3.18.53 \(\int \frac {2 x^2+e^4 (5+5 x-x^2)}{-2 x^2+e^4 (-5 x+x^2)} \, dx\)

Optimal. Leaf size=29 \[ -\log ^2(4)+\log \left (3 e^{-x} \left (\frac {1}{e^4}-\frac {-5+x}{2 x}\right )\right ) \]

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Rubi [A]  time = 0.07, antiderivative size = 24, normalized size of antiderivative = 0.83, number of steps used = 4, number of rules used = 3, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.075, Rules used = {1984, 1593, 893} \begin {gather*} -x-\log (x)+\log \left (\left (2-e^4\right ) x+5 e^4\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2*x^2 + E^4*(5 + 5*x - x^2))/(-2*x^2 + E^4*(-5*x + x^2)),x]

[Out]

-x - Log[x] + Log[5*E^4 + (2 - E^4)*x]

Rule 893

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Int[ExpandIntegrand[(d + e*x)^m*(f + g*x)^n*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] &
& NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IntegerQ[p] && ((EqQ[p, 1] && I
ntegersQ[m, n]) || (ILtQ[m, 0] && ILtQ[n, 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 1984

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
QuadraticQ[{u, v}, x] &&  !QuadraticMatchQ[{u, v}, x]

Rubi steps

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

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Mathematica [A]  time = 0.01, size = 23, normalized size = 0.79 \begin {gather*} -x-\log (x)+\log \left (-5 e^4-2 x+e^4 x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x^2 + E^4*(5 + 5*x - x^2))/(-2*x^2 + E^4*(-5*x + x^2)),x]

[Out]

-x - Log[x] + Log[-5*E^4 - 2*x + E^4*x]

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fricas [A]  time = 0.65, size = 19, normalized size = 0.66 \begin {gather*} -x + \log \left ({\left (x - 5\right )} e^{4} - 2 \, x\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+5*x+5)*exp(4)+2*x^2)/((x^2-5*x)*exp(4)-2*x^2),x, algorithm="fricas")

[Out]

-x + log((x - 5)*e^4 - 2*x) - log(x)

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giac [A]  time = 0.22, size = 36, normalized size = 1.24 \begin {gather*} -\frac {x e^{4} - 2 \, x}{e^{4} - 2} + \log \left ({\left | x e^{4} - 2 \, x - 5 \, e^{4} \right |}\right ) - \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+5*x+5)*exp(4)+2*x^2)/((x^2-5*x)*exp(4)-2*x^2),x, algorithm="giac")

[Out]

-(x*e^4 - 2*x)/(e^4 - 2) + log(abs(x*e^4 - 2*x - 5*e^4)) - log(abs(x))

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maple [A]  time = 0.35, size = 22, normalized size = 0.76

method	result	size
default	\(-x +\ln \left (x \,{\mathrm e}^{4}-5 \,{\mathrm e}^{4}-2 x \right )-\ln \relax (x )\)	\(22\)
norman	\(-x +\ln \left (x \,{\mathrm e}^{4}-5 \,{\mathrm e}^{4}-2 x \right )-\ln \relax (x )\)	\(22\)
risch	\(-x -\ln \relax (x )+\ln \left (\left (2-{\mathrm e}^{4}\right ) x +5 \,{\mathrm e}^{4}\right )\)	\(23\)
meijerg	\(\ln \left (1-\frac {x \left ({\mathrm e}^{4}-2\right ) {\mathrm e}^{-4}}{5}\right )-\ln \relax (x )+\ln \relax (5)-\ln \left ({\mathrm e}^{4}-2\right )+4-i \pi +\frac {25 \,{\mathrm e}^{4} \left (\frac {{\mathrm e}^{4}}{5}-\frac {2}{5}\right ) \left (-\frac {x \left ({\mathrm e}^{4}-2\right ) {\mathrm e}^{-4}}{5}-\ln \left (1-\frac {x \left ({\mathrm e}^{4}-2\right ) {\mathrm e}^{-4}}{5}\right )\right )}{\left ({\mathrm e}^{4}-2\right )^{2}}+\frac {5 \,{\mathrm e}^{4} \ln \left (1-\frac {x \left ({\mathrm e}^{4}-2\right ) {\mathrm e}^{-4}}{5}\right )}{{\mathrm e}^{4}-2}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x^2+5*x+5)*exp(4)+2*x^2)/((x^2-5*x)*exp(4)-2*x^2),x,method=_RETURNVERBOSE)

[Out]

-x+ln(x*exp(4)-5*exp(4)-2*x)-ln(x)

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maxima [A]  time = 0.37, size = 20, normalized size = 0.69 \begin {gather*} -x + \log \left (x {\left (e^{4} - 2\right )} - 5 \, e^{4}\right ) - \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x^2+5*x+5)*exp(4)+2*x^2)/((x^2-5*x)*exp(4)-2*x^2),x, algorithm="maxima")

[Out]

-x + log(x*(e^4 - 2) - 5*e^4) - log(x)

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mupad [B]  time = 1.16, size = 20, normalized size = 0.69 \begin {gather*} -x-2\,\mathrm {atanh}\left (\frac {x\,{\mathrm {e}}^{-4}\,\left (2\,{\mathrm {e}}^4-4\right )}{5}-1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(4)*(5*x - x^2 + 5) + 2*x^2)/(exp(4)*(5*x - x^2) + 2*x^2),x)

[Out]

- x - 2*atanh((x*exp(-4)*(2*exp(4) - 4))/5 - 1)

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sympy [A]  time = 0.34, size = 19, normalized size = 0.66 \begin {gather*} - x - \log {\relax (x )} + \log {\left (x - \frac {10 e^{4}}{-4 + 2 e^{4}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x**2+5*x+5)*exp(4)+2*x**2)/((x**2-5*x)*exp(4)-2*x**2),x)

[Out]

-x - log(x) + log(x - 10*exp(4)/(-4 + 2*exp(4)))

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