3.96.3 \(\int \frac {32-16 x+e^5 (12 x^2-8 x^3)}{-32+16 x+e^5 (32 x^2-16 x^3)+e^{10} (-8 x^4+4 x^5)+(e^5 (16 x-8 x^2)+e^{10} (-8 x^3+4 x^4)) \log (\frac {2-x}{2})+e^{10} (-2 x^2+x^3) \log ^2(\frac {2-x}{2})} \, dx\)

Optimal. Leaf size=32 \[ \frac {x^2}{-x+\frac {1}{4} e^5 x^2 \left (2 x+\log \left (1-\frac {x}{2}\right )\right )} \]

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Rubi [F]  time = 0.64, 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 {32-16 x+e^5 \left (12 x^2-8 x^3\right )}{-32+16 x+e^5 \left (32 x^2-16 x^3\right )+e^{10} \left (-8 x^4+4 x^5\right )+\left (e^5 \left (16 x-8 x^2\right )+e^{10} \left (-8 x^3+4 x^4\right )\right ) \log \left (\frac {2-x}{2}\right )+e^{10} \left (-2 x^2+x^3\right ) \log ^2\left (\frac {2-x}{2}\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(32 - 16*x + E^5*(12*x^2 - 8*x^3))/(-32 + 16*x + E^5*(32*x^2 - 16*x^3) + E^10*(-8*x^4 + 4*x^5) + (E^5*(16*
x - 8*x^2) + E^10*(-8*x^3 + 4*x^4))*Log[(2 - x)/2] + E^10*(-2*x^2 + x^3)*Log[(2 - x)/2]^2),x]

[Out]

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

Rubi steps

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

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Mathematica [A]  time = 1.08, size = 28, normalized size = 0.88 \begin {gather*} \frac {4 x}{-4+2 e^5 x^2+e^5 x \log \left (1-\frac {x}{2}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(32 - 16*x + E^5*(12*x^2 - 8*x^3))/(-32 + 16*x + E^5*(32*x^2 - 16*x^3) + E^10*(-8*x^4 + 4*x^5) + (E^
5*(16*x - 8*x^2) + E^10*(-8*x^3 + 4*x^4))*Log[(2 - x)/2] + E^10*(-2*x^2 + x^3)*Log[(2 - x)/2]^2),x]

[Out]

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

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fricas [A]  time = 0.56, size = 24, normalized size = 0.75 \begin {gather*} \frac {4 \, x}{2 \, x^{2} e^{5} + x e^{5} \log \left (-\frac {1}{2} \, x + 1\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*log(1-1/2*x)^2+((4*x^4-8*x^3)*exp(5)^2+(-8*x^
2+16*x)*exp(5))*log(1-1/2*x)+(4*x^5-8*x^4)*exp(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.33, size = 24, normalized size = 0.75 \begin {gather*} \frac {4 \, x}{2 \, x^{2} e^{5} + x e^{5} \log \left (-\frac {1}{2} \, x + 1\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*log(1-1/2*x)^2+((4*x^4-8*x^3)*exp(5)^2+(-8*x^
2+16*x)*exp(5))*log(1-1/2*x)+(4*x^5-8*x^4)*exp(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.27, size = 25, normalized size = 0.78

method	result	size
norman	\(\frac {4 x}{{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right ) x +2 x^{2} {\mathrm e}^{5}-4}\)	\(25\)
risch	\(\frac {4 x}{{\mathrm e}^{5} \ln \left (1-\frac {x}{2}\right ) x +2 x^{2} {\mathrm e}^{5}-4}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*ln(1-1/2*x)^2+((4*x^4-8*x^3)*exp(5)^2+(-8*x^2+16*x)
*exp(5))*ln(1-1/2*x)+(4*x^5-8*x^4)*exp(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.61, size = 31, normalized size = 0.97 \begin {gather*} \frac {4 \, x}{2 \, x^{2} e^{5} - x e^{5} \log \relax (2) + x e^{5} \log \left (-x + 2\right ) - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^3+12*x^2)*exp(5)-16*x+32)/((x^3-2*x^2)*exp(5)^2*log(1-1/2*x)^2+((4*x^4-8*x^3)*exp(5)^2+(-8*x^
2+16*x)*exp(5))*log(1-1/2*x)+(4*x^5-8*x^4)*exp(5)^2+(-16*x^3+32*x^2)*exp(5)+16*x-32),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^5\,\left (12\,x^2-8\,x^3\right )-16\,x+32}{-{\mathrm {e}}^{10}\,\left (2\,x^2-x^3\right )\,{\ln \left (1-\frac {x}{2}\right )}^2+\left ({\mathrm {e}}^5\,\left (16\,x-8\,x^2\right )-{\mathrm {e}}^{10}\,\left (8\,x^3-4\,x^4\right )\right )\,\ln \left (1-\frac {x}{2}\right )+16\,x-{\mathrm {e}}^{10}\,\left (8\,x^4-4\,x^5\right )+{\mathrm {e}}^5\,\left (32\,x^2-16\,x^3\right )-32} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(5)*(12*x^2 - 8*x^3) - 16*x + 32)/(16*x + log(1 - x/2)*(exp(5)*(16*x - 8*x^2) - exp(10)*(8*x^3 - 4*x^4
)) - exp(10)*(8*x^4 - 4*x^5) + exp(5)*(32*x^2 - 16*x^3) - exp(10)*log(1 - x/2)^2*(2*x^2 - x^3) - 32),x)

[Out]

int((exp(5)*(12*x^2 - 8*x^3) - 16*x + 32)/(16*x + log(1 - x/2)*(exp(5)*(16*x - 8*x^2) - exp(10)*(8*x^3 - 4*x^4
)) - exp(10)*(8*x^4 - 4*x^5) + exp(5)*(32*x^2 - 16*x^3) - exp(10)*log(1 - x/2)^2*(2*x^2 - x^3) - 32), x)

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sympy [A]  time = 0.19, size = 24, normalized size = 0.75 \begin {gather*} \frac {4 x}{2 x^{2} e^{5} + x e^{5} \log {\left (1 - \frac {x}{2} \right )} - 4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x**3+12*x**2)*exp(5)-16*x+32)/((x**3-2*x**2)*exp(5)**2*ln(1-1/2*x)**2+((4*x**4-8*x**3)*exp(5)**
2+(-8*x**2+16*x)*exp(5))*ln(1-1/2*x)+(4*x**5-8*x**4)*exp(5)**2+(-16*x**3+32*x**2)*exp(5)+16*x-32),x)

[Out]

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

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