3.10.62 \(\int \frac {12+16 x^3+e^6 (2-24 x^2)-8 \log (2)}{e^{12}-2 e^6 x+x^2} \, dx\)

Optimal. Leaf size=25 \[ \frac {2 \left (2-x+4 \left (-2+x^3+\log (2)\right )\right )}{-e^6+x} \]

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Rubi [A]  time = 0.04, antiderivative size = 37, normalized size of antiderivative = 1.48, number of steps used = 3, number of rules used = 2, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {27, 1850} \begin {gather*} 8 x^2+8 e^6 x+\frac {2 \left (6+e^6-4 e^{18}-\log (16)\right )}{e^6-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(12 + 16*x^3 + E^6*(2 - 24*x^2) - 8*Log[2])/(E^12 - 2*E^6*x + x^2),x]

[Out]

8*E^6*x + 8*x^2 + (2*(6 + E^6 - 4*E^18 - Log[16]))/(E^6 - x)

Rule 27

Int[(u_.)*((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[u*Cancel[(b/2 + c*x)^(2*p)/c^p], x] /; Fr
eeQ[{a, b, c}, x] && EqQ[b^2 - 4*a*c, 0] && IntegerQ[p]

Rule 1850

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {12+16 x^3+e^6 \left (2-24 x^2\right )-8 \log (2)}{\left (-e^6+x\right )^2} \, dx\\ &=\int \left (8 e^6+16 x+\frac {2 \left (6+e^6-4 e^{18}-\log (16)\right )}{\left (e^6-x\right )^2}\right ) \, dx\\ &=8 e^6 x+8 x^2+\frac {2 \left (6+e^6-4 e^{18}-\log (16)\right )}{e^6-x}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 36, normalized size = 1.44 \begin {gather*} \frac {2 \left (6+e^6-12 e^{18}+12 e^{12} x-4 x^3-4 \log (2)\right )}{e^6-x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + 16*x^3 + E^6*(2 - 24*x^2) - 8*Log[2])/(E^12 - 2*E^6*x + x^2),x]

[Out]

(2*(6 + E^6 - 12*E^18 + 12*E^12*x - 4*x^3 - 4*Log[2]))/(E^6 - x)

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fricas [A]  time = 0.59, size = 34, normalized size = 1.36 \begin {gather*} \frac {2 \, {\left (4 \, x^{3} - 4 \, x e^{12} + 4 \, e^{18} - e^{6} + 4 \, \log \relax (2) - 6\right )}}{x - e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*log(2)+(-24*x^2+2)*exp(3)^2+16*x^3+12)/(exp(3)^4-2*x*exp(3)^2+x^2),x, algorithm="fricas")

[Out]

2*(4*x^3 - 4*x*e^12 + 4*e^18 - e^6 + 4*log(2) - 6)/(x - e^6)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*log(2)+(-24*x^2+2)*exp(3)^2+16*x^3+12)/(exp(3)^4-2*x*exp(3)^2+x^2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 2*(4*sageVARx^2+4*sageVARx*exp(6)+(4*exp
(6)^2-4*exp(12))*ln(sageVARx^2-2*sageVARx*exp(6)+exp(12))+(8*exp(6)^3-12*exp(6)*exp(12)+exp(6)-4*ln(2)+6)*1/2/
sqrt(-exp(12)+exp(6)^

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maple [A]  time = 0.22, size = 28, normalized size = 1.12

method	result	size
gosper	\(\frac {-8 x^{3}+2 \,{\mathrm e}^{6}-8 \ln \relax (2)+12}{{\mathrm e}^{6}-x}\)	\(28\)
norman	\(\frac {-8 x^{3}+2 \,{\mathrm e}^{6}-8 \ln \relax (2)+12}{{\mathrm e}^{6}-x}\)	\(29\)
risch	\(8 x \,{\mathrm e}^{6}+8 x^{2}-\frac {8 \,{\mathrm e}^{18}}{{\mathrm e}^{6}-x}+\frac {2 \,{\mathrm e}^{6}}{{\mathrm e}^{6}-x}-\frac {8 \ln \relax (2)}{{\mathrm e}^{6}-x}+\frac {12}{{\mathrm e}^{6}-x}\)	\(58\)
meijerg	\(\frac {12 \,{\mathrm e}^{-12} x}{1-x \,{\mathrm e}^{-6}}-\frac {8 \ln \relax (2) {\mathrm e}^{-12} x}{1-x \,{\mathrm e}^{-6}}+24 \,{\mathrm e}^{12} \left (-\frac {x \,{\mathrm e}^{-6} \left (-3 x \,{\mathrm e}^{-6}+6\right )}{3 \left (1-x \,{\mathrm e}^{-6}\right )}-2 \ln \left (1-x \,{\mathrm e}^{-6}\right )\right )+\frac {2 \,{\mathrm e}^{-6} x}{1-x \,{\mathrm e}^{-6}}+16 \,{\mathrm e}^{12} \left (\frac {x \,{\mathrm e}^{-6} \left (-2 x^{2} {\mathrm e}^{-12}-6 x \,{\mathrm e}^{-6}+12\right )}{4-4 x \,{\mathrm e}^{-6}}+3 \ln \left (1-x \,{\mathrm e}^{-6}\right )\right )\)	\(125\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-8*ln(2)+(-24*x^2+2)*exp(3)^2+16*x^3+12)/(exp(3)^4-2*x*exp(3)^2+x^2),x,method=_RETURNVERBOSE)

[Out]

2*(-4*x^3+exp(3)^2-4*ln(2)+6)/(exp(3)^2-x)

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maxima [A]  time = 0.47, size = 35, normalized size = 1.40 \begin {gather*} 8 \, x^{2} + 8 \, x e^{6} + \frac {2 \, {\left (4 \, e^{18} - e^{6} + 4 \, \log \relax (2) - 6\right )}}{x - e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*log(2)+(-24*x^2+2)*exp(3)^2+16*x^3+12)/(exp(3)^4-2*x*exp(3)^2+x^2),x, algorithm="maxima")

[Out]

8*x^2 + 8*x*e^6 + 2*(4*e^18 - e^6 + 4*log(2) - 6)/(x - e^6)

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mupad [B]  time = 0.72, size = 34, normalized size = 1.36 \begin {gather*} -\frac {-8\,x^3+8\,{\mathrm {e}}^{12}\,x+2\,{\mathrm {e}}^6-8\,{\mathrm {e}}^{18}-\ln \left (256\right )+12}{x-{\mathrm {e}}^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(8*log(2) + exp(6)*(24*x^2 - 2) - 16*x^3 - 12)/(exp(12) - 2*x*exp(6) + x^2),x)

[Out]

-(2*exp(6) - 8*exp(18) - log(256) + 8*x*exp(12) - 8*x^3 + 12)/(x - exp(6))

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sympy [A]  time = 0.20, size = 32, normalized size = 1.28 \begin {gather*} 8 x^{2} + 8 x e^{6} + \frac {- 2 e^{6} - 12 + 8 \log {\relax (2 )} + 8 e^{18}}{x - e^{6}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-8*ln(2)+(-24*x**2+2)*exp(3)**2+16*x**3+12)/(exp(3)**4-2*x*exp(3)**2+x**2),x)

[Out]

8*x**2 + 8*x*exp(6) + (-2*exp(6) - 12 + 8*log(2) + 8*exp(18))/(x - exp(6))

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