3.42.17 \(\int \frac {e^x (64+16 x^4+x^8)+e^x (64+8 x^4) \log (x^2)+16 e^x \log ^2(x^2)+e^{\frac {5 x^3+x^4}{e^x (8+x^4)+4 e^x \log (x^2)}} (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+(60 x^2-4 x^3-4 x^4) \log (x^2))}{e^x (64+16 x^4+x^8)+e^x (64+8 x^4) \log (x^2)+16 e^x \log ^2(x^2)} \, dx\)

Optimal. Leaf size=28 \[ e^{\frac {e^{-x} (5+x)}{x+\frac {4 \left (2+\log \left (x^2\right )\right )}{x^3}}}+x \]

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Rubi [F]  time = 57.52, 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^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )+e^{\frac {5 x^3+x^4}{e^x \left (8+x^4\right )+4 e^x \log \left (x^2\right )}} \left (80 x^2-16 x^3-8 x^4-5 x^6-5 x^7-x^8+\left (60 x^2-4 x^3-4 x^4\right ) \log \left (x^2\right )\right )}{e^x \left (64+16 x^4+x^8\right )+e^x \left (64+8 x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(64 + 16*x^4 + x^8) + E^x*(64 + 8*x^4)*Log[x^2] + 16*E^x*Log[x^2]^2 + E^((5*x^3 + x^4)/(E^x*(8 + x^4)
+ 4*E^x*Log[x^2]))*(80*x^2 - 16*x^3 - 8*x^4 - 5*x^6 - 5*x^7 - x^8 + (60*x^2 - 4*x^3 - 4*x^4)*Log[x^2]))/(E^x*
(64 + 16*x^4 + x^8) + E^x*(64 + 8*x^4)*Log[x^2] + 16*E^x*Log[x^2]^2),x]

[Out]

x - 40*Defer[Int][(E^(-x + (x^3*(5 + x))/(E^x*(8 + x^4 + 4*Log[x^2])))*x^2)/(8 + x^4 + 4*Log[x^2])^2, x] - 8*D
efer[Int][(E^(-x + (x^3*(5 + x))/(E^x*(8 + x^4 + 4*Log[x^2])))*x^3)/(8 + x^4 + 4*Log[x^2])^2, x] - 20*Defer[In
t][(E^(-x + (x^3*(5 + x))/(E^x*(8 + x^4 + 4*Log[x^2])))*x^6)/(8 + x^4 + 4*Log[x^2])^2, x] - 4*Defer[Int][(E^(-
x + (x^3*(5 + x))/(E^x*(8 + x^4 + 4*Log[x^2])))*x^7)/(8 + x^4 + 4*Log[x^2])^2, x] + 15*Defer[Int][(E^(-x + (x^
3*(5 + x))/(E^x*(8 + x^4 + 4*Log[x^2])))*x^2)/(8 + x^4 + 4*Log[x^2]), x] - Defer[Int][(E^(-x + (x^3*(5 + x))/(
E^x*(8 + x^4 + 4*Log[x^2])))*x^3)/(8 + x^4 + 4*Log[x^2]), x] - Defer[Int][(E^(-x + (x^3*(5 + x))/(E^x*(8 + x^4
+ 4*Log[x^2])))*x^4)/(8 + x^4 + 4*Log[x^2]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{-x} \left (e^x \left (8+x^4\right )^2+8 e^x \left (8+x^4\right ) \log \left (x^2\right )+16 e^x \log ^2\left (x^2\right )-e^{\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-80+16 x+8 x^2+5 x^4+5 x^5+x^6+4 \left (-15+x+x^2\right ) \log \left (x^2\right )\right )\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx\\ &=\int \left (1-\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-80+16 x+8 x^2+5 x^4+5 x^5+x^6-60 \log \left (x^2\right )+4 x \log \left (x^2\right )+4 x^2 \log \left (x^2\right )\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}\right ) \, dx\\ &=x-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-80+16 x+8 x^2+5 x^4+5 x^5+x^6-60 \log \left (x^2\right )+4 x \log \left (x^2\right )+4 x^2 \log \left (x^2\right )\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx\\ &=x-\int \left (\frac {4 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (10+2 x+5 x^4+x^5\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-15+x+x^2\right )}{8+x^4+4 \log \left (x^2\right )}\right ) \, dx\\ &=x-4 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (10+2 x+5 x^4+x^5\right )}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2 \left (-15+x+x^2\right )}{8+x^4+4 \log \left (x^2\right )} \, dx\\ &=x-4 \int \left (\frac {10 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {2 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {5 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^6}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^7}{\left (8+x^4+4 \log \left (x^2\right )\right )^2}\right ) \, dx-\int \left (-\frac {15 e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{8+x^4+4 \log \left (x^2\right )}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{8+x^4+4 \log \left (x^2\right )}+\frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^4}{8+x^4+4 \log \left (x^2\right )}\right ) \, dx\\ &=x-4 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^7}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-8 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx+15 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{8+x^4+4 \log \left (x^2\right )} \, dx-20 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^6}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-40 \int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^2}{\left (8+x^4+4 \log \left (x^2\right )\right )^2} \, dx-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^3}{8+x^4+4 \log \left (x^2\right )} \, dx-\int \frac {e^{-x+\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}} x^4}{8+x^4+4 \log \left (x^2\right )} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.30, size = 29, normalized size = 1.04 \begin {gather*} e^{\frac {e^{-x} x^3 (5+x)}{8+x^4+4 \log \left (x^2\right )}}+x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(64 + 16*x^4 + x^8) + E^x*(64 + 8*x^4)*Log[x^2] + 16*E^x*Log[x^2]^2 + E^((5*x^3 + x^4)/(E^x*(8
+ x^4) + 4*E^x*Log[x^2]))*(80*x^2 - 16*x^3 - 8*x^4 - 5*x^6 - 5*x^7 - x^8 + (60*x^2 - 4*x^3 - 4*x^4)*Log[x^2]))
/(E^x*(64 + 16*x^4 + x^8) + E^x*(64 + 8*x^4)*Log[x^2] + 16*E^x*Log[x^2]^2),x]

[Out]

E^((x^3*(5 + x))/(E^x*(8 + x^4 + 4*Log[x^2]))) + x

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fricas [A]  time = 0.99, size = 32, normalized size = 1.14 \begin {gather*} x + e^{\left (\frac {x^{4} + 5 \, x^{3}}{{\left (x^{4} + 8\right )} e^{x} + 4 \, e^{x} \log \left (x^{2}\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^4-4*x^3+60*x^2)*log(x^2)-x^8-5*x^7-5*x^6-8*x^4-16*x^3+80*x^2)*exp((x^4+5*x^3)/(4*exp(x)*log(
x^2)+(x^4+8)*exp(x)))+16*exp(x)*log(x^2)^2+(8*x^4+64)*exp(x)*log(x^2)+(x^8+16*x^4+64)*exp(x))/(16*exp(x)*log(x
^2)^2+(8*x^4+64)*exp(x)*log(x^2)+(x^8+16*x^4+64)*exp(x)),x, algorithm="fricas")

[Out]

x + e^((x^4 + 5*x^3)/((x^4 + 8)*e^x + 4*e^x*log(x^2)))

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {8 \, {\left (x^{4} + 8\right )} e^{x} \log \left (x^{2}\right ) + 16 \, e^{x} \log \left (x^{2}\right )^{2} + {\left (x^{8} + 16 \, x^{4} + 64\right )} e^{x} - {\left (x^{8} + 5 \, x^{7} + 5 \, x^{6} + 8 \, x^{4} + 16 \, x^{3} - 80 \, x^{2} + 4 \, {\left (x^{4} + x^{3} - 15 \, x^{2}\right )} \log \left (x^{2}\right )\right )} e^{\left (\frac {x^{4} + 5 \, x^{3}}{{\left (x^{4} + 8\right )} e^{x} + 4 \, e^{x} \log \left (x^{2}\right )}\right )}}{8 \, {\left (x^{4} + 8\right )} e^{x} \log \left (x^{2}\right ) + 16 \, e^{x} \log \left (x^{2}\right )^{2} + {\left (x^{8} + 16 \, x^{4} + 64\right )} e^{x}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^4-4*x^3+60*x^2)*log(x^2)-x^8-5*x^7-5*x^6-8*x^4-16*x^3+80*x^2)*exp((x^4+5*x^3)/(4*exp(x)*log(
x^2)+(x^4+8)*exp(x)))+16*exp(x)*log(x^2)^2+(8*x^4+64)*exp(x)*log(x^2)+(x^8+16*x^4+64)*exp(x))/(16*exp(x)*log(x
^2)^2+(8*x^4+64)*exp(x)*log(x^2)+(x^8+16*x^4+64)*exp(x)),x, algorithm="giac")

[Out]

integrate((8*(x^4 + 8)*e^x*log(x^2) + 16*e^x*log(x^2)^2 + (x^8 + 16*x^4 + 64)*e^x - (x^8 + 5*x^7 + 5*x^6 + 8*x
^4 + 16*x^3 - 80*x^2 + 4*(x^4 + x^3 - 15*x^2)*log(x^2))*e^((x^4 + 5*x^3)/((x^4 + 8)*e^x + 4*e^x*log(x^2))))/(8
*(x^4 + 8)*e^x*log(x^2) + 16*e^x*log(x^2)^2 + (x^8 + 16*x^4 + 64)*e^x), x)

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maple [C]  time = 0.67, size = 75, normalized size = 2.68

method	result	size
risch	\(x +{\mathrm e}^{\frac {x^{3} \left (5+x \right ) {\mathrm e}^{-x}}{-2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}+4 i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}-2 i \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )+x^{4}+8 \ln \relax (x )+8}}\)	\(75\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((-4*x^4-4*x^3+60*x^2)*ln(x^2)-x^8-5*x^7-5*x^6-8*x^4-16*x^3+80*x^2)*exp((x^4+5*x^3)/(4*exp(x)*ln(x^2)+(x^
4+8)*exp(x)))+16*exp(x)*ln(x^2)^2+(8*x^4+64)*exp(x)*ln(x^2)+(x^8+16*x^4+64)*exp(x))/(16*exp(x)*ln(x^2)^2+(8*x^
4+64)*exp(x)*ln(x^2)+(x^8+16*x^4+64)*exp(x)),x,method=_RETURNVERBOSE)

[Out]

x+exp(x^3*(5+x)*exp(-x)/(-2*I*Pi*csgn(I*x^2)^3+4*I*Pi*csgn(I*x^2)^2*csgn(I*x)-2*I*Pi*csgn(I*x^2)*csgn(I*x)^2+x
^4+8*ln(x)+8))

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maxima [B]  time = 0.43, size = 116, normalized size = 4.14 \begin {gather*} {\left (x e^{\left (\frac {8 \, \log \relax (x)}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \relax (x)} + \frac {8}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \relax (x)}\right )} + e^{\left (\frac {5 \, x^{3}}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \relax (x)} + e^{\left (-x\right )}\right )}\right )} e^{\left (-\frac {8 \, \log \relax (x)}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \relax (x)} - \frac {8}{{\left (x^{4} + 8\right )} e^{x} + 8 \, e^{x} \log \relax (x)}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x^4-4*x^3+60*x^2)*log(x^2)-x^8-5*x^7-5*x^6-8*x^4-16*x^3+80*x^2)*exp((x^4+5*x^3)/(4*exp(x)*log(
x^2)+(x^4+8)*exp(x)))+16*exp(x)*log(x^2)^2+(8*x^4+64)*exp(x)*log(x^2)+(x^8+16*x^4+64)*exp(x))/(16*exp(x)*log(x
^2)^2+(8*x^4+64)*exp(x)*log(x^2)+(x^8+16*x^4+64)*exp(x)),x, algorithm="maxima")

[Out]

(x*e^(8*log(x)/((x^4 + 8)*e^x + 8*e^x*log(x)) + 8/((x^4 + 8)*e^x + 8*e^x*log(x))) + e^(5*x^3/((x^4 + 8)*e^x +
8*e^x*log(x)) + e^(-x)))*e^(-8*log(x)/((x^4 + 8)*e^x + 8*e^x*log(x)) - 8/((x^4 + 8)*e^x + 8*e^x*log(x)))

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mupad [B]  time = 3.76, size = 56, normalized size = 2.00 \begin {gather*} x+{\mathrm {e}}^{\frac {x^4}{8\,{\mathrm {e}}^x+x^4\,{\mathrm {e}}^x+4\,\ln \left (x^2\right )\,{\mathrm {e}}^x}}\,{\mathrm {e}}^{\frac {5\,x^3}{8\,{\mathrm {e}}^x+x^4\,{\mathrm {e}}^x+4\,\ln \left (x^2\right )\,{\mathrm {e}}^x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(16*x^4 + x^8 + 64) - exp((5*x^3 + x^4)/(exp(x)*(x^4 + 8) + 4*log(x^2)*exp(x)))*(log(x^2)*(4*x^3 -
60*x^2 + 4*x^4) - 80*x^2 + 16*x^3 + 8*x^4 + 5*x^6 + 5*x^7 + x^8) + 16*log(x^2)^2*exp(x) + log(x^2)*exp(x)*(8*
x^4 + 64))/(exp(x)*(16*x^4 + x^8 + 64) + 16*log(x^2)^2*exp(x) + log(x^2)*exp(x)*(8*x^4 + 64)),x)

[Out]

x + exp(x^4/(8*exp(x) + x^4*exp(x) + 4*log(x^2)*exp(x)))*exp((5*x^3)/(8*exp(x) + x^4*exp(x) + 4*log(x^2)*exp(x
)))

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sympy [A]  time = 1.20, size = 29, normalized size = 1.04 \begin {gather*} x + e^{\frac {x^{4} + 5 x^{3}}{\left (x^{4} + 8\right ) e^{x} + 4 e^{x} \log {\left (x^{2} \right )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((-4*x**4-4*x**3+60*x**2)*ln(x**2)-x**8-5*x**7-5*x**6-8*x**4-16*x**3+80*x**2)*exp((x**4+5*x**3)/(4*
exp(x)*ln(x**2)+(x**4+8)*exp(x)))+16*exp(x)*ln(x**2)**2+(8*x**4+64)*exp(x)*ln(x**2)+(x**8+16*x**4+64)*exp(x))/
(16*exp(x)*ln(x**2)**2+(8*x**4+64)*exp(x)*ln(x**2)+(x**8+16*x**4+64)*exp(x)),x)

[Out]

x + exp((x**4 + 5*x**3)/((x**4 + 8)*exp(x) + 4*exp(x)*log(x**2)))

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