3.94.32 \(\int \frac {e^{\frac {-9+2 x \log (3+e^x+4 x^2)}{x \log (3+e^x+4 x^2)}} (9 e^x x+72 x^2+(27+9 e^x+36 x^2) \log (3+e^x+4 x^2))}{(3 x^2+e^x x^2+4 x^4) \log ^2(3+e^x+4 x^2)} \, dx\)

Optimal. Leaf size=25 \[ e^{\frac {2 x-\frac {9}{\log \left (3+e^x+4 x^2\right )}}{x}} \]

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Rubi [F]  time = 5.82, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {\exp \left (\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}\right ) \left (9 e^x x+72 x^2+\left (27+9 e^x+36 x^2\right ) \log \left (3+e^x+4 x^2\right )\right )}{\left (3 x^2+e^x x^2+4 x^4\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-9 + 2*x*Log[3 + E^x + 4*x^2])/(x*Log[3 + E^x + 4*x^2]))*(9*E^x*x + 72*x^2 + (27 + 9*E^x + 36*x^2)*Lo
g[3 + E^x + 4*x^2]))/((3*x^2 + E^x*x^2 + 4*x^4)*Log[3 + E^x + 4*x^2]^2),x]

[Out]

9*Defer[Int][E^(2 - 9/(x*Log[3 + E^x + 4*x^2]))/(x*Log[3 + E^x + 4*x^2]^2), x] + 72*Defer[Int][E^(2 - 9/(x*Log
[3 + E^x + 4*x^2]))/((3 + E^x + 4*x^2)*Log[3 + E^x + 4*x^2]^2), x] - 27*Defer[Int][E^(2 - 9/(x*Log[3 + E^x + 4
*x^2]))/(x*(3 + E^x + 4*x^2)*Log[3 + E^x + 4*x^2]^2), x] - 36*Defer[Int][(E^(2 - 9/(x*Log[3 + E^x + 4*x^2]))*x
)/((3 + E^x + 4*x^2)*Log[3 + E^x + 4*x^2]^2), x] + 9*Defer[Int][E^(2 - 9/(x*Log[3 + E^x + 4*x^2]))/(x^2*Log[3
+ E^x + 4*x^2]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {9 \exp \left (\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}\right ) \left (3-8 x+4 x^2\right )}{x \left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )}+\frac {9 \exp \left (\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}\right ) \left (x+\log \left (3+e^x+4 x^2\right )\right )}{x^2 \log ^2\left (3+e^x+4 x^2\right )}\right ) \, dx\\ &=-\left (9 \int \frac {\exp \left (\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}\right ) \left (3-8 x+4 x^2\right )}{x \left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx\right )+9 \int \frac {\exp \left (\frac {-9+2 x \log \left (3+e^x+4 x^2\right )}{x \log \left (3+e^x+4 x^2\right )}\right ) \left (x+\log \left (3+e^x+4 x^2\right )\right )}{x^2 \log ^2\left (3+e^x+4 x^2\right )} \, dx\\ &=-\left (9 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}} \left (3-8 x+4 x^2\right )}{x \left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx\right )+9 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}} \left (x+\log \left (3+e^x+4 x^2\right )\right )}{x^2 \log ^2\left (3+e^x+4 x^2\right )} \, dx\\ &=-\left (9 \int \left (-\frac {8 e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{\left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )}+\frac {3 e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x \left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )}+\frac {4 e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}} x}{\left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )}\right ) \, dx\right )+9 \int \left (\frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x \log ^2\left (3+e^x+4 x^2\right )}+\frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x^2 \log \left (3+e^x+4 x^2\right )}\right ) \, dx\\ &=9 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x \log ^2\left (3+e^x+4 x^2\right )} \, dx+9 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x^2 \log \left (3+e^x+4 x^2\right )} \, dx-27 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{x \left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx-36 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}} x}{\left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx+72 \int \frac {e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}}}{\left (3+e^x+4 x^2\right ) \log ^2\left (3+e^x+4 x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 22, normalized size = 0.88 \begin {gather*} e^{2-\frac {9}{x \log \left (3+e^x+4 x^2\right )}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-9 + 2*x*Log[3 + E^x + 4*x^2])/(x*Log[3 + E^x + 4*x^2]))*(9*E^x*x + 72*x^2 + (27 + 9*E^x + 36*x
^2)*Log[3 + E^x + 4*x^2]))/((3*x^2 + E^x*x^2 + 4*x^4)*Log[3 + E^x + 4*x^2]^2),x]

[Out]

E^(2 - 9/(x*Log[3 + E^x + 4*x^2]))

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fricas [A]  time = 0.51, size = 32, normalized size = 1.28 \begin {gather*} e^{\left (\frac {2 \, x \log \left (4 \, x^{2} + e^{x} + 3\right ) - 9}{x \log \left (4 \, x^{2} + e^{x} + 3\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*exp(x)+36*x^2+27)*log(exp(x)+4*x^2+3)+9*exp(x)*x+72*x^2)*exp((2*x*log(exp(x)+4*x^2+3)-9)/x/log(e
xp(x)+4*x^2+3))/(exp(x)*x^2+4*x^4+3*x^2)/log(exp(x)+4*x^2+3)^2,x, algorithm="fricas")

[Out]

e^((2*x*log(4*x^2 + e^x + 3) - 9)/(x*log(4*x^2 + e^x + 3)))

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giac [A]  time = 3.51, size = 20, normalized size = 0.80 \begin {gather*} e^{\left (-\frac {9}{x \log \left (4 \, x^{2} + e^{x} + 3\right )} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*exp(x)+36*x^2+27)*log(exp(x)+4*x^2+3)+9*exp(x)*x+72*x^2)*exp((2*x*log(exp(x)+4*x^2+3)-9)/x/log(e
xp(x)+4*x^2+3))/(exp(x)*x^2+4*x^4+3*x^2)/log(exp(x)+4*x^2+3)^2,x, algorithm="giac")

[Out]

e^(-9/(x*log(4*x^2 + e^x + 3)) + 2)

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maple [A]  time = 0.02, size = 33, normalized size = 1.32

method	result	size
risch	\({\mathrm e}^{\frac {2 x \ln \left ({\mathrm e}^{x}+4 x^{2}+3\right )-9}{x \ln \left ({\mathrm e}^{x}+4 x^{2}+3\right )}}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((9*exp(x)+36*x^2+27)*ln(exp(x)+4*x^2+3)+9*exp(x)*x+72*x^2)*exp((2*x*ln(exp(x)+4*x^2+3)-9)/x/ln(exp(x)+4*x
^2+3))/(exp(x)*x^2+4*x^4+3*x^2)/ln(exp(x)+4*x^2+3)^2,x,method=_RETURNVERBOSE)

[Out]

exp((2*x*ln(exp(x)+4*x^2+3)-9)/x/ln(exp(x)+4*x^2+3))

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*exp(x)+36*x^2+27)*log(exp(x)+4*x^2+3)+9*exp(x)*x+72*x^2)*exp((2*x*log(exp(x)+4*x^2+3)-9)/x/log(e
xp(x)+4*x^2+3))/(exp(x)*x^2+4*x^4+3*x^2)/log(exp(x)+4*x^2+3)^2,x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 8.03, size = 21, normalized size = 0.84 \begin {gather*} {\mathrm {e}}^2\,{\mathrm {e}}^{-\frac {9}{x\,\ln \left ({\mathrm {e}}^x+4\,x^2+3\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp((2*x*log(exp(x) + 4*x^2 + 3) - 9)/(x*log(exp(x) + 4*x^2 + 3)))*(log(exp(x) + 4*x^2 + 3)*(9*exp(x) + 3
6*x^2 + 27) + 9*x*exp(x) + 72*x^2))/(log(exp(x) + 4*x^2 + 3)^2*(x^2*exp(x) + 3*x^2 + 4*x^4)),x)

[Out]

exp(2)*exp(-9/(x*log(exp(x) + 4*x^2 + 3)))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((9*exp(x)+36*x**2+27)*ln(exp(x)+4*x**2+3)+9*exp(x)*x+72*x**2)*exp((2*x*ln(exp(x)+4*x**2+3)-9)/x/ln(
exp(x)+4*x**2+3))/(exp(x)*x**2+4*x**4+3*x**2)/ln(exp(x)+4*x**2+3)**2,x)

[Out]

Timed out

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