3.30.82 \(\int \frac {e^x (-320 x-96 x^2+208 x^3-64 x^4)+e^x (128 x+64 x^2) \log (x)}{-216+540 x-882 x^2+845 x^3-588 x^4+240 x^5-64 x^6+(432-720 x+876 x^2-480 x^3+192 x^4) \log (x)+(-288+240 x-192 x^2) \log ^2(x)+64 \log ^3(x)} \, dx\)

Optimal. Leaf size=27 \[ 2+\frac {e^x}{\left (-1+x-\frac {-1+\frac {1}{4} (-2+x)+\log (x)}{x}\right )^2} \]

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Rubi [F]  time = 1.82, 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 (-320 x-96 x^2+208 x^3-64 x^4\right )+e^x \left (128 x+64 x^2\right ) \log (x)}{-216+540 x-882 x^2+845 x^3-588 x^4+240 x^5-64 x^6+\left (432-720 x+876 x^2-480 x^3+192 x^4\right ) \log (x)+\left (-288+240 x-192 x^2\right ) \log ^2(x)+64 \log ^3(x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^x*(-320*x - 96*x^2 + 208*x^3 - 64*x^4) + E^x*(128*x + 64*x^2)*Log[x])/(-216 + 540*x - 882*x^2 + 845*x^3
- 588*x^4 + 240*x^5 - 64*x^6 + (432 - 720*x + 876*x^2 - 480*x^3 + 192*x^4)*Log[x] + (-288 + 240*x - 192*x^2)*
Log[x]^2 + 64*Log[x]^3),x]

[Out]

128*Defer[Int][(E^x*x)/(6 - 5*x + 4*x^2 - 4*Log[x])^3, x] + 160*Defer[Int][(E^x*x^2)/(6 - 5*x + 4*x^2 - 4*Log[
x])^3, x] - 256*Defer[Int][(E^x*x^3)/(6 - 5*x + 4*x^2 - 4*Log[x])^3, x] + 32*Defer[Int][(E^x*x)/(6 - 5*x + 4*x
^2 - 4*Log[x])^2, x] + 16*Defer[Int][(E^x*x^2)/(6 - 5*x + 4*x^2 - 4*Log[x])^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {16 e^x x \left (20+6 x-13 x^2+4 x^3-4 (2+x) \log (x)\right )}{\left (6-5 x+4 x^2-4 \log (x)\right )^3} \, dx\\ &=16 \int \frac {e^x x \left (20+6 x-13 x^2+4 x^3-4 (2+x) \log (x)\right )}{\left (6-5 x+4 x^2-4 \log (x)\right )^3} \, dx\\ &=16 \int \left (-\frac {2 e^x x \left (-4-5 x+8 x^2\right )}{\left (6-5 x+4 x^2-4 \log (x)\right )^3}+\frac {e^x x (2+x)}{\left (6-5 x+4 x^2-4 \log (x)\right )^2}\right ) \, dx\\ &=16 \int \frac {e^x x (2+x)}{\left (6-5 x+4 x^2-4 \log (x)\right )^2} \, dx-32 \int \frac {e^x x \left (-4-5 x+8 x^2\right )}{\left (6-5 x+4 x^2-4 \log (x)\right )^3} \, dx\\ &=16 \int \left (\frac {2 e^x x}{\left (6-5 x+4 x^2-4 \log (x)\right )^2}+\frac {e^x x^2}{\left (6-5 x+4 x^2-4 \log (x)\right )^2}\right ) \, dx-32 \int \left (-\frac {4 e^x x}{\left (6-5 x+4 x^2-4 \log (x)\right )^3}-\frac {5 e^x x^2}{\left (6-5 x+4 x^2-4 \log (x)\right )^3}+\frac {8 e^x x^3}{\left (6-5 x+4 x^2-4 \log (x)\right )^3}\right ) \, dx\\ &=16 \int \frac {e^x x^2}{\left (6-5 x+4 x^2-4 \log (x)\right )^2} \, dx+32 \int \frac {e^x x}{\left (6-5 x+4 x^2-4 \log (x)\right )^2} \, dx+128 \int \frac {e^x x}{\left (6-5 x+4 x^2-4 \log (x)\right )^3} \, dx+160 \int \frac {e^x x^2}{\left (6-5 x+4 x^2-4 \log (x)\right )^3} \, dx-256 \int \frac {e^x x^3}{\left (6-5 x+4 x^2-4 \log (x)\right )^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.46, size = 24, normalized size = 0.89 \begin {gather*} \frac {16 e^x x^2}{\left (-6+5 x-4 x^2+4 \log (x)\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^x*(-320*x - 96*x^2 + 208*x^3 - 64*x^4) + E^x*(128*x + 64*x^2)*Log[x])/(-216 + 540*x - 882*x^2 + 8
45*x^3 - 588*x^4 + 240*x^5 - 64*x^6 + (432 - 720*x + 876*x^2 - 480*x^3 + 192*x^4)*Log[x] + (-288 + 240*x - 192
*x^2)*Log[x]^2 + 64*Log[x]^3),x]

[Out]

(16*E^x*x^2)/(-6 + 5*x - 4*x^2 + 4*Log[x])^2

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fricas [A]  time = 0.74, size = 49, normalized size = 1.81 \begin {gather*} \frac {16 \, x^{2} e^{x}}{16 \, x^{4} - 40 \, x^{3} + 73 \, x^{2} - 8 \, {\left (4 \, x^{2} - 5 \, x + 6\right )} \log \relax (x) + 16 \, \log \relax (x)^{2} - 60 \, x + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^2+128*x)*exp(x)*log(x)+(-64*x^4+208*x^3-96*x^2-320*x)*exp(x))/(64*log(x)^3+(-192*x^2+240*x-28
8)*log(x)^2+(192*x^4-480*x^3+876*x^2-720*x+432)*log(x)-64*x^6+240*x^5-588*x^4+845*x^3-882*x^2+540*x-216),x, al
gorithm="fricas")

[Out]

16*x^2*e^x/(16*x^4 - 40*x^3 + 73*x^2 - 8*(4*x^2 - 5*x + 6)*log(x) + 16*log(x)^2 - 60*x + 36)

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giac [B]  time = 0.23, size = 51, normalized size = 1.89 \begin {gather*} \frac {16 \, x^{2} e^{x}}{16 \, x^{4} - 40 \, x^{3} - 32 \, x^{2} \log \relax (x) + 73 \, x^{2} + 40 \, x \log \relax (x) + 16 \, \log \relax (x)^{2} - 60 \, x - 48 \, \log \relax (x) + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^2+128*x)*exp(x)*log(x)+(-64*x^4+208*x^3-96*x^2-320*x)*exp(x))/(64*log(x)^3+(-192*x^2+240*x-28
8)*log(x)^2+(192*x^4-480*x^3+876*x^2-720*x+432)*log(x)-64*x^6+240*x^5-588*x^4+845*x^3-882*x^2+540*x-216),x, al
gorithm="giac")

[Out]

16*x^2*e^x/(16*x^4 - 40*x^3 - 32*x^2*log(x) + 73*x^2 + 40*x*log(x) + 16*log(x)^2 - 60*x - 48*log(x) + 36)

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maple [A]  time = 0.03, size = 24, normalized size = 0.89

method	result	size
risch	\(\frac {16 x^{2} {\mathrm e}^{x}}{\left (4 x^{2}-5 x -4 \ln \relax (x )+6\right )^{2}}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((64*x^2+128*x)*exp(x)*ln(x)+(-64*x^4+208*x^3-96*x^2-320*x)*exp(x))/(64*ln(x)^3+(-192*x^2+240*x-288)*ln(x)
^2+(192*x^4-480*x^3+876*x^2-720*x+432)*ln(x)-64*x^6+240*x^5-588*x^4+845*x^3-882*x^2+540*x-216),x,method=_RETUR
NVERBOSE)

[Out]

16*x^2*exp(x)/(4*x^2-5*x-4*ln(x)+6)^2

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maxima [A]  time = 0.85, size = 49, normalized size = 1.81 \begin {gather*} \frac {16 \, x^{2} e^{x}}{16 \, x^{4} - 40 \, x^{3} + 73 \, x^{2} - 8 \, {\left (4 \, x^{2} - 5 \, x + 6\right )} \log \relax (x) + 16 \, \log \relax (x)^{2} - 60 \, x + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x^2+128*x)*exp(x)*log(x)+(-64*x^4+208*x^3-96*x^2-320*x)*exp(x))/(64*log(x)^3+(-192*x^2+240*x-28
8)*log(x)^2+(192*x^4-480*x^3+876*x^2-720*x+432)*log(x)-64*x^6+240*x^5-588*x^4+845*x^3-882*x^2+540*x-216),x, al
gorithm="maxima")

[Out]

16*x^2*e^x/(16*x^4 - 40*x^3 + 73*x^2 - 8*(4*x^2 - 5*x + 6)*log(x) + 16*log(x)^2 - 60*x + 36)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {{\mathrm {e}}^x\,\left (64\,x^4-208\,x^3+96\,x^2+320\,x\right )-{\mathrm {e}}^x\,\ln \relax (x)\,\left (64\,x^2+128\,x\right )}{540\,x-{\ln \relax (x)}^2\,\left (192\,x^2-240\,x+288\right )+\ln \relax (x)\,\left (192\,x^4-480\,x^3+876\,x^2-720\,x+432\right )+64\,{\ln \relax (x)}^3-882\,x^2+845\,x^3-588\,x^4+240\,x^5-64\,x^6-216} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(x)*(320*x + 96*x^2 - 208*x^3 + 64*x^4) - exp(x)*log(x)*(128*x + 64*x^2))/(540*x - log(x)^2*(192*x^2
- 240*x + 288) + log(x)*(876*x^2 - 720*x - 480*x^3 + 192*x^4 + 432) + 64*log(x)^3 - 882*x^2 + 845*x^3 - 588*x^
4 + 240*x^5 - 64*x^6 - 216),x)

[Out]

int(-(exp(x)*(320*x + 96*x^2 - 208*x^3 + 64*x^4) - exp(x)*log(x)*(128*x + 64*x^2))/(540*x - log(x)^2*(192*x^2
- 240*x + 288) + log(x)*(876*x^2 - 720*x - 480*x^3 + 192*x^4 + 432) + 64*log(x)^3 - 882*x^2 + 845*x^3 - 588*x^
4 + 240*x^5 - 64*x^6 - 216), x)

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sympy [B]  time = 0.40, size = 54, normalized size = 2.00 \begin {gather*} \frac {16 x^{2} e^{x}}{16 x^{4} - 40 x^{3} - 32 x^{2} \log {\relax (x )} + 73 x^{2} + 40 x \log {\relax (x )} - 60 x + 16 \log {\relax (x )}^{2} - 48 \log {\relax (x )} + 36} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((64*x**2+128*x)*exp(x)*ln(x)+(-64*x**4+208*x**3-96*x**2-320*x)*exp(x))/(64*ln(x)**3+(-192*x**2+240*
x-288)*ln(x)**2+(192*x**4-480*x**3+876*x**2-720*x+432)*ln(x)-64*x**6+240*x**5-588*x**4+845*x**3-882*x**2+540*x
-216),x)

[Out]

16*x**2*exp(x)/(16*x**4 - 40*x**3 - 32*x**2*log(x) + 73*x**2 + 40*x*log(x) - 60*x + 16*log(x)**2 - 48*log(x) +
36)

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