3.7.48 \(\int (-12+e^x (-14+e^5 (-2-2 x))-2 x-4 e^5 x+4 e^{x^4} x^3+(e^x (-2-2 x)-4 x) \log (x)) \, dx\)

Optimal. Leaf size=27 \[ 4+e^{x^4}-4 \left (e^x+x\right ) \left (3+\frac {1}{2} x \left (e^5+\log (x)\right )\right ) \]

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Rubi [B]  time = 0.08, antiderivative size = 75, normalized size of antiderivative = 2.78, number of steps used = 10, number of rules used = 7, integrand size = 52, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {6, 2187, 2176, 2194, 2209, 2554, 12} \begin {gather*} e^{x^4}-\left (1+2 e^5\right ) x^2+x^2-2 x^2 \log (x)-12 x+2 e^x+2 e^{x+5}-2 e^x \left (e^5 x+e^5+7\right )+2 e^x \log (x)-2 e^x (x+1) \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[-12 + E^x*(-14 + E^5*(-2 - 2*x)) - 2*x - 4*E^5*x + 4*E^x^4*x^3 + (E^x*(-2 - 2*x) - 4*x)*Log[x],x]

[Out]

2*E^x + E^x^4 + 2*E^(5 + x) - 12*x + x^2 - (1 + 2*E^5)*x^2 - 2*E^x*(7 + E^5 + E^5*x) + 2*E^x*Log[x] - 2*x^2*Lo
g[x] - 2*E^x*(1 + x)*Log[x]

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, 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 2187

Int[((a_.) + (b_.)*((F_)^((g_.)*(v_)))^(n_.))^(p_.)*(u_)^(m_.), x_Symbol] :> Int[NormalizePowerOfLinear[u, x]^
m*(a + b*(F^(g*ExpandToSum[v, x]))^n)^p, x] /; FreeQ[{F, a, b, g, n, p}, x] && LinearQ[v, x] && PowerOfLinearQ
[u, x] &&  !(LinearMatchQ[v, x] && PowerOfLinearMatchQ[u, x]) && IntegerQ[m]

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 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

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 \left (-12+e^x \left (-14+e^5 (-2-2 x)\right )+\left (-2-4 e^5\right ) x+4 e^{x^4} x^3+\left (e^x (-2-2 x)-4 x\right ) \log (x)\right ) \, dx\\ &=-12 x-\left (1+2 e^5\right ) x^2+4 \int e^{x^4} x^3 \, dx+\int e^x \left (-14+e^5 (-2-2 x)\right ) \, dx+\int \left (e^x (-2-2 x)-4 x\right ) \log (x) \, dx\\ &=e^{x^4}-12 x-\left (1+2 e^5\right ) x^2+2 e^x \log (x)-2 x^2 \log (x)-2 e^x (1+x) \log (x)-\int 2 \left (-e^x-x\right ) \, dx+\int e^x \left (-2 \left (7+e^5\right )-2 e^5 x\right ) \, dx\\ &=e^{x^4}-12 x-\left (1+2 e^5\right ) x^2-2 e^x \left (7+e^5+e^5 x\right )+2 e^x \log (x)-2 x^2 \log (x)-2 e^x (1+x) \log (x)-2 \int \left (-e^x-x\right ) \, dx+\left (2 e^5\right ) \int e^x \, dx\\ &=e^{x^4}+2 e^{5+x}-12 x+x^2-\left (1+2 e^5\right ) x^2-2 e^x \left (7+e^5+e^5 x\right )+2 e^x \log (x)-2 x^2 \log (x)-2 e^x (1+x) \log (x)+2 \int e^x \, dx\\ &=2 e^x+e^{x^4}+2 e^{5+x}-12 x+x^2-\left (1+2 e^5\right ) x^2-2 e^x \left (7+e^5+e^5 x\right )+2 e^x \log (x)-2 x^2 \log (x)-2 e^x (1+x) \log (x)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 43, normalized size = 1.59 \begin {gather*} -2 \left (6 e^x-\frac {e^{x^4}}{2}+6 x+e^{5+x} x+e^5 x^2+x \left (e^x+x\right ) \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[-12 + E^x*(-14 + E^5*(-2 - 2*x)) - 2*x - 4*E^5*x + 4*E^x^4*x^3 + (E^x*(-2 - 2*x) - 4*x)*Log[x],x]

[Out]

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

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fricas [A]  time = 0.73, size = 37, normalized size = 1.37 \begin {gather*} -2 \, x^{2} e^{5} - 2 \, {\left (x e^{5} + 6\right )} e^{x} - 2 \, {\left (x^{2} + x e^{x}\right )} \log \relax (x) - 12 \, x + e^{\left (x^{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-2)*exp(x)-4*x)*log(x)+4*x^3*exp(x^4)+((-2*x-2)*exp(5)-14)*exp(x)-4*x*exp(5)-2*x-12,x, algorit
hm="fricas")

[Out]

-2*x^2*e^5 - 2*(x*e^5 + 6)*e^x - 2*(x^2 + x*e^x)*log(x) - 12*x + e^(x^4)

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giac [A]  time = 0.39, size = 38, normalized size = 1.41 \begin {gather*} -2 \, x^{2} e^{5} - 2 \, x e^{\left (x + 5\right )} - 2 \, {\left (x^{2} + x e^{x}\right )} \log \relax (x) - 12 \, x + e^{\left (x^{4}\right )} - 12 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-2)*exp(x)-4*x)*log(x)+4*x^3*exp(x^4)+((-2*x-2)*exp(5)-14)*exp(x)-4*x*exp(5)-2*x-12,x, algorit
hm="giac")

[Out]

-2*x^2*e^5 - 2*x*e^(x + 5) - 2*(x^2 + x*e^x)*log(x) - 12*x + e^(x^4) - 12*e^x

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maple [A]  time = 0.05, size = 41, normalized size = 1.52

method	result	size
risch	\(-2 x \,{\mathrm e}^{5+x}-2 x^{2} {\mathrm e}^{5}-2 x \,{\mathrm e}^{x} \ln \relax (x )-2 x^{2} \ln \relax (x )-12 \,{\mathrm e}^{x}+{\mathrm e}^{x^{4}}-12 x\)	\(41\)
default	\(-12 x -2 \,{\mathrm e}^{5} {\mathrm e}^{x}-2 \,{\mathrm e}^{5} \left ({\mathrm e}^{x} x -{\mathrm e}^{x}\right )-12 \,{\mathrm e}^{x}-2 x \,{\mathrm e}^{x} \ln \relax (x )-2 x^{2} \ln \relax (x )-2 x^{2} {\mathrm e}^{5}+{\mathrm e}^{x^{4}}\)	\(53\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*x-2)*exp(x)-4*x)*ln(x)+4*x^3*exp(x^4)+((-2*x-2)*exp(5)-14)*exp(x)-4*x*exp(5)-2*x-12,x,method=_RETURNV
ERBOSE)

[Out]

-2*x*exp(5+x)-2*x^2*exp(5)-2*x*exp(x)*ln(x)-2*x^2*ln(x)-12*exp(x)+exp(x^4)-12*x

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maxima [A]  time = 0.52, size = 41, normalized size = 1.52 \begin {gather*} -2 \, x^{2} e^{5} - 2 \, {\left (x e^{5} + 7\right )} e^{x} - 2 \, {\left (x^{2} + x e^{x}\right )} \log \relax (x) - 12 \, x + e^{\left (x^{4}\right )} + 2 \, e^{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-2)*exp(x)-4*x)*log(x)+4*x^3*exp(x^4)+((-2*x-2)*exp(5)-14)*exp(x)-4*x*exp(5)-2*x-12,x, algorit
hm="maxima")

[Out]

-2*x^2*e^5 - 2*(x*e^5 + 7)*e^x - 2*(x^2 + x*e^x)*log(x) - 12*x + e^(x^4) + 2*e^x

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mupad [B]  time = 0.72, size = 40, normalized size = 1.48 \begin {gather*} {\mathrm {e}}^{x^4}-12\,x-12\,{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^{x+5}-2\,x^2\,\ln \relax (x)-2\,x^2\,{\mathrm {e}}^5-2\,x\,{\mathrm {e}}^x\,\ln \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(4*x^3*exp(x^4) - 4*x*exp(5) - 2*x - exp(x)*(exp(5)*(2*x + 2) + 14) - log(x)*(4*x + exp(x)*(2*x + 2)) - 12,
x)

[Out]

exp(x^4) - 12*x - 12*exp(x) - 2*x*exp(x + 5) - 2*x^2*log(x) - 2*x^2*exp(5) - 2*x*exp(x)*log(x)

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sympy [A]  time = 0.44, size = 44, normalized size = 1.63 \begin {gather*} - 2 x^{2} \log {\relax (x )} - 2 x^{2} e^{5} - 12 x + \left (- 2 x \log {\relax (x )} - 2 x e^{5} - 12\right ) e^{x} + e^{x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*x-2)*exp(x)-4*x)*ln(x)+4*x**3*exp(x**4)+((-2*x-2)*exp(5)-14)*exp(x)-4*x*exp(5)-2*x-12,x)

[Out]

-2*x**2*log(x) - 2*x**2*exp(5) - 12*x + (-2*x*log(x) - 2*x*exp(5) - 12)*exp(x) + exp(x**4)

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