3.94.65 \(\int \frac {20+8 x+8 x \log (x)+e^{14-x} (25 x+20 x^2+4 x^3) \log ^2(x)}{(25 x+20 x^2+4 x^3) \log ^2(x)} \, dx\)

Optimal. Leaf size=24 \[ 14-e^{14-x}-\frac {4}{(5+2 x) \log (x)} \]

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Rubi [F]  time = 1.23, 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 {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{\left (25 x+20 x^2+4 x^3\right ) \log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(20 + 8*x + 8*x*Log[x] + E^(14 - x)*(25*x + 20*x^2 + 4*x^3)*Log[x]^2)/((25*x + 20*x^2 + 4*x^3)*Log[x]^2),x
]

[Out]

-E^(14 - x) + 4*Defer[Int][1/(x*(5 + 2*x)*Log[x]^2), x] + 8*Defer[Int][1/((5 + 2*x)^2*Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{x \left (25+20 x+4 x^2\right ) \log ^2(x)} \, dx\\ &=\int \frac {20+8 x+8 x \log (x)+e^{14-x} \left (25 x+20 x^2+4 x^3\right ) \log ^2(x)}{x (5+2 x)^2 \log ^2(x)} \, dx\\ &=\int \left (e^{14-x}+\frac {4 (5+2 x+2 x \log (x))}{x (5+2 x)^2 \log ^2(x)}\right ) \, dx\\ &=4 \int \frac {5+2 x+2 x \log (x)}{x (5+2 x)^2 \log ^2(x)} \, dx+\int e^{14-x} \, dx\\ &=-e^{14-x}+4 \int \left (\frac {1}{x (5+2 x) \log ^2(x)}+\frac {2}{(5+2 x)^2 \log (x)}\right ) \, dx\\ &=-e^{14-x}+4 \int \frac {1}{x (5+2 x) \log ^2(x)} \, dx+8 \int \frac {1}{(5+2 x)^2 \log (x)} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.58, size = 23, normalized size = 0.96 \begin {gather*} -e^{14-x}-\frac {4}{(5+2 x) \log (x)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(20 + 8*x + 8*x*Log[x] + E^(14 - x)*(25*x + 20*x^2 + 4*x^3)*Log[x]^2)/((25*x + 20*x^2 + 4*x^3)*Log[x
]^2),x]

[Out]

-E^(14 - x) - 4/((5 + 2*x)*Log[x])

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fricas [A]  time = 0.60, size = 29, normalized size = 1.21 \begin {gather*} -\frac {{\left (2 \, x + 5\right )} e^{\left (-x + 14\right )} \log \relax (x) + 4}{{\left (2 \, x + 5\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+20*x^2+25*x)*exp(-x+14)*log(x)^2+8*x*log(x)+8*x+20)/(4*x^3+20*x^2+25*x)/log(x)^2,x, algorith
m="fricas")

[Out]

-((2*x + 5)*e^(-x + 14)*log(x) + 4)/((2*x + 5)*log(x))

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giac [A]  time = 0.18, size = 37, normalized size = 1.54 \begin {gather*} -\frac {2 \, x e^{\left (-x + 14\right )} \log \relax (x) + 5 \, e^{\left (-x + 14\right )} \log \relax (x) + 4}{2 \, x \log \relax (x) + 5 \, \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+20*x^2+25*x)*exp(-x+14)*log(x)^2+8*x*log(x)+8*x+20)/(4*x^3+20*x^2+25*x)/log(x)^2,x, algorith
m="giac")

[Out]

-(2*x*e^(-x + 14)*log(x) + 5*e^(-x + 14)*log(x) + 4)/(2*x*log(x) + 5*log(x))

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maple [A]  time = 0.04, size = 23, normalized size = 0.96

method	result	size
risch	\(-{\mathrm e}^{-x +14}-\frac {4}{\left (5+2 x \right ) \ln \relax (x )}\)	\(23\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((4*x^3+20*x^2+25*x)*exp(-x+14)*ln(x)^2+8*x*ln(x)+8*x+20)/(4*x^3+20*x^2+25*x)/ln(x)^2,x,method=_RETURNVERB
OSE)

[Out]

-exp(-x+14)-4/(5+2*x)/ln(x)

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maxima [A]  time = 0.40, size = 35, normalized size = 1.46 \begin {gather*} -\frac {{\left ({\left (2 \, x e^{14} + 5 \, e^{14}\right )} \log \relax (x) + 4 \, e^{x}\right )} e^{\left (-x\right )}}{{\left (2 \, x + 5\right )} \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x^3+20*x^2+25*x)*exp(-x+14)*log(x)^2+8*x*log(x)+8*x+20)/(4*x^3+20*x^2+25*x)/log(x)^2,x, algorith
m="maxima")

[Out]

-((2*x*e^14 + 5*e^14)*log(x) + 4*e^x)*e^(-x)/((2*x + 5)*log(x))

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mupad [B]  time = 6.34, size = 45, normalized size = 1.88 \begin {gather*} -\frac {4}{\ln \relax (x)\,\left (2\,x+5\right )}-\frac {5\,{\mathrm {e}}^{14-x}}{2\,x+5}-\frac {2\,x\,{\mathrm {e}}^{14-x}}{2\,x+5} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((8*x + 8*x*log(x) + exp(14 - x)*log(x)^2*(25*x + 20*x^2 + 4*x^3) + 20)/(log(x)^2*(25*x + 20*x^2 + 4*x^3)),
x)

[Out]

- 4/(log(x)*(2*x + 5)) - (5*exp(14 - x))/(2*x + 5) - (2*x*exp(14 - x))/(2*x + 5)

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sympy [A]  time = 0.32, size = 15, normalized size = 0.62 \begin {gather*} - e^{14 - x} - \frac {4}{\left (2 x + 5\right ) \log {\relax (x )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((4*x**3+20*x**2+25*x)*exp(-x+14)*ln(x)**2+8*x*ln(x)+8*x+20)/(4*x**3+20*x**2+25*x)/ln(x)**2,x)

[Out]

-exp(14 - x) - 4/((2*x + 5)*log(x))

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