3.87.61 \(\int \frac {-5 x-70 x^2+e^{2 x} x^2+70 x^3+5 x^4+e^x (-1-4 x^2+5 x^3)}{e^{2 x} x^2+e^x (x+15 x^2+x^3)} \, dx\)

Optimal. Leaf size=25 \[ 2-5 e^{-x} x+\log \left (3+\frac {1}{5} \left (e^x+\frac {1}{x}+x\right )\right ) \]

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Rubi [F]  time = 3.32, 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 {-5 x-70 x^2+e^{2 x} x^2+70 x^3+5 x^4+e^x \left (-1-4 x^2+5 x^3\right )}{e^{2 x} x^2+e^x \left (x+15 x^2+x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5*x - 70*x^2 + E^(2*x)*x^2 + 70*x^3 + 5*x^4 + E^x*(-1 - 4*x^2 + 5*x^3))/(E^(2*x)*x^2 + E^x*(x + 15*x^2 +
x^3)),x]

[Out]

15/E^x + 1/(E^x*x) + x - (4*x)/E^x + 30*Defer[Int][1/(E^x*(1 + 15*x + E^x*x + x^2)), x] + Defer[Int][1/(E^x*x^
2*(1 + 15*x + E^x*x + x^2)), x] + 16*Defer[Int][1/(E^x*x*(1 + 15*x + E^x*x + x^2)), x] + 212*Defer[Int][x/(E^x
*(1 + 15*x + E^x*x + x^2)), x] + 29*Defer[Int][x^2/(E^x*(1 + 15*x + E^x*x + x^2)), x] + Defer[Int][x^3/(E^x*(1
+ 15*x + E^x*x + x^2)), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (-5 x-70 x^2+e^{2 x} x^2+70 x^3+5 x^4+e^x \left (-1-4 x^2+5 x^3\right )\right )}{x \left (1+15 x+e^x x+x^2\right )} \, dx\\ &=\int \left (1+\frac {e^{-x} \left (-1-x-19 x^2+4 x^3\right )}{x^2}+\frac {e^{-x} \left (1+16 x+30 x^2+212 x^3+29 x^4+x^5\right )}{x^2 \left (1+15 x+e^x x+x^2\right )}\right ) \, dx\\ &=x+\int \frac {e^{-x} \left (-1-x-19 x^2+4 x^3\right )}{x^2} \, dx+\int \frac {e^{-x} \left (1+16 x+30 x^2+212 x^3+29 x^4+x^5\right )}{x^2 \left (1+15 x+e^x x+x^2\right )} \, dx\\ &=x+\int \left (-19 e^{-x}-\frac {e^{-x}}{x^2}-\frac {e^{-x}}{x}+4 e^{-x} x\right ) \, dx+\int \left (\frac {30 e^{-x}}{1+15 x+e^x x+x^2}+\frac {e^{-x}}{x^2 \left (1+15 x+e^x x+x^2\right )}+\frac {16 e^{-x}}{x \left (1+15 x+e^x x+x^2\right )}+\frac {212 e^{-x} x}{1+15 x+e^x x+x^2}+\frac {29 e^{-x} x^2}{1+15 x+e^x x+x^2}+\frac {e^{-x} x^3}{1+15 x+e^x x+x^2}\right ) \, dx\\ &=x+4 \int e^{-x} x \, dx+16 \int \frac {e^{-x}}{x \left (1+15 x+e^x x+x^2\right )} \, dx-19 \int e^{-x} \, dx+29 \int \frac {e^{-x} x^2}{1+15 x+e^x x+x^2} \, dx+30 \int \frac {e^{-x}}{1+15 x+e^x x+x^2} \, dx+212 \int \frac {e^{-x} x}{1+15 x+e^x x+x^2} \, dx-\int \frac {e^{-x}}{x^2} \, dx-\int \frac {e^{-x}}{x} \, dx+\int \frac {e^{-x}}{x^2 \left (1+15 x+e^x x+x^2\right )} \, dx+\int \frac {e^{-x} x^3}{1+15 x+e^x x+x^2} \, dx\\ &=19 e^{-x}+\frac {e^{-x}}{x}+x-4 e^{-x} x-\text {Ei}(-x)+4 \int e^{-x} \, dx+16 \int \frac {e^{-x}}{x \left (1+15 x+e^x x+x^2\right )} \, dx+29 \int \frac {e^{-x} x^2}{1+15 x+e^x x+x^2} \, dx+30 \int \frac {e^{-x}}{1+15 x+e^x x+x^2} \, dx+212 \int \frac {e^{-x} x}{1+15 x+e^x x+x^2} \, dx+\int \frac {e^{-x}}{x} \, dx+\int \frac {e^{-x}}{x^2 \left (1+15 x+e^x x+x^2\right )} \, dx+\int \frac {e^{-x} x^3}{1+15 x+e^x x+x^2} \, dx\\ &=15 e^{-x}+\frac {e^{-x}}{x}+x-4 e^{-x} x+16 \int \frac {e^{-x}}{x \left (1+15 x+e^x x+x^2\right )} \, dx+29 \int \frac {e^{-x} x^2}{1+15 x+e^x x+x^2} \, dx+30 \int \frac {e^{-x}}{1+15 x+e^x x+x^2} \, dx+212 \int \frac {e^{-x} x}{1+15 x+e^x x+x^2} \, dx+\int \frac {e^{-x}}{x^2 \left (1+15 x+e^x x+x^2\right )} \, dx+\int \frac {e^{-x} x^3}{1+15 x+e^x x+x^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.02, size = 33, normalized size = 1.32 \begin {gather*} x-5 e^{-x} x-\log (x)+\log \left (e^{-x} \left (1+\left (15+e^x\right ) x+x^2\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*x - 70*x^2 + E^(2*x)*x^2 + 70*x^3 + 5*x^4 + E^x*(-1 - 4*x^2 + 5*x^3))/(E^(2*x)*x^2 + E^x*(x + 15
*x^2 + x^3)),x]

[Out]

x - (5*x)/E^x - Log[x] + Log[(1 + (15 + E^x)*x + x^2)/E^x]

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fricas [A]  time = 0.55, size = 29, normalized size = 1.16 \begin {gather*} {\left (e^{x} \log \left (\frac {x^{2} + x e^{x} + 15 \, x + 1}{x}\right ) - 5 \, x\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)^2*x^2+(5*x^3-4*x^2-1)*exp(x)+5*x^4+70*x^3-70*x^2-5*x)/(exp(x)^2*x^2+(x^3+15*x^2+x)*exp(x)),x
, algorithm="fricas")

[Out]

(e^x*log((x^2 + x*e^x + 15*x + 1)/x) - 5*x)*e^(-x)

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giac [A]  time = 0.20, size = 31, normalized size = 1.24 \begin {gather*} {\left (e^{x} \log \left (x^{2} + x e^{x} + 15 \, x + 1\right ) - e^{x} \log \relax (x) - 5 \, x\right )} e^{\left (-x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)^2*x^2+(5*x^3-4*x^2-1)*exp(x)+5*x^4+70*x^3-70*x^2-5*x)/(exp(x)^2*x^2+(x^3+15*x^2+x)*exp(x)),x
, algorithm="giac")

[Out]

(e^x*log(x^2 + x*e^x + 15*x + 1) - e^x*log(x) - 5*x)*e^(-x)

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maple [A]  time = 0.05, size = 25, normalized size = 1.00

method	result	size
risch	\(-5 x \,{\mathrm e}^{-x}+\ln \left ({\mathrm e}^{x}+\frac {x^{2}+15 x +1}{x}\right )\)	\(25\)
norman	\(-5 x \,{\mathrm e}^{-x}-\ln \relax (x )+\ln \left ({\mathrm e}^{x} x +x^{2}+15 x +1\right )\)	\(26\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)^2*x^2+(5*x^3-4*x^2-1)*exp(x)+5*x^4+70*x^3-70*x^2-5*x)/(exp(x)^2*x^2+(x^3+15*x^2+x)*exp(x)),x,metho
d=_RETURNVERBOSE)

[Out]

-5*x*exp(-x)+ln(exp(x)+(x^2+15*x+1)/x)

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maxima [A]  time = 0.43, size = 25, normalized size = 1.00 \begin {gather*} -5 \, x e^{\left (-x\right )} + \log \left (\frac {x^{2} + x e^{x} + 15 \, x + 1}{x}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)^2*x^2+(5*x^3-4*x^2-1)*exp(x)+5*x^4+70*x^3-70*x^2-5*x)/(exp(x)^2*x^2+(x^3+15*x^2+x)*exp(x)),x
, algorithm="maxima")

[Out]

-5*x*e^(-x) + log((x^2 + x*e^x + 15*x + 1)/x)

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mupad [B]  time = 0.20, size = 25, normalized size = 1.00 \begin {gather*} \ln \left (15\,x+x\,{\mathrm {e}}^x+x^2+1\right )-\ln \relax (x)-5\,x\,{\mathrm {e}}^{-x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(5*x + exp(x)*(4*x^2 - 5*x^3 + 1) - x^2*exp(2*x) + 70*x^2 - 70*x^3 - 5*x^4)/(exp(x)*(x + 15*x^2 + x^3) +
x^2*exp(2*x)),x)

[Out]

log(15*x + x*exp(x) + x^2 + 1) - log(x) - 5*x*exp(-x)

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sympy [A]  time = 0.22, size = 20, normalized size = 0.80 \begin {gather*} - 5 x e^{- x} + \log {\left (e^{x} + \frac {x^{2} + 15 x + 1}{x} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((exp(x)**2*x**2+(5*x**3-4*x**2-1)*exp(x)+5*x**4+70*x**3-70*x**2-5*x)/(exp(x)**2*x**2+(x**3+15*x**2+x
)*exp(x)),x)

[Out]

-5*x*exp(-x) + log(exp(x) + (x**2 + 15*x + 1)/x)

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