3.26.72 \(\int \frac {-10+e^{2 x^2} (64 x^3-6 x^4+64 x^5-8 x^6)}{40-5 x+e^{2 x^2} (-8 x^4+x^5)} \, dx\)

Optimal. Leaf size=22 \[ \log \left (\frac {(-8+x)^2}{\left (-5+e^{2 x^2} x^4\right )^2}\right ) \]

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Rubi [F]  time = 1.03, 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 {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{40-5 x+e^{2 x^2} \left (-8 x^4+x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-10 + E^(2*x^2)*(64*x^3 - 6*x^4 + 64*x^5 - 8*x^6))/(40 - 5*x + E^(2*x^2)*(-8*x^4 + x^5)),x]

[Out]

-4*x^2 + 2*Log[8 - x] - 8*Log[x] - 20*Defer[Subst][Defer[Int][(-5 + E^(2*x)*x^2)^(-1), x], x, x^2] - 20*Defer[
Subst][Defer[Int][1/(x*(-5 + E^(2*x)*x^2)), x], x, x^2]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10+e^{2 x^2} \left (64 x^3-6 x^4+64 x^5-8 x^6\right )}{(8-x) \left (5-e^{2 x^2} x^4\right )} \, dx\\ &=\int \left (-\frac {2 \left (-32+3 x-32 x^2+4 x^3\right )}{(-8+x) x}-\frac {40 \left (1+x^2\right )}{x \left (-5+e^{2 x^2} x^4\right )}\right ) \, dx\\ &=-\left (2 \int \frac {-32+3 x-32 x^2+4 x^3}{(-8+x) x} \, dx\right )-40 \int \frac {1+x^2}{x \left (-5+e^{2 x^2} x^4\right )} \, dx\\ &=-\left (2 \int \left (\frac {1}{8-x}+\frac {4}{x}+4 x\right ) \, dx\right )-20 \operatorname {Subst}\left (\int \frac {1+x}{x \left (-5+e^{2 x} x^2\right )} \, dx,x,x^2\right )\\ &=-4 x^2+2 \log (8-x)-8 \log (x)-20 \operatorname {Subst}\left (\int \left (\frac {1}{-5+e^{2 x} x^2}+\frac {1}{x \left (-5+e^{2 x} x^2\right )}\right ) \, dx,x,x^2\right )\\ &=-4 x^2+2 \log (8-x)-8 \log (x)-20 \operatorname {Subst}\left (\int \frac {1}{-5+e^{2 x} x^2} \, dx,x,x^2\right )-20 \operatorname {Subst}\left (\int \frac {1}{x \left (-5+e^{2 x} x^2\right )} \, dx,x,x^2\right )\\ \end {aligned} \end {gather*}

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Mathematica [B]  time = 0.20, size = 120, normalized size = 5.45 \begin {gather*} -2 \log \left (5-4096 e^{128+32 (-8+x)+2 (-8+x)^2}-2048 e^{128+32 (-8+x)+2 (-8+x)^2} (-8+x)-384 e^{128+32 (-8+x)+2 (-8+x)^2} (-8+x)^2-32 e^{128+32 (-8+x)+2 (-8+x)^2} (-8+x)^3-e^{128+32 (-8+x)+2 (-8+x)^2} (-8+x)^4\right )+2 \log (-8+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-10 + E^(2*x^2)*(64*x^3 - 6*x^4 + 64*x^5 - 8*x^6))/(40 - 5*x + E^(2*x^2)*(-8*x^4 + x^5)),x]

[Out]

-2*Log[5 - 4096*E^(128 + 32*(-8 + x) + 2*(-8 + x)^2) - 2048*E^(128 + 32*(-8 + x) + 2*(-8 + x)^2)*(-8 + x) - 38
4*E^(128 + 32*(-8 + x) + 2*(-8 + x)^2)*(-8 + x)^2 - 32*E^(128 + 32*(-8 + x) + 2*(-8 + x)^2)*(-8 + x)^3 - E^(12
8 + 32*(-8 + x) + 2*(-8 + x)^2)*(-8 + x)^4] + 2*Log[-8 + x]

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fricas [A]  time = 0.61, size = 30, normalized size = 1.36 \begin {gather*} 2 \, \log \left (x - 8\right ) - 8 \, \log \relax (x) - 2 \, \log \left (\frac {x^{4} e^{\left (2 \, x^{2}\right )} - 5}{x^{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^6+64*x^5-6*x^4+64*x^3)*exp(x^2)^2-10)/((x^5-8*x^4)*exp(x^2)^2-5*x+40),x, algorithm="fricas")

[Out]

2*log(x - 8) - 8*log(x) - 2*log((x^4*e^(2*x^2) - 5)/x^4)

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giac [A]  time = 0.25, size = 22, normalized size = 1.00 \begin {gather*} -2 \, \log \left (x^{4} e^{\left (2 \, x^{2}\right )} - 5\right ) + 2 \, \log \left (x - 8\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^6+64*x^5-6*x^4+64*x^3)*exp(x^2)^2-10)/((x^5-8*x^4)*exp(x^2)^2-5*x+40),x, algorithm="giac")

[Out]

-2*log(x^4*e^(2*x^2) - 5) + 2*log(x - 8)

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maple [A]  time = 0.06, size = 23, normalized size = 1.05

method	result	size
norman	\(2 \ln \left (-8+x \right )-2 \ln \left (x^{4} {\mathrm e}^{2 x^{2}}-5\right )\)	\(23\)
risch	\(-8 \ln \relax (x )+2 \ln \left (-8+x \right )-2 \ln \left ({\mathrm e}^{2 x^{2}}-\frac {5}{x^{4}}\right )\)	\(27\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-8*x^6+64*x^5-6*x^4+64*x^3)*exp(x^2)^2-10)/((x^5-8*x^4)*exp(x^2)^2-5*x+40),x,method=_RETURNVERBOSE)

[Out]

2*ln(-8+x)-2*ln(x^4*exp(x^2)^2-5)

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maxima [A]  time = 0.76, size = 30, normalized size = 1.36 \begin {gather*} 2 \, \log \left (x - 8\right ) - 8 \, \log \relax (x) - 2 \, \log \left (\frac {x^{4} e^{\left (2 \, x^{2}\right )} - 5}{x^{4}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x^6+64*x^5-6*x^4+64*x^3)*exp(x^2)^2-10)/((x^5-8*x^4)*exp(x^2)^2-5*x+40),x, algorithm="maxima")

[Out]

2*log(x - 8) - 8*log(x) - 2*log((x^4*e^(2*x^2) - 5)/x^4)

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mupad [B]  time = 0.15, size = 22, normalized size = 1.00 \begin {gather*} 2\,\ln \left (x-8\right )-2\,\ln \left (x^4\,{\mathrm {e}}^{2\,x^2}-5\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(2*x^2)*(64*x^3 - 6*x^4 + 64*x^5 - 8*x^6) - 10)/(5*x + exp(2*x^2)*(8*x^4 - x^5) - 40),x)

[Out]

2*log(x - 8) - 2*log(x^4*exp(2*x^2) - 5)

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sympy [A]  time = 0.23, size = 26, normalized size = 1.18 \begin {gather*} - 8 \log {\relax (x )} + 2 \log {\left (x - 8 \right )} - 2 \log {\left (e^{2 x^{2}} - \frac {5}{x^{4}} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-8*x**6+64*x**5-6*x**4+64*x**3)*exp(x**2)**2-10)/((x**5-8*x**4)*exp(x**2)**2-5*x+40),x)

[Out]

-8*log(x) + 2*log(x - 8) - 2*log(exp(2*x**2) - 5/x**4)

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