3.32.90 \(\int \frac {-30+3 e^{2 x}+e^{9 x+x^2} (1-e^{2 x}-2 x)+(-6+2 e^{9 x+x^2}) \log (-3+e^{9 x+x^2})}{-15-3 e^{2 x} x+e^{9 x+x^2} (5+e^{2 x} x)+(-3+e^{9 x+x^2}) \log (-3+e^{9 x+x^2})} \, dx\)

Optimal. Leaf size=25 \[ \log \left (\frac {3}{x+e^{-2 x} \left (5+\log \left (-3+e^{x (9+x)}\right )\right )}\right ) \]

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Rubi [F]  time = 3.98, 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 {-30+3 e^{2 x}+e^{9 x+x^2} \left (1-e^{2 x}-2 x\right )+\left (-6+2 e^{9 x+x^2}\right ) \log \left (-3+e^{9 x+x^2}\right )}{-15-3 e^{2 x} x+e^{9 x+x^2} \left (5+e^{2 x} x\right )+\left (-3+e^{9 x+x^2}\right ) \log \left (-3+e^{9 x+x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-30 + 3*E^(2*x) + E^(9*x + x^2)*(1 - E^(2*x) - 2*x) + (-6 + 2*E^(9*x + x^2))*Log[-3 + E^(9*x + x^2)])/(-1
5 - 3*E^(2*x)*x + E^(9*x + x^2)*(5 + E^(2*x)*x) + (-3 + E^(9*x + x^2))*Log[-3 + E^(9*x + x^2)]),x]

[Out]

Defer[Int][E^(11*x + x^2)/((3 - E^(x*(9 + x)))*(5 + E^(2*x)*x + Log[-3 + E^(x*(9 + x))])), x] - 30*Defer[Int][
1/((-3 + E^(x*(9 + x)))*(5 + E^(2*x)*x + Log[-3 + E^(x*(9 + x))])), x] + 3*Defer[Int][E^(2*x)/((-3 + E^(x*(9 +
x)))*(5 + E^(2*x)*x + Log[-3 + E^(x*(9 + x))])), x] + Defer[Int][E^(9*x + x^2)/((-3 + E^(x*(9 + x)))*(5 + E^(
2*x)*x + Log[-3 + E^(x*(9 + x))])), x] + 2*Defer[Int][(E^(9*x + x^2)*x)/((3 - E^(x*(9 + x)))*(5 + E^(2*x)*x +
Log[-3 + E^(x*(9 + x))])), x] - 6*Defer[Int][Log[-3 + E^(x*(9 + x))]/((-3 + E^(x*(9 + x)))*(5 + E^(2*x)*x + Lo
g[-3 + E^(x*(9 + x))])), x] + 2*Defer[Int][(E^(9*x + x^2)*Log[-3 + E^(x*(9 + x))])/((-3 + E^(x*(9 + x)))*(5 +
E^(2*x)*x + Log[-3 + E^(x*(9 + x))])), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {30-3 e^{2 x}-e^{9 x+x^2} \left (1-e^{2 x}-2 x\right )-\left (-6+2 e^{9 x+x^2}\right ) \log \left (-3+e^{9 x+x^2}\right )}{\left (3-e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx\\ &=\int \left (\frac {e^{11 x+x^2}}{\left (3-e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}-\frac {30}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}+\frac {3 e^{2 x}}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}+\frac {e^{9 x+x^2}}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}+\frac {2 e^{9 x+x^2} x}{\left (3-e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}-\frac {6 \log \left (-3+e^{x (9+x)}\right )}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}+\frac {2 e^{9 x+x^2} \log \left (-3+e^{x (9+x)}\right )}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )}\right ) \, dx\\ &=2 \int \frac {e^{9 x+x^2} x}{\left (3-e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx+2 \int \frac {e^{9 x+x^2} \log \left (-3+e^{x (9+x)}\right )}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx+3 \int \frac {e^{2 x}}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx-6 \int \frac {\log \left (-3+e^{x (9+x)}\right )}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx-30 \int \frac {1}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx+\int \frac {e^{11 x+x^2}}{\left (3-e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx+\int \frac {e^{9 x+x^2}}{\left (-3+e^{x (9+x)}\right ) \left (5+e^{2 x} x+\log \left (-3+e^{x (9+x)}\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [F]  time = 0.56, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {-30+3 e^{2 x}+e^{9 x+x^2} \left (1-e^{2 x}-2 x\right )+\left (-6+2 e^{9 x+x^2}\right ) \log \left (-3+e^{9 x+x^2}\right )}{-15-3 e^{2 x} x+e^{9 x+x^2} \left (5+e^{2 x} x\right )+\left (-3+e^{9 x+x^2}\right ) \log \left (-3+e^{9 x+x^2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(-30 + 3*E^(2*x) + E^(9*x + x^2)*(1 - E^(2*x) - 2*x) + (-6 + 2*E^(9*x + x^2))*Log[-3 + E^(9*x + x^2)
])/(-15 - 3*E^(2*x)*x + E^(9*x + x^2)*(5 + E^(2*x)*x) + (-3 + E^(9*x + x^2))*Log[-3 + E^(9*x + x^2)]),x]

[Out]

Integrate[(-30 + 3*E^(2*x) + E^(9*x + x^2)*(1 - E^(2*x) - 2*x) + (-6 + 2*E^(9*x + x^2))*Log[-3 + E^(9*x + x^2)
])/(-15 - 3*E^(2*x)*x + E^(9*x + x^2)*(5 + E^(2*x)*x) + (-3 + E^(9*x + x^2))*Log[-3 + E^(9*x + x^2)]), x]

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fricas [A]  time = 0.52, size = 26, normalized size = 1.04 \begin {gather*} 2 \, x - \log \left (x e^{\left (2 \, x\right )} + \log \left (e^{\left (x^{2} + 9 \, x\right )} - 3\right ) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x^2+9*x)-6)*log(exp(x^2+9*x)-3)+(-exp(2*x)+1-2*x)*exp(x^2+9*x)+3*exp(2*x)-30)/((exp(x^2+9*x)
-3)*log(exp(x^2+9*x)-3)+(x*exp(2*x)+5)*exp(x^2+9*x)-3*x*exp(2*x)-15),x, algorithm="fricas")

[Out]

2*x - log(x*e^(2*x) + log(e^(x^2 + 9*x) - 3) + 5)

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giac [A]  time = 0.34, size = 26, normalized size = 1.04 \begin {gather*} 2 \, x - \log \left (x e^{\left (2 \, x\right )} + \log \left (e^{\left (x^{2} + 9 \, x\right )} - 3\right ) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x^2+9*x)-6)*log(exp(x^2+9*x)-3)+(-exp(2*x)+1-2*x)*exp(x^2+9*x)+3*exp(2*x)-30)/((exp(x^2+9*x)
-3)*log(exp(x^2+9*x)-3)+(x*exp(2*x)+5)*exp(x^2+9*x)-3*x*exp(2*x)-15),x, algorithm="giac")

[Out]

2*x - log(x*e^(2*x) + log(e^(x^2 + 9*x) - 3) + 5)

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maple [A]  time = 0.08, size = 25, normalized size = 1.00

method	result	size
risch	\(2 x -\ln \left (x \,{\mathrm e}^{2 x}+\ln \left ({\mathrm e}^{\left (x +9\right ) x}-3\right )+5\right )\)	\(25\)
norman	\(2 x -\ln \left (x \,{\mathrm e}^{2 x}+\ln \left ({\mathrm e}^{x^{2}+9 x}-3\right )+5\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*exp(x^2+9*x)-6)*ln(exp(x^2+9*x)-3)+(-exp(2*x)+1-2*x)*exp(x^2+9*x)+3*exp(2*x)-30)/((exp(x^2+9*x)-3)*ln(
exp(x^2+9*x)-3)+(x*exp(2*x)+5)*exp(x^2+9*x)-3*x*exp(2*x)-15),x,method=_RETURNVERBOSE)

[Out]

2*x-ln(x*exp(2*x)+ln(exp((x+9)*x)-3)+5)

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maxima [A]  time = 0.69, size = 26, normalized size = 1.04 \begin {gather*} 2 \, x - \log \left (x e^{\left (2 \, x\right )} + \log \left (e^{\left (x^{2} + 9 \, x\right )} - 3\right ) + 5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x^2+9*x)-6)*log(exp(x^2+9*x)-3)+(-exp(2*x)+1-2*x)*exp(x^2+9*x)+3*exp(2*x)-30)/((exp(x^2+9*x)
-3)*log(exp(x^2+9*x)-3)+(x*exp(2*x)+5)*exp(x^2+9*x)-3*x*exp(2*x)-15),x, algorithm="maxima")

[Out]

2*x - log(x*e^(2*x) + log(e^(x^2 + 9*x) - 3) + 5)

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mupad [B]  time = 0.57, size = 24, normalized size = 0.96 \begin {gather*} 2\,x-\ln \left (\ln \left ({\mathrm {e}}^{x\,\left (x+9\right )}-3\right )+x\,{\mathrm {e}}^{2\,x}+5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*exp(2*x) - exp(9*x + x^2)*(2*x + exp(2*x) - 1) + log(exp(9*x + x^2) - 3)*(2*exp(9*x + x^2) - 6) - 30)/
(3*x*exp(2*x) - exp(9*x + x^2)*(x*exp(2*x) + 5) - log(exp(9*x + x^2) - 3)*(exp(9*x + x^2) - 3) + 15),x)

[Out]

2*x - log(log(exp(x*(x + 9)) - 3) + x*exp(2*x) + 5)

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sympy [A]  time = 0.49, size = 24, normalized size = 0.96 \begin {gather*} 2 x - \log {\left (x e^{2 x} + \log {\left (e^{x^{2} + 9 x} - 3 \right )} + 5 \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*exp(x**2+9*x)-6)*ln(exp(x**2+9*x)-3)+(-exp(2*x)+1-2*x)*exp(x**2+9*x)+3*exp(2*x)-30)/((exp(x**2+9
*x)-3)*ln(exp(x**2+9*x)-3)+(x*exp(2*x)+5)*exp(x**2+9*x)-3*x*exp(2*x)-15),x)

[Out]

2*x - log(x*exp(2*x) + log(exp(x**2 + 9*x) - 3) + 5)

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