3.8.52 \(\int \frac {1}{4} (1+32 x^3-64 e^2 x^3+32 e^4 x^3+(-64 e^2 x^3+64 e^4 x^3) \log (3)+32 e^4 x^3 \log ^2(3)) \, dx\)

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

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Rubi [A]  time = 0.03, antiderivative size = 45, normalized size of antiderivative = 1.55, number of steps used = 8, number of rules used = 3, integrand size = 59, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.051, Rules used = {6, 12, 30} \begin {gather*} 2 x^4 \left (1-2 e^2+e^4 \left (1+\log ^2(3)\right )\right )-4 e^2 \left (1-e^2\right ) x^4 \log (3)+\frac {x}{4} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(1 + 32*x^3 - 64*E^2*x^3 + 32*E^4*x^3 + (-64*E^2*x^3 + 64*E^4*x^3)*Log[3] + 32*E^4*x^3*Log[3]^2)/4,x]

[Out]

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

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]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {1}{4} \left (1+32 e^4 x^3+\left (32-64 e^2\right ) x^3+\left (-64 e^2 x^3+64 e^4 x^3\right ) \log (3)+32 e^4 x^3 \log ^2(3)\right ) \, dx\\ &=\int \frac {1}{4} \left (1+\left (32-64 e^2+32 e^4\right ) x^3+\left (-64 e^2 x^3+64 e^4 x^3\right ) \log (3)+32 e^4 x^3 \log ^2(3)\right ) \, dx\\ &=\int \frac {1}{4} \left (1+\left (-64 e^2 x^3+64 e^4 x^3\right ) \log (3)+x^3 \left (32-64 e^2+32 e^4+32 e^4 \log ^2(3)\right )\right ) \, dx\\ &=\frac {1}{4} \int \left (1+\left (-64 e^2 x^3+64 e^4 x^3\right ) \log (3)+x^3 \left (32-64 e^2+32 e^4+32 e^4 \log ^2(3)\right )\right ) \, dx\\ &=\frac {x}{4}+2 x^4 \left (1-2 e^2+e^4 \left (1+\log ^2(3)\right )\right )+\frac {1}{4} \log (3) \int \left (-64 e^2 x^3+64 e^4 x^3\right ) \, dx\\ &=\frac {x}{4}+2 x^4 \left (1-2 e^2+e^4 \left (1+\log ^2(3)\right )\right )+\frac {1}{4} \log (3) \int \left (-64 e^2+64 e^4\right ) x^3 \, dx\\ &=\frac {x}{4}+2 x^4 \left (1-2 e^2+e^4 \left (1+\log ^2(3)\right )\right )-\left (16 e^2 \left (1-e^2\right ) \log (3)\right ) \int x^3 \, dx\\ &=\frac {x}{4}-4 e^2 \left (1-e^2\right ) x^4 \log (3)+2 x^4 \left (1-2 e^2+e^4 \left (1+\log ^2(3)\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 23, normalized size = 0.79 \begin {gather*} \frac {x}{4}+2 x^4 \left (-1+e^2 (1+\log (3))\right )^2 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(1 + 32*x^3 - 64*E^2*x^3 + 32*E^4*x^3 + (-64*E^2*x^3 + 64*E^4*x^3)*Log[3] + 32*E^4*x^3*Log[3]^2)/4,x
]

[Out]

x/4 + 2*x^4*(-1 + E^2*(1 + Log[3]))^2

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fricas [B]  time = 0.64, size = 52, normalized size = 1.79 \begin {gather*} 2 \, x^{4} e^{4} \log \relax (3)^{2} + 2 \, x^{4} e^{4} - 4 \, x^{4} e^{2} + 2 \, x^{4} + 4 \, {\left (x^{4} e^{4} - x^{4} e^{2}\right )} \log \relax (3) + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x^3*exp(2)^2*log(3)^2+1/4*(64*x^3*exp(2)^2-64*x^3*exp(2))*log(3)+8*x^3*exp(2)^2-16*x^3*exp(2)+8*x^
3+1/4,x, algorithm="fricas")

[Out]

2*x^4*e^4*log(3)^2 + 2*x^4*e^4 - 4*x^4*e^2 + 2*x^4 + 4*(x^4*e^4 - x^4*e^2)*log(3) + 1/4*x

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giac [B]  time = 0.28, size = 52, normalized size = 1.79 \begin {gather*} 2 \, x^{4} e^{4} \log \relax (3)^{2} + 2 \, x^{4} e^{4} - 4 \, x^{4} e^{2} + 2 \, x^{4} + 4 \, {\left (x^{4} e^{4} - x^{4} e^{2}\right )} \log \relax (3) + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x^3*exp(2)^2*log(3)^2+1/4*(64*x^3*exp(2)^2-64*x^3*exp(2))*log(3)+8*x^3*exp(2)^2-16*x^3*exp(2)+8*x^
3+1/4,x, algorithm="giac")

[Out]

2*x^4*e^4*log(3)^2 + 2*x^4*e^4 - 4*x^4*e^2 + 2*x^4 + 4*(x^4*e^4 - x^4*e^2)*log(3) + 1/4*x

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maple [A]  time = 0.04, size = 45, normalized size = 1.55

method	result	size
norman	\(\left (2 \,{\mathrm e}^{4} \ln \relax (3)^{2}+4 \,{\mathrm e}^{4} \ln \relax (3)+2 \,{\mathrm e}^{4}-4 \,{\mathrm e}^{2} \ln \relax (3)-4 \,{\mathrm e}^{2}+2\right ) x^{4}+\frac {x}{4}\)	\(45\)
risch	\(2 x^{4} {\mathrm e}^{4} \ln \relax (3)^{2}+4 \ln \relax (3) x^{4} {\mathrm e}^{4}-4 \ln \relax (3) x^{4} {\mathrm e}^{2}+2 x^{4} {\mathrm e}^{4}-4 x^{4} {\mathrm e}^{2}+2 x^{4}+\frac {x}{4}\)	\(55\)
gosper	\(\frac {x \left (8 x^{3} {\mathrm e}^{4} \ln \relax (3)^{2}+16 \,{\mathrm e}^{4} \ln \relax (3) x^{3}+8 x^{3} {\mathrm e}^{4}-16 \,{\mathrm e}^{2} \ln \relax (3) x^{3}-16 x^{3} {\mathrm e}^{2}+8 x^{3}+1\right )}{4}\)	\(60\)
default	\(2 x^{4} {\mathrm e}^{4} \ln \relax (3)^{2}+\frac {\ln \relax (3) \left (16 x^{4} {\mathrm e}^{4}-16 x^{4} {\mathrm e}^{2}\right )}{4}+2 x^{4} {\mathrm e}^{4}-4 x^{4} {\mathrm e}^{2}+2 x^{4}+\frac {x}{4}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*x^3*exp(2)^2*ln(3)^2+1/4*(64*x^3*exp(2)^2-64*x^3*exp(2))*ln(3)+8*x^3*exp(2)^2-16*x^3*exp(2)+8*x^3+1/4,x,
method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.69, size = 52, normalized size = 1.79 \begin {gather*} 2 \, x^{4} e^{4} \log \relax (3)^{2} + 2 \, x^{4} e^{4} - 4 \, x^{4} e^{2} + 2 \, x^{4} + 4 \, {\left (x^{4} e^{4} - x^{4} e^{2}\right )} \log \relax (3) + \frac {1}{4} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x^3*exp(2)^2*log(3)^2+1/4*(64*x^3*exp(2)^2-64*x^3*exp(2))*log(3)+8*x^3*exp(2)^2-16*x^3*exp(2)+8*x^
3+1/4,x, algorithm="maxima")

[Out]

2*x^4*e^4*log(3)^2 + 2*x^4*e^4 - 4*x^4*e^2 + 2*x^4 + 4*(x^4*e^4 - x^4*e^2)*log(3) + 1/4*x

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mupad [B]  time = 0.56, size = 39, normalized size = 1.34 \begin {gather*} \left (2\,{\mathrm {e}}^4-4\,{\mathrm {e}}^2+2\,{\mathrm {e}}^4\,{\ln \relax (3)}^2-\frac {\ln \relax (3)\,\left (64\,{\mathrm {e}}^2-64\,{\mathrm {e}}^4\right )}{16}+2\right )\,x^4+\frac {x}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(8*x^3*exp(4) - 16*x^3*exp(2) - (log(3)*(64*x^3*exp(2) - 64*x^3*exp(4)))/4 + 8*x^3 + 8*x^3*exp(4)*log(3)^2
+ 1/4,x)

[Out]

x/4 + x^4*(2*exp(4) - 4*exp(2) + 2*exp(4)*log(3)^2 - (log(3)*(64*exp(2) - 64*exp(4)))/16 + 2)

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sympy [A]  time = 0.08, size = 44, normalized size = 1.52 \begin {gather*} x^{4} \left (- 4 e^{2} \log {\relax (3 )} - 4 e^{2} + 2 + 2 e^{4} + 2 e^{4} \log {\relax (3 )}^{2} + 4 e^{4} \log {\relax (3 )}\right ) + \frac {x}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(8*x**3*exp(2)**2*ln(3)**2+1/4*(64*x**3*exp(2)**2-64*x**3*exp(2))*ln(3)+8*x**3*exp(2)**2-16*x**3*exp(
2)+8*x**3+1/4,x)

[Out]

x**4*(-4*exp(2)*log(3) - 4*exp(2) + 2 + 2*exp(4) + 2*exp(4)*log(3)**2 + 4*exp(4)*log(3)) + x/4

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