3.3.84 \(\int \frac {e^{\frac {x}{4+2 x+2 \log (x^2)}} (-16+8 x+20 x^2+6 x^3+(-16+17 x+11 x^2) \log (x^2)+(-4+6 x) \log ^2(x^2))}{8 x^3-8 x^4-6 x^5+4 x^6+2 x^7+(8 x^3-12 x^4+4 x^6) \log (x^2)+(2 x^3-4 x^4+2 x^5) \log ^2(x^2)} \, dx\)

Optimal. Leaf size=29 \[ \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x \left (x-x^2\right )} \]

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Rubi [F]  time = 4.75, 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 {e^{\frac {x}{4+2 x+2 \log \left (x^2\right )}} \left (-16+8 x+20 x^2+6 x^3+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{8 x^3-8 x^4-6 x^5+4 x^6+2 x^7+\left (8 x^3-12 x^4+4 x^6\right ) \log \left (x^2\right )+\left (2 x^3-4 x^4+2 x^5\right ) \log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(x/(4 + 2*x + 2*Log[x^2]))*(-16 + 8*x + 20*x^2 + 6*x^3 + (-16 + 17*x + 11*x^2)*Log[x^2] + (-4 + 6*x)*Lo
g[x^2]^2))/(8*x^3 - 8*x^4 - 6*x^5 + 4*x^6 + 2*x^7 + (8*x^3 - 12*x^4 + 4*x^6)*Log[x^2] + (2*x^3 - 4*x^4 + 2*x^5
)*Log[x^2]^2),x]

[Out]

Defer[Int][E^(x/(2*(2 + x + Log[x^2])))/(-1 + x)^2, x] - 2*Defer[Int][E^(x/(2*(2 + x + Log[x^2])))/x^3, x] - D
efer[Int][E^(x/(2*(2 + x + Log[x^2])))/x^2, x] + (3*Defer[Int][E^(x/(2*(2 + x + Log[x^2])))/((-1 + x)*(2 + x +
Log[x^2])^2), x])/2 - Defer[Int][E^(x/(2*(2 + x + Log[x^2])))/(x^2*(2 + x + Log[x^2])^2), x] - (3*Defer[Int][
E^(x/(2*(2 + x + Log[x^2])))/(x*(2 + x + Log[x^2])^2), x])/2 - Defer[Int][E^(x/(2*(2 + x + Log[x^2])))/((-1 +
x)*(2 + x + Log[x^2])), x]/2 + Defer[Int][E^(x/(2*(2 + x + Log[x^2])))/(x^2*(2 + x + Log[x^2])), x]/2 + Defer[
Int][E^(x/(2*(2 + x + Log[x^2])))/(x*(2 + x + Log[x^2])), x]/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} \left (2 (2+x)^2 (-2+3 x)+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{2 (1-x)^2 x^3 \left (2+x+\log \left (x^2\right )\right )^2} \, dx\\ &=\frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} \left (2 (2+x)^2 (-2+3 x)+\left (-16+17 x+11 x^2\right ) \log \left (x^2\right )+(-4+6 x) \log ^2\left (x^2\right )\right )}{(1-x)^2 x^3 \left (2+x+\log \left (x^2\right )\right )^2} \, dx\\ &=\frac {1}{2} \int \left (\frac {2 e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} (-2+3 x)}{(-1+x)^2 x^3}+\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} (2+x)}{(-1+x) x^2 \left (2+x+\log \left (x^2\right )\right )^2}-\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) x^2 \left (2+x+\log \left (x^2\right )\right )}\right ) \, dx\\ &=\frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} (2+x)}{(-1+x) x^2 \left (2+x+\log \left (x^2\right )\right )^2} \, dx-\frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) x^2 \left (2+x+\log \left (x^2\right )\right )} \, dx+\int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}} (-2+3 x)}{(-1+x)^2 x^3} \, dx\\ &=\frac {1}{2} \int \left (\frac {3 e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) \left (2+x+\log \left (x^2\right )\right )^2}-\frac {2 e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2 \left (2+x+\log \left (x^2\right )\right )^2}-\frac {3 e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x \left (2+x+\log \left (x^2\right )\right )^2}\right ) \, dx-\frac {1}{2} \int \left (\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) \left (2+x+\log \left (x^2\right )\right )}-\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2 \left (2+x+\log \left (x^2\right )\right )}-\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x \left (2+x+\log \left (x^2\right )\right )}\right ) \, dx+\int \left (\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x)^2}-\frac {2 e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^3}-\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2}\right ) \, dx\\ &=-\left (\frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) \left (2+x+\log \left (x^2\right )\right )} \, dx\right )+\frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2 \left (2+x+\log \left (x^2\right )\right )} \, dx+\frac {1}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x \left (2+x+\log \left (x^2\right )\right )} \, dx+\frac {3}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) \left (2+x+\log \left (x^2\right )\right )^2} \, dx-\frac {3}{2} \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x \left (2+x+\log \left (x^2\right )\right )^2} \, dx-2 \int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^3} \, dx+\int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x)^2} \, dx-\int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2} \, dx-\int \frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{x^2 \left (2+x+\log \left (x^2\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.68, size = 26, normalized size = 0.90 \begin {gather*} -\frac {e^{\frac {x}{2 \left (2+x+\log \left (x^2\right )\right )}}}{(-1+x) x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x/(4 + 2*x + 2*Log[x^2]))*(-16 + 8*x + 20*x^2 + 6*x^3 + (-16 + 17*x + 11*x^2)*Log[x^2] + (-4 + 6
*x)*Log[x^2]^2))/(8*x^3 - 8*x^4 - 6*x^5 + 4*x^6 + 2*x^7 + (8*x^3 - 12*x^4 + 4*x^6)*Log[x^2] + (2*x^3 - 4*x^4 +
2*x^5)*Log[x^2]^2),x]

[Out]

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

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fricas [A]  time = 0.60, size = 26, normalized size = 0.90 \begin {gather*} -\frac {e^{\left (\frac {x}{2 \, {\left (x + \log \left (x^{2}\right ) + 2\right )}}\right )}}{x^{3} - x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-4)*log(x^2)^2+(11*x^2+17*x-16)*log(x^2)+6*x^3+20*x^2+8*x-16)*exp(x/(2*log(x^2)+2*x+4))/((2*x^5
-4*x^4+2*x^3)*log(x^2)^2+(4*x^6-12*x^4+8*x^3)*log(x^2)+2*x^7+4*x^6-6*x^5-8*x^4+8*x^3),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.59, size = 26, normalized size = 0.90 \begin {gather*} -\frac {e^{\left (\frac {x}{2 \, {\left (x + \log \left (x^{2}\right ) + 2\right )}}\right )}}{x^{3} - x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-4)*log(x^2)^2+(11*x^2+17*x-16)*log(x^2)+6*x^3+20*x^2+8*x-16)*exp(x/(2*log(x^2)+2*x+4))/((2*x^5
-4*x^4+2*x^3)*log(x^2)^2+(4*x^6-12*x^4+8*x^3)*log(x^2)+2*x^7+4*x^6-6*x^5-8*x^4+8*x^3),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.09, size = 24, normalized size = 0.83

method	result	size
risch	\(-\frac {{\mathrm e}^{\frac {x}{2 \ln \left (x^{2}\right )+2 x +4}}}{x^{2} \left (x -1\right )}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((6*x-4)*ln(x^2)^2+(11*x^2+17*x-16)*ln(x^2)+6*x^3+20*x^2+8*x-16)*exp(x/(2*ln(x^2)+2*x+4))/((2*x^5-4*x^4+2*
x^3)*ln(x^2)^2+(4*x^6-12*x^4+8*x^3)*ln(x^2)+2*x^7+4*x^6-6*x^5-8*x^4+8*x^3),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: RuntimeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-4)*log(x^2)^2+(11*x^2+17*x-16)*log(x^2)+6*x^3+20*x^2+8*x-16)*exp(x/(2*log(x^2)+2*x+4))/((2*x^5
-4*x^4+2*x^3)*log(x^2)^2+(4*x^6-12*x^4+8*x^3)*log(x^2)+2*x^7+4*x^6-6*x^5-8*x^4+8*x^3),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: In function CAR, the value of the first argument is  0which is not
of the expected type LIST

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mupad [B]  time = 0.96, size = 26, normalized size = 0.90 \begin {gather*} \frac {{\mathrm {e}}^{\frac {x}{2\,x+\ln \left (x^4\right )+4}}}{x^2-x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x/(2*x + 2*log(x^2) + 4))*(8*x + log(x^2)*(17*x + 11*x^2 - 16) + log(x^2)^2*(6*x - 4) + 20*x^2 + 6*x^
3 - 16))/(log(x^2)^2*(2*x^3 - 4*x^4 + 2*x^5) + log(x^2)*(8*x^3 - 12*x^4 + 4*x^6) + 8*x^3 - 8*x^4 - 6*x^5 + 4*x
^6 + 2*x^7),x)

[Out]

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

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sympy [A]  time = 0.48, size = 22, normalized size = 0.76 \begin {gather*} - \frac {e^{\frac {x}{2 x + 2 \log {\left (x^{2} \right )} + 4}}}{x^{3} - x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((6*x-4)*ln(x**2)**2+(11*x**2+17*x-16)*ln(x**2)+6*x**3+20*x**2+8*x-16)*exp(x/(2*ln(x**2)+2*x+4))/((2
*x**5-4*x**4+2*x**3)*ln(x**2)**2+(4*x**6-12*x**4+8*x**3)*ln(x**2)+2*x**7+4*x**6-6*x**5-8*x**4+8*x**3),x)

[Out]

-exp(x/(2*x + 2*log(x**2) + 4))/(x**3 - x**2)

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