3.24.47 \(\int \frac {-50 x+e^{e^x} (10 x+e^x (-2 x+5 x^2))+(-10+e^{e^x} (2-5 x)+25 x) \log (20-100 x+125 x^2+e^{e^x} (-4+20 x-25 x^2))}{10 x^2-25 x^3+e^{e^x} (-2 x^2+5 x^3)} \, dx\)

Optimal. Leaf size=24 \[ 1+\frac {\log \left (\left (5-e^{e^x}\right ) (2-5 x)^2\right )}{x} \]

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Rubi [A]  time = 1.46, antiderivative size = 22, normalized size of antiderivative = 0.92, number of steps used = 15, number of rules used = 7, integrand size = 102, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.069, Rules used = {6688, 14, 6742, 36, 29, 31, 2551} \begin {gather*} \frac {\log \left (\left (5-e^{e^x}\right ) (2-5 x)^2\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-50*x + E^E^x*(10*x + E^x*(-2*x + 5*x^2)) + (-10 + E^E^x*(2 - 5*x) + 25*x)*Log[20 - 100*x + 125*x^2 + E^E
^x*(-4 + 20*x - 25*x^2)])/(10*x^2 - 25*x^3 + E^E^x*(-2*x^2 + 5*x^3)),x]

[Out]

Log[(5 - E^E^x)*(2 - 5*x)^2]/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 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 {\frac {x \left (-50+10 e^{e^x}+e^{e^x+x} (-2+5 x)\right )}{\left (-5+e^{e^x}\right ) (-2+5 x)}-\log \left (-\left (\left (-5+e^{e^x}\right ) (-2+5 x)^2\right )\right )}{x^2} \, dx\\ &=\int \left (\frac {e^{e^x+x}}{\left (-5+e^{e^x}\right ) x}+\frac {10 x+2 \log \left (-\left (\left (-5+e^{e^x}\right ) (-2+5 x)^2\right )\right )-5 x \log \left (-\left (\left (-5+e^{e^x}\right ) (-2+5 x)^2\right )\right )}{x^2 (-2+5 x)}\right ) \, dx\\ &=\int \frac {e^{e^x+x}}{\left (-5+e^{e^x}\right ) x} \, dx+\int \frac {10 x+2 \log \left (-\left (\left (-5+e^{e^x}\right ) (-2+5 x)^2\right )\right )-5 x \log \left (-\left (\left (-5+e^{e^x}\right ) (-2+5 x)^2\right )\right )}{x^2 (-2+5 x)} \, dx\\ &=\int \frac {e^{e^x+x}}{\left (-5+e^{e^x}\right ) x} \, dx+\int \frac {-10 x-(2-5 x) \log \left (-\left (\left (-5+e^{e^x}\right ) (-2+5 x)^2\right )\right )}{(2-5 x) x^2} \, dx\\ &=\int \frac {e^{e^x+x}}{\left (-5+e^{e^x}\right ) x} \, dx+\int \left (\frac {10}{x (-2+5 x)}-\frac {\log \left (-\left (\left (-5+e^{e^x}\right ) (-2+5 x)^2\right )\right )}{x^2}\right ) \, dx\\ &=10 \int \frac {1}{x (-2+5 x)} \, dx+\int \frac {e^{e^x+x}}{\left (-5+e^{e^x}\right ) x} \, dx-\int \frac {\log \left (-\left (\left (-5+e^{e^x}\right ) (-2+5 x)^2\right )\right )}{x^2} \, dx\\ &=\frac {\log \left (\left (5-e^{e^x}\right ) (2-5 x)^2\right )}{x}-5 \int \frac {1}{x} \, dx+25 \int \frac {1}{-2+5 x} \, dx+\int \frac {e^{e^x+x}}{\left (-5+e^{e^x}\right ) x} \, dx-\int \frac {-50+10 e^{e^x}+e^{e^x+x} (-2+5 x)}{\left (5-e^{e^x}\right ) (2-5 x) x} \, dx\\ &=5 \log (2-5 x)+\frac {\log \left (\left (5-e^{e^x}\right ) (2-5 x)^2\right )}{x}-5 \log (x)+\int \frac {e^{e^x+x}}{\left (-5+e^{e^x}\right ) x} \, dx-\int \left (\frac {e^{e^x+x}}{\left (-5+e^{e^x}\right ) x}+\frac {10}{x (-2+5 x)}\right ) \, dx\\ &=5 \log (2-5 x)+\frac {\log \left (\left (5-e^{e^x}\right ) (2-5 x)^2\right )}{x}-5 \log (x)-10 \int \frac {1}{x (-2+5 x)} \, dx\\ &=5 \log (2-5 x)+\frac {\log \left (\left (5-e^{e^x}\right ) (2-5 x)^2\right )}{x}-5 \log (x)+5 \int \frac {1}{x} \, dx-25 \int \frac {1}{-2+5 x} \, dx\\ &=\frac {\log \left (\left (5-e^{e^x}\right ) (2-5 x)^2\right )}{x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.31, size = 21, normalized size = 0.88 \begin {gather*} \frac {\log \left (-\left (\left (-5+e^{e^x}\right ) (-2+5 x)^2\right )\right )}{x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-50*x + E^E^x*(10*x + E^x*(-2*x + 5*x^2)) + (-10 + E^E^x*(2 - 5*x) + 25*x)*Log[20 - 100*x + 125*x^2
+ E^E^x*(-4 + 20*x - 25*x^2)])/(10*x^2 - 25*x^3 + E^E^x*(-2*x^2 + 5*x^3)),x]

[Out]

Log[-((-5 + E^E^x)*(-2 + 5*x)^2)]/x

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fricas [A]  time = 0.47, size = 30, normalized size = 1.25 \begin {gather*} \frac {\log \left (125 \, x^{2} - {\left (25 \, x^{2} - 20 \, x + 4\right )} e^{\left (e^{x}\right )} - 100 \, x + 20\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x+2)*exp(exp(x))+25*x-10)*log((-25*x^2+20*x-4)*exp(exp(x))+125*x^2-100*x+20)+((5*x^2-2*x)*exp(
x)+10*x)*exp(exp(x))-50*x)/((5*x^3-2*x^2)*exp(exp(x))-25*x^3+10*x^2),x, algorithm="fricas")

[Out]

log(125*x^2 - (25*x^2 - 20*x + 4)*e^(e^x) - 100*x + 20)/x

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giac [B]  time = 0.47, size = 53, normalized size = 2.21 \begin {gather*} \frac {\log \left (-{\left (25 \, x^{2} e^{\left (x + e^{x}\right )} - 125 \, x^{2} e^{x} - 20 \, x e^{\left (x + e^{x}\right )} + 100 \, x e^{x} + 4 \, e^{\left (x + e^{x}\right )} - 20 \, e^{x}\right )} e^{\left (-x\right )}\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x+2)*exp(exp(x))+25*x-10)*log((-25*x^2+20*x-4)*exp(exp(x))+125*x^2-100*x+20)+((5*x^2-2*x)*exp(
x)+10*x)*exp(exp(x))-50*x)/((5*x^3-2*x^2)*exp(exp(x))-25*x^3+10*x^2),x, algorithm="giac")

[Out]

log(-(25*x^2*e^(x + e^x) - 125*x^2*e^x - 20*x*e^(x + e^x) + 100*x*e^x + 4*e^(x + e^x) - 20*e^x)*e^(-x))/x

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maple [C]  time = 0.19, size = 221, normalized size = 9.21

method	result	size
risch	\(\frac {\ln \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )}{x}+\frac {-i \pi \mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )\right )^{2} \mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )\right ) \mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )^{2}\right )^{3}-i \pi \,\mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )^{2}\right ) \mathrm {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right ) \mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )+i \pi \,\mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )^{2}\right ) \mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )^{2}+i \pi \,\mathrm {csgn}\left (i \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right ) \mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )^{2}+i \pi \mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )^{3}-2 i \pi \mathrm {csgn}\left (i \left (x -\frac {2}{5}\right )^{2} \left ({\mathrm e}^{{\mathrm e}^{x}}-5\right )\right )^{2}+2 i \pi +4 \ln \left (x -\frac {2}{5}\right )}{2 x}\)	\(221\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x+2)*exp(exp(x))+25*x-10)*ln((-25*x^2+20*x-4)*exp(exp(x))+125*x^2-100*x+20)+((5*x^2-2*x)*exp(x)+10*x
)*exp(exp(x))-50*x)/((5*x^3-2*x^2)*exp(exp(x))-25*x^3+10*x^2),x,method=_RETURNVERBOSE)

[Out]

1/x*ln(exp(exp(x))-5)+1/2*(-I*Pi*csgn(I*(x-2/5))^2*csgn(I*(x-2/5)^2)+2*I*Pi*csgn(I*(x-2/5))*csgn(I*(x-2/5)^2)^
2-I*Pi*csgn(I*(x-2/5)^2)^3-I*Pi*csgn(I*(x-2/5)^2)*csgn(I*(exp(exp(x))-5))*csgn(I*(x-2/5)^2*(exp(exp(x))-5))+I*
Pi*csgn(I*(x-2/5)^2)*csgn(I*(x-2/5)^2*(exp(exp(x))-5))^2+I*Pi*csgn(I*(exp(exp(x))-5))*csgn(I*(x-2/5)^2*(exp(ex
p(x))-5))^2+I*Pi*csgn(I*(x-2/5)^2*(exp(exp(x))-5))^3-2*I*Pi*csgn(I*(x-2/5)^2*(exp(exp(x))-5))^2+2*I*Pi+4*ln(x-
2/5))/x

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maxima [A]  time = 0.50, size = 21, normalized size = 0.88 \begin {gather*} \frac {2 \, \log \left (5 \, x - 2\right ) + \log \left (-e^{\left (e^{x}\right )} + 5\right )}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x+2)*exp(exp(x))+25*x-10)*log((-25*x^2+20*x-4)*exp(exp(x))+125*x^2-100*x+20)+((5*x^2-2*x)*exp(
x)+10*x)*exp(exp(x))-50*x)/((5*x^3-2*x^2)*exp(exp(x))-25*x^3+10*x^2),x, algorithm="maxima")

[Out]

(2*log(5*x - 2) + log(-e^(e^x) + 5))/x

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mupad [B]  time = 0.30, size = 47, normalized size = 1.96 \begin {gather*} -\frac {\ln \left (125\,x^2-{\mathrm {e}}^{{\mathrm {e}}^x}\,\left (25\,x^2-20\,x+4\right )-100\,x+20\right )\,\left (2\,x-5\,x^2\right )}{x^2\,\left (5\,x-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((50*x + log(125*x^2 - exp(exp(x))*(25*x^2 - 20*x + 4) - 100*x + 20)*(exp(exp(x))*(5*x - 2) - 25*x + 10) -
exp(exp(x))*(10*x - exp(x)*(2*x - 5*x^2)))/(exp(exp(x))*(2*x^2 - 5*x^3) - 10*x^2 + 25*x^3),x)

[Out]

-(log(125*x^2 - exp(exp(x))*(25*x^2 - 20*x + 4) - 100*x + 20)*(2*x - 5*x^2))/(x^2*(5*x - 2))

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sympy [A]  time = 0.70, size = 27, normalized size = 1.12 \begin {gather*} \frac {\log {\left (125 x^{2} - 100 x + \left (- 25 x^{2} + 20 x - 4\right ) e^{e^{x}} + 20 \right )}}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x+2)*exp(exp(x))+25*x-10)*ln((-25*x**2+20*x-4)*exp(exp(x))+125*x**2-100*x+20)+((5*x**2-2*x)*ex
p(x)+10*x)*exp(exp(x))-50*x)/((5*x**3-2*x**2)*exp(exp(x))-25*x**3+10*x**2),x)

[Out]

log(125*x**2 - 100*x + (-25*x**2 + 20*x - 4)*exp(exp(x)) + 20)/x

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