3.49.9 \(\int (e^{3+x} (4-16 x+16 x^2+32 x^3-32 x^4-32 x^5-4 x^6)+e^{3+x} (4-12 x-20 x^2+28 x^3+28 x^4+4 x^5) \log (x)) \, dx\)

Optimal. Leaf size=25 \[ e^{3+x} x (2-x (2+2 x))^2 (-x+\log (x)) \]

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Rubi [B]  time = 0.75, antiderivative size = 109, normalized size of antiderivative = 4.36, number of steps used = 50, number of rules used = 5, integrand size = 70, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.071, Rules used = {2196, 2194, 2176, 2554, 12} \begin {gather*} -4 e^{x+3} x^6-8 e^{x+3} x^5+4 e^{x+3} x^5 \log (x)+4 e^{x+3} x^4+8 e^{x+3} x^4 \log (x)+8 e^{x+3} x^3-4 e^{x+3} x^3 \log (x)-4 e^{x+3} x^2-8 e^{x+3} x^2 \log (x)+4 e^{x+3} x \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(3 + x)*(4 - 16*x + 16*x^2 + 32*x^3 - 32*x^4 - 32*x^5 - 4*x^6) + E^(3 + x)*(4 - 12*x - 20*x^2 + 28*x^3 +
28*x^4 + 4*x^5)*Log[x],x]

[Out]

-4*E^(3 + x)*x^2 + 8*E^(3 + x)*x^3 + 4*E^(3 + x)*x^4 - 8*E^(3 + x)*x^5 - 4*E^(3 + x)*x^6 + 4*E^(3 + x)*x*Log[x
] - 8*E^(3 + x)*x^2*Log[x] - 4*E^(3 + x)*x^3*Log[x] + 8*E^(3 + x)*x^4*Log[x] + 4*E^(3 + x)*x^5*Log[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 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

Rule 2554

Int[Log[u_]*(v_), x_Symbol] :> With[{w = IntHide[v, x]}, Dist[Log[u], w, x] - Int[SimplifyIntegrand[(w*D[u, x]
)/u, x], x] /; InverseFunctionFreeQ[w, x]] /; InverseFunctionFreeQ[u, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int e^{3+x} \left (4-16 x+16 x^2+32 x^3-32 x^4-32 x^5-4 x^6\right ) \, dx+\int e^{3+x} \left (4-12 x-20 x^2+28 x^3+28 x^4+4 x^5\right ) \log (x) \, dx\\ &=4 e^{3+x} x \log (x)-8 e^{3+x} x^2 \log (x)-4 e^{3+x} x^3 \log (x)+8 e^{3+x} x^4 \log (x)+4 e^{3+x} x^5 \log (x)-\int 4 e^{3+x} \left (1-x-x^2\right )^2 \, dx+\int \left (4 e^{3+x}-16 e^{3+x} x+16 e^{3+x} x^2+32 e^{3+x} x^3-32 e^{3+x} x^4-32 e^{3+x} x^5-4 e^{3+x} x^6\right ) \, dx\\ &=4 e^{3+x} x \log (x)-8 e^{3+x} x^2 \log (x)-4 e^{3+x} x^3 \log (x)+8 e^{3+x} x^4 \log (x)+4 e^{3+x} x^5 \log (x)+4 \int e^{3+x} \, dx-4 \int e^{3+x} x^6 \, dx-4 \int e^{3+x} \left (1-x-x^2\right )^2 \, dx-16 \int e^{3+x} x \, dx+16 \int e^{3+x} x^2 \, dx+32 \int e^{3+x} x^3 \, dx-32 \int e^{3+x} x^4 \, dx-32 \int e^{3+x} x^5 \, dx\\ &=4 e^{3+x}-16 e^{3+x} x+16 e^{3+x} x^2+32 e^{3+x} x^3-32 e^{3+x} x^4-32 e^{3+x} x^5-4 e^{3+x} x^6+4 e^{3+x} x \log (x)-8 e^{3+x} x^2 \log (x)-4 e^{3+x} x^3 \log (x)+8 e^{3+x} x^4 \log (x)+4 e^{3+x} x^5 \log (x)-4 \int \left (e^{3+x}-2 e^{3+x} x-e^{3+x} x^2+2 e^{3+x} x^3+e^{3+x} x^4\right ) \, dx+16 \int e^{3+x} \, dx+24 \int e^{3+x} x^5 \, dx-32 \int e^{3+x} x \, dx-96 \int e^{3+x} x^2 \, dx+128 \int e^{3+x} x^3 \, dx+160 \int e^{3+x} x^4 \, dx\\ &=20 e^{3+x}-48 e^{3+x} x-80 e^{3+x} x^2+160 e^{3+x} x^3+128 e^{3+x} x^4-8 e^{3+x} x^5-4 e^{3+x} x^6+4 e^{3+x} x \log (x)-8 e^{3+x} x^2 \log (x)-4 e^{3+x} x^3 \log (x)+8 e^{3+x} x^4 \log (x)+4 e^{3+x} x^5 \log (x)-4 \int e^{3+x} \, dx+4 \int e^{3+x} x^2 \, dx-4 \int e^{3+x} x^4 \, dx+8 \int e^{3+x} x \, dx-8 \int e^{3+x} x^3 \, dx+32 \int e^{3+x} \, dx-120 \int e^{3+x} x^4 \, dx+192 \int e^{3+x} x \, dx-384 \int e^{3+x} x^2 \, dx-640 \int e^{3+x} x^3 \, dx\\ &=48 e^{3+x}+152 e^{3+x} x-460 e^{3+x} x^2-488 e^{3+x} x^3+4 e^{3+x} x^4-8 e^{3+x} x^5-4 e^{3+x} x^6+4 e^{3+x} x \log (x)-8 e^{3+x} x^2 \log (x)-4 e^{3+x} x^3 \log (x)+8 e^{3+x} x^4 \log (x)+4 e^{3+x} x^5 \log (x)-8 \int e^{3+x} \, dx-8 \int e^{3+x} x \, dx+16 \int e^{3+x} x^3 \, dx+24 \int e^{3+x} x^2 \, dx-192 \int e^{3+x} \, dx+480 \int e^{3+x} x^3 \, dx+768 \int e^{3+x} x \, dx+1920 \int e^{3+x} x^2 \, dx\\ &=-152 e^{3+x}+912 e^{3+x} x+1484 e^{3+x} x^2+8 e^{3+x} x^3+4 e^{3+x} x^4-8 e^{3+x} x^5-4 e^{3+x} x^6+4 e^{3+x} x \log (x)-8 e^{3+x} x^2 \log (x)-4 e^{3+x} x^3 \log (x)+8 e^{3+x} x^4 \log (x)+4 e^{3+x} x^5 \log (x)+8 \int e^{3+x} \, dx-48 \int e^{3+x} x \, dx-48 \int e^{3+x} x^2 \, dx-768 \int e^{3+x} \, dx-1440 \int e^{3+x} x^2 \, dx-3840 \int e^{3+x} x \, dx\\ &=-912 e^{3+x}-2976 e^{3+x} x-4 e^{3+x} x^2+8 e^{3+x} x^3+4 e^{3+x} x^4-8 e^{3+x} x^5-4 e^{3+x} x^6+4 e^{3+x} x \log (x)-8 e^{3+x} x^2 \log (x)-4 e^{3+x} x^3 \log (x)+8 e^{3+x} x^4 \log (x)+4 e^{3+x} x^5 \log (x)+48 \int e^{3+x} \, dx+96 \int e^{3+x} x \, dx+2880 \int e^{3+x} x \, dx+3840 \int e^{3+x} \, dx\\ &=2976 e^{3+x}-4 e^{3+x} x^2+8 e^{3+x} x^3+4 e^{3+x} x^4-8 e^{3+x} x^5-4 e^{3+x} x^6+4 e^{3+x} x \log (x)-8 e^{3+x} x^2 \log (x)-4 e^{3+x} x^3 \log (x)+8 e^{3+x} x^4 \log (x)+4 e^{3+x} x^5 \log (x)-96 \int e^{3+x} \, dx-2880 \int e^{3+x} \, dx\\ &=-4 e^{3+x} x^2+8 e^{3+x} x^3+4 e^{3+x} x^4-8 e^{3+x} x^5-4 e^{3+x} x^6+4 e^{3+x} x \log (x)-8 e^{3+x} x^2 \log (x)-4 e^{3+x} x^3 \log (x)+8 e^{3+x} x^4 \log (x)+4 e^{3+x} x^5 \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.28, size = 22, normalized size = 0.88 \begin {gather*} -4 e^{3+x} x \left (-1+x+x^2\right )^2 (x-\log (x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(3 + x)*(4 - 16*x + 16*x^2 + 32*x^3 - 32*x^4 - 32*x^5 - 4*x^6) + E^(3 + x)*(4 - 12*x - 20*x^2 + 28
*x^3 + 28*x^4 + 4*x^5)*Log[x],x]

[Out]

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

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fricas [B]  time = 0.58, size = 57, normalized size = 2.28 \begin {gather*} 4 \, {\left (x^{5} + 2 \, x^{4} - x^{3} - 2 \, x^{2} + x\right )} e^{\left (x + 3\right )} \log \relax (x) - 4 \, {\left (x^{6} + 2 \, x^{5} - x^{4} - 2 \, x^{3} + x^{2}\right )} e^{\left (x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^5+28*x^4+28*x^3-20*x^2-12*x+4)*exp(2)*exp(x+1)*log(x)+(-4*x^6-32*x^5-32*x^4+32*x^3+16*x^2-16*x+
4)*exp(2)*exp(x+1),x, algorithm="fricas")

[Out]

4*(x^5 + 2*x^4 - x^3 - 2*x^2 + x)*e^(x + 3)*log(x) - 4*(x^6 + 2*x^5 - x^4 - 2*x^3 + x^2)*e^(x + 3)

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giac [B]  time = 0.12, size = 130, normalized size = 5.20 \begin {gather*} 4 \, {\left (x^{5} + 2 \, x^{4} - x^{3} - 2 \, x^{2} + x\right )} e^{\left (x + 3\right )} \log \relax (x) - 4 \, {\left (x^{6} + 2 \, x^{5} - 2 \, x^{4} - 4 \, x^{2} + 12 \, x - 13\right )} e^{\left (x + 3\right )} - 4 \, {\left (x^{4} - 4 \, x^{3} + 12 \, x^{2} - 24 \, x + 24\right )} e^{\left (x + 3\right )} - 8 \, {\left (x^{3} - 3 \, x^{2} + 6 \, x - 6\right )} e^{\left (x + 3\right )} + 4 \, {\left (x^{2} - 2 \, x + 2\right )} e^{\left (x + 3\right )} + 8 \, {\left (x - 1\right )} e^{\left (x + 3\right )} - 4 \, e^{\left (x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^5+28*x^4+28*x^3-20*x^2-12*x+4)*exp(2)*exp(x+1)*log(x)+(-4*x^6-32*x^5-32*x^4+32*x^3+16*x^2-16*x+
4)*exp(2)*exp(x+1),x, algorithm="giac")

[Out]

4*(x^5 + 2*x^4 - x^3 - 2*x^2 + x)*e^(x + 3)*log(x) - 4*(x^6 + 2*x^5 - 2*x^4 - 4*x^2 + 12*x - 13)*e^(x + 3) - 4
*(x^4 - 4*x^3 + 12*x^2 - 24*x + 24)*e^(x + 3) - 8*(x^3 - 3*x^2 + 6*x - 6)*e^(x + 3) + 4*(x^2 - 2*x + 2)*e^(x +
3) + 8*(x - 1)*e^(x + 3) - 4*e^(x + 3)

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maple [B]  time = 0.10, size = 83, normalized size = 3.32

method	result	size
risch	\(4 \ln \relax (x ) x \left (x^{4}+2 x^{3}-x^{2}-2 x +1\right ) {\mathrm e}^{3+x}-4 \left (x^{4}-2 x^{3}+5 x^{2}-12 x +13\right ) {\mathrm e}^{3+x}+\left (-4 x^{6}-8 x^{5}+8 x^{4}+16 x^{2}-48 x +52\right ) {\mathrm e}^{3+x}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x^5+28*x^4+28*x^3-20*x^2-12*x+4)*exp(2)*exp(x+1)*ln(x)+(-4*x^6-32*x^5-32*x^4+32*x^3+16*x^2-16*x+4)*exp(
2)*exp(x+1),x,method=_RETURNVERBOSE)

[Out]

4*ln(x)*x*(x^4+2*x^3-x^2-2*x+1)*exp(3+x)-4*(x^4-2*x^3+5*x^2-12*x+13)*exp(3+x)+(-4*x^6-8*x^5+8*x^4+16*x^2-48*x+
52)*exp(3+x)

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maxima [B]  time = 0.37, size = 262, normalized size = 10.48 \begin {gather*} 4 \, {\left (x^{5} e^{3} + 2 \, x^{4} e^{3} - x^{3} e^{3} - 2 \, x^{2} e^{3} + x e^{3}\right )} e^{x} \log \relax (x) - 4 \, {\left (x^{6} e^{3} - 6 \, x^{5} e^{3} + 30 \, x^{4} e^{3} - 120 \, x^{3} e^{3} + 360 \, x^{2} e^{3} - 720 \, x e^{3} + 720 \, e^{3}\right )} e^{x} - 32 \, {\left (x^{5} e^{3} - 5 \, x^{4} e^{3} + 20 \, x^{3} e^{3} - 60 \, x^{2} e^{3} + 120 \, x e^{3} - 120 \, e^{3}\right )} e^{x} - 4 \, {\left (x^{4} e^{3} - 2 \, x^{3} e^{3} + 5 \, x^{2} e^{3} - 12 \, x e^{3} + 13 \, e^{3}\right )} e^{x} - 32 \, {\left (x^{4} e^{3} - 4 \, x^{3} e^{3} + 12 \, x^{2} e^{3} - 24 \, x e^{3} + 24 \, e^{3}\right )} e^{x} + 32 \, {\left (x^{3} e^{3} - 3 \, x^{2} e^{3} + 6 \, x e^{3} - 6 \, e^{3}\right )} e^{x} + 16 \, {\left (x^{2} e^{3} - 2 \, x e^{3} + 2 \, e^{3}\right )} e^{x} - 16 \, {\left (x e^{3} - e^{3}\right )} e^{x} + 4 \, e^{\left (x + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x^5+28*x^4+28*x^3-20*x^2-12*x+4)*exp(2)*exp(x+1)*log(x)+(-4*x^6-32*x^5-32*x^4+32*x^3+16*x^2-16*x+
4)*exp(2)*exp(x+1),x, algorithm="maxima")

[Out]

4*(x^5*e^3 + 2*x^4*e^3 - x^3*e^3 - 2*x^2*e^3 + x*e^3)*e^x*log(x) - 4*(x^6*e^3 - 6*x^5*e^3 + 30*x^4*e^3 - 120*x
^3*e^3 + 360*x^2*e^3 - 720*x*e^3 + 720*e^3)*e^x - 32*(x^5*e^3 - 5*x^4*e^3 + 20*x^3*e^3 - 60*x^2*e^3 + 120*x*e^
3 - 120*e^3)*e^x - 4*(x^4*e^3 - 2*x^3*e^3 + 5*x^2*e^3 - 12*x*e^3 + 13*e^3)*e^x - 32*(x^4*e^3 - 4*x^3*e^3 + 12*
x^2*e^3 - 24*x*e^3 + 24*e^3)*e^x + 32*(x^3*e^3 - 3*x^2*e^3 + 6*x*e^3 - 6*e^3)*e^x + 16*(x^2*e^3 - 2*x*e^3 + 2*
e^3)*e^x - 16*(x*e^3 - e^3)*e^x + 4*e^(x + 3)

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mupad [B]  time = 3.59, size = 21, normalized size = 0.84 \begin {gather*} -4\,x\,{\mathrm {e}}^{x+3}\,\left (x-\ln \relax (x)\right )\,{\left (x^2+x-1\right )}^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x + 1)*exp(2)*log(x)*(28*x^3 - 20*x^2 - 12*x + 28*x^4 + 4*x^5 + 4) - exp(x + 1)*exp(2)*(16*x - 16*x^2
- 32*x^3 + 32*x^4 + 32*x^5 + 4*x^6 - 4),x)

[Out]

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

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sympy [B]  time = 0.45, size = 104, normalized size = 4.16 \begin {gather*} \left (- 4 x^{6} e^{2} + 4 x^{5} e^{2} \log {\relax (x )} - 8 x^{5} e^{2} + 8 x^{4} e^{2} \log {\relax (x )} + 4 x^{4} e^{2} - 4 x^{3} e^{2} \log {\relax (x )} + 8 x^{3} e^{2} - 8 x^{2} e^{2} \log {\relax (x )} - 4 x^{2} e^{2} + 4 x e^{2} \log {\relax (x )}\right ) e^{x + 1} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x**5+28*x**4+28*x**3-20*x**2-12*x+4)*exp(2)*exp(x+1)*ln(x)+(-4*x**6-32*x**5-32*x**4+32*x**3+16*x*
*2-16*x+4)*exp(2)*exp(x+1),x)

[Out]

(-4*x**6*exp(2) + 4*x**5*exp(2)*log(x) - 8*x**5*exp(2) + 8*x**4*exp(2)*log(x) + 4*x**4*exp(2) - 4*x**3*exp(2)*
log(x) + 8*x**3*exp(2) - 8*x**2*exp(2)*log(x) - 4*x**2*exp(2) + 4*x*exp(2)*log(x))*exp(x + 1)

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