3.93.71 \(\int \frac {-5 e^{2 x}+e^x (-45+15 x+10 x^2-3 x^3)+e^x (-5+5 x+3 x^2) \log (x)+e^x (-5+5 x+3 x^2) \log (81 x^4)}{16 x^2+e^{2 x} x^2-8 x^3+x^4+e^x (8 x^2-2 x^3)+(8 x^2+2 e^x x^2-2 x^3) \log (x)+x^2 \log ^2(x)+(8 x^2+2 e^x x^2-2 x^3+2 x^2 \log (x)) \log (81 x^4)+x^2 \log ^2(81 x^4)} \, dx\)

Optimal. Leaf size=29 \[ \frac {e^x \left (3+\frac {5}{x}\right )}{4+e^x-x+\log (x)+\log \left (81 x^4\right )} \]

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Rubi [F]  time = 13.46, 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 {-5 e^{2 x}+e^x \left (-45+15 x+10 x^2-3 x^3\right )+e^x \left (-5+5 x+3 x^2\right ) \log (x)+e^x \left (-5+5 x+3 x^2\right ) \log \left (81 x^4\right )}{16 x^2+e^{2 x} x^2-8 x^3+x^4+e^x \left (8 x^2-2 x^3\right )+\left (8 x^2+2 e^x x^2-2 x^3\right ) \log (x)+x^2 \log ^2(x)+\left (8 x^2+2 e^x x^2-2 x^3+2 x^2 \log (x)\right ) \log \left (81 x^4\right )+x^2 \log ^2\left (81 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-5*E^(2*x) + E^x*(-45 + 15*x + 10*x^2 - 3*x^3) + E^x*(-5 + 5*x + 3*x^2)*Log[x] + E^x*(-5 + 5*x + 3*x^2)*L
og[81*x^4])/(16*x^2 + E^(2*x)*x^2 - 8*x^3 + x^4 + E^x*(8*x^2 - 2*x^3) + (8*x^2 + 2*E^x*x^2 - 2*x^3)*Log[x] + x
^2*Log[x]^2 + (8*x^2 + 2*E^x*x^2 - 2*x^3 + 2*x^2*Log[x])*Log[81*x^4] + x^2*Log[81*x^4]^2),x]

[Out]

5*Defer[Int][E^x/(x^2*(-4 - E^x + x - Log[81*x] - Log[x^4])), x] - 25*Defer[Int][E^x/(x^2*(-4 - E^x + x - Log[
x] - Log[81*x^4])^2), x] + 10*Defer[Int][E^x/(x*(-4 - E^x + x - Log[x] - Log[81*x^4])^2), x] - 3*Defer[Int][(E
^x*x)/(-4 - E^x + x - Log[x] - Log[81*x^4])^2, x] + 5*Defer[Int][(E^x*Log[x])/(x*(-4 - E^x + x - Log[x] - Log[
81*x^4])^2), x] + 5*Defer[Int][(E^x*Log[81*x^4])/(x*(-4 - E^x + x - Log[x] - Log[81*x^4])^2), x] + 10*Defer[In
t][E^x/(4 + E^x - x + Log[x] + Log[81*x^4])^2, x] + 3*Defer[Int][(E^x*Log[x])/(4 + E^x - x + Log[x] + Log[81*x
^4])^2, x] + 3*Defer[Int][(E^x*Log[81*x^4])/(4 + E^x - x + Log[x] + Log[81*x^4])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^x \left (-45-5 e^x+15 x+10 x^2-3 x^3-\left (5-5 x-3 x^2\right ) \log (x)-\left (5-5 x-3 x^2\right ) \log \left (81 x^4\right )\right )}{x^2 \left (4+e^x-x+\log (x)+\log \left (81 x^4\right )\right )^2} \, dx\\ &=\int \left (\frac {5 e^x}{x^2 \left (-4-e^x+x-\log (81 x)-\log \left (x^4\right )\right )}-\frac {e^x (5+3 x) \left (5-5 x+x^2-x \log (x)-x \log \left (81 x^4\right )\right )}{x^2 \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2}\right ) \, dx\\ &=5 \int \frac {e^x}{x^2 \left (-4-e^x+x-\log (81 x)-\log \left (x^4\right )\right )} \, dx-\int \frac {e^x (5+3 x) \left (5-5 x+x^2-x \log (x)-x \log \left (81 x^4\right )\right )}{x^2 \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2} \, dx\\ &=5 \int \frac {e^x}{x^2 \left (-4-e^x+x-\log (81 x)-\log \left (x^4\right )\right )} \, dx-\int \left (\frac {5 e^x \left (5-5 x+x^2-x \log (x)-x \log \left (81 x^4\right )\right )}{x^2 \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2}+\frac {3 e^x \left (5-5 x+x^2-x \log (x)-x \log \left (81 x^4\right )\right )}{x \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2}\right ) \, dx\\ &=-\left (3 \int \frac {e^x \left (5-5 x+x^2-x \log (x)-x \log \left (81 x^4\right )\right )}{x \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2} \, dx\right )+5 \int \frac {e^x}{x^2 \left (-4-e^x+x-\log (81 x)-\log \left (x^4\right )\right )} \, dx-5 \int \frac {e^x \left (5-5 x+x^2-x \log (x)-x \log \left (81 x^4\right )\right )}{x^2 \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2} \, dx\\ &=-\left (3 \int \left (\frac {5 e^x}{x \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2}+\frac {e^x x}{\left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2}-\frac {5 e^x}{\left (4+e^x-x+\log (x)+\log \left (81 x^4\right )\right )^2}-\frac {e^x \log (x)}{\left (4+e^x-x+\log (x)+\log \left (81 x^4\right )\right )^2}-\frac {e^x \log \left (81 x^4\right )}{\left (4+e^x-x+\log (x)+\log \left (81 x^4\right )\right )^2}\right ) \, dx\right )+5 \int \frac {e^x}{x^2 \left (-4-e^x+x-\log (81 x)-\log \left (x^4\right )\right )} \, dx-5 \int \left (\frac {5 e^x}{x^2 \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2}-\frac {5 e^x}{x \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2}-\frac {e^x \log (x)}{x \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2}-\frac {e^x \log \left (81 x^4\right )}{x \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2}+\frac {e^x}{\left (4+e^x-x+\log (x)+\log \left (81 x^4\right )\right )^2}\right ) \, dx\\ &=-\left (3 \int \frac {e^x x}{\left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2} \, dx\right )+3 \int \frac {e^x \log (x)}{\left (4+e^x-x+\log (x)+\log \left (81 x^4\right )\right )^2} \, dx+3 \int \frac {e^x \log \left (81 x^4\right )}{\left (4+e^x-x+\log (x)+\log \left (81 x^4\right )\right )^2} \, dx+5 \int \frac {e^x}{x^2 \left (-4-e^x+x-\log (81 x)-\log \left (x^4\right )\right )} \, dx+5 \int \frac {e^x \log (x)}{x \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2} \, dx+5 \int \frac {e^x \log \left (81 x^4\right )}{x \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2} \, dx-5 \int \frac {e^x}{\left (4+e^x-x+\log (x)+\log \left (81 x^4\right )\right )^2} \, dx-15 \int \frac {e^x}{x \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2} \, dx+15 \int \frac {e^x}{\left (4+e^x-x+\log (x)+\log \left (81 x^4\right )\right )^2} \, dx-25 \int \frac {e^x}{x^2 \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2} \, dx+25 \int \frac {e^x}{x \left (-4-e^x+x-\log (x)-\log \left (81 x^4\right )\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 5.08, size = 30, normalized size = 1.03 \begin {gather*} \frac {e^x (5+3 x)}{x \left (4+e^x-x+\log (x)+\log \left (81 x^4\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-5*E^(2*x) + E^x*(-45 + 15*x + 10*x^2 - 3*x^3) + E^x*(-5 + 5*x + 3*x^2)*Log[x] + E^x*(-5 + 5*x + 3*
x^2)*Log[81*x^4])/(16*x^2 + E^(2*x)*x^2 - 8*x^3 + x^4 + E^x*(8*x^2 - 2*x^3) + (8*x^2 + 2*E^x*x^2 - 2*x^3)*Log[
x] + x^2*Log[x]^2 + (8*x^2 + 2*E^x*x^2 - 2*x^3 + 2*x^2*Log[x])*Log[81*x^4] + x^2*Log[81*x^4]^2),x]

[Out]

(E^x*(5 + 3*x))/(x*(4 + E^x - x + Log[x] + Log[81*x^4]))

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fricas [A]  time = 0.91, size = 33, normalized size = 1.14 \begin {gather*} -\frac {{\left (3 \, x + 5\right )} e^{x}}{x^{2} - x e^{x} - 4 \, x \log \relax (3) - 5 \, x \log \relax (x) - 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+5*x-5)*exp(x)*log(81*x^4)+(3*x^2+5*x-5)*exp(x)*log(x)-5*exp(x)^2+(-3*x^3+10*x^2+15*x-45)*exp
(x))/(x^2*log(81*x^4)^2+(2*x^2*log(x)+2*exp(x)*x^2-2*x^3+8*x^2)*log(81*x^4)+x^2*log(x)^2+(2*exp(x)*x^2-2*x^3+8
*x^2)*log(x)+exp(x)^2*x^2+(-2*x^3+8*x^2)*exp(x)+x^4-8*x^3+16*x^2),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.37, size = 36, normalized size = 1.24 \begin {gather*} -\frac {3 \, x e^{x} + 5 \, e^{x}}{x^{2} - x e^{x} - 4 \, x \log \relax (3) - 5 \, x \log \relax (x) - 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+5*x-5)*exp(x)*log(81*x^4)+(3*x^2+5*x-5)*exp(x)*log(x)-5*exp(x)^2+(-3*x^3+10*x^2+15*x-45)*exp
(x))/(x^2*log(81*x^4)^2+(2*x^2*log(x)+2*exp(x)*x^2-2*x^3+8*x^2)*log(81*x^4)+x^2*log(x)^2+(2*exp(x)*x^2-2*x^3+8
*x^2)*log(x)+exp(x)^2*x^2+(-2*x^3+8*x^2)*exp(x)+x^4-8*x^3+16*x^2),x, algorithm="giac")

[Out]

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

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maple [C]  time = 0.30, size = 229, normalized size = 7.90

method	result	size
risch	\(-\frac {2 \left (3 x +5\right ) {\mathrm e}^{x}}{x \left (-i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+i \pi \mathrm {csgn}\left (i x^{3}\right )^{3}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )+i \pi \mathrm {csgn}\left (i x^{4}\right )^{3}+i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{4}\right )^{2}-2 i \pi \mathrm {csgn}\left (i x^{2}\right )^{2} \mathrm {csgn}\left (i x \right )+i \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x \right )^{2}+i \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-i \pi \,\mathrm {csgn}\left (i x^{3}\right ) \mathrm {csgn}\left (i x^{4}\right )^{2}-i \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}-8-8 \ln \relax (3)+2 x -2 \,{\mathrm e}^{x}-10 \ln \relax (x )\right )}\)	\(229\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((3*x^2+5*x-5)*exp(x)*ln(81*x^4)+(3*x^2+5*x-5)*exp(x)*ln(x)-5*exp(x)^2+(-3*x^3+10*x^2+15*x-45)*exp(x))/(x^
2*ln(81*x^4)^2+(2*x^2*ln(x)+2*exp(x)*x^2-2*x^3+8*x^2)*ln(81*x^4)+x^2*ln(x)^2+(2*exp(x)*x^2-2*x^3+8*x^2)*ln(x)+
exp(x)^2*x^2+(-2*x^3+8*x^2)*exp(x)+x^4-8*x^3+16*x^2),x,method=_RETURNVERBOSE)

[Out]

-2*(3*x+5)*exp(x)/x/(-I*Pi*csgn(I*x^2)*csgn(I*x^3)^2+I*Pi*csgn(I*x^3)^3+I*Pi*csgn(I*x)*csgn(I*x^3)*csgn(I*x^4)
+I*Pi*csgn(I*x^4)^3+I*Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-I*Pi*csgn(I*x)*csgn(I*x^4)^2-2*I*Pi*csgn(I*x^2)^2*c
sgn(I*x)+I*Pi*csgn(I*x^2)*csgn(I*x)^2+I*Pi*csgn(I*x^2)^3-I*Pi*csgn(I*x^3)*csgn(I*x^4)^2-I*Pi*csgn(I*x)*csgn(I*
x^3)^2-8-8*ln(3)+2*x-2*exp(x)-10*ln(x))

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maxima [A]  time = 0.52, size = 47, normalized size = 1.62 \begin {gather*} -\frac {3 \, x^{2} - 12 \, x {\left (\log \relax (3) + 1\right )} - 15 \, x \log \relax (x) + 5 \, e^{x}}{x^{2} - 4 \, x {\left (\log \relax (3) + 1\right )} - x e^{x} - 5 \, x \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x^2+5*x-5)*exp(x)*log(81*x^4)+(3*x^2+5*x-5)*exp(x)*log(x)-5*exp(x)^2+(-3*x^3+10*x^2+15*x-45)*exp
(x))/(x^2*log(81*x^4)^2+(2*x^2*log(x)+2*exp(x)*x^2-2*x^3+8*x^2)*log(81*x^4)+x^2*log(x)^2+(2*exp(x)*x^2-2*x^3+8
*x^2)*log(x)+exp(x)^2*x^2+(-2*x^3+8*x^2)*exp(x)+x^4-8*x^3+16*x^2),x, algorithm="maxima")

[Out]

-(3*x^2 - 12*x*(log(3) + 1) - 15*x*log(x) + 5*e^x)/(x^2 - 4*x*(log(3) + 1) - x*e^x - 5*x*log(x))

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mupad [B]  time = 6.57, size = 49, normalized size = 1.69 \begin {gather*} -\frac {12\,x-5\,{\mathrm {e}}^x+3\,x\,\ln \left (81\,x^4\right )+3\,x\,\ln \relax (x)-3\,x^2}{x\,\left (\ln \left (81\,x^4\right )-x+{\mathrm {e}}^x+\ln \relax (x)+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(15*x + 10*x^2 - 3*x^3 - 45) - 5*exp(2*x) + exp(x)*log(81*x^4)*(5*x + 3*x^2 - 5) + exp(x)*log(x)*(
5*x + 3*x^2 - 5))/(exp(x)*(8*x^2 - 2*x^3) + log(x)*(2*x^2*exp(x) + 8*x^2 - 2*x^3) + x^2*log(81*x^4)^2 + x^2*ex
p(2*x) + x^2*log(x)^2 + 16*x^2 - 8*x^3 + x^4 + log(81*x^4)*(2*x^2*exp(x) + 2*x^2*log(x) + 8*x^2 - 2*x^3)),x)

[Out]

-(12*x - 5*exp(x) + 3*x*log(81*x^4) + 3*x*log(x) - 3*x^2)/(x*(log(81*x^4) - x + exp(x) + log(x) + 4))

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sympy [B]  time = 0.43, size = 61, normalized size = 2.10 \begin {gather*} \frac {3 x^{2} - 15 x \log {\relax (x )} - 12 x \log {\relax (3 )} - 7 x - 25 \log {\relax (x )} - 20 \log {\relax (3 )} - 20}{- x^{2} + x e^{x} + 5 x \log {\relax (x )} + 4 x + 4 x \log {\relax (3 )}} + \frac {5}{x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((3*x**2+5*x-5)*exp(x)*ln(81*x**4)+(3*x**2+5*x-5)*exp(x)*ln(x)-5*exp(x)**2+(-3*x**3+10*x**2+15*x-45)
*exp(x))/(x**2*ln(81*x**4)**2+(2*x**2*ln(x)+2*exp(x)*x**2-2*x**3+8*x**2)*ln(81*x**4)+x**2*ln(x)**2+(2*exp(x)*x
**2-2*x**3+8*x**2)*ln(x)+exp(x)**2*x**2+(-2*x**3+8*x**2)*exp(x)+x**4-8*x**3+16*x**2),x)

[Out]

(3*x**2 - 15*x*log(x) - 12*x*log(3) - 7*x - 25*log(x) - 20*log(3) - 20)/(-x**2 + x*exp(x) + 5*x*log(x) + 4*x +
4*x*log(3)) + 5/x

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