3.35.73 \(\int e^{4 e^x+e^5 x^2} (4 e^{5+x}+2 e^{10} x+(-4 e^x-2 e^5 x) \log ^2(1+\log (4))) \, dx\)

Optimal. Leaf size=29 \[ e^{4 e^x+e^5 x^2} \left (e^5-\log ^2(1+\log (4))\right ) \]

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Rubi [A]  time = 0.14, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, integrand size = 50, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6706} \begin {gather*} e^{e^5 x^2+4 e^x} \left (e^5-\log ^2(1+\log (4))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^(4*E^x + E^5*x^2)*(4*E^(5 + x) + 2*E^10*x + (-4*E^x - 2*E^5*x)*Log[1 + Log[4]]^2),x]

[Out]

E^(4*E^x + E^5*x^2)*(E^5 - Log[1 + Log[4]]^2)

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=e^{4 e^x+e^5 x^2} \left (e^5-\log ^2(1+\log (4))\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.07, size = 29, normalized size = 1.00 \begin {gather*} e^{4 e^x+e^5 x^2} \left (e^5-\log ^2(1+\log (4))\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^(4*E^x + E^5*x^2)*(4*E^(5 + x) + 2*E^10*x + (-4*E^x - 2*E^5*x)*Log[1 + Log[4]]^2),x]

[Out]

E^(4*E^x + E^5*x^2)*(E^5 - Log[1 + Log[4]]^2)

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fricas [A]  time = 1.26, size = 33, normalized size = 1.14 \begin {gather*} -{\left (\log \left (2 \, \log \relax (2) + 1\right )^{2} - e^{5}\right )} e^{\left ({\left (x^{2} e^{10} + 4 \, e^{\left (x + 5\right )}\right )} e^{\left (-5\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)-2*x*exp(5))*log(1+2*log(2))^2+4*exp(5)*exp(x)+2*x*exp(5)^2)*exp(4*exp(x)+x^2*exp(5)),x,
algorithm="fricas")

[Out]

-(log(2*log(2) + 1)^2 - e^5)*e^((x^2*e^10 + 4*e^(x + 5))*e^(-5))

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giac [A]  time = 0.29, size = 37, normalized size = 1.28 \begin {gather*} -e^{\left (x^{2} e^{5} + 4 \, e^{x}\right )} \log \left (2 \, \log \relax (2) + 1\right )^{2} + e^{\left (x^{2} e^{5} + 4 \, e^{x} + 5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)-2*x*exp(5))*log(1+2*log(2))^2+4*exp(5)*exp(x)+2*x*exp(5)^2)*exp(4*exp(x)+x^2*exp(5)),x,
algorithm="giac")

[Out]

-e^(x^2*e^5 + 4*e^x)*log(2*log(2) + 1)^2 + e^(x^2*e^5 + 4*e^x + 5)

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maple [A]  time = 0.05, size = 28, normalized size = 0.97

method	result	size
norman	\(\left ({\mathrm e}^{5}-\ln \left (1+2 \ln \relax (2)\right )^{2}\right ) {\mathrm e}^{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{5}}\)	\(28\)
risch	\(-\ln \left (1+2 \ln \relax (2)\right )^{2} {\mathrm e}^{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{5}}+{\mathrm e}^{5} {\mathrm e}^{4 \,{\mathrm e}^{x}+x^{2} {\mathrm e}^{5}}\)	\(40\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*exp(x)-2*x*exp(5))*ln(1+2*ln(2))^2+4*exp(5)*exp(x)+2*x*exp(5)^2)*exp(4*exp(x)+x^2*exp(5)),x,method=_R
ETURNVERBOSE)

[Out]

(exp(5)-ln(1+2*ln(2))^2)*exp(4*exp(x)+x^2*exp(5))

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maxima [A]  time = 0.75, size = 28, normalized size = 0.97 \begin {gather*} -{\left (\log \left (2 \, \log \relax (2) + 1\right )^{2} - e^{5}\right )} e^{\left (x^{2} e^{5} + 4 \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)-2*x*exp(5))*log(1+2*log(2))^2+4*exp(5)*exp(x)+2*x*exp(5)^2)*exp(4*exp(x)+x^2*exp(5)),x,
algorithm="maxima")

[Out]

-(log(2*log(2) + 1)^2 - e^5)*e^(x^2*e^5 + 4*e^x)

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mupad [B]  time = 0.10, size = 25, normalized size = 0.86 \begin {gather*} {\mathrm {e}}^{4\,{\mathrm {e}}^x+x^2\,{\mathrm {e}}^5}\,\left ({\mathrm {e}}^5-{\ln \left (\ln \relax (4)+1\right )}^2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(4*exp(x) + x^2*exp(5))*(2*x*exp(10) - log(2*log(2) + 1)^2*(4*exp(x) + 2*x*exp(5)) + 4*exp(5)*exp(x)),x
)

[Out]

exp(4*exp(x) + x^2*exp(5))*(exp(5) - log(log(4) + 1)^2)

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sympy [A]  time = 0.21, size = 26, normalized size = 0.90 \begin {gather*} \left (- \log {\left (1 + 2 \log {\relax (2 )} \right )}^{2} + e^{5}\right ) e^{x^{2} e^{5} + 4 e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*exp(x)-2*x*exp(5))*ln(1+2*ln(2))**2+4*exp(5)*exp(x)+2*x*exp(5)**2)*exp(4*exp(x)+x**2*exp(5)),x)

[Out]

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

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