3.56.66 \(\int \frac {-e^3+8 e^5 x+64 x^3+64 e x^3+e^2 (2 x+16 x^3)-e^2 \log (4)}{e^2} \, dx\)

Optimal. Leaf size=28 \[ x \left (-e+x+4 x \left (e^3+\left (x+\frac {2 x}{e}\right )^2\right )-\log (4)\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 35, normalized size of antiderivative = 1.25, number of steps used = 4, number of rules used = 2, integrand size = 47, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.043, Rules used = {6, 12} \begin {gather*} \frac {16 (1+e) x^4}{e^2}+4 x^4+4 e^3 x^2+x^2-x (e+\log (4)) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-E^3 + 8*E^5*x + 64*x^3 + 64*E*x^3 + E^2*(2*x + 16*x^3) - E^2*Log[4])/E^2,x]

[Out]

x^2 + 4*E^3*x^2 + 4*x^4 + (16*(1 + E)*x^4)/E^2 - x*(E + Log[4])

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^3+8 e^5 x+(64+64 e) x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)}{e^2} \, dx\\ &=\frac {\int \left (-e^3+8 e^5 x+(64+64 e) x^3+e^2 \left (2 x+16 x^3\right )-e^2 \log (4)\right ) \, dx}{e^2}\\ &=4 e^3 x^2+\frac {16 (1+e) x^4}{e^2}-x (e+\log (4))+\int \left (2 x+16 x^3\right ) \, dx\\ &=x^2+4 e^3 x^2+4 x^4+\frac {16 (1+e) x^4}{e^2}-x (e+\log (4))\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 42, normalized size = 1.50 \begin {gather*} -e x+x^2+4 e^3 x^2+4 x^4+\frac {16 x^4}{e^2}+\frac {16 x^4}{e}-x \log (4) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-E^3 + 8*E^5*x + 64*x^3 + 64*E*x^3 + E^2*(2*x + 16*x^3) - E^2*Log[4])/E^2,x]

[Out]

-(E*x) + x^2 + 4*E^3*x^2 + 4*x^4 + (16*x^4)/E^2 + (16*x^4)/E - x*Log[4]

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fricas [A]  time = 0.78, size = 47, normalized size = 1.68 \begin {gather*} {\left (16 \, x^{4} e + 16 \, x^{4} + 4 \, x^{2} e^{5} - 2 \, x e^{2} \log \relax (2) - x e^{3} + {\left (4 \, x^{4} + x^{2}\right )} e^{2}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(1)^2*log(2)+8*x*exp(1)^2*exp(3)-exp(1)^3+(16*x^3+2*x)*exp(1)^2+64*x^3*exp(1)+64*x^3)/exp(1)^
2,x, algorithm="fricas")

[Out]

(16*x^4*e + 16*x^4 + 4*x^2*e^5 - 2*x*e^2*log(2) - x*e^3 + (4*x^4 + x^2)*e^2)*e^(-2)

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giac [A]  time = 0.16, size = 47, normalized size = 1.68 \begin {gather*} {\left (16 \, x^{4} e + 16 \, x^{4} + 4 \, x^{2} e^{5} - 2 \, x e^{2} \log \relax (2) - x e^{3} + {\left (4 \, x^{4} + x^{2}\right )} e^{2}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(1)^2*log(2)+8*x*exp(1)^2*exp(3)-exp(1)^3+(16*x^3+2*x)*exp(1)^2+64*x^3*exp(1)+64*x^3)/exp(1)^
2,x, algorithm="giac")

[Out]

(16*x^4*e + 16*x^4 + 4*x^2*e^5 - 2*x*e^2*log(2) - x*e^3 + (4*x^4 + x^2)*e^2)*e^(-2)

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maple [A]  time = 0.04, size = 41, normalized size = 1.46

method	result	size
risch	\(4 x^{2} {\mathrm e}^{3}+4 x^{4}+16 x^{4} {\mathrm e}^{-1}-x \,{\mathrm e}-2 x \ln \relax (2)+x^{2}+16 x^{4} {\mathrm e}^{-2}\)	\(41\)
norman	\(\left ({\mathrm e} \left (4 \,{\mathrm e}^{3}+1\right ) x^{2}+4 \left ({\mathrm e}^{2}+4 \,{\mathrm e}+4\right ) {\mathrm e}^{-1} x^{4}-{\mathrm e} \left ({\mathrm e}+2 \ln \relax (2)\right ) x \right ) {\mathrm e}^{-1}\)	\(50\)
gosper	\(-x \left (-4 x^{3} {\mathrm e}^{2}-4 x \,{\mathrm e}^{2} {\mathrm e}^{3}-16 x^{3} {\mathrm e}+{\mathrm e}^{3}+2 \,{\mathrm e}^{2} \ln \relax (2)-{\mathrm e}^{2} x -16 x^{3}\right ) {\mathrm e}^{-2}\)	\(58\)
default	\({\mathrm e}^{-2} \left (-2 x \,{\mathrm e}^{2} \ln \relax (2)+4 x^{2} {\mathrm e}^{2} {\mathrm e}^{3}-x \,{\mathrm e}^{3}+{\mathrm e}^{2} \left (4 x^{4}+x^{2}\right )+16 x^{4} {\mathrm e}+16 x^{4}\right )\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-2*exp(1)^2*ln(2)+8*x*exp(1)^2*exp(3)-exp(1)^3+(16*x^3+2*x)*exp(1)^2+64*x^3*exp(1)+64*x^3)/exp(1)^2,x,met
hod=_RETURNVERBOSE)

[Out]

4*x^2*exp(3)+4*x^4+16*x^4*exp(-1)-x*exp(1)-2*x*ln(2)+x^2+16*x^4*exp(-2)

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maxima [A]  time = 0.37, size = 47, normalized size = 1.68 \begin {gather*} {\left (16 \, x^{4} e + 16 \, x^{4} + 4 \, x^{2} e^{5} - 2 \, x e^{2} \log \relax (2) - x e^{3} + {\left (4 \, x^{4} + x^{2}\right )} e^{2}\right )} e^{\left (-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(1)^2*log(2)+8*x*exp(1)^2*exp(3)-exp(1)^3+(16*x^3+2*x)*exp(1)^2+64*x^3*exp(1)+64*x^3)/exp(1)^
2,x, algorithm="maxima")

[Out]

(16*x^4*e + 16*x^4 + 4*x^2*e^5 - 2*x*e^2*log(2) - x*e^3 + (4*x^4 + x^2)*e^2)*e^(-2)

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mupad [B]  time = 0.06, size = 33, normalized size = 1.18 \begin {gather*} \left (16\,{\mathrm {e}}^{-1}+16\,{\mathrm {e}}^{-2}+4\right )\,x^4+\left (4\,{\mathrm {e}}^3+1\right )\,x^2+\left (-\mathrm {e}-\ln \relax (4)\right )\,x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(-2)*(exp(2)*(2*x + 16*x^3) - 2*exp(2)*log(2) - exp(3) + 8*x*exp(5) + 64*x^3*exp(1) + 64*x^3),x)

[Out]

x^4*(16*exp(-1) + 16*exp(-2) + 4) - x*(exp(1) + log(4)) + x^2*(4*exp(3) + 1)

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sympy [A]  time = 0.06, size = 39, normalized size = 1.39 \begin {gather*} \frac {x^{4} \left (16 + 4 e^{2} + 16 e\right )}{e^{2}} + x^{2} \left (1 + 4 e^{3}\right ) + x \left (- e - 2 \log {\relax (2 )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-2*exp(1)**2*ln(2)+8*x*exp(1)**2*exp(3)-exp(1)**3+(16*x**3+2*x)*exp(1)**2+64*x**3*exp(1)+64*x**3)/e
xp(1)**2,x)

[Out]

x**4*(16 + 4*exp(2) + 16*E)*exp(-2) + x**2*(1 + 4*exp(3)) + x*(-E - 2*log(2))

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