3.70.50 \(\int \frac {16-8 x^3+e (-64+16 x+4 x^2+32 x^3-8 x^4)+(4-2 x^3+e (-16+4 x+x^2+8 x^3-2 x^4)) \log ^2(5)+e^x (e (-16-4 x^2+8 x^3)+e (-4-x^2+2 x^3) \log ^2(5))}{-x^2+e^{1+x} x^2+e (4 x^2-x^3)} \, dx\)

Optimal. Leaf size=30 \[ \left (4+\log ^2(5)\right ) \left (\frac {4}{x}+x^2-\log \left (-4+\frac {1}{e}-e^x+x\right )\right ) \]

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Rubi [F]  time = 0.86, 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 {16-8 x^3+e \left (-64+16 x+4 x^2+32 x^3-8 x^4\right )+\left (4-2 x^3+e \left (-16+4 x+x^2+8 x^3-2 x^4\right )\right ) \log ^2(5)+e^x \left (e \left (-16-4 x^2+8 x^3\right )+e \left (-4-x^2+2 x^3\right ) \log ^2(5)\right )}{-x^2+e^{1+x} x^2+e \left (4 x^2-x^3\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(16 - 8*x^3 + E*(-64 + 16*x + 4*x^2 + 32*x^3 - 8*x^4) + (4 - 2*x^3 + E*(-16 + 4*x + x^2 + 8*x^3 - 2*x^4))*
Log[5]^2 + E^x*(E*(-16 - 4*x^2 + 8*x^3) + E*(-4 - x^2 + 2*x^3)*Log[5]^2))/(-x^2 + E^(1 + x)*x^2 + E*(4*x^2 - x
^3)),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\left (e^{1+x} \left (4+x^2-2 x^3\right )+2 \left (-2+x^3\right )+e \left (16-4 x-x^2-8 x^3+2 x^4\right )\right ) \left (4+\log ^2(5)\right )}{\left (1-e^{1+x}+e (-4+x)\right ) x^2} \, dx\\ &=\left (4+\log ^2(5)\right ) \int \frac {e^{1+x} \left (4+x^2-2 x^3\right )+2 \left (-2+x^3\right )+e \left (16-4 x-x^2-8 x^3+2 x^4\right )}{\left (1-e^{1+x}+e (-4+x)\right ) x^2} \, dx\\ &=\left (4+\log ^2(5)\right ) \int \left (\frac {1-5 e+e x}{1-4 e-e^{1+x}+e x}+\frac {-4-x^2+2 x^3}{x^2}\right ) \, dx\\ &=\left (4+\log ^2(5)\right ) \int \frac {1-5 e+e x}{1-4 e-e^{1+x}+e x} \, dx+\left (4+\log ^2(5)\right ) \int \frac {-4-x^2+2 x^3}{x^2} \, dx\\ &=\left (4+\log ^2(5)\right ) \int \left (-1-\frac {4}{x^2}+2 x\right ) \, dx+\left (4+\log ^2(5)\right ) \int \left (-\frac {1-5 e}{-1+4 e+e^{1+x}-e x}+\frac {e x}{1-4 e-e^{1+x}+e x}\right ) \, dx\\ &=\frac {4 \left (4+\log ^2(5)\right )}{x}-x \left (4+\log ^2(5)\right )+x^2 \left (4+\log ^2(5)\right )-\left ((1-5 e) \left (4+\log ^2(5)\right )\right ) \int \frac {1}{-1+4 e+e^{1+x}-e x} \, dx+\left (e \left (4+\log ^2(5)\right )\right ) \int \frac {x}{1-4 e-e^{1+x}+e x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.19, size = 34, normalized size = 1.13 \begin {gather*} \left (4+\log ^2(5)\right ) \left (\frac {4}{x}+x^2-\log \left (1-4 e-e^{1+x}+e x\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(16 - 8*x^3 + E*(-64 + 16*x + 4*x^2 + 32*x^3 - 8*x^4) + (4 - 2*x^3 + E*(-16 + 4*x + x^2 + 8*x^3 - 2*
x^4))*Log[5]^2 + E^x*(E*(-16 - 4*x^2 + 8*x^3) + E*(-4 - x^2 + 2*x^3)*Log[5]^2))/(-x^2 + E^(1 + x)*x^2 + E*(4*x
^2 - x^3)),x]

[Out]

(4 + Log[5]^2)*(4/x + x^2 - Log[1 - 4*E - E^(1 + x) + E*x])

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fricas [A]  time = 0.57, size = 47, normalized size = 1.57 \begin {gather*} \frac {4 \, x^{3} + {\left (x^{3} + 4\right )} \log \relax (5)^{2} - {\left (x \log \relax (5)^{2} + 4 \, x\right )} \log \left (-{\left (x - 4\right )} e + e^{\left (x + 1\right )} - 1\right ) + 16}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-x^2-4)*exp(1)*log(5)^2+(8*x^3-4*x^2-16)*exp(1))*exp(x)+((-2*x^4+8*x^3+x^2+4*x-16)*exp(1)-2*
x^3+4)*log(5)^2+(-8*x^4+32*x^3+4*x^2+16*x-64)*exp(1)-8*x^3+16)/(x^2*exp(1)*exp(x)+(-x^3+4*x^2)*exp(1)-x^2),x,
algorithm="fricas")

[Out]

(4*x^3 + (x^3 + 4)*log(5)^2 - (x*log(5)^2 + 4*x)*log(-(x - 4)*e + e^(x + 1) - 1) + 16)/x

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giac [B]  time = 0.23, size = 67, normalized size = 2.23 \begin {gather*} \frac {x^{3} \log \relax (5)^{2} - x \log \relax (5)^{2} \log \left (-x e + 4 \, e + e^{\left (x + 1\right )} - 1\right ) + 4 \, x^{3} + 4 \, \log \relax (5)^{2} - 4 \, x \log \left (-x e + 4 \, e + e^{\left (x + 1\right )} - 1\right ) + 16}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-x^2-4)*exp(1)*log(5)^2+(8*x^3-4*x^2-16)*exp(1))*exp(x)+((-2*x^4+8*x^3+x^2+4*x-16)*exp(1)-2*
x^3+4)*log(5)^2+(-8*x^4+32*x^3+4*x^2+16*x-64)*exp(1)-8*x^3+16)/(x^2*exp(1)*exp(x)+(-x^3+4*x^2)*exp(1)-x^2),x,
algorithm="giac")

[Out]

(x^3*log(5)^2 - x*log(5)^2*log(-x*e + 4*e + e^(x + 1) - 1) + 4*x^3 + 4*log(5)^2 - 4*x*log(-x*e + 4*e + e^(x +
1) - 1) + 16)/x

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maple [A]  time = 0.11, size = 50, normalized size = 1.67

method	result	size
norman	\(\frac {\left (4+\ln \relax (5)^{2}\right ) x^{3}+4 \ln \relax (5)^{2}+16}{x}+\left (-\ln \relax (5)^{2}-4\right ) \ln \left (x \,{\mathrm e}-{\mathrm e} \,{\mathrm e}^{x}-4 \,{\mathrm e}+1\right )\)	\(50\)
risch	\(x^{2} \ln \relax (5)^{2}+4 x^{2}+\frac {4 \ln \relax (5)^{2}}{x}+\frac {16}{x}-\ln \relax (5)^{2} \ln \left ({\mathrm e}^{x}-\left (x \,{\mathrm e}-4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1}\right )-4 \ln \left ({\mathrm e}^{x}-\left (x \,{\mathrm e}-4 \,{\mathrm e}+1\right ) {\mathrm e}^{-1}\right )\)	\(73\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((2*x^3-x^2-4)*exp(1)*ln(5)^2+(8*x^3-4*x^2-16)*exp(1))*exp(x)+((-2*x^4+8*x^3+x^2+4*x-16)*exp(1)-2*x^3+4)*
ln(5)^2+(-8*x^4+32*x^3+4*x^2+16*x-64)*exp(1)-8*x^3+16)/(x^2*exp(1)*exp(x)+(-x^3+4*x^2)*exp(1)-x^2),x,method=_R
ETURNVERBOSE)

[Out]

((4+ln(5)^2)*x^3+4*ln(5)^2+16)/x+(-ln(5)^2-4)*ln(x*exp(1)-exp(1)*exp(x)-4*exp(1)+1)

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maxima [A]  time = 0.50, size = 52, normalized size = 1.73 \begin {gather*} -{\left (\log \relax (5)^{2} + 4\right )} \log \left (-{\left (x e - 4 \, e - e^{\left (x + 1\right )} + 1\right )} e^{\left (-1\right )}\right ) + \frac {{\left (\log \relax (5)^{2} + 4\right )} x^{3} + 4 \, \log \relax (5)^{2} + 16}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x^3-x^2-4)*exp(1)*log(5)^2+(8*x^3-4*x^2-16)*exp(1))*exp(x)+((-2*x^4+8*x^3+x^2+4*x-16)*exp(1)-2*
x^3+4)*log(5)^2+(-8*x^4+32*x^3+4*x^2+16*x-64)*exp(1)-8*x^3+16)/(x^2*exp(1)*exp(x)+(-x^3+4*x^2)*exp(1)-x^2),x,
algorithm="maxima")

[Out]

-(log(5)^2 + 4)*log(-(x*e - 4*e - e^(x + 1) + 1)*e^(-1)) + ((log(5)^2 + 4)*x^3 + 4*log(5)^2 + 16)/x

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mupad [B]  time = 0.27, size = 41, normalized size = 1.37 \begin {gather*} \frac {\left ({\ln \relax (5)}^2+4\right )\,x^3+4\,{\ln \relax (5)}^2+16}{x}-\ln \left (x+{\mathrm {e}}^{-1}-{\mathrm {e}}^x-4\right )\,\left ({\ln \relax (5)}^2+4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(5)^2*(exp(1)*(4*x + x^2 + 8*x^3 - 2*x^4 - 16) - 2*x^3 + 4) - exp(x)*(exp(1)*(4*x^2 - 8*x^3 + 16) + ex
p(1)*log(5)^2*(x^2 - 2*x^3 + 4)) + exp(1)*(16*x + 4*x^2 + 32*x^3 - 8*x^4 - 64) - 8*x^3 + 16)/(exp(1)*(4*x^2 -
x^3) - x^2 + x^2*exp(1)*exp(x)),x)

[Out]

(x^3*(log(5)^2 + 4) + 4*log(5)^2 + 16)/x - log(x + exp(-1) - exp(x) - 4)*(log(5)^2 + 4)

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sympy [A]  time = 0.25, size = 48, normalized size = 1.60 \begin {gather*} x^{2} \left (\log {\relax (5 )}^{2} + 4\right ) + \left (-4 - \log {\relax (5 )}^{2}\right ) \log {\left (\frac {- e x - 1 + 4 e}{e} + e^{x} \right )} + \frac {4 \log {\relax (5 )}^{2} + 16}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((2*x**3-x**2-4)*exp(1)*ln(5)**2+(8*x**3-4*x**2-16)*exp(1))*exp(x)+((-2*x**4+8*x**3+x**2+4*x-16)*ex
p(1)-2*x**3+4)*ln(5)**2+(-8*x**4+32*x**3+4*x**2+16*x-64)*exp(1)-8*x**3+16)/(x**2*exp(1)*exp(x)+(-x**3+4*x**2)*
exp(1)-x**2),x)

[Out]

x**2*(log(5)**2 + 4) + (-4 - log(5)**2)*log((-E*x - 1 + 4*E)*exp(-1) + exp(x)) + (4*log(5)**2 + 16)/x

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