3.20.87 \(\int \frac {64 e^3+e^3 (384+192 x) \log (2)+e^3 (576+576 x+108 x^2) \log ^2(2)+e^x (-64-64 x+(-384-576 x-96 x^2) \log (2)+(-576-1152 x-396 x^2-36 x^3) \log ^2(2))}{e^3 \log ^2(2)} \, dx\)

Optimal. Leaf size=25 \[ \left (x-e^{-3+x} x\right ) \left (6 (4+x)+\frac {8}{\log (2)}\right )^2 \]

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Rubi [B]  time = 0.22, antiderivative size = 98, normalized size of antiderivative = 3.92, number of steps used = 28, number of rules used = 4, integrand size = 85, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.047, Rules used = {12, 2196, 2194, 2176} \begin {gather*} -36 e^{x-3} x^3+36 x^3-288 e^{x-3} x^2+288 x^2-\frac {96 e^{x-3} x^2}{\log (2)}-576 e^{x-3} x+576 x-\frac {64 e^{x-3} x}{\log ^2(2)}+\frac {64 x}{\log ^2(2)}-\frac {384 e^{x-3} x}{\log (2)}+\frac {96 (x+2)^2}{\log (2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(64*E^3 + E^3*(384 + 192*x)*Log[2] + E^3*(576 + 576*x + 108*x^2)*Log[2]^2 + E^x*(-64 - 64*x + (-384 - 576*
x - 96*x^2)*Log[2] + (-576 - 1152*x - 396*x^2 - 36*x^3)*Log[2]^2))/(E^3*Log[2]^2),x]

[Out]

576*x - 576*E^(-3 + x)*x + 288*x^2 - 288*E^(-3 + x)*x^2 + 36*x^3 - 36*E^(-3 + x)*x^3 + (64*x)/Log[2]^2 - (64*E
^(-3 + x)*x)/Log[2]^2 - (384*E^(-3 + x)*x)/Log[2] - (96*E^(-3 + x)*x^2)/Log[2] + (96*(2 + x)^2)/Log[2]

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 {\int \left (64 e^3+e^3 (384+192 x) \log (2)+e^3 \left (576+576 x+108 x^2\right ) \log ^2(2)+e^x \left (-64-64 x+\left (-384-576 x-96 x^2\right ) \log (2)+\left (-576-1152 x-396 x^2-36 x^3\right ) \log ^2(2)\right )\right ) \, dx}{e^3 \log ^2(2)}\\ &=\frac {64 x}{\log ^2(2)}+\frac {96 (2+x)^2}{\log (2)}+\frac {\int e^x \left (-64-64 x+\left (-384-576 x-96 x^2\right ) \log (2)+\left (-576-1152 x-396 x^2-36 x^3\right ) \log ^2(2)\right ) \, dx}{e^3 \log ^2(2)}+\int \left (576+576 x+108 x^2\right ) \, dx\\ &=576 x+288 x^2+36 x^3+\frac {64 x}{\log ^2(2)}+\frac {96 (2+x)^2}{\log (2)}+\frac {\int \left (-64 e^x-64 e^x x-96 e^x \left (4+6 x+x^2\right ) \log (2)-36 e^x \left (16+32 x+11 x^2+x^3\right ) \log ^2(2)\right ) \, dx}{e^3 \log ^2(2)}\\ &=576 x+288 x^2+36 x^3+\frac {64 x}{\log ^2(2)}+\frac {96 (2+x)^2}{\log (2)}-\frac {36 \int e^x \left (16+32 x+11 x^2+x^3\right ) \, dx}{e^3}-\frac {64 \int e^x \, dx}{e^3 \log ^2(2)}-\frac {64 \int e^x x \, dx}{e^3 \log ^2(2)}-\frac {96 \int e^x \left (4+6 x+x^2\right ) \, dx}{e^3 \log (2)}\\ &=576 x+288 x^2+36 x^3-\frac {64 e^{-3+x}}{\log ^2(2)}+\frac {64 x}{\log ^2(2)}-\frac {64 e^{-3+x} x}{\log ^2(2)}+\frac {96 (2+x)^2}{\log (2)}-\frac {36 \int \left (16 e^x+32 e^x x+11 e^x x^2+e^x x^3\right ) \, dx}{e^3}+\frac {64 \int e^x \, dx}{e^3 \log ^2(2)}-\frac {96 \int \left (4 e^x+6 e^x x+e^x x^2\right ) \, dx}{e^3 \log (2)}\\ &=576 x+288 x^2+36 x^3+\frac {64 x}{\log ^2(2)}-\frac {64 e^{-3+x} x}{\log ^2(2)}+\frac {96 (2+x)^2}{\log (2)}-\frac {36 \int e^x x^3 \, dx}{e^3}-\frac {396 \int e^x x^2 \, dx}{e^3}-\frac {576 \int e^x \, dx}{e^3}-\frac {1152 \int e^x x \, dx}{e^3}-\frac {96 \int e^x x^2 \, dx}{e^3 \log (2)}-\frac {384 \int e^x \, dx}{e^3 \log (2)}-\frac {576 \int e^x x \, dx}{e^3 \log (2)}\\ &=-576 e^{-3+x}+576 x-1152 e^{-3+x} x+288 x^2-396 e^{-3+x} x^2+36 x^3-36 e^{-3+x} x^3+\frac {64 x}{\log ^2(2)}-\frac {64 e^{-3+x} x}{\log ^2(2)}-\frac {384 e^{-3+x}}{\log (2)}-\frac {576 e^{-3+x} x}{\log (2)}-\frac {96 e^{-3+x} x^2}{\log (2)}+\frac {96 (2+x)^2}{\log (2)}+\frac {108 \int e^x x^2 \, dx}{e^3}+\frac {792 \int e^x x \, dx}{e^3}+\frac {1152 \int e^x \, dx}{e^3}+\frac {192 \int e^x x \, dx}{e^3 \log (2)}+\frac {576 \int e^x \, dx}{e^3 \log (2)}\\ &=576 e^{-3+x}+576 x-360 e^{-3+x} x+288 x^2-288 e^{-3+x} x^2+36 x^3-36 e^{-3+x} x^3+\frac {64 x}{\log ^2(2)}-\frac {64 e^{-3+x} x}{\log ^2(2)}+\frac {192 e^{-3+x}}{\log (2)}-\frac {384 e^{-3+x} x}{\log (2)}-\frac {96 e^{-3+x} x^2}{\log (2)}+\frac {96 (2+x)^2}{\log (2)}-\frac {216 \int e^x x \, dx}{e^3}-\frac {792 \int e^x \, dx}{e^3}-\frac {192 \int e^x \, dx}{e^3 \log (2)}\\ &=-216 e^{-3+x}+576 x-576 e^{-3+x} x+288 x^2-288 e^{-3+x} x^2+36 x^3-36 e^{-3+x} x^3+\frac {64 x}{\log ^2(2)}-\frac {64 e^{-3+x} x}{\log ^2(2)}-\frac {384 e^{-3+x} x}{\log (2)}-\frac {96 e^{-3+x} x^2}{\log (2)}+\frac {96 (2+x)^2}{\log (2)}+\frac {216 \int e^x \, dx}{e^3}\\ &=576 x-576 e^{-3+x} x+288 x^2-288 e^{-3+x} x^2+36 x^3-36 e^{-3+x} x^3+\frac {64 x}{\log ^2(2)}-\frac {64 e^{-3+x} x}{\log ^2(2)}-\frac {384 e^{-3+x} x}{\log (2)}-\frac {96 e^{-3+x} x^2}{\log (2)}+\frac {96 (2+x)^2}{\log (2)}\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.27, size = 89, normalized size = 3.56 \begin {gather*} -\frac {4 \left (-\frac {1}{2} e^3 x \left (2 x^2 \log ^2(8)+x \log (4096) (4+\log (4096))+2 (4+\log (4096))^2\right )+e^x x \left (16+6 \log ^2(8)+x^2 \log ^2(8)+21 \log (2) (4+\log (64))+\log (4096)-\log (8) \log (4096)+x \log (8) (8+\log (16777216))\right )\right )}{e^3 \log ^2(2)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(64*E^3 + E^3*(384 + 192*x)*Log[2] + E^3*(576 + 576*x + 108*x^2)*Log[2]^2 + E^x*(-64 - 64*x + (-384
- 576*x - 96*x^2)*Log[2] + (-576 - 1152*x - 396*x^2 - 36*x^3)*Log[2]^2))/(E^3*Log[2]^2),x]

[Out]

(-4*(-1/2*(E^3*x*(2*x^2*Log[8]^2 + x*Log[4096]*(4 + Log[4096]) + 2*(4 + Log[4096])^2)) + E^x*x*(16 + 6*Log[8]^
2 + x^2*Log[8]^2 + 21*Log[2]*(4 + Log[64]) + Log[4096] - Log[8]*Log[4096] + x*Log[8]*(8 + Log[16777216]))))/(E
^3*Log[2]^2)

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fricas [B]  time = 0.79, size = 84, normalized size = 3.36 \begin {gather*} \frac {4 \, {\left (9 \, {\left (x^{3} + 8 \, x^{2} + 16 \, x\right )} e^{3} \log \relax (2)^{2} + 24 \, {\left (x^{2} + 4 \, x\right )} e^{3} \log \relax (2) + 16 \, x e^{3} - {\left (9 \, {\left (x^{3} + 8 \, x^{2} + 16 \, x\right )} \log \relax (2)^{2} + 24 \, {\left (x^{2} + 4 \, x\right )} \log \relax (2) + 16 \, x\right )} e^{x}\right )} e^{\left (-3\right )}}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-36*x^3-396*x^2-1152*x-576)*log(2)^2+(-96*x^2-576*x-384)*log(2)-64*x-64)*exp(x)+(108*x^2+576*x+57
6)*exp(3)*log(2)^2+(192*x+384)*exp(3)*log(2)+64*exp(3))/exp(3)/log(2)^2,x, algorithm="fricas")

[Out]

4*(9*(x^3 + 8*x^2 + 16*x)*e^3*log(2)^2 + 24*(x^2 + 4*x)*e^3*log(2) + 16*x*e^3 - (9*(x^3 + 8*x^2 + 16*x)*log(2)
^2 + 24*(x^2 + 4*x)*log(2) + 16*x)*e^x)*e^(-3)/log(2)^2

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giac [B]  time = 0.15, size = 92, normalized size = 3.68 \begin {gather*} \frac {4 \, {\left (9 \, {\left (x^{3} + 8 \, x^{2} + 16 \, x\right )} e^{3} \log \relax (2)^{2} + 24 \, {\left (x^{2} + 4 \, x\right )} e^{3} \log \relax (2) + 16 \, x e^{3} - {\left (9 \, x^{3} \log \relax (2)^{2} + 72 \, x^{2} \log \relax (2)^{2} + 24 \, x^{2} \log \relax (2) + 144 \, x \log \relax (2)^{2} + 96 \, x \log \relax (2) + 16 \, x\right )} e^{x}\right )} e^{\left (-3\right )}}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-36*x^3-396*x^2-1152*x-576)*log(2)^2+(-96*x^2-576*x-384)*log(2)-64*x-64)*exp(x)+(108*x^2+576*x+57
6)*exp(3)*log(2)^2+(192*x+384)*exp(3)*log(2)+64*exp(3))/exp(3)/log(2)^2,x, algorithm="giac")

[Out]

4*(9*(x^3 + 8*x^2 + 16*x)*e^3*log(2)^2 + 24*(x^2 + 4*x)*e^3*log(2) + 16*x*e^3 - (9*x^3*log(2)^2 + 72*x^2*log(2
)^2 + 24*x^2*log(2) + 144*x*log(2)^2 + 96*x*log(2) + 16*x)*e^x)*e^(-3)/log(2)^2

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maple [B]  time = 0.05, size = 88, normalized size = 3.52

method	result	size
risch	\(36 x^{3}+288 x^{2}+576 x +\frac {96 x^{2}}{\ln \relax (2)}+\frac {384 x}{\ln \relax (2)}+\frac {64 x}{\ln \relax (2)^{2}}+\frac {\left (-36 x^{3} \ln \relax (2)^{2}-288 x^{2} \ln \relax (2)^{2}-576 x \ln \relax (2)^{2}-96 x^{2} \ln \relax (2)-384 x \ln \relax (2)-64 x \right ) {\mathrm e}^{x -3}}{\ln \relax (2)^{2}}\)	\(88\)
norman	\(\frac {\left (288 \ln \relax (2)+96\right ) x^{2}+36 x^{3} \ln \relax (2)+\frac {64 \left (9 \ln \relax (2)^{2}+6 \ln \relax (2)+1\right ) x}{\ln \relax (2)}-96 \left (3 \ln \relax (2)+1\right ) {\mathrm e}^{-3} x^{2} {\mathrm e}^{x}-36 \ln \relax (2) {\mathrm e}^{-3} x^{3} {\mathrm e}^{x}-\frac {64 \left (9 \ln \relax (2)^{2}+6 \ln \relax (2)+1\right ) {\mathrm e}^{-3} x \,{\mathrm e}^{x}}{\ln \relax (2)}}{\ln \relax (2)}\)	\(98\)
default	\(\frac {{\mathrm e}^{-3} \left (-64 \,{\mathrm e}^{x} x -96 x^{2} \ln \relax (2) {\mathrm e}^{x}-384 x \ln \relax (2) {\mathrm e}^{x}-288 \,{\mathrm e}^{x} \ln \relax (2)^{2} x^{2}-576 x \ln \relax (2)^{2} {\mathrm e}^{x}-36 \,{\mathrm e}^{x} \ln \relax (2)^{2} x^{3}+96 \,{\mathrm e}^{3} \ln \relax (2) x^{2}+384 \,{\mathrm e}^{3} \ln \relax (2) x +36 \,{\mathrm e}^{3} \ln \relax (2)^{2} x^{3}+288 \,{\mathrm e}^{3} \ln \relax (2)^{2} x^{2}+576 \,{\mathrm e}^{3} \ln \relax (2)^{2} x +64 x \,{\mathrm e}^{3}\right )}{\ln \relax (2)^{2}}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-36*x^3-396*x^2-1152*x-576)*ln(2)^2+(-96*x^2-576*x-384)*ln(2)-64*x-64)*exp(x)+(108*x^2+576*x+576)*exp(3
)*ln(2)^2+(192*x+384)*exp(3)*ln(2)+64*exp(3))/exp(3)/ln(2)^2,x,method=_RETURNVERBOSE)

[Out]

36*x^3+288*x^2+576*x+96*x^2/ln(2)+384*x/ln(2)+64*x/ln(2)^2+(-36*x^3*ln(2)^2-288*x^2*ln(2)^2-576*x*ln(2)^2-96*x
^2*ln(2)-384*x*ln(2)-64*x)/ln(2)^2*exp(x-3)

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maxima [B]  time = 1.19, size = 90, normalized size = 3.60 \begin {gather*} \frac {4 \, {\left (9 \, {\left (x^{3} + 8 \, x^{2} + 16 \, x\right )} e^{3} \log \relax (2)^{2} + 24 \, {\left (x^{2} + 4 \, x\right )} e^{3} \log \relax (2) + 16 \, x e^{3} - {\left (9 \, x^{3} \log \relax (2)^{2} + 24 \, {\left (3 \, \log \relax (2)^{2} + \log \relax (2)\right )} x^{2} + 16 \, {\left (9 \, \log \relax (2)^{2} + 6 \, \log \relax (2) + 1\right )} x\right )} e^{x}\right )} e^{\left (-3\right )}}{\log \relax (2)^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-36*x^3-396*x^2-1152*x-576)*log(2)^2+(-96*x^2-576*x-384)*log(2)-64*x-64)*exp(x)+(108*x^2+576*x+57
6)*exp(3)*log(2)^2+(192*x+384)*exp(3)*log(2)+64*exp(3))/exp(3)/log(2)^2,x, algorithm="maxima")

[Out]

4*(9*(x^3 + 8*x^2 + 16*x)*e^3*log(2)^2 + 24*(x^2 + 4*x)*e^3*log(2) + 16*x*e^3 - (9*x^3*log(2)^2 + 24*(3*log(2)
^2 + log(2))*x^2 + 16*(9*log(2)^2 + 6*log(2) + 1)*x)*e^x)*e^(-3)/log(2)^2

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mupad [B]  time = 0.11, size = 90, normalized size = 3.60 \begin {gather*} \frac {36\,x^3\,{\ln \relax (2)}^2+x\,\left (384\,\ln \relax (2)+576\,{\ln \relax (2)}^2+64\right )-64\,x\,{\mathrm {e}}^{x-3}\,\left (\ln \left (64\right )+9\,{\ln \relax (2)}^2+1\right )-36\,x^3\,{\mathrm {e}}^{x-3}\,{\ln \relax (2)}^2-96\,x^2\,{\mathrm {e}}^{x-3}\,\left (\ln \relax (2)+3\,{\ln \relax (2)}^2\right )+96\,x^2\,\ln \relax (2)\,\left (3\,\ln \relax (2)+1\right )}{{\ln \relax (2)}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(-3)*(64*exp(3) - exp(x)*(64*x + log(2)*(576*x + 96*x^2 + 384) + log(2)^2*(1152*x + 396*x^2 + 36*x^3 +
576) + 64) + exp(3)*log(2)*(192*x + 384) + exp(3)*log(2)^2*(576*x + 108*x^2 + 576)))/log(2)^2,x)

[Out]

(36*x^3*log(2)^2 + x*(384*log(2) + 576*log(2)^2 + 64) - 64*x*exp(x - 3)*(log(64) + 9*log(2)^2 + 1) - 36*x^3*ex
p(x - 3)*log(2)^2 - 96*x^2*exp(x - 3)*(log(2) + 3*log(2)^2) + 96*x^2*log(2)*(3*log(2) + 1))/log(2)^2

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sympy [B]  time = 0.22, size = 99, normalized size = 3.96 \begin {gather*} 36 x^{3} + \frac {x^{2} \left (96 + 288 \log {\relax (2 )}\right )}{\log {\relax (2 )}} + \frac {x \left (64 + 384 \log {\relax (2 )} + 576 \log {\relax (2 )}^{2}\right )}{\log {\relax (2 )}^{2}} + \frac {\left (- 36 x^{3} \log {\relax (2 )}^{2} - 288 x^{2} \log {\relax (2 )}^{2} - 96 x^{2} \log {\relax (2 )} - 576 x \log {\relax (2 )}^{2} - 384 x \log {\relax (2 )} - 64 x\right ) e^{x}}{e^{3} \log {\relax (2 )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-36*x**3-396*x**2-1152*x-576)*ln(2)**2+(-96*x**2-576*x-384)*ln(2)-64*x-64)*exp(x)+(108*x**2+576*x
+576)*exp(3)*ln(2)**2+(192*x+384)*exp(3)*ln(2)+64*exp(3))/exp(3)/ln(2)**2,x)

[Out]

36*x**3 + x**2*(96 + 288*log(2))/log(2) + x*(64 + 384*log(2) + 576*log(2)**2)/log(2)**2 + (-36*x**3*log(2)**2
- 288*x**2*log(2)**2 - 96*x**2*log(2) - 576*x*log(2)**2 - 384*x*log(2) - 64*x)*exp(-3)*exp(x)/log(2)**2

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