3.83.5 \(\int \frac {e^{4+x} (-150-50 x^2)+e^4 (300+100 x^2)+(-200 e^4 x^2+e^{4+x} (150 x+100 x^2+50 x^3)) \log (x)}{36 x+24 x^3+4 x^5+e^x (-36 x-24 x^3-4 x^5)+e^{2 x} (9 x+6 x^3+x^5)} \, dx\)

Optimal. Leaf size=25 \[ \frac {25 e^4 \log (x)}{\left (1-\frac {e^x}{2}\right ) \left (3+x^2\right )} \]

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Rubi [F]  time = 8.92, 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^{4+x} \left (-150-50 x^2\right )+e^4 \left (300+100 x^2\right )+\left (-200 e^4 x^2+e^{4+x} \left (150 x+100 x^2+50 x^3\right )\right ) \log (x)}{36 x+24 x^3+4 x^5+e^x \left (-36 x-24 x^3-4 x^5\right )+e^{2 x} \left (9 x+6 x^3+x^5\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(4 + x)*(-150 - 50*x^2) + E^4*(300 + 100*x^2) + (-200*E^4*x^2 + E^(4 + x)*(150*x + 100*x^2 + 50*x^3))*L
og[x])/(36*x + 24*x^3 + 4*x^5 + E^x*(-36*x - 24*x^3 - 4*x^5) + E^(2*x)*(9*x + 6*x^3 + x^5)),x]

[Out]

((50*I)*E^4*Log[x]*Defer[Int][1/((-2 + E^x)^2*(I*Sqrt[3] - x)), x])/Sqrt[3] - (25*E^4*Defer[Int][1/((-2 + E^x)
*(I*Sqrt[3] - x)), x])/3 + ((25*I)*E^4*Log[x]*Defer[Int][1/((-2 + E^x)*(I*Sqrt[3] - x)), x])/Sqrt[3] - (50*E^4
*Defer[Int][1/((-2 + E^x)*x), x])/3 + ((50*I)*E^4*Log[x]*Defer[Int][1/((-2 + E^x)^2*(I*Sqrt[3] + x)), x])/Sqrt
[3] + (25*E^4*Defer[Int][1/((-2 + E^x)*(I*Sqrt[3] + x)), x])/3 + ((25*I)*E^4*Log[x]*Defer[Int][1/((-2 + E^x)*(
I*Sqrt[3] + x)), x])/Sqrt[3] + 100*E^4*Log[x]*Defer[Int][x/((-2 + E^x)*(3 + x^2)^2), x] - ((50*I)*E^4*Defer[In
t][Defer[Int][1/((-2 + E^x)^2*(I*Sqrt[3] - x)), x]/x, x])/Sqrt[3] - ((25*I)*E^4*Defer[Int][Defer[Int][1/((-2 +
E^x)*(I*Sqrt[3] - x)), x]/x, x])/Sqrt[3] - ((50*I)*E^4*Defer[Int][Defer[Int][1/((-2 + E^x)^2*(I*Sqrt[3] + x))
, x]/x, x])/Sqrt[3] - ((25*I)*E^4*Defer[Int][Defer[Int][1/((-2 + E^x)*(I*Sqrt[3] + x)), x]/x, x])/Sqrt[3] - 10
0*E^4*Defer[Int][Defer[Int][x/((-2 + E^x)*(3 + x^2)^2), x]/x, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {50 e^4 \left (-\left (\left (-2+e^x\right ) \left (3+x^2\right )\right )+x \left (-4 x+e^x \left (3+2 x+x^2\right )\right ) \log (x)\right )}{\left (2-e^x\right )^2 x \left (3+x^2\right )^2} \, dx\\ &=\left (50 e^4\right ) \int \frac {-\left (\left (-2+e^x\right ) \left (3+x^2\right )\right )+x \left (-4 x+e^x \left (3+2 x+x^2\right )\right ) \log (x)}{\left (2-e^x\right )^2 x \left (3+x^2\right )^2} \, dx\\ &=\left (50 e^4\right ) \int \left (\frac {2 \log (x)}{\left (-2+e^x\right )^2 \left (3+x^2\right )}+\frac {-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)}{\left (-2+e^x\right ) x \left (3+x^2\right )^2}\right ) \, dx\\ &=\left (50 e^4\right ) \int \frac {-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)}{\left (-2+e^x\right ) x \left (3+x^2\right )^2} \, dx+\left (100 e^4\right ) \int \frac {\log (x)}{\left (-2+e^x\right )^2 \left (3+x^2\right )} \, dx\\ &=\left (50 e^4\right ) \int \left (\frac {-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)}{9 \left (-2+e^x\right ) x}-\frac {x \left (-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)\right )}{3 \left (-2+e^x\right ) \left (3+x^2\right )^2}-\frac {x \left (-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)\right )}{9 \left (-2+e^x\right ) \left (3+x^2\right )}\right ) \, dx-\left (100 e^4\right ) \int \frac {i \left (\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx+\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx\right )}{2 \sqrt {3} x} \, dx+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{\sqrt {3}}\\ &=\frac {1}{9} \left (50 e^4\right ) \int \frac {-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)}{\left (-2+e^x\right ) x} \, dx-\frac {1}{9} \left (50 e^4\right ) \int \frac {x \left (-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)\right )}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx-\frac {1}{3} \left (50 e^4\right ) \int \frac {x \left (-3-x^2+3 x \log (x)+2 x^2 \log (x)+x^3 \log (x)\right )}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx-\frac {\left (50 i e^4\right ) \int \frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx+\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{x} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{\sqrt {3}}\\ &=\frac {1}{9} \left (50 e^4\right ) \int \left (-\frac {3}{\left (-2+e^x\right ) x}-\frac {x}{-2+e^x}+\frac {3 \log (x)}{-2+e^x}+\frac {2 x \log (x)}{-2+e^x}+\frac {x^2 \log (x)}{-2+e^x}\right ) \, dx-\frac {1}{9} \left (50 e^4\right ) \int \left (-\frac {3 x}{\left (-2+e^x\right ) \left (3+x^2\right )}-\frac {x^3}{\left (-2+e^x\right ) \left (3+x^2\right )}+\frac {3 x^2 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )}+\frac {2 x^3 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )}+\frac {x^4 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )}\right ) \, dx-\frac {1}{3} \left (50 e^4\right ) \int \left (-\frac {3 x}{\left (-2+e^x\right ) \left (3+x^2\right )^2}-\frac {x^3}{\left (-2+e^x\right ) \left (3+x^2\right )^2}+\frac {3 x^2 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2}+\frac {2 x^3 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2}+\frac {x^4 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2}\right ) \, dx-\frac {\left (50 i e^4\right ) \int \left (\frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{x}+\frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{x}\right ) \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{\sqrt {3}}\\ &=-\left (\frac {1}{9} \left (50 e^4\right ) \int \frac {x}{-2+e^x} \, dx\right )+\frac {1}{9} \left (50 e^4\right ) \int \frac {x^3}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx+\frac {1}{9} \left (50 e^4\right ) \int \frac {x^2 \log (x)}{-2+e^x} \, dx-\frac {1}{9} \left (50 e^4\right ) \int \frac {x^4 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx+\frac {1}{9} \left (100 e^4\right ) \int \frac {x \log (x)}{-2+e^x} \, dx-\frac {1}{9} \left (100 e^4\right ) \int \frac {x^3 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx-\frac {1}{3} \left (50 e^4\right ) \int \frac {1}{\left (-2+e^x\right ) x} \, dx+\frac {1}{3} \left (50 e^4\right ) \int \frac {x^3}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx+\frac {1}{3} \left (50 e^4\right ) \int \frac {x}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx+\frac {1}{3} \left (50 e^4\right ) \int \frac {\log (x)}{-2+e^x} \, dx-\frac {1}{3} \left (50 e^4\right ) \int \frac {x^4 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx-\frac {1}{3} \left (50 e^4\right ) \int \frac {x^2 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )} \, dx-\frac {1}{3} \left (100 e^4\right ) \int \frac {x^3 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx+\left (50 e^4\right ) \int \frac {x}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx-\left (50 e^4\right ) \int \frac {x^2 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )^2} \, dx-\frac {\left (50 i e^4\right ) \int \frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{x} \, dx}{\sqrt {3}}-\frac {\left (50 i e^4\right ) \int \frac {\int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{x} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}-x\right )} \, dx}{\sqrt {3}}+\frac {\left (50 i e^4 \log (x)\right ) \int \frac {1}{\left (-2+e^x\right )^2 \left (i \sqrt {3}+x\right )} \, dx}{\sqrt {3}}\\ &=\text {Rest of rules removed due to large latex content} \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.14, size = 21, normalized size = 0.84 \begin {gather*} -\frac {50 e^4 \log (x)}{\left (-2+e^x\right ) \left (3+x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(4 + x)*(-150 - 50*x^2) + E^4*(300 + 100*x^2) + (-200*E^4*x^2 + E^(4 + x)*(150*x + 100*x^2 + 50*x
^3))*Log[x])/(36*x + 24*x^3 + 4*x^5 + E^x*(-36*x - 24*x^3 - 4*x^5) + E^(2*x)*(9*x + 6*x^3 + x^5)),x]

[Out]

(-50*E^4*Log[x])/((-2 + E^x)*(3 + x^2))

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fricas [A]  time = 0.77, size = 29, normalized size = 1.16 \begin {gather*} \frac {50 \, e^{8} \log \relax (x)}{2 \, {\left (x^{2} + 3\right )} e^{4} - {\left (x^{2} + 3\right )} e^{\left (x + 4\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x^3+100*x^2+150*x)*exp(4)*exp(x)-200*x^2*exp(4))*log(x)+(-50*x^2-150)*exp(4)*exp(x)+(100*x^2+3
00)*exp(4))/((x^5+6*x^3+9*x)*exp(x)^2+(-4*x^5-24*x^3-36*x)*exp(x)+4*x^5+24*x^3+36*x),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.22, size = 25, normalized size = 1.00 \begin {gather*} -\frac {50 \, e^{4} \log \relax (x)}{x^{2} e^{x} - 2 \, x^{2} + 3 \, e^{x} - 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x^3+100*x^2+150*x)*exp(4)*exp(x)-200*x^2*exp(4))*log(x)+(-50*x^2-150)*exp(4)*exp(x)+(100*x^2+3
00)*exp(4))/((x^5+6*x^3+9*x)*exp(x)^2+(-4*x^5-24*x^3-36*x)*exp(x)+4*x^5+24*x^3+36*x),x, algorithm="giac")

[Out]

-50*e^4*log(x)/(x^2*e^x - 2*x^2 + 3*e^x - 6)

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maple [A]  time = 0.05, size = 20, normalized size = 0.80

method	result	size
risch	\(-\frac {50 \,{\mathrm e}^{4} \ln \relax (x )}{\left (x^{2}+3\right ) \left ({\mathrm e}^{x}-2\right )}\)	\(20\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((50*x^3+100*x^2+150*x)*exp(4)*exp(x)-200*x^2*exp(4))*ln(x)+(-50*x^2-150)*exp(4)*exp(x)+(100*x^2+300)*exp
(4))/((x^5+6*x^3+9*x)*exp(x)^2+(-4*x^5-24*x^3-36*x)*exp(x)+4*x^5+24*x^3+36*x),x,method=_RETURNVERBOSE)

[Out]

-50*exp(4)/(x^2+3)/(exp(x)-2)*ln(x)

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maxima [A]  time = 0.42, size = 24, normalized size = 0.96 \begin {gather*} \frac {50 \, e^{4} \log \relax (x)}{2 \, x^{2} - {\left (x^{2} + 3\right )} e^{x} + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x^3+100*x^2+150*x)*exp(4)*exp(x)-200*x^2*exp(4))*log(x)+(-50*x^2-150)*exp(4)*exp(x)+(100*x^2+3
00)*exp(4))/((x^5+6*x^3+9*x)*exp(x)^2+(-4*x^5-24*x^3-36*x)*exp(x)+4*x^5+24*x^3+36*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 5.55, size = 19, normalized size = 0.76 \begin {gather*} -\frac {50\,{\mathrm {e}}^4\,\ln \relax (x)}{\left (x^2+3\right )\,\left ({\mathrm {e}}^x-2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(log(x)*(200*x^2*exp(4) - exp(4)*exp(x)*(150*x + 100*x^2 + 50*x^3)) - exp(4)*(100*x^2 + 300) + exp(4)*exp
(x)*(50*x^2 + 150))/(36*x + 24*x^3 + 4*x^5 + exp(2*x)*(9*x + 6*x^3 + x^5) - exp(x)*(36*x + 24*x^3 + 4*x^5)),x)

[Out]

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

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sympy [A]  time = 0.29, size = 24, normalized size = 0.96 \begin {gather*} - \frac {50 e^{4} \log {\relax (x )}}{- 2 x^{2} + \left (x^{2} + 3\right ) e^{x} - 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((50*x**3+100*x**2+150*x)*exp(4)*exp(x)-200*x**2*exp(4))*ln(x)+(-50*x**2-150)*exp(4)*exp(x)+(100*x*
*2+300)*exp(4))/((x**5+6*x**3+9*x)*exp(x)**2+(-4*x**5-24*x**3-36*x)*exp(x)+4*x**5+24*x**3+36*x),x)

[Out]

-50*exp(4)*log(x)/(-2*x**2 + (x**2 + 3)*exp(x) - 6)

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