3.44.9 \(\int \frac {-36-6 x-3 x^2+e^x (-9+3 x-3 x^2)+(12+e^x (3-3 x)) \log (x^2)}{64-96 x+4 x^2+24 x^3+4 x^4+e^{2 x} (4-8 x+4 x^2)+e^x (32-56 x+16 x^2+8 x^3)+(-128+64 x+56 x^2+8 x^3+e^{2 x} (-8+8 x)+e^x (-64+48 x+16 x^2)) \log (x^2)+(64+4 e^{2 x}+32 x+4 x^2+e^x (32+8 x)) \log ^2(x^2)} \, dx\)

Optimal. Leaf size=22 \[ \frac {3 x}{4 \left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )} \]

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Rubi [F]  time = 2.65, 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 {-36-6 x-3 x^2+e^x \left (-9+3 x-3 x^2\right )+\left (12+e^x (3-3 x)\right ) \log \left (x^2\right )}{64-96 x+4 x^2+24 x^3+4 x^4+e^{2 x} \left (4-8 x+4 x^2\right )+e^x \left (32-56 x+16 x^2+8 x^3\right )+\left (-128+64 x+56 x^2+8 x^3+e^{2 x} (-8+8 x)+e^x \left (-64+48 x+16 x^2\right )\right ) \log \left (x^2\right )+\left (64+4 e^{2 x}+32 x+4 x^2+e^x (32+8 x)\right ) \log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-36 - 6*x - 3*x^2 + E^x*(-9 + 3*x - 3*x^2) + (12 + E^x*(3 - 3*x))*Log[x^2])/(64 - 96*x + 4*x^2 + 24*x^3 +
4*x^4 + E^(2*x)*(4 - 8*x + 4*x^2) + E^x*(32 - 56*x + 16*x^2 + 8*x^3) + (-128 + 64*x + 56*x^2 + 8*x^3 + E^(2*x
)*(-8 + 8*x) + E^x*(-64 + 48*x + 16*x^2))*Log[x^2] + (64 + 4*E^(2*x) + 32*x + 4*x^2 + E^x*(32 + 8*x))*Log[x^2]
^2),x]

[Out]

(-9*Defer[Int][1/((4 + E^x + x)*(-1 + x + Log[x^2])^2), x])/4 + (3*Defer[Int][x/((4 + E^x + x)*(-1 + x + Log[x
^2])^2), x])/4 - (3*Defer[Int][x^2/((4 + E^x + x)*(-1 + x + Log[x^2])^2), x])/4 + (3*Defer[Int][Log[x^2]/((4 +
E^x + x)*(-1 + x + Log[x^2])^2), x])/4 - (3*Defer[Int][(x*Log[x^2])/((4 + E^x + x)*(-1 + x + Log[x^2])^2), x]
)/4 + (9*Defer[Int][x/((4 + E^x + x)^2*(-1 + x + Log[x^2])), x])/4 + (3*Defer[Int][x^2/((4 + E^x + x)^2*(-1 +
x + Log[x^2])), x])/4

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {3 \left (-12-2 x-x^2-e^x \left (3-x+x^2\right )-\left (-4+e^x (-1+x)\right ) \log \left (x^2\right )\right )}{4 \left (4+e^x+x\right )^2 \left (1-x-\log \left (x^2\right )\right )^2} \, dx\\ &=\frac {3}{4} \int \frac {-12-2 x-x^2-e^x \left (3-x+x^2\right )-\left (-4+e^x (-1+x)\right ) \log \left (x^2\right )}{\left (4+e^x+x\right )^2 \left (1-x-\log \left (x^2\right )\right )^2} \, dx\\ &=\frac {3}{4} \int \left (\frac {x (3+x)}{\left (4+e^x+x\right )^2 \left (-1+x+\log \left (x^2\right )\right )}-\frac {3-x+x^2-\log \left (x^2\right )+x \log \left (x^2\right )}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2}\right ) \, dx\\ &=\frac {3}{4} \int \frac {x (3+x)}{\left (4+e^x+x\right )^2 \left (-1+x+\log \left (x^2\right )\right )} \, dx-\frac {3}{4} \int \frac {3-x+x^2-\log \left (x^2\right )+x \log \left (x^2\right )}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2} \, dx\\ &=-\left (\frac {3}{4} \int \left (\frac {3}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2}-\frac {x}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2}+\frac {x^2}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2}-\frac {\log \left (x^2\right )}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2}+\frac {x \log \left (x^2\right )}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2}\right ) \, dx\right )+\frac {3}{4} \int \left (\frac {3 x}{\left (4+e^x+x\right )^2 \left (-1+x+\log \left (x^2\right )\right )}+\frac {x^2}{\left (4+e^x+x\right )^2 \left (-1+x+\log \left (x^2\right )\right )}\right ) \, dx\\ &=\frac {3}{4} \int \frac {x}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2} \, dx-\frac {3}{4} \int \frac {x^2}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2} \, dx+\frac {3}{4} \int \frac {\log \left (x^2\right )}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2} \, dx-\frac {3}{4} \int \frac {x \log \left (x^2\right )}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2} \, dx+\frac {3}{4} \int \frac {x^2}{\left (4+e^x+x\right )^2 \left (-1+x+\log \left (x^2\right )\right )} \, dx-\frac {9}{4} \int \frac {1}{\left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )^2} \, dx+\frac {9}{4} \int \frac {x}{\left (4+e^x+x\right )^2 \left (-1+x+\log \left (x^2\right )\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.43, size = 22, normalized size = 1.00 \begin {gather*} \frac {3 x}{4 \left (4+e^x+x\right ) \left (-1+x+\log \left (x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-36 - 6*x - 3*x^2 + E^x*(-9 + 3*x - 3*x^2) + (12 + E^x*(3 - 3*x))*Log[x^2])/(64 - 96*x + 4*x^2 + 24
*x^3 + 4*x^4 + E^(2*x)*(4 - 8*x + 4*x^2) + E^x*(32 - 56*x + 16*x^2 + 8*x^3) + (-128 + 64*x + 56*x^2 + 8*x^3 +
E^(2*x)*(-8 + 8*x) + E^x*(-64 + 48*x + 16*x^2))*Log[x^2] + (64 + 4*E^(2*x) + 32*x + 4*x^2 + E^x*(32 + 8*x))*Lo
g[x^2]^2),x]

[Out]

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

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fricas [A]  time = 0.57, size = 29, normalized size = 1.32 \begin {gather*} \frac {3 \, x}{4 \, {\left (x^{2} + {\left (x - 1\right )} e^{x} + {\left (x + e^{x} + 4\right )} \log \left (x^{2}\right ) + 3 \, x - 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x+3)*exp(x)+12)*log(x^2)+(-3*x^2+3*x-9)*exp(x)-3*x^2-6*x-36)/((4*exp(x)^2+(8*x+32)*exp(x)+4*x^
2+32*x+64)*log(x^2)^2+((8*x-8)*exp(x)^2+(16*x^2+48*x-64)*exp(x)+8*x^3+56*x^2+64*x-128)*log(x^2)+(4*x^2-8*x+4)*
exp(x)^2+(8*x^3+16*x^2-56*x+32)*exp(x)+4*x^4+24*x^3+4*x^2-96*x+64),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.37, size = 40, normalized size = 1.82 \begin {gather*} \frac {3 \, x}{4 \, {\left (x^{2} + x e^{x} + x \log \left (x^{2}\right ) + e^{x} \log \left (x^{2}\right ) + 3 \, x - e^{x} + 4 \, \log \left (x^{2}\right ) - 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x+3)*exp(x)+12)*log(x^2)+(-3*x^2+3*x-9)*exp(x)-3*x^2-6*x-36)/((4*exp(x)^2+(8*x+32)*exp(x)+4*x^
2+32*x+64)*log(x^2)^2+((8*x-8)*exp(x)^2+(16*x^2+48*x-64)*exp(x)+8*x^3+56*x^2+64*x-128)*log(x^2)+(4*x^2-8*x+4)*
exp(x)^2+(8*x^3+16*x^2-56*x+32)*exp(x)+4*x^4+24*x^3+4*x^2-96*x+64),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.12, size = 71, normalized size = 3.23

method	result	size
risch	\(\frac {3 x}{2 \left (4+x +{\mathrm e}^{x}\right ) \left (-i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+2 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+2 x +4 \ln \relax (x )-2\right )}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-3*x+3)*exp(x)+12)*ln(x^2)+(-3*x^2+3*x-9)*exp(x)-3*x^2-6*x-36)/((4*exp(x)^2+(8*x+32)*exp(x)+4*x^2+32*x+
64)*ln(x^2)^2+((8*x-8)*exp(x)^2+(16*x^2+48*x-64)*exp(x)+8*x^3+56*x^2+64*x-128)*ln(x^2)+(4*x^2-8*x+4)*exp(x)^2+
(8*x^3+16*x^2-56*x+32)*exp(x)+4*x^4+24*x^3+4*x^2-96*x+64),x,method=_RETURNVERBOSE)

[Out]

3/2*x/(4+x+exp(x))/(-I*Pi*csgn(I*x)^2*csgn(I*x^2)+2*I*Pi*csgn(I*x)*csgn(I*x^2)^2-I*Pi*csgn(I*x^2)^3+2*x+4*ln(x
)-2)

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maxima [A]  time = 0.50, size = 30, normalized size = 1.36 \begin {gather*} \frac {3 \, x}{4 \, {\left (x^{2} + {\left (x + 2 \, \log \relax (x) - 1\right )} e^{x} + 2 \, {\left (x + 4\right )} \log \relax (x) + 3 \, x - 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x+3)*exp(x)+12)*log(x^2)+(-3*x^2+3*x-9)*exp(x)-3*x^2-6*x-36)/((4*exp(x)^2+(8*x+32)*exp(x)+4*x^
2+32*x+64)*log(x^2)^2+((8*x-8)*exp(x)^2+(16*x^2+48*x-64)*exp(x)+8*x^3+56*x^2+64*x-128)*log(x^2)+(4*x^2-8*x+4)*
exp(x)^2+(8*x^3+16*x^2-56*x+32)*exp(x)+4*x^4+24*x^3+4*x^2-96*x+64),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.05 \begin {gather*} \int -\frac {6\,x+\ln \left (x^2\right )\,\left ({\mathrm {e}}^x\,\left (3\,x-3\right )-12\right )+{\mathrm {e}}^x\,\left (3\,x^2-3\,x+9\right )+3\,x^2+36}{{\mathrm {e}}^{2\,x}\,\left (4\,x^2-8\,x+4\right )-96\,x+{\ln \left (x^2\right )}^2\,\left (32\,x+4\,{\mathrm {e}}^{2\,x}+{\mathrm {e}}^x\,\left (8\,x+32\right )+4\,x^2+64\right )+\ln \left (x^2\right )\,\left (64\,x+{\mathrm {e}}^x\,\left (16\,x^2+48\,x-64\right )+{\mathrm {e}}^{2\,x}\,\left (8\,x-8\right )+56\,x^2+8\,x^3-128\right )+4\,x^2+24\,x^3+4\,x^4+{\mathrm {e}}^x\,\left (8\,x^3+16\,x^2-56\,x+32\right )+64} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(6*x + log(x^2)*(exp(x)*(3*x - 3) - 12) + exp(x)*(3*x^2 - 3*x + 9) + 3*x^2 + 36)/(exp(2*x)*(4*x^2 - 8*x +
4) - 96*x + log(x^2)^2*(32*x + 4*exp(2*x) + exp(x)*(8*x + 32) + 4*x^2 + 64) + log(x^2)*(64*x + exp(x)*(48*x +
16*x^2 - 64) + exp(2*x)*(8*x - 8) + 56*x^2 + 8*x^3 - 128) + 4*x^2 + 24*x^3 + 4*x^4 + exp(x)*(16*x^2 - 56*x +
8*x^3 + 32) + 64),x)

[Out]

int(-(6*x + log(x^2)*(exp(x)*(3*x - 3) - 12) + exp(x)*(3*x^2 - 3*x + 9) + 3*x^2 + 36)/(exp(2*x)*(4*x^2 - 8*x +
4) - 96*x + log(x^2)^2*(32*x + 4*exp(2*x) + exp(x)*(8*x + 32) + 4*x^2 + 64) + log(x^2)*(64*x + exp(x)*(48*x +
16*x^2 - 64) + exp(2*x)*(8*x - 8) + 56*x^2 + 8*x^3 - 128) + 4*x^2 + 24*x^3 + 4*x^4 + exp(x)*(16*x^2 - 56*x +
8*x^3 + 32) + 64), x)

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sympy [B]  time = 0.36, size = 42, normalized size = 1.91 \begin {gather*} \frac {3 x}{4 x^{2} + 4 x \log {\left (x^{2} \right )} + 12 x + \left (4 x + 4 \log {\left (x^{2} \right )} - 4\right ) e^{x} + 16 \log {\left (x^{2} \right )} - 16} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-3*x+3)*exp(x)+12)*ln(x**2)+(-3*x**2+3*x-9)*exp(x)-3*x**2-6*x-36)/((4*exp(x)**2+(8*x+32)*exp(x)+4
*x**2+32*x+64)*ln(x**2)**2+((8*x-8)*exp(x)**2+(16*x**2+48*x-64)*exp(x)+8*x**3+56*x**2+64*x-128)*ln(x**2)+(4*x*
*2-8*x+4)*exp(x)**2+(8*x**3+16*x**2-56*x+32)*exp(x)+4*x**4+24*x**3+4*x**2-96*x+64),x)

[Out]

3*x/(4*x**2 + 4*x*log(x**2) + 12*x + (4*x + 4*log(x**2) - 4)*exp(x) + 16*log(x**2) - 16)

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