3.89.1 \(\int \frac {e^2 (5-5 x)+10 x-5 x^2}{e^2 x+x^2} \, dx\)

Optimal. Leaf size=21 \[ 1+5 \left (-x+\log \left (x \left (e^2+x\right )\right )\right )+\log (6+\log (4)) \]

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Rubi [A]  time = 0.04, antiderivative size = 16, normalized size of antiderivative = 0.76, number of steps used = 3, number of rules used = 2, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.067, Rules used = {1593, 1820} \begin {gather*} -5 x+5 \log (x)+5 \log \left (x+e^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^2*(5 - 5*x) + 10*x - 5*x^2)/(E^2*x + x^2),x]

[Out]

-5*x + 5*Log[x] + 5*Log[E^2 + x]

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 1820

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a +
b*x^n)^p, x], x] /; FreeQ[{a, b, c, m, n}, x] && PolyQ[Pq, x] && (IGtQ[p, 0] || EqQ[n, 1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^2 (5-5 x)+10 x-5 x^2}{x \left (e^2+x\right )} \, dx\\ &=\int \left (-5+\frac {5}{x}+\frac {5}{e^2+x}\right ) \, dx\\ &=-5 x+5 \log (x)+5 \log \left (e^2+x\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 16, normalized size = 0.76 \begin {gather*} -5 \left (x-\log (x)-\log \left (e^2+x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^2*(5 - 5*x) + 10*x - 5*x^2)/(E^2*x + x^2),x]

[Out]

-5*(x - Log[x] - Log[E^2 + x])

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fricas [A]  time = 0.75, size = 15, normalized size = 0.71 \begin {gather*} -5 \, x + 5 \, \log \left (x^{2} + x e^{2}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*exp(2)-5*x^2+10*x)/(exp(2)*x+x^2),x, algorithm="fricas")

[Out]

-5*x + 5*log(x^2 + x*e^2)

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giac [A]  time = 0.20, size = 17, normalized size = 0.81 \begin {gather*} -5 \, x + 5 \, \log \left ({\left | x + e^{2} \right |}\right ) + 5 \, \log \left ({\left | x \right |}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*exp(2)-5*x^2+10*x)/(exp(2)*x+x^2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.47, size = 16, normalized size = 0.76

method	result	size
default	\(-5 x +5 \ln \relax (x )+5 \ln \left (x +{\mathrm e}^{2}\right )\)	\(16\)
norman	\(-5 x +5 \ln \relax (x )+5 \ln \left (x +{\mathrm e}^{2}\right )\)	\(16\)
risch	\(-5 x +5 \ln \left ({\mathrm e}^{2} x +x^{2}\right )\)	\(16\)
meijerg	\(5 \ln \relax (x )-10-5 \ln \left (1+x \,{\mathrm e}^{-2}\right )+\left (-5 \,{\mathrm e}^{2}+10\right ) \ln \left (1+x \,{\mathrm e}^{-2}\right )-5 \,{\mathrm e}^{2} \left (x \,{\mathrm e}^{-2}-\ln \left (1+x \,{\mathrm e}^{-2}\right )\right )\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-5*x+5)*exp(2)-5*x^2+10*x)/(exp(2)*x+x^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.36, size = 15, normalized size = 0.71 \begin {gather*} -5 \, x + 5 \, \log \left (x + e^{2}\right ) + 5 \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*exp(2)-5*x^2+10*x)/(exp(2)*x+x^2),x, algorithm="maxima")

[Out]

-5*x + 5*log(x + e^2) + 5*log(x)

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mupad [B]  time = 0.10, size = 13, normalized size = 0.62 \begin {gather*} 5\,\ln \left (x\,\left (x+{\mathrm {e}}^2\right )\right )-5\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x^2 - 10*x + exp(2)*(5*x - 5))/(x*exp(2) + x^2),x)

[Out]

5*log(x*(x + exp(2))) - 5*x

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sympy [A]  time = 0.13, size = 14, normalized size = 0.67 \begin {gather*} - 5 x + 5 \log {\left (x^{2} + x e^{2} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-5*x+5)*exp(2)-5*x**2+10*x)/(exp(2)*x+x**2),x)

[Out]

-5*x + 5*log(x**2 + x*exp(2))

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