3.89.82 \(\int \frac {e^{5+e^{2 x}} (6+e^{16+x} (-9+9 x-18 e^{2 x} x)+e^{16+x} (3-3 x+6 e^{2 x} x) \log (x^2))+e^{5+e^{2 x}} (9+18 e^{2 x} x+(-3-6 e^{2 x} x) \log (x^2)) \log (3-\log (x^2))}{-3 e^{32+2 x}+e^{32+2 x} \log (x^2)+(6 e^{16+x}-2 e^{16+x} \log (x^2)) \log (3-\log (x^2))+(-3+\log (x^2)) \log ^2(3-\log (x^2))} \, dx\)

Optimal. Leaf size=31 \[ \frac {3 e^{5+e^{2 x}} x}{e^{16+x}-\log \left (3-\log \left (x^2\right )\right )} \]

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Rubi [F]  time = 5.87, 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^{5+e^{2 x}} \left (6+e^{16+x} \left (-9+9 x-18 e^{2 x} x\right )+e^{16+x} \left (3-3 x+6 e^{2 x} x\right ) \log \left (x^2\right )\right )+e^{5+e^{2 x}} \left (9+18 e^{2 x} x+\left (-3-6 e^{2 x} x\right ) \log \left (x^2\right )\right ) \log \left (3-\log \left (x^2\right )\right )}{-3 e^{32+2 x}+e^{32+2 x} \log \left (x^2\right )+\left (6 e^{16+x}-2 e^{16+x} \log \left (x^2\right )\right ) \log \left (3-\log \left (x^2\right )\right )+\left (-3+\log \left (x^2\right )\right ) \log ^2\left (3-\log \left (x^2\right )\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(5 + E^(2*x))*(6 + E^(16 + x)*(-9 + 9*x - 18*E^(2*x)*x) + E^(16 + x)*(3 - 3*x + 6*E^(2*x)*x)*Log[x^2])
+ E^(5 + E^(2*x))*(9 + 18*E^(2*x)*x + (-3 - 6*E^(2*x)*x)*Log[x^2])*Log[3 - Log[x^2]])/(-3*E^(32 + 2*x) + E^(32
+ 2*x)*Log[x^2] + (6*E^(16 + x) - 2*E^(16 + x)*Log[x^2])*Log[3 - Log[x^2]] + (-3 + Log[x^2])*Log[3 - Log[x^2]
]^2),x]

[Out]

6*Defer[Int][E^(-11 + E^(2*x) + x)*x, x] + 6*Defer[Int][E^(5 + E^(2*x))/((-3 + Log[x^2])*(E^(16 + x) - Log[3 -
Log[x^2]])^2), x] + 3*Defer[Int][E^(5 + E^(2*x))/(E^(16 + x) - Log[3 - Log[x^2]]), x] - 3*Defer[Int][(E^(5 +
E^(2*x))*x)/(E^(16 + x) - Log[3 - Log[x^2]]), x] + 6*Defer[Int][E^(-27 + E^(2*x))*x*Log[3 - Log[x^2]], x] + 9*
Defer[Int][(E^(5 + E^(2*x))*x*Log[3 - Log[x^2]])/((-3 + Log[x^2])*(E^(16 + x) - Log[3 - Log[x^2]])^2), x] - 3*
Defer[Int][(E^(5 + E^(2*x))*x*Log[x^2]*Log[3 - Log[x^2]])/((-3 + Log[x^2])*(E^(16 + x) - Log[3 - Log[x^2]])^2)
, x] - 6*Defer[Int][(E^E^(2*x)*x*Log[3 - Log[x^2]]^2)/(-E^(43 + x) + E^27*Log[3 - Log[x^2]]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-e^{5+e^{2 x}} \left (6+e^{16+x} \left (-9+9 x-18 e^{2 x} x\right )+e^{16+x} \left (3-3 x+6 e^{2 x} x\right ) \log \left (x^2\right )\right )-e^{5+e^{2 x}} \left (9+18 e^{2 x} x+\left (-3-6 e^{2 x} x\right ) \log \left (x^2\right )\right ) \log \left (3-\log \left (x^2\right )\right )}{\left (3-\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2} \, dx\\ &=\int \left (6 e^{-11+e^{2 x}+x} x+6 e^{-27+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right )-\frac {3 e^{5+e^{2 x}} \left (-2-3 x \log \left (3-\log \left (x^2\right )\right )+x \log \left (x^2\right ) \log \left (3-\log \left (x^2\right )\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2}-\frac {3 e^{e^{2 x}} \left (-e^{32}+e^{32} x-2 x \log ^2\left (3-\log \left (x^2\right )\right )\right )}{e^{43+x}-e^{27} \log \left (3-\log \left (x^2\right )\right )}\right ) \, dx\\ &=-\left (3 \int \frac {e^{5+e^{2 x}} \left (-2-3 x \log \left (3-\log \left (x^2\right )\right )+x \log \left (x^2\right ) \log \left (3-\log \left (x^2\right )\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2} \, dx\right )-3 \int \frac {e^{e^{2 x}} \left (-e^{32}+e^{32} x-2 x \log ^2\left (3-\log \left (x^2\right )\right )\right )}{e^{43+x}-e^{27} \log \left (3-\log \left (x^2\right )\right )} \, dx+6 \int e^{-11+e^{2 x}+x} x \, dx+6 \int e^{-27+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right ) \, dx\\ &=-\left (3 \int \left (-\frac {2 e^{5+e^{2 x}}}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2}-\frac {3 e^{5+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2}+\frac {e^{5+e^{2 x}} x \log \left (x^2\right ) \log \left (3-\log \left (x^2\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2}\right ) \, dx\right )-3 \int \left (-\frac {e^{5+e^{2 x}}}{e^{16+x}-\log \left (3-\log \left (x^2\right )\right )}+\frac {e^{5+e^{2 x}} x}{e^{16+x}-\log \left (3-\log \left (x^2\right )\right )}+\frac {2 e^{e^{2 x}} x \log ^2\left (3-\log \left (x^2\right )\right )}{-e^{43+x}+e^{27} \log \left (3-\log \left (x^2\right )\right )}\right ) \, dx+6 \int e^{-11+e^{2 x}+x} x \, dx+6 \int e^{-27+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right ) \, dx\\ &=3 \int \frac {e^{5+e^{2 x}}}{e^{16+x}-\log \left (3-\log \left (x^2\right )\right )} \, dx-3 \int \frac {e^{5+e^{2 x}} x}{e^{16+x}-\log \left (3-\log \left (x^2\right )\right )} \, dx-3 \int \frac {e^{5+e^{2 x}} x \log \left (x^2\right ) \log \left (3-\log \left (x^2\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2} \, dx+6 \int e^{-11+e^{2 x}+x} x \, dx+6 \int \frac {e^{5+e^{2 x}}}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2} \, dx+6 \int e^{-27+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right ) \, dx-6 \int \frac {e^{e^{2 x}} x \log ^2\left (3-\log \left (x^2\right )\right )}{-e^{43+x}+e^{27} \log \left (3-\log \left (x^2\right )\right )} \, dx+9 \int \frac {e^{5+e^{2 x}} x \log \left (3-\log \left (x^2\right )\right )}{\left (-3+\log \left (x^2\right )\right ) \left (e^{16+x}-\log \left (3-\log \left (x^2\right )\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.21, size = 31, normalized size = 1.00 \begin {gather*} -\frac {3 e^{5+e^{2 x}} x}{-e^{16+x}+\log \left (3-\log \left (x^2\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(5 + E^(2*x))*(6 + E^(16 + x)*(-9 + 9*x - 18*E^(2*x)*x) + E^(16 + x)*(3 - 3*x + 6*E^(2*x)*x)*Log[
x^2]) + E^(5 + E^(2*x))*(9 + 18*E^(2*x)*x + (-3 - 6*E^(2*x)*x)*Log[x^2])*Log[3 - Log[x^2]])/(-3*E^(32 + 2*x) +
E^(32 + 2*x)*Log[x^2] + (6*E^(16 + x) - 2*E^(16 + x)*Log[x^2])*Log[3 - Log[x^2]] + (-3 + Log[x^2])*Log[3 - Lo
g[x^2]]^2),x]

[Out]

(-3*E^(5 + E^(2*x))*x)/(-E^(16 + x) + Log[3 - Log[x^2]])

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fricas [A]  time = 0.49, size = 36, normalized size = 1.16 \begin {gather*} \frac {3 \, x e^{\left ({\left (5 \, e^{32} + e^{\left (2 \, x + 32\right )}\right )} e^{\left (-32\right )}\right )}}{e^{\left (x + 16\right )} - \log \left (-\log \left (x^{2}\right ) + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x*exp(x)^2-3)*log(x^2)+18*x*exp(x)^2+9)*exp(exp(x)^2+5)*log(3-log(x^2))+((6*x*exp(x)^2-3*x+3)*
exp(x+16)*log(x^2)+(-18*x*exp(x)^2+9*x-9)*exp(x+16)+6)*exp(exp(x)^2+5))/((log(x^2)-3)*log(3-log(x^2))^2+(-2*ex
p(x+16)*log(x^2)+6*exp(x+16))*log(3-log(x^2))+exp(x+16)^2*log(x^2)-3*exp(x+16)^2),x, algorithm="fricas")

[Out]

3*x*e^((5*e^32 + e^(2*x + 32))*e^(-32))/(e^(x + 16) - log(-log(x^2) + 3))

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giac [B]  time = 0.49, size = 127, normalized size = 4.10 \begin {gather*} -\frac {3 \, {\left (2 \, x^{2} e^{\left (x + e^{\left (2 \, x\right )} + 21\right )} \log \left (x \mathrm {sgn}\relax (x)\right ) - 3 \, x^{2} e^{\left (x + e^{\left (2 \, x\right )} + 21\right )} - 2 \, x e^{\left (e^{\left (2 \, x\right )} + 5\right )}\right )}}{2 \, x e^{\left (x + 16\right )} \log \left (x \mathrm {sgn}\relax (x)\right ) \log \left (-2 \, \log \left (x \mathrm {sgn}\relax (x)\right ) + 3\right ) - 2 \, x e^{\left (2 \, x + 32\right )} \log \left (x \mathrm {sgn}\relax (x)\right ) - 3 \, x e^{\left (x + 16\right )} \log \left (-2 \, \log \left (x \mathrm {sgn}\relax (x)\right ) + 3\right ) + 3 \, x e^{\left (2 \, x + 32\right )} + 2 \, e^{\left (x + 16\right )} - 2 \, \log \left (-2 \, \log \left (x \mathrm {sgn}\relax (x)\right ) + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x*exp(x)^2-3)*log(x^2)+18*x*exp(x)^2+9)*exp(exp(x)^2+5)*log(3-log(x^2))+((6*x*exp(x)^2-3*x+3)*
exp(x+16)*log(x^2)+(-18*x*exp(x)^2+9*x-9)*exp(x+16)+6)*exp(exp(x)^2+5))/((log(x^2)-3)*log(3-log(x^2))^2+(-2*ex
p(x+16)*log(x^2)+6*exp(x+16))*log(3-log(x^2))+exp(x+16)^2*log(x^2)-3*exp(x+16)^2),x, algorithm="giac")

[Out]

-3*(2*x^2*e^(x + e^(2*x) + 21)*log(x*sgn(x)) - 3*x^2*e^(x + e^(2*x) + 21) - 2*x*e^(e^(2*x) + 5))/(2*x*e^(x + 1
6)*log(x*sgn(x))*log(-2*log(x*sgn(x)) + 3) - 2*x*e^(2*x + 32)*log(x*sgn(x)) - 3*x*e^(x + 16)*log(-2*log(x*sgn(
x)) + 3) + 3*x*e^(2*x + 32) + 2*e^(x + 16) - 2*log(-2*log(x*sgn(x)) + 3))

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maple [C]  time = 0.65, size = 55, normalized size = 1.77

method	result	size
risch	\(\frac {3 \,{\mathrm e}^{5+{\mathrm e}^{2 x}} x}{{\mathrm e}^{x +16}-\ln \left (3-2 \ln \relax (x )+\frac {i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \left (-\mathrm {csgn}\left (i x^{2}\right )+\mathrm {csgn}\left (i x \right )\right )^{2}}{2}\right )}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-6*x*exp(x)^2-3)*ln(x^2)+18*x*exp(x)^2+9)*exp(exp(x)^2+5)*ln(3-ln(x^2))+((6*x*exp(x)^2-3*x+3)*exp(x+16)
*ln(x^2)+(-18*x*exp(x)^2+9*x-9)*exp(x+16)+6)*exp(exp(x)^2+5))/((ln(x^2)-3)*ln(3-ln(x^2))^2+(-2*exp(x+16)*ln(x^
2)+6*exp(x+16))*ln(3-ln(x^2))+exp(x+16)^2*ln(x^2)-3*exp(x+16)^2),x,method=_RETURNVERBOSE)

[Out]

3*exp(5+exp(2*x))*x/(exp(x+16)-ln(3-2*ln(x)+1/2*I*Pi*csgn(I*x^2)*(-csgn(I*x^2)+csgn(I*x))^2))

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maxima [A]  time = 0.49, size = 26, normalized size = 0.84 \begin {gather*} \frac {3 \, x e^{\left (e^{\left (2 \, x\right )} + 5\right )}}{e^{\left (x + 16\right )} - \log \left (-2 \, \log \relax (x) + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x*exp(x)^2-3)*log(x^2)+18*x*exp(x)^2+9)*exp(exp(x)^2+5)*log(3-log(x^2))+((6*x*exp(x)^2-3*x+3)*
exp(x+16)*log(x^2)+(-18*x*exp(x)^2+9*x-9)*exp(x+16)+6)*exp(exp(x)^2+5))/((log(x^2)-3)*log(3-log(x^2))^2+(-2*ex
p(x+16)*log(x^2)+6*exp(x+16))*log(3-log(x^2))+exp(x+16)^2*log(x^2)-3*exp(x+16)^2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {{\mathrm {e}}^{{\mathrm {e}}^{2\,x}+5}\,\left (\ln \left (x^2\right )\,{\mathrm {e}}^{x+16}\,\left (6\,x\,{\mathrm {e}}^{2\,x}-3\,x+3\right )-{\mathrm {e}}^{x+16}\,\left (18\,x\,{\mathrm {e}}^{2\,x}-9\,x+9\right )+6\right )+{\mathrm {e}}^{{\mathrm {e}}^{2\,x}+5}\,\ln \left (3-\ln \left (x^2\right )\right )\,\left (18\,x\,{\mathrm {e}}^{2\,x}-\ln \left (x^2\right )\,\left (6\,x\,{\mathrm {e}}^{2\,x}+3\right )+9\right )}{\left (\ln \left (x^2\right )-3\right )\,{\ln \left (3-\ln \left (x^2\right )\right )}^2+\left (6\,{\mathrm {e}}^{x+16}-2\,\ln \left (x^2\right )\,{\mathrm {e}}^{x+16}\right )\,\ln \left (3-\ln \left (x^2\right )\right )-3\,{\mathrm {e}}^{2\,x+32}+\ln \left (x^2\right )\,{\mathrm {e}}^{2\,x+32}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(2*x) + 5)*(log(x^2)*exp(x + 16)*(6*x*exp(2*x) - 3*x + 3) - exp(x + 16)*(18*x*exp(2*x) - 9*x + 9)
+ 6) + exp(exp(2*x) + 5)*log(3 - log(x^2))*(18*x*exp(2*x) - log(x^2)*(6*x*exp(2*x) + 3) + 9))/(log(x^2)*exp(2*
x + 32) - 3*exp(2*x + 32) + log(3 - log(x^2))*(6*exp(x + 16) - 2*log(x^2)*exp(x + 16)) + log(3 - log(x^2))^2*(
log(x^2) - 3)),x)

[Out]

int((exp(exp(2*x) + 5)*(log(x^2)*exp(x + 16)*(6*x*exp(2*x) - 3*x + 3) - exp(x + 16)*(18*x*exp(2*x) - 9*x + 9)
+ 6) + exp(exp(2*x) + 5)*log(3 - log(x^2))*(18*x*exp(2*x) - log(x^2)*(6*x*exp(2*x) + 3) + 9))/(log(x^2)*exp(2*
x + 32) - 3*exp(2*x + 32) + log(3 - log(x^2))*(6*exp(x + 16) - 2*log(x^2)*exp(x + 16)) + log(3 - log(x^2))^2*(
log(x^2) - 3)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-6*x*exp(x)**2-3)*ln(x**2)+18*x*exp(x)**2+9)*exp(exp(x)**2+5)*ln(3-ln(x**2))+((6*x*exp(x)**2-3*x+
3)*exp(x+16)*ln(x**2)+(-18*x*exp(x)**2+9*x-9)*exp(x+16)+6)*exp(exp(x)**2+5))/((ln(x**2)-3)*ln(3-ln(x**2))**2+(
-2*exp(x+16)*ln(x**2)+6*exp(x+16))*ln(3-ln(x**2))+exp(x+16)**2*ln(x**2)-3*exp(x+16)**2),x)

[Out]

Timed out

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