-5/(E^3 - 2*x)])*Log[x])/(4*E^3*x^2 -
8*x^3),x]

[Out]

(-15*(3 - Log[2])*Log[E^3 - 2*x])/(2*E^3) - (45*Log[x])/(2*E^3) + (15*Log[-5/(E^3 - 2*x)]*Log[x])/(4*x) + (15*
Log[E^3 - 2*x]*Log[x])/(2*E^3) + (15*PolyLog[2, (2*x)/E^3])/(2*E^3) + (15*PolyLog[2, 1 - (2*x)/E^3])/(2*E^3)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

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 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]

Rule 2316

Int[((a_.) + Log[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[((a + b*Log[-((c*d)/e)])*Log[d + e*
x])/e, x] + Dist[b, Int[Log[-((e*x)/d)]/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[-((c*d)/e), 0]

Rule 2344

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x_Symbol] :> Dist[1/d, Int[(a + b*
Log[c*x^n])^p/x, x], x] - Dist[e/d, Int[(a + b*Log[c*x^n])^p/(d + e*x), x], x] /; FreeQ[{a, b, c, d, e, n}, x]
&& IGtQ[p, 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 2392

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/(x_), x_Symbol] :> Simp[(a + b*Log[c*d])*Log[x], x] + Dist[
b, Int[Log[1 + (e*x)/d]/x, x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[c*d, 0]

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 6688

Int[u_, x_Symbol] :> With[{v = SimplifyIntegrand[u, x]}, Int[v, x] /; SimplerIntegrandQ[v, u, x]]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (15 e^3-30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )+\left (30 x+\left (-15 e^3+30 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{\left (4 e^3-8 x\right ) x^2} \, dx\\ &=\int \frac {15 \left (-\log \left (-\frac {5}{e^3-2 x}\right ) (-1+\log (x))+\frac {2 x \log (x)}{e^3-2 x}\right )}{4 x^2} \, dx\\ &=\frac {15}{4} \int \frac {-\log \left (-\frac {5}{e^3-2 x}\right ) (-1+\log (x))+\frac {2 x \log (x)}{e^3-2 x}}{x^2} \, dx\\ &=\frac {15}{4} \int \left (\frac {\log \left (-\frac {5}{e^3-2 x}\right )}{x^2}-\frac {\left (2 x-e^3 \log \left (-\frac {5}{e^3-2 x}\right )+2 x \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{x^2 \left (-e^3+2 x\right )}\right ) \, dx\\ &=\frac {15}{4} \int \frac {\log \left (-\frac {5}{e^3-2 x}\right )}{x^2} \, dx-\frac {15}{4} \int \frac {\left (2 x-e^3 \log \left (-\frac {5}{e^3-2 x}\right )+2 x \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{x^2 \left (-e^3+2 x\right )} \, dx\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15}{4} \int \frac {\left (-2 x+\left (e^3-2 x\right ) \log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{\left (e^3-2 x\right ) x^2} \, dx+\frac {15}{2} \int \frac {1}{\left (e^3-2 x\right ) x} \, dx\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15}{4} \int \frac {\left (-\frac {2 x}{e^3-2 x}+\log \left (-\frac {5}{e^3-2 x}\right )\right ) \log (x)}{x^2} \, dx+\frac {15 \int \frac {1}{x} \, dx}{2 e^3}+\frac {15 \int \frac {1}{e^3-2 x} \, dx}{e^3}\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}+\frac {15 \log (x)}{2 e^3}-\frac {15}{4} \int \left (-\frac {2 \log (x)}{\left (e^3-2 x\right ) x}+\frac {\log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{x^2}\right ) \, dx\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}+\frac {15 \log (x)}{2 e^3}-\frac {15}{4} \int \frac {\log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{x^2} \, dx+\frac {15}{2} \int \frac {\log (x)}{\left (e^3-2 x\right ) x} \, dx\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}+\frac {15 \log (x)}{2 e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}+\frac {15 \log \left (e^3-2 x\right ) \log (x)}{2 e^3}-\frac {15 \log ^2(x)}{2 e^3}+\frac {15}{4} \int \left (-\frac {\log \left (-\frac {5}{e^3-2 x}\right )}{x^2}-\frac {2 \log \left (e^3-2 x\right )}{e^3 x}+\frac {2 \log (x)}{e^3 x}\right ) \, dx+\frac {15 \int \frac {\log (x)}{x} \, dx}{2 e^3}+\frac {15 \int \frac {\log (x)}{e^3-2 x} \, dx}{e^3}\\ &=-\frac {15 \log \left (-\frac {5}{e^3-2 x}\right )}{4 x}-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}-\frac {15 (3-\log (2)) \log \left (e^3-2 x\right )}{2 e^3}+\frac {15 \log (x)}{2 e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}+\frac {15 \log \left (e^3-2 x\right ) \log (x)}{2 e^3}-\frac {15 \log ^2(x)}{4 e^3}-\frac {15}{4} \int \frac {\log \left (-\frac {5}{e^3-2 x}\right )}{x^2} \, dx-\frac {15 \int \frac {\log \left (e^3-2 x\right )}{x} \, dx}{2 e^3}+\frac {15 \int \frac {\log (x)}{x} \, dx}{2 e^3}+\frac {15 \int \frac {\log \left (\frac {2 x}{e^3}\right )}{e^3-2 x} \, dx}{e^3}\\ &=-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}-\frac {15 (3-\log (2)) \log \left (e^3-2 x\right )}{2 e^3}-\frac {15 \log (x)}{e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}+\frac {15 \log \left (e^3-2 x\right ) \log (x)}{2 e^3}+\frac {15 \text {Li}_2\left (1-\frac {2 x}{e^3}\right )}{2 e^3}-\frac {15}{2} \int \frac {1}{\left (e^3-2 x\right ) x} \, dx-\frac {15 \int \frac {\log \left (1-\frac {2 x}{e^3}\right )}{x} \, dx}{2 e^3}\\ &=-\frac {15 \log \left (e^3-2 x\right )}{2 e^3}-\frac {15 (3-\log (2)) \log \left (e^3-2 x\right )}{2 e^3}-\frac {15 \log (x)}{e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}+\frac {15 \log \left (e^3-2 x\right ) \log (x)}{2 e^3}+\frac {15 \text {Li}_2\left (\frac {2 x}{e^3}\right )}{2 e^3}+\frac {15 \text {Li}_2\left (1-\frac {2 x}{e^3}\right )}{2 e^3}-\frac {15 \int \frac {1}{x} \, dx}{2 e^3}-\frac {15 \int \frac {1}{e^3-2 x} \, dx}{e^3}\\ &=-\frac {15 (3-\log (2)) \log \left (e^3-2 x\right )}{2 e^3}-\frac {45 \log (x)}{2 e^3}+\frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x}+\frac {15 \log \left (e^3-2 x\right ) \log (x)}{2 e^3}+\frac {15 \text {Li}_2\left (\frac {2 x}{e^3}\right )}{2 e^3}+\frac {15 \text {Li}_2\left (1-\frac {2 x}{e^3}\right )}{2 e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.08, size = 21, normalized size = 0.91 \begin {gather*} \frac {15 \log \left (-\frac {5}{e^3-2 x}\right ) \log (x)}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((15*E^3 - 30*x)*Log[-5/(E^3 - 2*x)] + (30*x + (-15*E^3 + 30*x)*Log[-5/(E^3 - 2*x)])*Log[x])/(4*E^3*
x^2 - 8*x^3),x]

[Out]

(15*Log[-5/(E^3 - 2*x)]*Log[x])/(4*x)

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fricas [A]  time = 0.57, size = 20, normalized size = 0.87 \begin {gather*} \frac {15 \, \log \relax (x) \log \left (\frac {5}{2 \, x - e^{3}}\right )}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(3)+30*x)*log(-5/(exp(3)-2*x))+30*x)*log(x)+(15*exp(3)-30*x)*log(-5/(exp(3)-2*x)))/(4*x^2*
exp(3)-8*x^3),x, algorithm="fricas")

[Out]

15/4*log(x)*log(5/(2*x - e^3))/x

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giac [B]  time = 1.14, size = 69, normalized size = 3.00 \begin {gather*} \frac {15 \, {\left (\pi ^{2} \mathrm {sgn}\left (2 \, x - e^{3}\right ) \mathrm {sgn}\relax (x) - \pi ^{2} \mathrm {sgn}\left (2 \, x - e^{3}\right ) + 3 \, \pi ^{2} \mathrm {sgn}\relax (x) - 3 \, \pi ^{2} + 4 \, \log \relax (5) \log \left ({\left | x \right |}\right ) - 4 \, \log \left ({\left | 2 \, x - e^{3} \right |}\right ) \log \left ({\left | x \right |}\right )\right )}}{16 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(3)+30*x)*log(-5/(exp(3)-2*x))+30*x)*log(x)+(15*exp(3)-30*x)*log(-5/(exp(3)-2*x)))/(4*x^2*
exp(3)-8*x^3),x, algorithm="giac")

[Out]

15/16*(pi^2*sgn(2*x - e^3)*sgn(x) - pi^2*sgn(2*x - e^3) + 3*pi^2*sgn(x) - 3*pi^2 + 4*log(5)*log(abs(x)) - 4*lo
g(abs(2*x - e^3))*log(abs(x)))/x

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maple [C]  time = 0.24, size = 68, normalized size = 2.96

method	result	size
risch	\(-\frac {15 \ln \relax (x ) \ln \left ({\mathrm e}^{3}-2 x \right )}{4 x}+\frac {15 \left (-2 i \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-2 x}\right )^{2}+2 i \pi \mathrm {csgn}\left (\frac {i}{{\mathrm e}^{3}-2 x}\right )^{3}+2 i \pi +2 \ln \relax (5)\right ) \ln \relax (x )}{8 x}\)	\(68\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-15*exp(3)+30*x)*ln(-5/(exp(3)-2*x))+30*x)*ln(x)+(15*exp(3)-30*x)*ln(-5/(exp(3)-2*x)))/(4*x^2*exp(3)-8*
x^3),x,method=_RETURNVERBOSE)

[Out]

-15/4/x*ln(x)*ln(exp(3)-2*x)+15/8*(-2*I*Pi*csgn(I/(exp(3)-2*x))^2+2*I*Pi*csgn(I/(exp(3)-2*x))^3+2*I*Pi+2*ln(5)
)/x*ln(x)

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maxima [A]  time = 0.49, size = 24, normalized size = 1.04 \begin {gather*} \frac {15 \, {\left (\log \relax (5) \log \relax (x) - \log \left (2 \, x - e^{3}\right ) \log \relax (x)\right )}}{4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(3)+30*x)*log(-5/(exp(3)-2*x))+30*x)*log(x)+(15*exp(3)-30*x)*log(-5/(exp(3)-2*x)))/(4*x^2*
exp(3)-8*x^3),x, algorithm="maxima")

[Out]

15/4*(log(5)*log(x) - log(2*x - e^3)*log(x))/x

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mupad [B]  time = 3.50, size = 21, normalized size = 0.91 \begin {gather*} \frac {15\,\ln \relax (x)\,\left (\ln \relax (5)+\ln \left (\frac {1}{2\,x-{\mathrm {e}}^3}\right )\right )}{4\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(5/(2*x - exp(3)))*(30*x - 15*exp(3)) - log(x)*(30*x + log(5/(2*x - exp(3)))*(30*x - 15*exp(3))))/(4*
x^2*exp(3) - 8*x^3),x)

[Out]

(15*log(x)*(log(5) + log(1/(2*x - exp(3)))))/(4*x)

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sympy [A]  time = 0.65, size = 19, normalized size = 0.83 \begin {gather*} \frac {15 \log {\relax (x )} \log {\left (- \frac {5}{- 2 x + e^{3}} \right )}}{4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-15*exp(3)+30*x)*ln(-5/(exp(3)-2*x))+30*x)*ln(x)+(15*exp(3)-30*x)*ln(-5/(exp(3)-2*x)))/(4*x**2*ex
p(3)-8*x**3),x)

[Out]

15*log(x)*log(-5/(-2*x + exp(3)))/(4*x)

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