3.79.23 \(\int \frac {1}{3} e^{2 x} (4-50 x-6 x^2) \log ^4(4) \, dx\)

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

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Rubi [A]  time = 0.05, antiderivative size = 42, normalized size of antiderivative = 1.45, number of steps used = 9, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {12, 2196, 2194, 2176} \begin {gather*} -e^{2 x} x^2 \log ^4(4)-\frac {22}{3} e^{2 x} x \log ^4(4)+\frac {13}{3} e^{2 x} \log ^4(4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(2*x)*(4 - 50*x - 6*x^2)*Log[4]^4)/3,x]

[Out]

(13*E^(2*x)*Log[4]^4)/3 - (22*E^(2*x)*x*Log[4]^4)/3 - E^(2*x)*x^2*Log[4]^4

Rule 12

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

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} &=\frac {1}{3} \log ^4(4) \int e^{2 x} \left (4-50 x-6 x^2\right ) \, dx\\ &=\frac {1}{3} \log ^4(4) \int \left (4 e^{2 x}-50 e^{2 x} x-6 e^{2 x} x^2\right ) \, dx\\ &=\frac {1}{3} \left (4 \log ^4(4)\right ) \int e^{2 x} \, dx-\left (2 \log ^4(4)\right ) \int e^{2 x} x^2 \, dx-\frac {1}{3} \left (50 \log ^4(4)\right ) \int e^{2 x} x \, dx\\ &=\frac {2}{3} e^{2 x} \log ^4(4)-\frac {25}{3} e^{2 x} x \log ^4(4)-e^{2 x} x^2 \log ^4(4)+\left (2 \log ^4(4)\right ) \int e^{2 x} x \, dx+\frac {1}{3} \left (25 \log ^4(4)\right ) \int e^{2 x} \, dx\\ &=\frac {29}{6} e^{2 x} \log ^4(4)-\frac {22}{3} e^{2 x} x \log ^4(4)-e^{2 x} x^2 \log ^4(4)-\log ^4(4) \int e^{2 x} \, dx\\ &=\frac {13}{3} e^{2 x} \log ^4(4)-\frac {22}{3} e^{2 x} x \log ^4(4)-e^{2 x} x^2 \log ^4(4)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 27, normalized size = 0.93 \begin {gather*} -\frac {2}{3} e^{2 x} \left (-\frac {13}{2}+11 x+\frac {3 x^2}{2}\right ) \log ^4(4) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(2*x)*(4 - 50*x - 6*x^2)*Log[4]^4)/3,x]

[Out]

(-2*E^(2*x)*(-13/2 + 11*x + (3*x^2)/2)*Log[4]^4)/3

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fricas [A]  time = 0.64, size = 20, normalized size = 0.69 \begin {gather*} -\frac {16}{3} \, {\left (3 \, x^{2} + 22 \, x - 13\right )} e^{\left (2 \, x\right )} \log \relax (2)^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16/3*(-6*x^2-50*x+4)*log(2)^4*exp(x)^2,x, algorithm="fricas")

[Out]

-16/3*(3*x^2 + 22*x - 13)*e^(2*x)*log(2)^4

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giac [A]  time = 0.15, size = 20, normalized size = 0.69 \begin {gather*} -\frac {16}{3} \, {\left (3 \, x^{2} + 22 \, x - 13\right )} e^{\left (2 \, x\right )} \log \relax (2)^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16/3*(-6*x^2-50*x+4)*log(2)^4*exp(x)^2,x, algorithm="giac")

[Out]

-16/3*(3*x^2 + 22*x - 13)*e^(2*x)*log(2)^4

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maple [A]  time = 0.03, size = 21, normalized size = 0.72

method	result	size
gosper	\(-\frac {16 \,{\mathrm e}^{2 x} \ln \relax (2)^{4} \left (3 x^{2}+22 x -13\right )}{3}\)	\(21\)
risch	\(\frac {16 \ln \relax (2)^{4} \left (-3 x^{2}-22 x +13\right ) {\mathrm e}^{2 x}}{3}\)	\(21\)
default	\(\frac {32 \ln \relax (2)^{4} \left (\frac {13 \,{\mathrm e}^{2 x}}{2}-11 x \,{\mathrm e}^{2 x}-\frac {3 \,{\mathrm e}^{2 x} x^{2}}{2}\right )}{3}\)	\(30\)
norman	\(\frac {208 \ln \relax (2)^{4} {\mathrm e}^{2 x}}{3}-\frac {352 x \ln \relax (2)^{4} {\mathrm e}^{2 x}}{3}-16 x^{2} \ln \relax (2)^{4} {\mathrm e}^{2 x}\)	\(36\)
meijerg	\(-\frac {32 \ln \relax (2)^{4} \left (1-{\mathrm e}^{2 x}\right )}{3}+4 \ln \relax (2)^{4} \left (2-\frac {\left (12 x^{2}-12 x +6\right ) {\mathrm e}^{2 x}}{3}\right )-\frac {200 \ln \relax (2)^{4} \left (1-\frac {\left (-4 x +2\right ) {\mathrm e}^{2 x}}{2}\right )}{3}\)	\(59\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(16/3*(-6*x^2-50*x+4)*ln(2)^4*exp(x)^2,x,method=_RETURNVERBOSE)

[Out]

-16/3*exp(x)^2*ln(2)^4*(3*x^2+22*x-13)

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maxima [A]  time = 0.37, size = 40, normalized size = 1.38 \begin {gather*} -\frac {8}{3} \, {\left (3 \, {\left (2 \, x^{2} - 2 \, x + 1\right )} e^{\left (2 \, x\right )} + 25 \, {\left (2 \, x - 1\right )} e^{\left (2 \, x\right )} - 4 \, e^{\left (2 \, x\right )}\right )} \log \relax (2)^{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16/3*(-6*x^2-50*x+4)*log(2)^4*exp(x)^2,x, algorithm="maxima")

[Out]

-8/3*(3*(2*x^2 - 2*x + 1)*e^(2*x) + 25*(2*x - 1)*e^(2*x) - 4*e^(2*x))*log(2)^4

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mupad [B]  time = 5.58, size = 20, normalized size = 0.69 \begin {gather*} -\frac {16\,{\mathrm {e}}^{2\,x}\,{\ln \relax (2)}^4\,\left (3\,x^2+22\,x-13\right )}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(16*exp(2*x)*log(2)^4*(50*x + 6*x^2 - 4))/3,x)

[Out]

-(16*exp(2*x)*log(2)^4*(22*x + 3*x^2 - 13))/3

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sympy [A]  time = 0.15, size = 31, normalized size = 1.07 \begin {gather*} \frac {\left (- 48 x^{2} \log {\relax (2 )}^{4} - 352 x \log {\relax (2 )}^{4} + 208 \log {\relax (2 )}^{4}\right ) e^{2 x}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(16/3*(-6*x**2-50*x+4)*ln(2)**4*exp(x)**2,x)

[Out]

(-48*x**2*log(2)**4 - 352*x*log(2)**4 + 208*log(2)**4)*exp(2*x)/3

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