3.86.47 \(\int (3 x^2+e^{e^{e^x}} (5+5 e^{e^x+x} x)+5 \log (3)) \, dx\)

Optimal. Leaf size=22 \[ x \left (x^2-5 \left (-e^{e^{e^x}}-\log (3)\right )\right ) \]

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Rubi [A]  time = 0.02, antiderivative size = 19, normalized size of antiderivative = 0.86, number of steps used = 2, number of rules used = 1, integrand size = 30, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.033, Rules used = {2288} \begin {gather*} x^3+5 e^{e^{e^x}} x+5 x \log (3) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[3*x^2 + E^E^E^x*(5 + 5*E^(E^x + x)*x) + 5*Log[3],x]

[Out]

5*E^E^E^x*x + x^3 + 5*x*Log[3]

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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x^3+5 x \log (3)+\int e^{e^{e^x}} \left (5+5 e^{e^x+x} x\right ) \, dx\\ &=5 e^{e^{e^x}} x+x^3+5 x \log (3)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 18, normalized size = 0.82 \begin {gather*} 5 e^{e^{e^x}} x+x^3+x \log (243) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[3*x^2 + E^E^E^x*(5 + 5*E^(E^x + x)*x) + 5*Log[3],x]

[Out]

5*E^E^E^x*x + x^3 + x*Log[243]

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fricas [A]  time = 0.53, size = 16, normalized size = 0.73 \begin {gather*} x^{3} + 5 \, x e^{\left (e^{\left (e^{x}\right )}\right )} + 5 \, x \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x)*exp(exp(x))+5)*exp(exp(exp(x)))+5*log(3)+3*x^2,x, algorithm="fricas")

[Out]

x^3 + 5*x*e^(e^(e^x)) + 5*x*log(3)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int 3 \, x^{2} + 5 \, {\left (x e^{\left (x + e^{x}\right )} + 1\right )} e^{\left (e^{\left (e^{x}\right )}\right )} + 5 \, \log \relax (3)\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x)*exp(exp(x))+5)*exp(exp(exp(x)))+5*log(3)+3*x^2,x, algorithm="giac")

[Out]

integrate(3*x^2 + 5*(x*e^(x + e^x) + 1)*e^(e^(e^x)) + 5*log(3), x)

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maple [A]  time = 0.07, size = 17, normalized size = 0.77

method	result	size
default	\(5 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}} x +x^{3}+5 x \ln \relax (3)\)	\(17\)
norman	\(5 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}} x +x^{3}+5 x \ln \relax (3)\)	\(17\)
risch	\(5 \,{\mathrm e}^{{\mathrm e}^{{\mathrm e}^{x}}} x +x^{3}+5 x \ln \relax (3)\)	\(17\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((5*x*exp(x)*exp(exp(x))+5)*exp(exp(exp(x)))+5*ln(3)+3*x^2,x,method=_RETURNVERBOSE)

[Out]

5*exp(exp(exp(x)))*x+x^3+5*x*ln(3)

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maxima [A]  time = 0.38, size = 16, normalized size = 0.73 \begin {gather*} x^{3} + 5 \, x e^{\left (e^{\left (e^{x}\right )}\right )} + 5 \, x \log \relax (3) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x)*exp(exp(x))+5)*exp(exp(exp(x)))+5*log(3)+3*x^2,x, algorithm="maxima")

[Out]

x^3 + 5*x*e^(e^(e^x)) + 5*x*log(3)

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mupad [B]  time = 5.20, size = 14, normalized size = 0.64 \begin {gather*} x\,\left (\ln \left (243\right )+5\,{\mathrm {e}}^{{\mathrm {e}}^{{\mathrm {e}}^x}}+x^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(5*log(3) + exp(exp(exp(x)))*(5*x*exp(exp(x))*exp(x) + 5) + 3*x^2,x)

[Out]

x*(log(243) + 5*exp(exp(exp(x))) + x^2)

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sympy [A]  time = 0.97, size = 19, normalized size = 0.86 \begin {gather*} x^{3} + 5 x e^{e^{e^{x}}} + 5 x \log {\relax (3 )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((5*x*exp(x)*exp(exp(x))+5)*exp(exp(exp(x)))+5*ln(3)+3*x**2,x)

[Out]

x**3 + 5*x*exp(exp(exp(x))) + 5*x*log(3)

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