3.23.30 \(\int \frac {-4-4 e^6-2 x-4 x^2}{1+e^6+x^2} \, dx\)

Optimal. Leaf size=24 \[ 5-3 x-\log \left (e^x \left (-1-e^6-x^2\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 15, normalized size of antiderivative = 0.62, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {1810, 260} \begin {gather*} -\log \left (x^2+e^6+1\right )-4 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-4 - 4*E^6 - 2*x - 4*x^2)/(1 + E^6 + x^2),x]

[Out]

-4*x - Log[1 + E^6 + x^2]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 1810

Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[Pq*(a + b*x^2)^p, x], x] /; FreeQ[{a,
b}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-4-\frac {2 x}{1+e^6+x^2}\right ) \, dx\\ &=-4 x-2 \int \frac {x}{1+e^6+x^2} \, dx\\ &=-4 x-\log \left (1+e^6+x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 19, normalized size = 0.79 \begin {gather*} -2 \left (2 x+\frac {1}{2} \log \left (1+e^6+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 4*E^6 - 2*x - 4*x^2)/(1 + E^6 + x^2),x]

[Out]

-2*(2*x + Log[1 + E^6 + x^2]/2)

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fricas [A]  time = 0.81, size = 14, normalized size = 0.58 \begin {gather*} -4 \, x - \log \left (x^{2} + e^{6} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(3)^2-4*x^2-2*x-4)/(exp(3)^2+x^2+1),x, algorithm="fricas")

[Out]

-4*x - log(x^2 + e^6 + 1)

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giac [A]  time = 0.27, size = 14, normalized size = 0.58 \begin {gather*} -4 \, x - \log \left (x^{2} + e^{6} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(3)^2-4*x^2-2*x-4)/(exp(3)^2+x^2+1),x, algorithm="giac")

[Out]

-4*x - log(x^2 + e^6 + 1)

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maple [A]  time = 0.39, size = 15, normalized size = 0.62

method	result	size
default	\(-4 x -\ln \left ({\mathrm e}^{6}+x^{2}+1\right )\)	\(15\)
risch	\(-4 x -\ln \left ({\mathrm e}^{6}+x^{2}+1\right )\)	\(15\)
norman	\(-4 x -\ln \left ({\mathrm e}^{6}+x^{2}+1\right )\)	\(17\)
meijerg	\(-\frac {4 \,{\mathrm e}^{6} \arctan \left (\frac {x}{\sqrt {1+{\mathrm e}^{6}}}\right )}{\sqrt {1+{\mathrm e}^{6}}}-2 \sqrt {1+{\mathrm e}^{6}}\, \left (\frac {2 x}{\sqrt {1+{\mathrm e}^{6}}}-2 \arctan \left (\frac {x}{\sqrt {1+{\mathrm e}^{6}}}\right )\right )-\ln \left (1+\frac {x^{2}}{1+{\mathrm e}^{6}}\right )-\frac {4 \arctan \left (\frac {x}{\sqrt {1+{\mathrm e}^{6}}}\right )}{\sqrt {1+{\mathrm e}^{6}}}\)	\(82\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4*x-ln(exp(6)+x^2+1)

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maxima [A]  time = 0.46, size = 14, normalized size = 0.58 \begin {gather*} -4 \, x - \log \left (x^{2} + e^{6} + 1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(3)^2-4*x^2-2*x-4)/(exp(3)^2+x^2+1),x, algorithm="maxima")

[Out]

-4*x - log(x^2 + e^6 + 1)

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mupad [B]  time = 1.25, size = 14, normalized size = 0.58 \begin {gather*} -4\,x-\ln \left (x^2+{\mathrm {e}}^6+1\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(2*x + 4*exp(6) + 4*x^2 + 4)/(exp(6) + x^2 + 1),x)

[Out]

- 4*x - log(exp(6) + x^2 + 1)

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sympy [A]  time = 0.10, size = 14, normalized size = 0.58 \begin {gather*} - 4 x - \log {\left (x^{2} + 1 + e^{6} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*exp(3)**2-4*x**2-2*x-4)/(exp(3)**2+x**2+1),x)

[Out]

-4*x - log(x**2 + 1 + exp(6))

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