3.91.52 \(\int e^{4 e^{1+\log ^2(5)} x} (-4 x^3-4 e^{1+\log ^2(5)} x^4) \, dx\)

Optimal. Leaf size=20 \[ 2-e^{4 e^{1+\log ^2(5)} x} x^4 \]

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Rubi [A]  time = 0.40, antiderivative size = 18, normalized size of antiderivative = 0.90, number of steps used = 12, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {1593, 2196, 2176, 2194} \begin {gather*} x^4 \left (-e^{4 x e^{1+\log ^2(5)}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(4*E^(1 + Log[5]^2)*x)*(-4*x^3 - 4*E^(1 + Log[5]^2)*x^4),x]

[Out]

-(E^(4*E^(1 + Log[5]^2)*x)*x^4)

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 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int e^{4 e^{1+\log ^2(5)} x} x^3 \left (-4-4 e^{1+\log ^2(5)} x\right ) \, dx\\ &=\int \left (-4 e^{4 e^{1+\log ^2(5)} x} x^3-4 e^{1+4 e^{1+\log ^2(5)} x+\log ^2(5)} x^4\right ) \, dx\\ &=-\left (4 \int e^{4 e^{1+\log ^2(5)} x} x^3 \, dx\right )-4 \int e^{1+4 e^{1+\log ^2(5)} x+\log ^2(5)} x^4 \, dx\\ &=-e^{-1+4 e^{1+\log ^2(5)} x-\log ^2(5)} x^3-e^{4 e^{1+\log ^2(5)} x} x^4+\left (3 e^{-1-\log ^2(5)}\right ) \int e^{4 e^{1+\log ^2(5)} x} x^2 \, dx+\left (4 e^{-1-\log ^2(5)}\right ) \int e^{1+4 e^{1+\log ^2(5)} x+\log ^2(5)} x^3 \, dx\\ &=\frac {3}{4} e^{4 e^{1+\log ^2(5)} x-2 \left (1+\log ^2(5)\right )} x^2-e^{4 e^{1+\log ^2(5)} x} x^4-\frac {1}{2} \left (3 e^{-2 \left (1+\log ^2(5)\right )}\right ) \int e^{4 e^{1+\log ^2(5)} x} x \, dx-\left (3 e^{-2 \left (1+\log ^2(5)\right )}\right ) \int e^{1+4 e^{1+\log ^2(5)} x+\log ^2(5)} x^2 \, dx\\ &=-\frac {3}{8} e^{4 e^{1+\log ^2(5)} x-3 \left (1+\log ^2(5)\right )} x-e^{4 e^{1+\log ^2(5)} x} x^4+\frac {1}{8} \left (3 e^{-3 \left (1+\log ^2(5)\right )}\right ) \int e^{4 e^{1+\log ^2(5)} x} \, dx+\frac {1}{2} \left (3 e^{-3 \left (1+\log ^2(5)\right )}\right ) \int e^{1+4 e^{1+\log ^2(5)} x+\log ^2(5)} x \, dx\\ &=\frac {3}{32} e^{4 e^{1+\log ^2(5)} x-4 \left (1+\log ^2(5)\right )}-e^{4 e^{1+\log ^2(5)} x} x^4-\frac {1}{8} \left (3 e^{-4 \left (1+\log ^2(5)\right )}\right ) \int e^{1+4 e^{1+\log ^2(5)} x+\log ^2(5)} \, dx\\ &=-e^{4 e^{1+\log ^2(5)} x} x^4\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 18, normalized size = 0.90 \begin {gather*} -e^{4 e^{1+\log ^2(5)} x} x^4 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(4*E^(1 + Log[5]^2)*x)*(-4*x^3 - 4*E^(1 + Log[5]^2)*x^4),x]

[Out]

-(E^(4*E^(1 + Log[5]^2)*x)*x^4)

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fricas [A]  time = 0.60, size = 16, normalized size = 0.80 \begin {gather*} -x^{4} e^{\left (4 \, x e^{\left (\log \relax (5)^{2} + 1\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^4*exp(1)*exp(log(5)^2)-4*x^3)*exp(x*exp(1)*exp(log(5)^2))^4,x, algorithm="fricas")

[Out]

-x^4*e^(4*x*e^(log(5)^2 + 1))

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giac [B]  time = 0.17, size = 137, normalized size = 6.85 \begin {gather*} -\frac {1}{32} \, {\left (32 \, x^{4} e^{\left (4 \, \log \relax (5)^{2} + 4\right )} - 32 \, x^{3} e^{\left (3 \, \log \relax (5)^{2} + 3\right )} + 24 \, x^{2} e^{\left (2 \, \log \relax (5)^{2} + 2\right )} - 12 \, x e^{\left (\log \relax (5)^{2} + 1\right )} + 3\right )} e^{\left (4 \, x e^{\left (\log \relax (5)^{2} + 1\right )} - 4 \, \log \relax (5)^{2} - 4\right )} - \frac {1}{32} \, {\left (32 \, x^{3} e^{\left (3 \, \log \relax (5)^{2} + 3\right )} - 24 \, x^{2} e^{\left (2 \, \log \relax (5)^{2} + 2\right )} + 12 \, x e^{\left (\log \relax (5)^{2} + 1\right )} - 3\right )} e^{\left (4 \, x e^{\left (\log \relax (5)^{2} + 1\right )} - 4 \, \log \relax (5)^{2} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^4*exp(1)*exp(log(5)^2)-4*x^3)*exp(x*exp(1)*exp(log(5)^2))^4,x, algorithm="giac")

[Out]

-1/32*(32*x^4*e^(4*log(5)^2 + 4) - 32*x^3*e^(3*log(5)^2 + 3) + 24*x^2*e^(2*log(5)^2 + 2) - 12*x*e^(log(5)^2 +
1) + 3)*e^(4*x*e^(log(5)^2 + 1) - 4*log(5)^2 - 4) - 1/32*(32*x^3*e^(3*log(5)^2 + 3) - 24*x^2*e^(2*log(5)^2 + 2
) + 12*x*e^(log(5)^2 + 1) - 3)*e^(4*x*e^(log(5)^2 + 1) - 4*log(5)^2 - 4)

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maple [A]  time = 0.15, size = 17, normalized size = 0.85

method	result	size
risch	\(-{\mathrm e}^{4 x \,{\mathrm e}^{1+\ln \relax (5)^{2}}} x^{4}\)	\(17\)
gosper	\(-{\mathrm e}^{4 x \,{\mathrm e}^{1+\ln \relax (5)^{2}}} x^{4}\)	\(18\)
derivativedivides	\(-{\mathrm e}^{4 x \,{\mathrm e}^{1+\ln \relax (5)^{2}}} x^{4}\)	\(18\)
default	\(-{\mathrm e}^{4 x \,{\mathrm e}^{1+\ln \relax (5)^{2}}} x^{4}\)	\(18\)
norman	\(-{\mathrm e}^{4 x \,{\mathrm e}^{1+\ln \relax (5)^{2}}} x^{4}\)	\(18\)
meijerg	\(\frac {{\mathrm e}^{-4-4 \ln \relax (5)^{2}} \left (24-\frac {{\mathrm e}^{4 x \,{\mathrm e}^{1+\ln \relax (5)^{2}}} \left (1280 x^{4} {\mathrm e}^{4+4 \ln \relax (5)^{2}}-1280 x^{3} {\mathrm e}^{3+3 \ln \relax (5)^{2}}+960 x^{2} {\mathrm e}^{2+2 \ln \relax (5)^{2}}-480 x \,{\mathrm e}^{1+\ln \relax (5)^{2}}+120\right )}{5}\right )}{256}-\frac {{\mathrm e}^{-4-4 \ln \relax (5)^{2}} \left (6-\frac {{\mathrm e}^{4 x \,{\mathrm e}^{1+\ln \relax (5)^{2}}} \left (-256 x^{3} {\mathrm e}^{3+3 \ln \relax (5)^{2}}+192 x^{2} {\mathrm e}^{2+2 \ln \relax (5)^{2}}-96 x \,{\mathrm e}^{1+\ln \relax (5)^{2}}+24\right )}{4}\right )}{64}\)	\(148\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-4*x^4*exp(1)*exp(ln(5)^2)-4*x^3)*exp(x*exp(1)*exp(ln(5)^2))^4,x,method=_RETURNVERBOSE)

[Out]

-exp(4*x*exp(1+ln(5)^2))*x^4

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maxima [B]  time = 0.45, size = 137, normalized size = 6.85 \begin {gather*} -\frac {1}{32} \, {\left (32 \, x^{4} e^{\left (4 \, \log \relax (5)^{2} + 4\right )} - 32 \, x^{3} e^{\left (3 \, \log \relax (5)^{2} + 3\right )} + 24 \, x^{2} e^{\left (2 \, \log \relax (5)^{2} + 2\right )} - 12 \, x e^{\left (\log \relax (5)^{2} + 1\right )} + 3\right )} e^{\left (4 \, x e^{\left (\log \relax (5)^{2} + 1\right )} - 4 \, \log \relax (5)^{2} - 4\right )} - \frac {1}{32} \, {\left (32 \, x^{3} e^{\left (3 \, \log \relax (5)^{2} + 3\right )} - 24 \, x^{2} e^{\left (2 \, \log \relax (5)^{2} + 2\right )} + 12 \, x e^{\left (\log \relax (5)^{2} + 1\right )} - 3\right )} e^{\left (4 \, x e^{\left (\log \relax (5)^{2} + 1\right )} - 4 \, \log \relax (5)^{2} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x^4*exp(1)*exp(log(5)^2)-4*x^3)*exp(x*exp(1)*exp(log(5)^2))^4,x, algorithm="maxima")

[Out]

-1/32*(32*x^4*e^(4*log(5)^2 + 4) - 32*x^3*e^(3*log(5)^2 + 3) + 24*x^2*e^(2*log(5)^2 + 2) - 12*x*e^(log(5)^2 +
1) + 3)*e^(4*x*e^(log(5)^2 + 1) - 4*log(5)^2 - 4) - 1/32*(32*x^3*e^(3*log(5)^2 + 3) - 24*x^2*e^(2*log(5)^2 + 2
) + 12*x*e^(log(5)^2 + 1) - 3)*e^(4*x*e^(log(5)^2 + 1) - 4*log(5)^2 - 4)

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mupad [B]  time = 0.13, size = 16, normalized size = 0.80 \begin {gather*} -x^4\,{\mathrm {e}}^{4\,x\,{\mathrm {e}}^{{\ln \relax (5)}^2}\,\mathrm {e}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-exp(4*x*exp(log(5)^2)*exp(1))*(4*x^3 + 4*x^4*exp(log(5)^2)*exp(1)),x)

[Out]

-x^4*exp(4*x*exp(log(5)^2)*exp(1))

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sympy [A]  time = 0.11, size = 19, normalized size = 0.95 \begin {gather*} - x^{4} e^{4 e x e^{\log {\relax (5 )}^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-4*x**4*exp(1)*exp(ln(5)**2)-4*x**3)*exp(x*exp(1)*exp(ln(5)**2))**4,x)

[Out]

-x**4*exp(4*E*x*exp(log(5)**2))

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