3.61.65 \(\int (18 x^2-32 x^3+(-12 x+36 x^2) \log (x)-12 x \log ^2(x)+e^{e^{e^{e^x}}} (-16 x+24 x^2+(12-24 x) \log (x)+6 \log ^2(x)+e^{e^{e^x}+e^x} (e^x (-2 x^2+8 x^3)-12 e^x x^2 \log (x)+6 e^x x \log ^2(x)))) \, dx\)

Optimal. Leaf size=33 \[ 2 \left (e^{e^{e^{e^x}}}-x\right ) x \left (-x+x^2+3 (-x+\log (x))^2\right ) \]

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Rubi [B]  time = 0.25, antiderivative size = 89, normalized size of antiderivative = 2.70, number of steps used = 9, number of rules used = 7, integrand size = 111, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.063, Rules used = {1593, 43, 2334, 12, 2305, 2304, 2288} \begin {gather*} -8 x^4+2 x^3-6 x^2 \log ^2(x)+6 x^2 \log (x)-2 e^{e^{e^{e^x}}-x} \left (6 e^x x^2 \log (x)+e^x \left (x^2-4 x^3\right )-3 e^x x \log ^2(x)\right )-6 \left (x^2-2 x^3\right ) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[18*x^2 - 32*x^3 + (-12*x + 36*x^2)*Log[x] - 12*x*Log[x]^2 + E^E^E^E^x*(-16*x + 24*x^2 + (12 - 24*x)*Log[x]
+ 6*Log[x]^2 + E^(E^E^x + E^x)*(E^x*(-2*x^2 + 8*x^3) - 12*E^x*x^2*Log[x] + 6*E^x*x*Log[x]^2)),x]

[Out]

2*x^3 - 8*x^4 + 6*x^2*Log[x] - 6*(x^2 - 2*x^3)*Log[x] - 6*x^2*Log[x]^2 - 2*E^(E^E^E^x - x)*(E^x*(x^2 - 4*x^3)
+ 6*E^x*x^2*Log[x] - 3*E^x*x*Log[x]^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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 2304

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Log[c*x^
n]))/(d*(m + 1)), x] - Simp[(b*n*(d*x)^(m + 1))/(d*(m + 1)^2), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1
]

Rule 2305

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*Lo
g[c*x^n])^p)/(d*(m + 1)), x] - Dist[(b*n*p)/(m + 1), Int[(d*x)^m*(a + b*Log[c*x^n])^(p - 1), x], x] /; FreeQ[{
a, b, c, d, m, n}, x] && NeQ[m, -1] && GtQ[p, 0]

Rule 2334

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(x_)^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> With[{u = I
ntHide[x^m*(d + e*x^r)^q, x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[SimplifyIntegrand[u/x, x], x], x]
] /; FreeQ[{a, b, c, d, e, n, r}, x] && IGtQ[q, 0] && IntegerQ[m] &&  !(EqQ[q, 1] && EqQ[m, -1])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=6 x^3-8 x^4-12 \int x \log ^2(x) \, dx+\int \left (-12 x+36 x^2\right ) \log (x) \, dx+\int e^{e^{e^{e^x}}} \left (-16 x+24 x^2+(12-24 x) \log (x)+6 \log ^2(x)+e^{e^{e^x}+e^x} \left (e^x \left (-2 x^2+8 x^3\right )-12 e^x x^2 \log (x)+6 e^x x \log ^2(x)\right )\right ) \, dx\\ &=6 x^3-8 x^4-6 x^2 \log ^2(x)-2 e^{e^{e^{e^x}}-x} \left (e^x \left (x^2-4 x^3\right )+6 e^x x^2 \log (x)-3 e^x x \log ^2(x)\right )+12 \int x \log (x) \, dx+\int x (-12+36 x) \log (x) \, dx\\ &=-3 x^2+6 x^3-8 x^4+6 x^2 \log (x)-6 \left (x^2-2 x^3\right ) \log (x)-6 x^2 \log ^2(x)-2 e^{e^{e^{e^x}}-x} \left (e^x \left (x^2-4 x^3\right )+6 e^x x^2 \log (x)-3 e^x x \log ^2(x)\right )-\int 6 x (-1+2 x) \, dx\\ &=-3 x^2+6 x^3-8 x^4+6 x^2 \log (x)-6 \left (x^2-2 x^3\right ) \log (x)-6 x^2 \log ^2(x)-2 e^{e^{e^{e^x}}-x} \left (e^x \left (x^2-4 x^3\right )+6 e^x x^2 \log (x)-3 e^x x \log ^2(x)\right )-6 \int x (-1+2 x) \, dx\\ &=-3 x^2+6 x^3-8 x^4+6 x^2 \log (x)-6 \left (x^2-2 x^3\right ) \log (x)-6 x^2 \log ^2(x)-2 e^{e^{e^{e^x}}-x} \left (e^x \left (x^2-4 x^3\right )+6 e^x x^2 \log (x)-3 e^x x \log ^2(x)\right )-6 \int \left (-x+2 x^2\right ) \, dx\\ &=2 x^3-8 x^4+6 x^2 \log (x)-6 \left (x^2-2 x^3\right ) \log (x)-6 x^2 \log ^2(x)-2 e^{e^{e^{e^x}}-x} \left (e^x \left (x^2-4 x^3\right )+6 e^x x^2 \log (x)-3 e^x x \log ^2(x)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.25, size = 35, normalized size = 1.06 \begin {gather*} 2 \left (e^{e^{e^{e^x}}}-x\right ) x \left (x (-1+4 x)-6 x \log (x)+3 \log ^2(x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[18*x^2 - 32*x^3 + (-12*x + 36*x^2)*Log[x] - 12*x*Log[x]^2 + E^E^E^E^x*(-16*x + 24*x^2 + (12 - 24*x)*
Log[x] + 6*Log[x]^2 + E^(E^E^x + E^x)*(E^x*(-2*x^2 + 8*x^3) - 12*E^x*x^2*Log[x] + 6*E^x*x*Log[x]^2)),x]

[Out]

2*(E^E^E^E^x - x)*x*(x*(-1 + 4*x) - 6*x*Log[x] + 3*Log[x]^2)

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fricas [B]  time = 0.52, size = 59, normalized size = 1.79 \begin {gather*} -8 \, x^{4} + 12 \, x^{3} \log \relax (x) - 6 \, x^{2} \log \relax (x)^{2} + 2 \, x^{3} + 2 \, {\left (4 \, x^{3} - 6 \, x^{2} \log \relax (x) + 3 \, x \log \relax (x)^{2} - x^{2}\right )} e^{\left (e^{\left (e^{\left (e^{x}\right )}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x*exp(x)*log(x)^2-12*x^2*exp(x)*log(x)+(8*x^3-2*x^2)*exp(x))*exp(exp(x))*exp(exp(exp(x)))+6*log(
x)^2+(-24*x+12)*log(x)+24*x^2-16*x)*exp(exp(exp(exp(x))))-12*x*log(x)^2+(36*x^2-12*x)*log(x)-32*x^3+18*x^2,x,
algorithm="fricas")

[Out]

-8*x^4 + 12*x^3*log(x) - 6*x^2*log(x)^2 + 2*x^3 + 2*(4*x^3 - 6*x^2*log(x) + 3*x*log(x)^2 - x^2)*e^(e^(e^(e^x))
)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -32 \, x^{3} - 12 \, x \log \relax (x)^{2} + 18 \, x^{2} + 2 \, {\left (12 \, x^{2} - {\left (6 \, x^{2} e^{x} \log \relax (x) - 3 \, x e^{x} \log \relax (x)^{2} - {\left (4 \, x^{3} - x^{2}\right )} e^{x}\right )} e^{\left (e^{x} + e^{\left (e^{x}\right )}\right )} - 6 \, {\left (2 \, x - 1\right )} \log \relax (x) + 3 \, \log \relax (x)^{2} - 8 \, x\right )} e^{\left (e^{\left (e^{\left (e^{x}\right )}\right )}\right )} + 12 \, {\left (3 \, x^{2} - x\right )} \log \relax (x)\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x*exp(x)*log(x)^2-12*x^2*exp(x)*log(x)+(8*x^3-2*x^2)*exp(x))*exp(exp(x))*exp(exp(exp(x)))+6*log(
x)^2+(-24*x+12)*log(x)+24*x^2-16*x)*exp(exp(exp(exp(x))))-12*x*log(x)^2+(36*x^2-12*x)*log(x)-32*x^3+18*x^2,x,
algorithm="giac")

[Out]

integrate(-32*x^3 - 12*x*log(x)^2 + 18*x^2 + 2*(12*x^2 - (6*x^2*e^x*log(x) - 3*x*e^x*log(x)^2 - (4*x^3 - x^2)*
e^x)*e^(e^x + e^(e^x)) - 6*(2*x - 1)*log(x) + 3*log(x)^2 - 8*x)*e^(e^(e^(e^x))) + 12*(3*x^2 - x)*log(x), x)

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maple [B]  time = 0.09, size = 70, normalized size = 2.12

method	result	size
risch	\(2 x \left (4 x^{2}-6 x \ln \relax (x )+3 \ln \relax (x )^{2}-x \right ) {\mathrm e}^{{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}}}-6 x^{2} \ln \relax (x )^{2}+6 x^{2} \ln \relax (x )+\left (12 x^{3}-6 x^{2}\right ) \ln \relax (x )+2 x^{3}-8 x^{4}\)	\(70\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x*exp(x)*ln(x)^2-12*x^2*exp(x)*ln(x)+(8*x^3-2*x^2)*exp(x))*exp(exp(x))*exp(exp(exp(x)))+6*ln(x)^2+(-24
*x+12)*ln(x)+24*x^2-16*x)*exp(exp(exp(exp(x))))-12*x*ln(x)^2+(36*x^2-12*x)*ln(x)-32*x^3+18*x^2,x,method=_RETUR
NVERBOSE)

[Out]

2*x*(4*x^2-6*x*ln(x)+3*ln(x)^2-x)*exp(exp(exp(exp(x))))-6*x^2*ln(x)^2+6*x^2*ln(x)+(12*x^3-6*x^2)*ln(x)+2*x^3-8
*x^4

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maxima [B]  time = 0.37, size = 80, normalized size = 2.42 \begin {gather*} -8 \, x^{4} - 3 \, {\left (2 \, \log \relax (x)^{2} - 2 \, \log \relax (x) + 1\right )} x^{2} + 2 \, x^{3} + 3 \, x^{2} + 2 \, {\left (4 \, x^{3} - 6 \, x^{2} \log \relax (x) + 3 \, x \log \relax (x)^{2} - x^{2}\right )} e^{\left (e^{\left (e^{\left (e^{x}\right )}\right )}\right )} + 6 \, {\left (2 \, x^{3} - x^{2}\right )} \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x*exp(x)*log(x)^2-12*x^2*exp(x)*log(x)+(8*x^3-2*x^2)*exp(x))*exp(exp(x))*exp(exp(exp(x)))+6*log(
x)^2+(-24*x+12)*log(x)+24*x^2-16*x)*exp(exp(exp(exp(x))))-12*x*log(x)^2+(36*x^2-12*x)*log(x)-32*x^3+18*x^2,x,
algorithm="maxima")

[Out]

-8*x^4 - 3*(2*log(x)^2 - 2*log(x) + 1)*x^2 + 2*x^3 + 3*x^2 + 2*(4*x^3 - 6*x^2*log(x) + 3*x*log(x)^2 - x^2)*e^(
e^(e^(e^x))) + 6*(2*x^3 - x^2)*log(x)

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mupad [B]  time = 4.54, size = 58, normalized size = 1.76 \begin {gather*} 12\,x^3\,\ln \relax (x)+{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}}\,\left (8\,x^3-12\,x^2\,\ln \relax (x)-2\,x^2+6\,x\,{\ln \relax (x)}^2\right )-6\,x^2\,{\ln \relax (x)}^2+2\,x^3-8\,x^4 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(18*x^2 - exp(exp(exp(exp(x))))*(16*x - 6*log(x)^2 + log(x)*(24*x - 12) - 24*x^2 + exp(exp(x))*exp(exp(exp(
x)))*(exp(x)*(2*x^2 - 8*x^3) - 6*x*exp(x)*log(x)^2 + 12*x^2*exp(x)*log(x))) - log(x)*(12*x - 36*x^2) - 12*x*lo
g(x)^2 - 32*x^3,x)

[Out]

12*x^3*log(x) + exp(exp(exp(exp(x))))*(6*x*log(x)^2 - 12*x^2*log(x) - 2*x^2 + 8*x^3) - 6*x^2*log(x)^2 + 2*x^3
- 8*x^4

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sympy [B]  time = 13.56, size = 63, normalized size = 1.91 \begin {gather*} - 8 x^{4} + 12 x^{3} \log {\relax (x )} + 2 x^{3} - 6 x^{2} \log {\relax (x )}^{2} + \left (8 x^{3} - 12 x^{2} \log {\relax (x )} - 2 x^{2} + 6 x \log {\relax (x )}^{2}\right ) e^{e^{e^{e^{x}}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x*exp(x)*ln(x)**2-12*x**2*exp(x)*ln(x)+(8*x**3-2*x**2)*exp(x))*exp(exp(x))*exp(exp(exp(x)))+6*ln
(x)**2+(-24*x+12)*ln(x)+24*x**2-16*x)*exp(exp(exp(exp(x))))-12*x*ln(x)**2+(36*x**2-12*x)*ln(x)-32*x**3+18*x**2
,x)

[Out]

-8*x**4 + 12*x**3*log(x) + 2*x**3 - 6*x**2*log(x)**2 + (8*x**3 - 12*x**2*log(x) - 2*x**2 + 6*x*log(x)**2)*exp(
exp(exp(exp(x))))

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