3.90.67 \(\int \frac {e^{\frac {1}{9} (-1-e^{2 x+x \log (2+x)}-10 x)} (90-55 x-50 x^2+e^{2 x+x \log (2+x)} (-20 x-15 x^2+(-10 x-5 x^2) \log (2+x)))}{18+9 x} \, dx\)

Optimal. Leaf size=30 \[ 5 e^{\frac {1}{9} \left (-1-e^{x (2+\log (2+x))}-x\right )-x} x \]

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Rubi [F]  time = 4.64, 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 {e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \left (90-55 x-50 x^2+e^{2 x+x \log (2+x)} \left (-20 x-15 x^2+\left (-10 x-5 x^2\right ) \log (2+x)\right )\right )}{18+9 x} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((-1 - E^(2*x + x*Log[2 + x]) - 10*x)/9)*(90 - 55*x - 50*x^2 + E^(2*x + x*Log[2 + x])*(-20*x - 15*x^2 +
(-10*x - 5*x^2)*Log[2 + x])))/(18 + 9*x),x]

[Out]

5*Defer[Int][E^((-1 - E^(2*x + x*Log[2 + x]) - 10*x)/9), x] - (50*Defer[Int][E^((-1 - E^(2*x + x*Log[2 + x]) -
10*x)/9)*x, x])/9 - (20*Defer[Int][E^((-1 + 8*x - E^(2*x)*(2 + x)^x)/9)*x*(2 + x)^(-1 + x), x])/9 - (5*Defer[
Int][E^((-1 + 8*x - E^(2*x)*(2 + x)^x)/9)*x^2*(2 + x)^(-1 + x), x])/3 - (5*Log[2 + x]*Defer[Int][E^((-1 + 8*x
- E^(2*x)*(2 + x)^x)/9)*x*(2 + x)^x, x])/9 + (5*Defer[Int][Defer[Int][E^((-1 + 8*x - E^(2*x)*(2 + x)^x)/9)*x*(
2 + x)^x, x]/(2 + x), x])/9

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {5}{9} e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} (-9+10 x)-\frac {5}{9} \exp \left (\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )+2 x\right ) x (2+x)^{-1+x} (4+3 x+2 \log (2+x)+x \log (2+x))\right ) \, dx\\ &=-\left (\frac {5}{9} \int e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} (-9+10 x) \, dx\right )-\frac {5}{9} \int \exp \left (\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )+2 x\right ) x (2+x)^{-1+x} (4+3 x+2 \log (2+x)+x \log (2+x)) \, dx\\ &=-\left (\frac {5}{9} \int \left (-9 e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )}+10 e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} x\right ) \, dx\right )-\frac {5}{9} \int e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x (2+x)^{-1+x} (4+3 x+2 \log (2+x)+x \log (2+x)) \, dx\\ &=-\left (\frac {5}{9} \int \left (e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x (2+x)^{-1+x} (4+3 x)+e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x (2+x)^x \log (2+x)\right ) \, dx\right )+5 \int e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \, dx-\frac {50}{9} \int e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} x \, dx\\ &=-\left (\frac {5}{9} \int e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x (2+x)^{-1+x} (4+3 x) \, dx\right )-\frac {5}{9} \int e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x (2+x)^x \log (2+x) \, dx+5 \int e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \, dx-\frac {50}{9} \int e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} x \, dx\\ &=-\left (\frac {5}{9} \int \left (4 e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x (2+x)^{-1+x}+3 e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x^2 (2+x)^{-1+x}\right ) \, dx\right )+\frac {5}{9} \int \frac {\int e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x (2+x)^x \, dx}{2+x} \, dx+5 \int e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \, dx-\frac {50}{9} \int e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} x \, dx-\frac {1}{9} (5 \log (2+x)) \int e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x (2+x)^x \, dx\\ &=\frac {5}{9} \int \frac {\int e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x (2+x)^x \, dx}{2+x} \, dx-\frac {5}{3} \int e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x^2 (2+x)^{-1+x} \, dx-\frac {20}{9} \int e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x (2+x)^{-1+x} \, dx+5 \int e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} \, dx-\frac {50}{9} \int e^{\frac {1}{9} \left (-1-e^{2 x+x \log (2+x)}-10 x\right )} x \, dx-\frac {1}{9} (5 \log (2+x)) \int e^{\frac {1}{9} \left (-1+8 x-e^{2 x} (2+x)^x\right )} x (2+x)^x \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 28, normalized size = 0.93 \begin {gather*} 5 e^{-\frac {1}{9}-\frac {10 x}{9}-\frac {1}{9} e^{2 x} (2+x)^x} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((-1 - E^(2*x + x*Log[2 + x]) - 10*x)/9)*(90 - 55*x - 50*x^2 + E^(2*x + x*Log[2 + x])*(-20*x - 15
*x^2 + (-10*x - 5*x^2)*Log[2 + x])))/(18 + 9*x),x]

[Out]

5*E^(-1/9 - (10*x)/9 - (E^(2*x)*(2 + x)^x)/9)*x

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fricas [A]  time = 0.51, size = 22, normalized size = 0.73 \begin {gather*} 5 \, x e^{\left (-\frac {10}{9} \, x - \frac {1}{9} \, e^{\left (x \log \left (x + 2\right ) + 2 \, x\right )} - \frac {1}{9}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2-10*x)*log(2+x)-15*x^2-20*x)*exp(x*log(2+x)+2*x)-50*x^2-55*x+90)/(9*x+18)/exp(1/9*exp(x*log
(2+x)+2*x)+10/9*x+1/9),x, algorithm="fricas")

[Out]

5*x*e^(-10/9*x - 1/9*e^(x*log(x + 2) + 2*x) - 1/9)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int -\frac {5 \, {\left (10 \, x^{2} + {\left (3 \, x^{2} + {\left (x^{2} + 2 \, x\right )} \log \left (x + 2\right ) + 4 \, x\right )} e^{\left (x \log \left (x + 2\right ) + 2 \, x\right )} + 11 \, x - 18\right )} e^{\left (-\frac {10}{9} \, x - \frac {1}{9} \, e^{\left (x \log \left (x + 2\right ) + 2 \, x\right )} - \frac {1}{9}\right )}}{9 \, {\left (x + 2\right )}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2-10*x)*log(2+x)-15*x^2-20*x)*exp(x*log(2+x)+2*x)-50*x^2-55*x+90)/(9*x+18)/exp(1/9*exp(x*log
(2+x)+2*x)+10/9*x+1/9),x, algorithm="giac")

[Out]

integrate(-5/9*(10*x^2 + (3*x^2 + (x^2 + 2*x)*log(x + 2) + 4*x)*e^(x*log(x + 2) + 2*x) + 11*x - 18)*e^(-10/9*x
- 1/9*e^(x*log(x + 2) + 2*x) - 1/9)/(x + 2), x)

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maple [A]  time = 0.47, size = 21, normalized size = 0.70

method	result	size
risch	\(5 x \,{\mathrm e}^{-\frac {\left (2+x \right )^{x} {\mathrm e}^{2 x}}{9}-\frac {10 x}{9}-\frac {1}{9}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x^2-10*x)*ln(2+x)-15*x^2-20*x)*exp(x*ln(2+x)+2*x)-50*x^2-55*x+90)/(9*x+18)/exp(1/9*exp(x*ln(2+x)+2*x
)+10/9*x+1/9),x,method=_RETURNVERBOSE)

[Out]

5*x*exp(-1/9*(2+x)^x*exp(2*x)-10/9*x-1/9)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} 5 \, x e^{\left (-\frac {10}{9} \, x - \frac {1}{9} \, e^{\left (x \log \left (x + 2\right ) + 2 \, x\right )} - \frac {1}{9}\right )} + \frac {5}{9} \, \int 0\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2-10*x)*log(2+x)-15*x^2-20*x)*exp(x*log(2+x)+2*x)-50*x^2-55*x+90)/(9*x+18)/exp(1/9*exp(x*log
(2+x)+2*x)+10/9*x+1/9),x, algorithm="maxima")

[Out]

5*x*e^(-10/9*x - 1/9*e^(x*log(x + 2) + 2*x) - 1/9) + 5/9*integrate(0, x)

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mupad [B]  time = 8.16, size = 21, normalized size = 0.70 \begin {gather*} 5\,x\,{\mathrm {e}}^{-\frac {10\,x}{9}}\,{\mathrm {e}}^{-\frac {1}{9}}\,{\mathrm {e}}^{-\frac {{\mathrm {e}}^{2\,x}\,{\left (x+2\right )}^x}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- (10*x)/9 - exp(2*x + x*log(x + 2))/9 - 1/9)*(55*x + exp(2*x + x*log(x + 2))*(20*x + log(x + 2)*(10
*x + 5*x^2) + 15*x^2) + 50*x^2 - 90))/(9*x + 18),x)

[Out]

5*x*exp(-(10*x)/9)*exp(-1/9)*exp(-(exp(2*x)*(x + 2)^x)/9)

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sympy [A]  time = 53.75, size = 27, normalized size = 0.90 \begin {gather*} 5 x e^{- \frac {10 x}{9} - \frac {e^{x \log {\left (x + 2 \right )} + 2 x}}{9} - \frac {1}{9}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x**2-10*x)*ln(2+x)-15*x**2-20*x)*exp(x*ln(2+x)+2*x)-50*x**2-55*x+90)/(9*x+18)/exp(1/9*exp(x*ln
(2+x)+2*x)+10/9*x+1/9),x)

[Out]

5*x*exp(-10*x/9 - exp(x*log(x + 2) + 2*x)/9 - 1/9)

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