3.3.94 \(\int \frac {2 e^x x-4 x^3+(-4 x+4 x^2) \log (16)+(-4 x^2+e^x (2 x-2 \log (16))+4 x \log (16)) \log (\frac {e^{-x} (e^x-2 x)}{-2 x+2 \log (16)})}{2 x^4-2 x^3 \log (16)+e^x (-x^3+x^2 \log (16))} \, dx\)

Optimal. Leaf size=31 \[ \frac {2 \log \left (\frac {e^{-x} \left (\frac {e^x}{2}-x\right )}{-x+\log (16)}\right )}{x} \]

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Rubi [B]  time = 2.70, antiderivative size = 75, normalized size of antiderivative = 2.42, number of steps used = 20, number of rules used = 9, integrand size = 104, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {6741, 6742, 6688, 14, 36, 29, 31, 2551, 12} \begin {gather*} -\frac {4 \log (x)}{\log (256)}+\frac {2 \log (x)}{\log (16)}-\frac {2 \log (x-\log (16))}{\log (16)}+\frac {2 \log \left (-\frac {e^{-x} \left (e^x-2 x\right )}{2 x-\log (256)}\right )}{x}+\frac {4 \log (2 x-\log (256))}{\log (256)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*E^x*x - 4*x^3 + (-4*x + 4*x^2)*Log[16] + (-4*x^2 + E^x*(2*x - 2*Log[16]) + 4*x*Log[16])*Log[(E^x - 2*x)
/(E^x*(-2*x + 2*Log[16]))])/(2*x^4 - 2*x^3*Log[16] + E^x*(-x^3 + x^2*Log[16])),x]

[Out]

(2*Log[x])/Log[16] - (4*Log[x])/Log[256] - (2*Log[x - Log[16]])/Log[16] + (2*Log[-((E^x - 2*x)/(E^x*(2*x - Log
[256])))])/x + (4*Log[2*x - Log[256]])/Log[256]

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 2551

Int[Log[u_]*((a_.) + (b_.)*(x_))^(m_.), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Log[u])/(b*(m + 1)), x] - Dist[1/
(b*(m + 1)), Int[SimplifyIntegrand[((a + b*x)^(m + 1)*D[u, x])/u, x], x], x] /; FreeQ[{a, b, m}, x] && Inverse
FunctionFreeQ[u, x] && NeQ[m, -1]

Rule 6688

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

Rule 6741

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

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 {-2 e^x x+4 x^3-\left (-4 x+4 x^2\right ) \log (16)-\left (-4 x^2+e^x (2 x-2 \log (16))+4 x \log (16)\right ) \log \left (\frac {e^{-x} \left (e^x-2 x\right )}{-2 x+2 \log (16)}\right )}{\left (e^x-2 x\right ) x^2 (x-\log (16))} \, dx\\ &=\int \left (\frac {4 (-1+x)}{\left (e^x-2 x\right ) x}-\frac {2 \left (x+x \log \left (\frac {e^{-x} \left (e^x-2 x\right )}{-2 x+\log (256)}\right )-\log (16) \log \left (\frac {e^{-x} \left (e^x-2 x\right )}{-2 x+\log (256)}\right )\right )}{x^2 (x-\log (16))}\right ) \, dx\\ &=-\left (2 \int \frac {x+x \log \left (\frac {e^{-x} \left (e^x-2 x\right )}{-2 x+\log (256)}\right )-\log (16) \log \left (\frac {e^{-x} \left (e^x-2 x\right )}{-2 x+\log (256)}\right )}{x^2 (x-\log (16))} \, dx\right )+4 \int \frac {-1+x}{\left (e^x-2 x\right ) x} \, dx\\ &=-\left (2 \int \frac {\frac {x}{x-\log (16)}+\log \left (\frac {e^{-x} \left (e^x-2 x\right )}{-2 x+\log (256)}\right )}{x^2} \, dx\right )+4 \int \left (\frac {1}{e^x-2 x}-\frac {1}{\left (e^x-2 x\right ) x}\right ) \, dx\\ &=-\left (2 \int \left (\frac {1}{x (x-\log (16))}+\frac {\log \left (\frac {e^{-x} \left (e^x-2 x\right )}{-2 x+\log (256)}\right )}{x^2}\right ) \, dx\right )+4 \int \frac {1}{e^x-2 x} \, dx-4 \int \frac {1}{\left (e^x-2 x\right ) x} \, dx\\ &=-\left (2 \int \frac {1}{x (x-\log (16))} \, dx\right )-2 \int \frac {\log \left (\frac {e^{-x} \left (e^x-2 x\right )}{-2 x+\log (256)}\right )}{x^2} \, dx+4 \int \frac {1}{e^x-2 x} \, dx-4 \int \frac {1}{\left (e^x-2 x\right ) x} \, dx\\ &=\frac {2 \log \left (-\frac {e^{-x} \left (e^x-2 x\right )}{2 x-\log (256)}\right )}{x}-2 \int \frac {2 \left (-e^x+2 x^2+\log (256)-x \log (256)\right )}{\left (e^x-2 x\right ) x (2 x-\log (256))} \, dx+4 \int \frac {1}{e^x-2 x} \, dx-4 \int \frac {1}{\left (e^x-2 x\right ) x} \, dx+\frac {2 \int \frac {1}{x} \, dx}{\log (16)}-\frac {2 \int \frac {1}{x-\log (16)} \, dx}{\log (16)}\\ &=\frac {2 \log (x)}{\log (16)}-\frac {2 \log (x-\log (16))}{\log (16)}+\frac {2 \log \left (-\frac {e^{-x} \left (e^x-2 x\right )}{2 x-\log (256)}\right )}{x}+4 \int \frac {1}{e^x-2 x} \, dx-4 \int \frac {1}{\left (e^x-2 x\right ) x} \, dx-4 \int \frac {-e^x+2 x^2+\log (256)-x \log (256)}{\left (e^x-2 x\right ) x (2 x-\log (256))} \, dx\\ &=\frac {2 \log (x)}{\log (16)}-\frac {2 \log (x-\log (16))}{\log (16)}+\frac {2 \log \left (-\frac {e^{-x} \left (e^x-2 x\right )}{2 x-\log (256)}\right )}{x}+4 \int \frac {1}{e^x-2 x} \, dx-4 \int \frac {1}{\left (e^x-2 x\right ) x} \, dx-4 \int \left (\frac {-1+x}{\left (e^x-2 x\right ) x}-\frac {1}{x (2 x-\log (256))}\right ) \, dx\\ &=\frac {2 \log (x)}{\log (16)}-\frac {2 \log (x-\log (16))}{\log (16)}+\frac {2 \log \left (-\frac {e^{-x} \left (e^x-2 x\right )}{2 x-\log (256)}\right )}{x}+4 \int \frac {1}{e^x-2 x} \, dx-4 \int \frac {1}{\left (e^x-2 x\right ) x} \, dx-4 \int \frac {-1+x}{\left (e^x-2 x\right ) x} \, dx+4 \int \frac {1}{x (2 x-\log (256))} \, dx\\ &=\frac {2 \log (x)}{\log (16)}-\frac {2 \log (x-\log (16))}{\log (16)}+\frac {2 \log \left (-\frac {e^{-x} \left (e^x-2 x\right )}{2 x-\log (256)}\right )}{x}-4 \int \left (\frac {1}{e^x-2 x}-\frac {1}{\left (e^x-2 x\right ) x}\right ) \, dx+4 \int \frac {1}{e^x-2 x} \, dx-4 \int \frac {1}{\left (e^x-2 x\right ) x} \, dx-\frac {4 \int \frac {1}{x} \, dx}{\log (256)}+\frac {8 \int \frac {1}{2 x-\log (256)} \, dx}{\log (256)}\\ &=\frac {2 \log (x)}{\log (16)}-\frac {4 \log (x)}{\log (256)}-\frac {2 \log (x-\log (16))}{\log (16)}+\frac {2 \log \left (-\frac {e^{-x} \left (e^x-2 x\right )}{2 x-\log (256)}\right )}{x}+\frac {4 \log (2 x-\log (256))}{\log (256)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.36, size = 75, normalized size = 2.42 \begin {gather*} \frac {2 \log (x)}{\log (16)}-\frac {4 \log (x)}{\log (256)}-\frac {2 \log (x-\log (16))}{\log (16)}+\frac {2 \log \left (-\frac {e^{-x} \left (e^x-2 x\right )}{2 x-\log (256)}\right )}{x}+\frac {4 \log (2 x-\log (256))}{\log (256)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*E^x*x - 4*x^3 + (-4*x + 4*x^2)*Log[16] + (-4*x^2 + E^x*(2*x - 2*Log[16]) + 4*x*Log[16])*Log[(E^x
- 2*x)/(E^x*(-2*x + 2*Log[16]))])/(2*x^4 - 2*x^3*Log[16] + E^x*(-x^3 + x^2*Log[16])),x]

[Out]

(2*Log[x])/Log[16] - (4*Log[x])/Log[256] - (2*Log[x - Log[16]])/Log[16] + (2*Log[-((E^x - 2*x)/(E^x*(2*x - Log
[256])))])/x + (4*Log[2*x - Log[256]])/Log[256]

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fricas [A]  time = 0.60, size = 28, normalized size = 0.90 \begin {gather*} \frac {2 \, \log \left (\frac {{\left (2 \, x - e^{x}\right )} e^{\left (-x\right )}}{2 \, {\left (x - 4 \, \log \relax (2)\right )}}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*log(2)+2*x)*exp(x)+16*x*log(2)-4*x^2)*log((exp(x)-2*x)/(8*log(2)-2*x)/exp(x))+2*exp(x)*x+4*(4*
x^2-4*x)*log(2)-4*x^3)/((4*x^2*log(2)-x^3)*exp(x)-8*x^3*log(2)+2*x^4),x, algorithm="fricas")

[Out]

2*log(1/2*(2*x - e^x)*e^(-x)/(x - 4*log(2)))/x

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giac [A]  time = 0.43, size = 31, normalized size = 1.00 \begin {gather*} -\frac {2 \, {\left (\log \relax (2) - \log \left ({\left (2 \, x - e^{x}\right )} e^{\left (-x\right )}\right ) + \log \left (x - 4 \, \log \relax (2)\right )\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*log(2)+2*x)*exp(x)+16*x*log(2)-4*x^2)*log((exp(x)-2*x)/(8*log(2)-2*x)/exp(x))+2*exp(x)*x+4*(4*
x^2-4*x)*log(2)-4*x^3)/((4*x^2*log(2)-x^3)*exp(x)-8*x^3*log(2)+2*x^4),x, algorithm="giac")

[Out]

-2*(log(2) - log((2*x - e^x)*e^(-x)) + log(x - 4*log(2)))/x

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maple [A]  time = 0.50, size = 28, normalized size = 0.90

method	result	size
norman	\(\frac {2 \ln \left (\frac {\left ({\mathrm e}^{x}-2 x \right ) {\mathrm e}^{-x}}{8 \ln \relax (2)-2 x}\right )}{x}\)	\(28\)
risch	\(-\frac {2 \ln \left ({\mathrm e}^{x}\right )}{x}-\frac {-i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x} \left (\frac {{\mathrm e}^{x}}{2}-x \right )}{\ln \relax (2)-\frac {x}{4}}\right )^{2}-i \pi \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{x}}{2}-x \right )}{\ln \relax (2)-\frac {x}{4}}\right )^{3}-i \pi \,\mathrm {csgn}\left (\frac {i}{\ln \relax (2)-\frac {x}{4}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{x}}{2}-x \right )}{\ln \relax (2)-\frac {x}{4}}\right )^{2}+2 i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x} \left (\frac {{\mathrm e}^{x}}{2}-x \right )}{\ln \relax (2)-\frac {x}{4}}\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (\frac {{\mathrm e}^{x}}{2}-x \right )\right ) \mathrm {csgn}\left (\frac {i}{\ln \relax (2)-\frac {x}{4}}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{x}}{2}-x \right )}{\ln \relax (2)-\frac {x}{4}}\right )+i \pi \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x} \left (\frac {{\mathrm e}^{x}}{2}-x \right )}{\ln \relax (2)-\frac {x}{4}}\right )^{3}+i \pi \,\mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{x}}{2}-x \right )}{\ln \relax (2)-\frac {x}{4}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x} \left (\frac {{\mathrm e}^{x}}{2}-x \right )}{\ln \relax (2)-\frac {x}{4}}\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left (\frac {{\mathrm e}^{x}}{2}-x \right )\right ) \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{x}}{2}-x \right )}{\ln \relax (2)-\frac {x}{4}}\right )^{2}-2 i \pi +i \pi \,\mathrm {csgn}\left (i {\mathrm e}^{-x}\right ) \mathrm {csgn}\left (\frac {i \left (\frac {{\mathrm e}^{x}}{2}-x \right )}{\ln \relax (2)-\frac {x}{4}}\right ) \mathrm {csgn}\left (\frac {i {\mathrm e}^{-x} \left (\frac {{\mathrm e}^{x}}{2}-x \right )}{\ln \relax (2)-\frac {x}{4}}\right )+4 \ln \relax (2)+2 \ln \left (\ln \relax (2)-\frac {x}{4}\right )-2 \ln \left (x -\frac {{\mathrm e}^{x}}{2}\right )}{x}\)	\(396\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-8*ln(2)+2*x)*exp(x)+16*x*ln(2)-4*x^2)*ln((exp(x)-2*x)/(8*ln(2)-2*x)/exp(x))+2*exp(x)*x+4*(4*x^2-4*x)*l
n(2)-4*x^3)/((4*x^2*ln(2)-x^3)*exp(x)-8*x^3*ln(2)+2*x^4),x,method=_RETURNVERBOSE)

[Out]

2*ln((exp(x)-2*x)/(8*ln(2)-2*x)/exp(x))/x

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maxima [A]  time = 0.61, size = 26, normalized size = 0.84 \begin {gather*} -\frac {2 \, {\left (\log \relax (2) - \log \left (2 \, x - e^{x}\right ) + \log \left (x - 4 \, \log \relax (2)\right )\right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*log(2)+2*x)*exp(x)+16*x*log(2)-4*x^2)*log((exp(x)-2*x)/(8*log(2)-2*x)/exp(x))+2*exp(x)*x+4*(4*
x^2-4*x)*log(2)-4*x^3)/((4*x^2*log(2)-x^3)*exp(x)-8*x^3*log(2)+2*x^4),x, algorithm="maxima")

[Out]

-2*(log(2) - log(2*x - e^x) + log(x - 4*log(2)))/x

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} -\int \frac {4\,\ln \relax (2)\,\left (4\,x-4\,x^2\right )-\ln \left (\frac {{\mathrm {e}}^{-x}\,\left (2\,x-{\mathrm {e}}^x\right )}{2\,x-8\,\ln \relax (2)}\right )\,\left ({\mathrm {e}}^x\,\left (2\,x-8\,\ln \relax (2)\right )+16\,x\,\ln \relax (2)-4\,x^2\right )-2\,x\,{\mathrm {e}}^x+4\,x^3}{{\mathrm {e}}^x\,\left (4\,x^2\,\ln \relax (2)-x^3\right )-8\,x^3\,\ln \relax (2)+2\,x^4} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(4*log(2)*(4*x - 4*x^2) - log((exp(-x)*(2*x - exp(x)))/(2*x - 8*log(2)))*(exp(x)*(2*x - 8*log(2)) + 16*x*
log(2) - 4*x^2) - 2*x*exp(x) + 4*x^3)/(exp(x)*(4*x^2*log(2) - x^3) - 8*x^3*log(2) + 2*x^4),x)

[Out]

-int((4*log(2)*(4*x - 4*x^2) - log((exp(-x)*(2*x - exp(x)))/(2*x - 8*log(2)))*(exp(x)*(2*x - 8*log(2)) + 16*x*
log(2) - 4*x^2) - 2*x*exp(x) + 4*x^3)/(exp(x)*(4*x^2*log(2) - x^3) - 8*x^3*log(2) + 2*x^4), x)

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sympy [A]  time = 0.52, size = 22, normalized size = 0.71 \begin {gather*} \frac {2 \log {\left (\frac {\left (- 2 x + e^{x}\right ) e^{- x}}{- 2 x + 8 \log {\relax (2 )}} \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-8*ln(2)+2*x)*exp(x)+16*x*ln(2)-4*x**2)*ln((exp(x)-2*x)/(8*ln(2)-2*x)/exp(x))+2*exp(x)*x+4*(4*x**
2-4*x)*ln(2)-4*x**3)/((4*x**2*ln(2)-x**3)*exp(x)-8*x**3*ln(2)+2*x**4),x)

[Out]

2*log((-2*x + exp(x))*exp(-x)/(-2*x + 8*log(2)))/x

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