3.94.55 \(\int \frac {e^{x+x \log (\frac {x^2}{(9+x) \log ^2(4)})} (27+2 x+(9+x) \log (\frac {x^2}{(9+x) \log ^2(4)}))}{9+x} \, dx\)

Optimal. Leaf size=20 \[ e^{x+x \log \left (\frac {x^2}{(9+x) \log ^2(4)}\right )} \]

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Rubi [A]  time = 0.29, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 1, number of rules used = 1, integrand size = 49, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.020, Rules used = {6706} \begin {gather*} e^x \left (\frac {x^2}{x+9}\right )^x \log ^{-2 x}(4) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^(x + x*Log[x^2/((9 + x)*Log[4]^2)])*(27 + 2*x + (9 + x)*Log[x^2/((9 + x)*Log[4]^2)]))/(9 + x),x]

[Out]

(E^x*(x^2/(9 + x))^x)/Log[4]^(2*x)

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^x \left (\frac {x^2}{9+x}\right )^x \log ^{-2 x}(4)\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.20, size = 21, normalized size = 1.05 \begin {gather*} e^x \left (\frac {x^2}{9+x}\right )^x \log ^{-2 x}(4) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x + x*Log[x^2/((9 + x)*Log[4]^2)])*(27 + 2*x + (9 + x)*Log[x^2/((9 + x)*Log[4]^2)]))/(9 + x),x]

[Out]

(E^x*(x^2/(9 + x))^x)/Log[4]^(2*x)

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fricas [A]  time = 0.61, size = 20, normalized size = 1.00 \begin {gather*} e^{\left (x \log \left (\frac {x^{2}}{4 \, {\left (x + 9\right )} \log \relax (2)^{2}}\right ) + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+9)*log(1/4*x^2/(x+9)/log(2)^2)+2*x+27)*exp(x*log(1/4*x^2/(x+9)/log(2)^2)+x)/(x+9),x, algorithm="
fricas")

[Out]

e^(x*log(1/4*x^2/((x + 9)*log(2)^2)) + x)

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giac [A]  time = 0.54, size = 26, normalized size = 1.30 \begin {gather*} e^{\left (x \log \left (\frac {x^{2}}{4 \, {\left (x \log \relax (2)^{2} + 9 \, \log \relax (2)^{2}\right )}}\right ) + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+9)*log(1/4*x^2/(x+9)/log(2)^2)+2*x+27)*exp(x*log(1/4*x^2/(x+9)/log(2)^2)+x)/(x+9),x, algorithm="
giac")

[Out]

e^(x*log(1/4*x^2/(x*log(2)^2 + 9*log(2)^2)) + x)

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maple [A]  time = 0.54, size = 20, normalized size = 1.00

method	result	size
risch	\(\left (\frac {x^{2}}{4 \left (x +9\right ) \ln \relax (2)^{2}}\right )^{x} {\mathrm e}^{x}\)	\(20\)
norman	\({\mathrm e}^{x \ln \left (\frac {x^{2}}{4 \left (x +9\right ) \ln \relax (2)^{2}}\right )+x}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x+9)*ln(1/4*x^2/(x+9)/ln(2)^2)+2*x+27)*exp(x*ln(1/4*x^2/(x+9)/ln(2)^2)+x)/(x+9),x,method=_RETURNVERBOSE)

[Out]

(1/4*x^2/(x+9)/ln(2)^2)^x*exp(x)

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maxima [A]  time = 0.59, size = 26, normalized size = 1.30 \begin {gather*} e^{\left (-2 \, x \log \relax (2) - x \log \left (x + 9\right ) + 2 \, x \log \relax (x) - 2 \, x \log \left (\log \relax (2)\right ) + x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+9)*log(1/4*x^2/(x+9)/log(2)^2)+2*x+27)*exp(x*log(1/4*x^2/(x+9)/log(2)^2)+x)/(x+9),x, algorithm="
maxima")

[Out]

e^(-2*x*log(2) - x*log(x + 9) + 2*x*log(x) - 2*x*log(log(2)) + x)

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mupad [B]  time = 8.25, size = 26, normalized size = 1.30 \begin {gather*} {\mathrm {e}}^x\,{\left (\frac {1}{4\,x\,{\ln \relax (2)}^2+36\,{\ln \relax (2)}^2}\right )}^x\,{\left (x^2\right )}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x + x*log(x^2/(4*log(2)^2*(x + 9))))*(2*x + log(x^2/(4*log(2)^2*(x + 9)))*(x + 9) + 27))/(x + 9),x)

[Out]

exp(x)*(1/(4*x*log(2)^2 + 36*log(2)^2))^x*(x^2)^x

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sympy [A]  time = 0.48, size = 19, normalized size = 0.95 \begin {gather*} e^{x \log {\left (\frac {x^{2}}{4 \left (x + 9\right ) \log {\relax (2 )}^{2}} \right )} + x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x+9)*ln(1/4*x**2/(x+9)/ln(2)**2)+2*x+27)*exp(x*ln(1/4*x**2/(x+9)/ln(2)**2)+x)/(x+9),x)

[Out]

exp(x*log(x**2/(4*(x + 9)*log(2)**2)) + x)

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