3.21.82 \(\int \frac {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x (5 x+20 x^2-50 x^3)+(5 x+e^x (-5 x-5 x^2)) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x (20 x^2-50 x^3)} \, dx\)

Optimal. Leaf size=18 \[ 4+x+\frac {\log (x)}{2+5 \left (-1+e^x\right ) x} \]

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Rubi [F]  time = 2.02, 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 {2-x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (5 x+20 x^2-50 x^3\right )+\left (5 x+e^x \left (-5 x-5 x^2\right )\right ) \log (x)}{4 x-20 x^2+25 x^3+25 e^{2 x} x^3+e^x \left (20 x^2-50 x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2 - x - 20*x^2 + 25*x^3 + 25*E^(2*x)*x^3 + E^x*(5*x + 20*x^2 - 50*x^3) + (5*x + E^x*(-5*x - 5*x^2))*Log[x
])/(4*x - 20*x^2 + 25*x^3 + 25*E^(2*x)*x^3 + E^x*(20*x^2 - 50*x^3)),x]

[Out]

x + 2*Log[x]*Defer[Int][(2 - 5*x + 5*E^x*x)^(-2), x] + 2*Log[x]*Defer[Int][1/(x*(2 - 5*x + 5*E^x*x)^2), x] - 5
*Log[x]*Defer[Int][x/(2 - 5*x + 5*E^x*x)^2, x] - Log[x]*Defer[Int][(2 - 5*x + 5*E^x*x)^(-1), x] + Defer[Int][1
/(x*(2 - 5*x + 5*E^x*x)), x] - Log[x]*Defer[Int][1/(x*(2 - 5*x + 5*E^x*x)), x] - 2*Defer[Int][Defer[Int][(2 +
5*(-1 + E^x)*x)^(-2), x]/x, x] - 2*Defer[Int][Defer[Int][1/(x*(2 + 5*(-1 + E^x)*x)^2), x]/x, x] + 5*Defer[Int]
[Defer[Int][x/(2 + 5*(-1 + E^x)*x)^2, x]/x, x] + Defer[Int][Defer[Int][(2 + 5*(-1 + E^x)*x)^(-1), x]/x, x] + D
efer[Int][Defer[Int][1/(x*(2 + 5*(-1 + E^x)*x)), x]/x, x]

Rubi steps

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

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Mathematica [A]  time = 0.42, size = 18, normalized size = 1.00 \begin {gather*} x+\frac {\log (x)}{2-5 x+5 e^x x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - x - 20*x^2 + 25*x^3 + 25*E^(2*x)*x^3 + E^x*(5*x + 20*x^2 - 50*x^3) + (5*x + E^x*(-5*x - 5*x^2))
*Log[x])/(4*x - 20*x^2 + 25*x^3 + 25*E^(2*x)*x^3 + E^x*(20*x^2 - 50*x^3)),x]

[Out]

x + Log[x]/(2 - 5*x + 5*E^x*x)

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fricas [A]  time = 0.54, size = 31, normalized size = 1.72 \begin {gather*} \frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \relax (x)}{5 \, x e^{x} - 5 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2-5*x)*exp(x)+5*x)*log(x)+25*exp(x)^2*x^3+(-50*x^3+20*x^2+5*x)*exp(x)+25*x^3-20*x^2-x+2)/(25
*exp(x)^2*x^3+(-50*x^3+20*x^2)*exp(x)+25*x^3-20*x^2+4*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 3.68, size = 31, normalized size = 1.72 \begin {gather*} \frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \relax (x)}{5 \, x e^{x} - 5 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2-5*x)*exp(x)+5*x)*log(x)+25*exp(x)^2*x^3+(-50*x^3+20*x^2+5*x)*exp(x)+25*x^3-20*x^2-x+2)/(25
*exp(x)^2*x^3+(-50*x^3+20*x^2)*exp(x)+25*x^3-20*x^2+4*x),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 18, normalized size = 1.00

method	result	size
risch	\(\frac {\ln \relax (x )}{5 \,{\mathrm e}^{x} x -5 x +2}+x\)	\(18\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-5*x^2-5*x)*exp(x)+5*x)*ln(x)+25*exp(x)^2*x^3+(-50*x^3+20*x^2+5*x)*exp(x)+25*x^3-20*x^2-x+2)/(25*exp(x)
^2*x^3+(-50*x^3+20*x^2)*exp(x)+25*x^3-20*x^2+4*x),x,method=_RETURNVERBOSE)

[Out]

1/(5*exp(x)*x-5*x+2)*ln(x)+x

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maxima [A]  time = 0.80, size = 31, normalized size = 1.72 \begin {gather*} \frac {5 \, x^{2} e^{x} - 5 \, x^{2} + 2 \, x + \log \relax (x)}{5 \, x e^{x} - 5 \, x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x^2-5*x)*exp(x)+5*x)*log(x)+25*exp(x)^2*x^3+(-50*x^3+20*x^2+5*x)*exp(x)+25*x^3-20*x^2-x+2)/(25
*exp(x)^2*x^3+(-50*x^3+20*x^2)*exp(x)+25*x^3-20*x^2+4*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.30, size = 17, normalized size = 0.94 \begin {gather*} x+\frac {\ln \relax (x)}{5\,x\,{\mathrm {e}}^x-5\,x+2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(x)*(5*x - exp(x)*(5*x + 5*x^2)) - x + 25*x^3*exp(2*x) - 20*x^2 + 25*x^3 + exp(x)*(5*x + 20*x^2 - 50*x
^3) + 2)/(4*x + exp(x)*(20*x^2 - 50*x^3) + 25*x^3*exp(2*x) - 20*x^2 + 25*x^3),x)

[Out]

x + log(x)/(5*x*exp(x) - 5*x + 2)

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sympy [A]  time = 0.29, size = 15, normalized size = 0.83 \begin {gather*} x + \frac {\log {\relax (x )}}{5 x e^{x} - 5 x + 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-5*x**2-5*x)*exp(x)+5*x)*ln(x)+25*exp(x)**2*x**3+(-50*x**3+20*x**2+5*x)*exp(x)+25*x**3-20*x**2-x+
2)/(25*exp(x)**2*x**3+(-50*x**3+20*x**2)*exp(x)+25*x**3-20*x**2+4*x),x)

[Out]

x + log(x)/(5*x*exp(x) - 5*x + 2)

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