3.89.96 \(\int \frac {e^{-3-x} (20-4 x+(56+2 x-4 x^2) \log (7+2 x))}{(175-20 x-13 x^2+2 x^3) \log ^2(7+2 x)} \, dx\)

Optimal. Leaf size=22 \[ \frac {2 e^{-3-x}}{(-5+x) \log (7+2 x)} \]

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Rubi [F]  time = 1.21, 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^{-3-x} \left (20-4 x+\left (56+2 x-4 x^2\right ) \log (7+2 x)\right )}{\left (175-20 x-13 x^2+2 x^3\right ) \log ^2(7+2 x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(-3 - x)*(20 - 4*x + (56 + 2*x - 4*x^2)*Log[7 + 2*x]))/((175 - 20*x - 13*x^2 + 2*x^3)*Log[7 + 2*x]^2),x
]

[Out]

(-4*Defer[Int][E^(-3 - x)/((-5 + x)*Log[7 + 2*x]^2), x])/17 + (8*Defer[Int][E^(-3 - x)/((7 + 2*x)*Log[7 + 2*x]
^2), x])/17 - 2*Defer[Int][E^(-3 - x)/((-5 + x)^2*Log[7 + 2*x]), x] - 2*Defer[Int][E^(-3 - x)/((-5 + x)*Log[7
+ 2*x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {4 e^{-3-x}}{(-5+x) (7+2 x) \log ^2(7+2 x)}-\frac {2 e^{-3-x} (-4+x)}{(-5+x)^2 \log (7+2 x)}\right ) \, dx\\ &=-\left (2 \int \frac {e^{-3-x} (-4+x)}{(-5+x)^2 \log (7+2 x)} \, dx\right )-4 \int \frac {e^{-3-x}}{(-5+x) (7+2 x) \log ^2(7+2 x)} \, dx\\ &=-\left (2 \int \left (\frac {e^{-3-x}}{(-5+x)^2 \log (7+2 x)}+\frac {e^{-3-x}}{(-5+x) \log (7+2 x)}\right ) \, dx\right )-4 \int \left (\frac {e^{-3-x}}{17 (-5+x) \log ^2(7+2 x)}-\frac {2 e^{-3-x}}{17 (7+2 x) \log ^2(7+2 x)}\right ) \, dx\\ &=-\left (\frac {4}{17} \int \frac {e^{-3-x}}{(-5+x) \log ^2(7+2 x)} \, dx\right )+\frac {8}{17} \int \frac {e^{-3-x}}{(7+2 x) \log ^2(7+2 x)} \, dx-2 \int \frac {e^{-3-x}}{(-5+x)^2 \log (7+2 x)} \, dx-2 \int \frac {e^{-3-x}}{(-5+x) \log (7+2 x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.38, size = 22, normalized size = 1.00 \begin {gather*} \frac {2 e^{-3-x}}{(-5+x) \log (7+2 x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(-3 - x)*(20 - 4*x + (56 + 2*x - 4*x^2)*Log[7 + 2*x]))/((175 - 20*x - 13*x^2 + 2*x^3)*Log[7 + 2*x
]^2),x]

[Out]

(2*E^(-3 - x))/((-5 + x)*Log[7 + 2*x])

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fricas [A]  time = 0.56, size = 21, normalized size = 0.95 \begin {gather*} \frac {2 \, e^{\left (-x - 3\right )}}{{\left (x - 5\right )} \log \left (2 \, x + 7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+2*x+56)*log(7+2*x)-4*x+20)/(2*x^3-13*x^2-20*x+175)/exp(3+x)/log(7+2*x)^2,x, algorithm="fric
as")

[Out]

2*e^(-x - 3)/((x - 5)*log(2*x + 7))

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giac [A]  time = 0.20, size = 29, normalized size = 1.32 \begin {gather*} \frac {2 \, e^{\left (-x\right )}}{x e^{3} \log \left (2 \, x + 7\right ) - 5 \, e^{3} \log \left (2 \, x + 7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+2*x+56)*log(7+2*x)-4*x+20)/(2*x^3-13*x^2-20*x+175)/exp(3+x)/log(7+2*x)^2,x, algorithm="giac
")

[Out]

2*e^(-x)/(x*e^3*log(2*x + 7) - 5*e^3*log(2*x + 7))

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maple [A]  time = 0.03, size = 22, normalized size = 1.00

method	result	size
risch	\(\frac {2 \,{\mathrm e}^{-3-x}}{\left (x -5\right ) \ln \left (7+2 x \right )}\)	\(22\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-4*x^2+2*x+56)*ln(7+2*x)-4*x+20)/(2*x^3-13*x^2-20*x+175)/exp(3+x)/ln(7+2*x)^2,x,method=_RETURNVERBOSE)

[Out]

2/(x-5)/ln(7+2*x)*exp(-3-x)

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maxima [A]  time = 0.44, size = 25, normalized size = 1.14 \begin {gather*} \frac {2 \, e^{\left (-x\right )}}{{\left (x e^{3} - 5 \, e^{3}\right )} \log \left (2 \, x + 7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x^2+2*x+56)*log(7+2*x)-4*x+20)/(2*x^3-13*x^2-20*x+175)/exp(3+x)/log(7+2*x)^2,x, algorithm="maxi
ma")

[Out]

2*e^(-x)/((x*e^3 - 5*e^3)*log(2*x + 7))

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mupad [B]  time = 5.81, size = 28, normalized size = 1.27 \begin {gather*} -\frac {2\,{\mathrm {e}}^{-x}\,{\mathrm {e}}^{-3}}{5\,\ln \left (2\,x+7\right )-x\,\ln \left (2\,x+7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(- x - 3)*(log(2*x + 7)*(2*x - 4*x^2 + 56) - 4*x + 20))/(log(2*x + 7)^2*(20*x + 13*x^2 - 2*x^3 - 175)
),x)

[Out]

-(2*exp(-x)*exp(-3))/(5*log(2*x + 7) - x*log(2*x + 7))

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sympy [A]  time = 0.32, size = 24, normalized size = 1.09 \begin {gather*} \frac {2 e^{- x - 3}}{x \log {\left (2 x + 7 \right )} - 5 \log {\left (2 x + 7 \right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-4*x**2+2*x+56)*ln(7+2*x)-4*x+20)/(2*x**3-13*x**2-20*x+175)/exp(3+x)/ln(7+2*x)**2,x)

[Out]

2*exp(-x - 3)/(x*log(2*x + 7) - 5*log(2*x + 7))

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