3.82.88 \(\int \frac {e^{\sqrt [338]{e}-x+2 \sqrt [676]{e} \log (\frac {4-5 x+x^2}{x})+\log ^2(\frac {4-5 x+x^2}{x})} (-4 x+5 x^2-x^3+\sqrt [676]{e} (-8+2 x^2)+(-8+2 x^2) \log (\frac {4-5 x+x^2}{x}))}{4 x-5 x^2+x^3} \, dx\)

Optimal. Leaf size=27 \[ e^{-x+\left (\sqrt [676]{e}+\log \left (-4+\frac {4-x}{x}+x\right )\right )^2} \]

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Rubi [B]  time = 2.47, antiderivative size = 146, normalized size of antiderivative = 5.41, number of steps used = 3, number of rules used = 3, integrand size = 109, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.028, Rules used = {1594, 2274, 2288} \begin {gather*} \frac {\left (\frac {x^2-5 x+4}{x}\right )^{2 \sqrt [676]{e}} e^{\log ^2\left (\frac {x^2-5 x+4}{x}\right )-x+\sqrt [338]{e}} \left (x^3-5 x^2+2 \left (4-x^2\right ) \log \left (\frac {x^2-5 x+4}{x}\right )+4 x\right )}{x \left (x^2-5 x+4\right ) \left (\frac {2 x \left (\frac {x^2-5 x+4}{x^2}+\frac {5-2 x}{x}\right ) \log \left (\frac {x^2-5 x+4}{x}\right )}{x^2-5 x+4}+1\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(E^(1/338) - x + 2*E^(1/676)*Log[(4 - 5*x + x^2)/x] + Log[(4 - 5*x + x^2)/x]^2)*(-4*x + 5*x^2 - x^3 + E
^(1/676)*(-8 + 2*x^2) + (-8 + 2*x^2)*Log[(4 - 5*x + x^2)/x]))/(4*x - 5*x^2 + x^3),x]

[Out]

(E^(E^(1/338) - x + Log[(4 - 5*x + x^2)/x]^2)*((4 - 5*x + x^2)/x)^(2*E^(1/676))*(4*x - 5*x^2 + x^3 + 2*(4 - x^
2)*Log[(4 - 5*x + x^2)/x]))/(x*(4 - 5*x + x^2)*(1 + (2*x*((5 - 2*x)/x + (4 - 5*x + x^2)/x^2)*Log[(4 - 5*x + x^
2)/x])/(4 - 5*x + x^2)))

Rule 1594

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

Rule 2274

Int[(u_.)*(F_)^((a_.)*(Log[z_]*(b_.) + (v_.))), x_Symbol] :> Int[u*F^(a*v)*z^(a*b*Log[F]), x] /; FreeQ[{F, a,
b}, x]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\sqrt [338]{e}-x+2 \sqrt [676]{e} \log \left (\frac {4-5 x+x^2}{x}\right )+\log ^2\left (\frac {4-5 x+x^2}{x}\right )\right ) \left (-4 x+5 x^2-x^3+\sqrt [676]{e} \left (-8+2 x^2\right )+\left (-8+2 x^2\right ) \log \left (\frac {4-5 x+x^2}{x}\right )\right )}{x \left (4-5 x+x^2\right )} \, dx\\ &=\int \frac {e^{\sqrt [338]{e}-x+\log ^2\left (\frac {4-5 x+x^2}{x}\right )} \left (\frac {4-5 x+x^2}{x}\right )^{2 \sqrt [676]{e}} \left (-4 x+5 x^2-x^3+\sqrt [676]{e} \left (-8+2 x^2\right )+\left (-8+2 x^2\right ) \log \left (\frac {4-5 x+x^2}{x}\right )\right )}{x \left (4-5 x+x^2\right )} \, dx\\ &=\frac {e^{\sqrt [338]{e}-x+\log ^2\left (\frac {4-5 x+x^2}{x}\right )} \left (\frac {4-5 x+x^2}{x}\right )^{2 \sqrt [676]{e}} \left (4 x-5 x^2+x^3+2 \left (4-x^2\right ) \log \left (\frac {4-5 x+x^2}{x}\right )\right )}{x \left (4-5 x+x^2\right ) \left (1+\frac {2 x \left (\frac {5-2 x}{x}+\frac {4-5 x+x^2}{x^2}\right ) \log \left (\frac {4-5 x+x^2}{x}\right )}{4-5 x+x^2}\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 39, normalized size = 1.44 \begin {gather*} e^{\sqrt [338]{e}-x+\log ^2\left (-5+\frac {4}{x}+x\right )} \left (-5+\frac {4}{x}+x\right )^{2 \sqrt [676]{e}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(E^(1/338) - x + 2*E^(1/676)*Log[(4 - 5*x + x^2)/x] + Log[(4 - 5*x + x^2)/x]^2)*(-4*x + 5*x^2 - x
^3 + E^(1/676)*(-8 + 2*x^2) + (-8 + 2*x^2)*Log[(4 - 5*x + x^2)/x]))/(4*x - 5*x^2 + x^3),x]

[Out]

E^(E^(1/338) - x + Log[-5 + 4/x + x]^2)*(-5 + 4/x + x)^(2*E^(1/676))

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fricas [A]  time = 0.59, size = 39, normalized size = 1.44 \begin {gather*} e^{\left (2 \, e^{\frac {1}{676}} \log \left (\frac {x^{2} - 5 \, x + 4}{x}\right ) + \log \left (\frac {x^{2} - 5 \, x + 4}{x}\right )^{2} - x + e^{\frac {1}{338}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-8)*log((x^2-5*x+4)/x)+(2*x^2-8)*exp(1/676)-x^3+5*x^2-4*x)*exp(log((x^2-5*x+4)/x)^2+2*exp(1/6
76)*log((x^2-5*x+4)/x)+exp(1/676)^2-x)/(x^3-5*x^2+4*x),x, algorithm="fricas")

[Out]

e^(2*e^(1/676)*log((x^2 - 5*x + 4)/x) + log((x^2 - 5*x + 4)/x)^2 - x + e^(1/338))

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giac [A]  time = 0.55, size = 31, normalized size = 1.15 \begin {gather*} e^{\left (2 \, e^{\frac {1}{676}} \log \left (x + \frac {4}{x} - 5\right ) + \log \left (x + \frac {4}{x} - 5\right )^{2} - x + e^{\frac {1}{338}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-8)*log((x^2-5*x+4)/x)+(2*x^2-8)*exp(1/676)-x^3+5*x^2-4*x)*exp(log((x^2-5*x+4)/x)^2+2*exp(1/6
76)*log((x^2-5*x+4)/x)+exp(1/676)^2-x)/(x^3-5*x^2+4*x),x, algorithm="giac")

[Out]

e^(2*e^(1/676)*log(x + 4/x - 5) + log(x + 4/x - 5)^2 - x + e^(1/338))

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maple [A]  time = 0.14, size = 41, normalized size = 1.52

method	result	size
risch	\(\left (\frac {x^{2}-5 x +4}{x}\right )^{2 \,{\mathrm e}^{\frac {1}{676}}} {\mathrm e}^{\ln \left (\frac {x^{2}-5 x +4}{x}\right )^{2}+{\mathrm e}^{\frac {1}{338}}-x}\)	\(41\)
norman	\({\mathrm e}^{\ln \left (\frac {x^{2}-5 x +4}{x}\right )^{2}+2 \,{\mathrm e}^{\frac {1}{676}} \ln \left (\frac {x^{2}-5 x +4}{x}\right )+{\mathrm e}^{\frac {1}{338}}-x}\)	\(42\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^2-8)*ln((x^2-5*x+4)/x)+(2*x^2-8)*exp(1/676)-x^3+5*x^2-4*x)*exp(ln((x^2-5*x+4)/x)^2+2*exp(1/676)*ln((
x^2-5*x+4)/x)+exp(1/676)^2-x)/(x^3-5*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

((x^2-5*x+4)/x)^(2*exp(1/676))*exp(ln((x^2-5*x+4)/x)^2+exp(1/338)-x)

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maxima [B]  time = 0.68, size = 71, normalized size = 2.63 \begin {gather*} e^{\left (2 \, e^{\frac {1}{676}} \log \left (x - 1\right ) + \log \left (x - 1\right )^{2} + 2 \, e^{\frac {1}{676}} \log \left (x - 4\right ) + 2 \, \log \left (x - 1\right ) \log \left (x - 4\right ) + \log \left (x - 4\right )^{2} - 2 \, e^{\frac {1}{676}} \log \relax (x) - 2 \, \log \left (x - 1\right ) \log \relax (x) - 2 \, \log \left (x - 4\right ) \log \relax (x) + \log \relax (x)^{2} - x + e^{\frac {1}{338}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^2-8)*log((x^2-5*x+4)/x)+(2*x^2-8)*exp(1/676)-x^3+5*x^2-4*x)*exp(log((x^2-5*x+4)/x)^2+2*exp(1/6
76)*log((x^2-5*x+4)/x)+exp(1/676)^2-x)/(x^3-5*x^2+4*x),x, algorithm="maxima")

[Out]

e^(2*e^(1/676)*log(x - 1) + log(x - 1)^2 + 2*e^(1/676)*log(x - 4) + 2*log(x - 1)*log(x - 4) + log(x - 4)^2 - 2
*e^(1/676)*log(x) - 2*log(x - 1)*log(x) - 2*log(x - 4)*log(x) + log(x)^2 - x + e^(1/338))

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mupad [B]  time = 5.87, size = 37, normalized size = 1.37 \begin {gather*} {\mathrm {e}}^{-x}\,{\mathrm {e}}^{{\ln \left (\frac {x^2-5\,x+4}{x}\right )}^2}\,{\mathrm {e}}^{{\mathrm {e}}^{1/338}}\,{\left (x+\frac {4}{x}-5\right )}^{2\,{\mathrm {e}}^{1/676}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(1/338) - x + log((x^2 - 5*x + 4)/x)^2 + 2*exp(1/676)*log((x^2 - 5*x + 4)/x))*(exp(1/676)*(2*x^2 -
8) - 4*x + log((x^2 - 5*x + 4)/x)*(2*x^2 - 8) + 5*x^2 - x^3))/(4*x - 5*x^2 + x^3),x)

[Out]

exp(-x)*exp(log((x^2 - 5*x + 4)/x)^2)*exp(exp(1/338))*(x + 4/x - 5)^(2*exp(1/676))

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sympy [F(-1)]  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(((2*x**2-8)*ln((x**2-5*x+4)/x)+(2*x**2-8)*exp(1/676)-x**3+5*x**2-4*x)*exp(ln((x**2-5*x+4)/x)**2+2*ex
p(1/676)*ln((x**2-5*x+4)/x)+exp(1/676)**2-x)/(x**3-5*x**2+4*x),x)

[Out]

Timed out

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