3.33.79 \(\int \frac {2 x-48 x^3+e (1-48 x^2)+e^{10+2 x} (1+2 x-48 x^2)}{e x+e^{10+2 x} x+x^2} \, dx\)

Optimal. Leaf size=20 \[ \log \left (e^{-24 x^2} x \left (e+e^{2 (5+x)}+x\right )\right ) \]

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Rubi [F]  time = 0.37, 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-48 x^3+e \left (1-48 x^2\right )+e^{10+2 x} \left (1+2 x-48 x^2\right )}{e x+e^{10+2 x} x+x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(2*x - 48*x^3 + E*(1 - 48*x^2) + E^(10 + 2*x)*(1 + 2*x - 48*x^2))/(E*x + E^(10 + 2*x)*x + x^2),x]

[Out]

2*x - 24*x^2 + Log[x] + (1 - 2*E)*Defer[Int][(E + E^(10 + 2*x) + x)^(-1), x] - 2*Defer[Int][x/(E + E^(10 + 2*x
) + x), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (-\frac {-1+2 e+2 x}{e+e^{10+2 x}+x}+\frac {1+2 x-48 x^2}{x}\right ) \, dx\\ &=-\int \frac {-1+2 e+2 x}{e+e^{10+2 x}+x} \, dx+\int \frac {1+2 x-48 x^2}{x} \, dx\\ &=\int \left (2+\frac {1}{x}-48 x\right ) \, dx-\int \left (-\frac {1-2 e}{e+e^{10+2 x}+x}+\frac {2 x}{e+e^{10+2 x}+x}\right ) \, dx\\ &=2 x-24 x^2+\log (x)-2 \int \frac {x}{e+e^{10+2 x}+x} \, dx-(-1+2 e) \int \frac {1}{e+e^{10+2 x}+x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.17, size = 19, normalized size = 0.95 \begin {gather*} -24 x^2+\log (x)+\log \left (e+e^{10+2 x}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2*x - 48*x^3 + E*(1 - 48*x^2) + E^(10 + 2*x)*(1 + 2*x - 48*x^2))/(E*x + E^(10 + 2*x)*x + x^2),x]

[Out]

-24*x^2 + Log[x] + Log[E + E^(10 + 2*x) + x]

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fricas [A]  time = 0.66, size = 19, normalized size = 0.95 \begin {gather*} -24 \, x^{2} + \log \left (x + e + e^{\left (2 \, x + 10\right )}\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*x^2+2*x+1)*exp(2*x+10)+(-48*x^2+1)*exp(1)-48*x^3+2*x)/(x*exp(2*x+10)+x*exp(1)+x^2),x, algorith
m="fricas")

[Out]

-24*x^2 + log(x + e + e^(2*x + 10)) + log(x)

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giac [A]  time = 0.21, size = 25, normalized size = 1.25 \begin {gather*} -24 \, x^{2} + \log \relax (x) + \log \left (-x - e - e^{\left (2 \, x + 10\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*x^2+2*x+1)*exp(2*x+10)+(-48*x^2+1)*exp(1)-48*x^3+2*x)/(x*exp(2*x+10)+x*exp(1)+x^2),x, algorith
m="giac")

[Out]

-24*x^2 + log(x) + log(-x - e - e^(2*x + 10))

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maple [A]  time = 0.10, size = 20, normalized size = 1.00

method	result	size
norman	\(-24 x^{2}+\ln \relax (x )+\ln \left ({\mathrm e}^{2 x +10}+x +{\mathrm e}\right )\)	\(20\)
risch	\(-24 x^{2}+\ln \relax (x )-10+\ln \left ({\mathrm e}^{2 x +10}+x +{\mathrm e}\right )\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-48*x^2+2*x+1)*exp(2*x+10)+(-48*x^2+1)*exp(1)-48*x^3+2*x)/(x*exp(2*x+10)+x*exp(1)+x^2),x,method=_RETURNV
ERBOSE)

[Out]

-24*x^2+ln(x)+ln(exp(2*x+10)+x+exp(1))

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maxima [A]  time = 0.77, size = 22, normalized size = 1.10 \begin {gather*} -24 \, x^{2} + \log \left ({\left (x + e + e^{\left (2 \, x + 10\right )}\right )} e^{\left (-10\right )}\right ) + \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*x^2+2*x+1)*exp(2*x+10)+(-48*x^2+1)*exp(1)-48*x^3+2*x)/(x*exp(2*x+10)+x*exp(1)+x^2),x, algorith
m="maxima")

[Out]

-24*x^2 + log((x + e + e^(2*x + 10))*e^(-10)) + log(x)

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mupad [B]  time = 0.16, size = 19, normalized size = 0.95 \begin {gather*} \ln \left (x+\mathrm {e}+{\mathrm {e}}^{2\,x+10}\right )+\ln \relax (x)-24\,x^2 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2*x - exp(1)*(48*x^2 - 1) + exp(2*x + 10)*(2*x - 48*x^2 + 1) - 48*x^3)/(x*exp(1) + x*exp(2*x + 10) + x^2)
,x)

[Out]

log(x + exp(1) + exp(2*x + 10)) + log(x) - 24*x^2

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sympy [A]  time = 0.16, size = 20, normalized size = 1.00 \begin {gather*} - 24 x^{2} + \log {\relax (x )} + \log {\left (x + e^{2 x + 10} + e \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-48*x**2+2*x+1)*exp(2*x+10)+(-48*x**2+1)*exp(1)-48*x**3+2*x)/(x*exp(2*x+10)+x*exp(1)+x**2),x)

[Out]

-24*x**2 + log(x) + log(x + exp(2*x + 10) + E)

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