3.87.22 \(\int \frac {e^6 (64 x^7-64 x^8+16 x^9)+e^x (4 e^3 x^2+e^6 (-24 x^6+8 x^7))+(e^{3+x} (-12 x^2-4 x^3)+e^3 (32 x^4-16 x^5)) \log (x)+4 x \log ^2(x)}{e^6 (16 x^6-16 x^7+4 x^8)+e^3 (8 x^3-4 x^4) \log (x)+\log ^2(x)} \, dx\)

Optimal. Leaf size=30 \[ 2 x \left (x+\frac {e^x}{-2 x+x^2-\frac {\log (x)}{2 e^3 x^2}}\right ) \]

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Rubi [F]  time = 3.61, 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^6 \left (64 x^7-64 x^8+16 x^9\right )+e^x \left (4 e^3 x^2+e^6 \left (-24 x^6+8 x^7\right )\right )+\left (e^{3+x} \left (-12 x^2-4 x^3\right )+e^3 \left (32 x^4-16 x^5\right )\right ) \log (x)+4 x \log ^2(x)}{e^6 \left (16 x^6-16 x^7+4 x^8\right )+e^3 \left (8 x^3-4 x^4\right ) \log (x)+\log ^2(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^6*(64*x^7 - 64*x^8 + 16*x^9) + E^x*(4*E^3*x^2 + E^6*(-24*x^6 + 8*x^7)) + (E^(3 + x)*(-12*x^2 - 4*x^3) +
E^3*(32*x^4 - 16*x^5))*Log[x] + 4*x*Log[x]^2)/(E^6*(16*x^6 - 16*x^7 + 4*x^8) + E^3*(8*x^3 - 4*x^4)*Log[x] + L
og[x]^2),x]

[Out]

2*x^2 + 4*Defer[Int][(E^(3 + x)*x^2)/(-4*E^3*x^3 + 2*E^3*x^4 - Log[x])^2, x] + 48*Defer[Int][(E^(6 + x)*x^5)/(
-4*E^3*x^3 + 2*E^3*x^4 - Log[x])^2, x] - 32*Defer[Int][(E^(6 + x)*x^6)/(-4*E^3*x^3 + 2*E^3*x^4 - Log[x])^2, x]
+ 12*Defer[Int][(E^(3 + x)*x^2)/(-4*E^3*x^3 + 2*E^3*x^4 - Log[x]), x] + 4*Defer[Int][(E^(3 + x)*x^3)/(-4*E^3*
x^3 + 2*E^3*x^4 - Log[x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {4 x \left (e^{3+x} x+2 e^{6+x} (-3+x) x^5+4 e^6 (-2+x)^2 x^6-e^3 x \left (4 (-2+x) x^2+e^x (3+x)\right ) \log (x)+\log ^2(x)\right )}{\left (2 e^3 (-2+x) x^3-\log (x)\right )^2} \, dx\\ &=4 \int \frac {x \left (e^{3+x} x+2 e^{6+x} (-3+x) x^5+4 e^6 (-2+x)^2 x^6-e^3 x \left (4 (-2+x) x^2+e^x (3+x)\right ) \log (x)+\log ^2(x)\right )}{\left (2 e^3 (-2+x) x^3-\log (x)\right )^2} \, dx\\ &=4 \int \left (x+\frac {e^{3+x} x^2 \left (1-6 e^3 x^4+2 e^3 x^5-3 \log (x)-x \log (x)\right )}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2}\right ) \, dx\\ &=2 x^2+4 \int \frac {e^{3+x} x^2 \left (1-6 e^3 x^4+2 e^3 x^5-3 \log (x)-x \log (x)\right )}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2} \, dx\\ &=2 x^2+4 \int \left (\frac {e^{3+x} x^2 (3+x)}{-4 e^3 x^3+2 e^3 x^4-\log (x)}+\frac {e^{3+x} x^2 \left (1+12 e^3 x^3-8 e^3 x^4\right )}{\left (4 e^3 x^3-2 e^3 x^4+\log (x)\right )^2}\right ) \, dx\\ &=2 x^2+4 \int \frac {e^{3+x} x^2 (3+x)}{-4 e^3 x^3+2 e^3 x^4-\log (x)} \, dx+4 \int \frac {e^{3+x} x^2 \left (1+12 e^3 x^3-8 e^3 x^4\right )}{\left (4 e^3 x^3-2 e^3 x^4+\log (x)\right )^2} \, dx\\ &=2 x^2+4 \int \left (\frac {e^{3+x} x^2}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2}+\frac {12 e^{6+x} x^5}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2}-\frac {8 e^{6+x} x^6}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2}\right ) \, dx+4 \int \left (\frac {3 e^{3+x} x^2}{-4 e^3 x^3+2 e^3 x^4-\log (x)}+\frac {e^{3+x} x^3}{-4 e^3 x^3+2 e^3 x^4-\log (x)}\right ) \, dx\\ &=2 x^2+4 \int \frac {e^{3+x} x^2}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2} \, dx+4 \int \frac {e^{3+x} x^3}{-4 e^3 x^3+2 e^3 x^4-\log (x)} \, dx+12 \int \frac {e^{3+x} x^2}{-4 e^3 x^3+2 e^3 x^4-\log (x)} \, dx-32 \int \frac {e^{6+x} x^6}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2} \, dx+48 \int \frac {e^{6+x} x^5}{\left (-4 e^3 x^3+2 e^3 x^4-\log (x)\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 33, normalized size = 1.10 \begin {gather*} 2 x^2 \left (1+\frac {2 e^{3+x} x}{2 e^3 (-2+x) x^3-\log (x)}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^6*(64*x^7 - 64*x^8 + 16*x^9) + E^x*(4*E^3*x^2 + E^6*(-24*x^6 + 8*x^7)) + (E^(3 + x)*(-12*x^2 - 4*
x^3) + E^3*(32*x^4 - 16*x^5))*Log[x] + 4*x*Log[x]^2)/(E^6*(16*x^6 - 16*x^7 + 4*x^8) + E^3*(8*x^3 - 4*x^4)*Log[
x] + Log[x]^2),x]

[Out]

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

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fricas [A]  time = 0.50, size = 52, normalized size = 1.73 \begin {gather*} \frac {2 \, {\left (2 \, x^{3} e^{\left (x + 3\right )} - x^{2} \log \relax (x) + 2 \, {\left (x^{6} - 2 \, x^{5}\right )} e^{3}\right )}}{2 \, {\left (x^{4} - 2 \, x^{3}\right )} e^{3} - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)^2+((-4*x^3-12*x^2)*exp(3)*exp(x)+(-16*x^5+32*x^4)*exp(3))*log(x)+((8*x^7-24*x^6)*exp(3)^
2+4*x^2*exp(3))*exp(x)+(16*x^9-64*x^8+64*x^7)*exp(3)^2)/(log(x)^2+(-4*x^4+8*x^3)*exp(3)*log(x)+(4*x^8-16*x^7+1
6*x^6)*exp(3)^2),x, algorithm="fricas")

[Out]

2*(2*x^3*e^(x + 3) - x^2*log(x) + 2*(x^6 - 2*x^5)*e^3)/(2*(x^4 - 2*x^3)*e^3 - log(x))

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giac [A]  time = 0.18, size = 54, normalized size = 1.80 \begin {gather*} \frac {2 \, {\left (2 \, x^{6} e^{3} - 4 \, x^{5} e^{3} + 2 \, x^{3} e^{\left (x + 3\right )} - x^{2} \log \relax (x)\right )}}{2 \, x^{4} e^{3} - 4 \, x^{3} e^{3} - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)^2+((-4*x^3-12*x^2)*exp(3)*exp(x)+(-16*x^5+32*x^4)*exp(3))*log(x)+((8*x^7-24*x^6)*exp(3)^
2+4*x^2*exp(3))*exp(x)+(16*x^9-64*x^8+64*x^7)*exp(3)^2)/(log(x)^2+(-4*x^4+8*x^3)*exp(3)*log(x)+(4*x^8-16*x^7+1
6*x^6)*exp(3)^2),x, algorithm="giac")

[Out]

2*(2*x^6*e^3 - 4*x^5*e^3 + 2*x^3*e^(x + 3) - x^2*log(x))/(2*x^4*e^3 - 4*x^3*e^3 - log(x))

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maple [A]  time = 0.11, size = 37, normalized size = 1.23

method	result	size
risch	\(2 x^{2}+\frac {4 x^{3} {\mathrm e}^{3+x}}{2 x^{4} {\mathrm e}^{3}-4 x^{3} {\mathrm e}^{3}-\ln \relax (x )}\)	\(37\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*ln(x)^2+((-4*x^3-12*x^2)*exp(3)*exp(x)+(-16*x^5+32*x^4)*exp(3))*ln(x)+((8*x^7-24*x^6)*exp(3)^2+4*x^2*
exp(3))*exp(x)+(16*x^9-64*x^8+64*x^7)*exp(3)^2)/(ln(x)^2+(-4*x^4+8*x^3)*exp(3)*ln(x)+(4*x^8-16*x^7+16*x^6)*exp
(3)^2),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [A]  time = 0.40, size = 54, normalized size = 1.80 \begin {gather*} \frac {2 \, {\left (2 \, x^{6} e^{3} - 4 \, x^{5} e^{3} + 2 \, x^{3} e^{\left (x + 3\right )} - x^{2} \log \relax (x)\right )}}{2 \, x^{4} e^{3} - 4 \, x^{3} e^{3} - \log \relax (x)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*log(x)^2+((-4*x^3-12*x^2)*exp(3)*exp(x)+(-16*x^5+32*x^4)*exp(3))*log(x)+((8*x^7-24*x^6)*exp(3)^
2+4*x^2*exp(3))*exp(x)+(16*x^9-64*x^8+64*x^7)*exp(3)^2)/(log(x)^2+(-4*x^4+8*x^3)*exp(3)*log(x)+(4*x^8-16*x^7+1
6*x^6)*exp(3)^2),x, algorithm="maxima")

[Out]

2*(2*x^6*e^3 - 4*x^5*e^3 + 2*x^3*e^(x + 3) - x^2*log(x))/(2*x^4*e^3 - 4*x^3*e^3 - log(x))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {4\,x\,{\ln \relax (x)}^2+\left ({\mathrm {e}}^3\,\left (32\,x^4-16\,x^5\right )-{\mathrm {e}}^{x+3}\,\left (4\,x^3+12\,x^2\right )\right )\,\ln \relax (x)+{\mathrm {e}}^6\,\left (16\,x^9-64\,x^8+64\,x^7\right )-{\mathrm {e}}^x\,\left ({\mathrm {e}}^6\,\left (24\,x^6-8\,x^7\right )-4\,x^2\,{\mathrm {e}}^3\right )}{{\ln \relax (x)}^2+{\mathrm {e}}^3\,\left (8\,x^3-4\,x^4\right )\,\ln \relax (x)+{\mathrm {e}}^6\,\left (4\,x^8-16\,x^7+16\,x^6\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*x*log(x)^2 + log(x)*(exp(3)*(32*x^4 - 16*x^5) - exp(3)*exp(x)*(12*x^2 + 4*x^3)) + exp(6)*(64*x^7 - 64*x
^8 + 16*x^9) - exp(x)*(exp(6)*(24*x^6 - 8*x^7) - 4*x^2*exp(3)))/(log(x)^2 + exp(6)*(16*x^6 - 16*x^7 + 4*x^8) +
exp(3)*log(x)*(8*x^3 - 4*x^4)),x)

[Out]

int((4*x*log(x)^2 - log(x)*(exp(x + 3)*(12*x^2 + 4*x^3) - exp(3)*(32*x^4 - 16*x^5)) + exp(6)*(64*x^7 - 64*x^8
+ 16*x^9) - exp(x)*(exp(6)*(24*x^6 - 8*x^7) - 4*x^2*exp(3)))/(log(x)^2 + exp(6)*(16*x^6 - 16*x^7 + 4*x^8) + ex
p(3)*log(x)*(8*x^3 - 4*x^4)), x)

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sympy [A]  time = 0.35, size = 36, normalized size = 1.20 \begin {gather*} \frac {4 x^{3} e^{3} e^{x}}{2 x^{4} e^{3} - 4 x^{3} e^{3} - \log {\relax (x )}} + 2 x^{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*x*ln(x)**2+((-4*x**3-12*x**2)*exp(3)*exp(x)+(-16*x**5+32*x**4)*exp(3))*ln(x)+((8*x**7-24*x**6)*ex
p(3)**2+4*x**2*exp(3))*exp(x)+(16*x**9-64*x**8+64*x**7)*exp(3)**2)/(ln(x)**2+(-4*x**4+8*x**3)*exp(3)*ln(x)+(4*
x**8-16*x**7+16*x**6)*exp(3)**2),x)

[Out]

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

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