3.8.9 \(\int \frac {250+10 e^{2 x}-100 x^2+10 x^4+e^2 (-75 x+45 x^2-15 x^3+9 x^4)+e^x (-100+20 x^2+e^2 (15 x-24 x^2+9 x^3))}{-750 x+450 x^2+300 x^3-180 x^4-30 x^5+18 x^6+e^{2 x} (-30 x+18 x^2)+e^x (300 x-180 x^2-60 x^3+36 x^4)} \, dx\)

Optimal. Leaf size=35 \[ \frac {e^2 x}{2 \left (5-e^x-x^2\right )}+\frac {1}{3} \log \left (-3+\frac {5}{x}\right ) \]

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Rubi [F]  time = 1.59, 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 {250+10 e^{2 x}-100 x^2+10 x^4+e^2 \left (-75 x+45 x^2-15 x^3+9 x^4\right )+e^x \left (-100+20 x^2+e^2 \left (15 x-24 x^2+9 x^3\right )\right )}{-750 x+450 x^2+300 x^3-180 x^4-30 x^5+18 x^6+e^{2 x} \left (-30 x+18 x^2\right )+e^x \left (300 x-180 x^2-60 x^3+36 x^4\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(250 + 10*E^(2*x) - 100*x^2 + 10*x^4 + E^2*(-75*x + 45*x^2 - 15*x^3 + 9*x^4) + E^x*(-100 + 20*x^2 + E^2*(1
5*x - 24*x^2 + 9*x^3)))/(-750*x + 450*x^2 + 300*x^3 - 180*x^4 - 30*x^5 + 18*x^6 + E^(2*x)*(-30*x + 18*x^2) + E
^x*(300*x - 180*x^2 - 60*x^3 + 36*x^4)),x]

[Out]

Log[5 - 3*x]/3 - Log[x]/3 + (5*E^2*Defer[Int][x/(-5 + E^x + x^2)^2, x])/2 + E^2*Defer[Int][x^2/(-5 + E^x + x^2
)^2, x] - (E^2*Defer[Int][x^3/(-5 + E^x + x^2)^2, x])/2 - (E^2*Defer[Int][(-5 + E^x + x^2)^(-1), x])/2 + (E^2*
Defer[Int][x/(-5 + E^x + x^2), x])/2

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-10 e^{2 x}-20 e^x \left (-5+x^2\right )-10 \left (-5+x^2\right )^2-3 e^{2+x} x \left (5-8 x+3 x^2\right )-3 e^2 x \left (-25+15 x-5 x^2+3 x^3\right )}{6 (5-3 x) x \left (5-e^x-x^2\right )^2} \, dx\\ &=\frac {1}{6} \int \frac {-10 e^{2 x}-20 e^x \left (-5+x^2\right )-10 \left (-5+x^2\right )^2-3 e^{2+x} x \left (5-8 x+3 x^2\right )-3 e^2 x \left (-25+15 x-5 x^2+3 x^3\right )}{(5-3 x) x \left (5-e^x-x^2\right )^2} \, dx\\ &=\frac {1}{6} \int \left (\frac {10}{x (-5+3 x)}+\frac {3 e^2 (-1+x)}{-5+e^x+x^2}-\frac {3 e^2 x \left (-5-2 x+x^2\right )}{\left (-5+e^x+x^2\right )^2}\right ) \, dx\\ &=\frac {5}{3} \int \frac {1}{x (-5+3 x)} \, dx+\frac {1}{2} e^2 \int \frac {-1+x}{-5+e^x+x^2} \, dx-\frac {1}{2} e^2 \int \frac {x \left (-5-2 x+x^2\right )}{\left (-5+e^x+x^2\right )^2} \, dx\\ &=-\left (\frac {1}{3} \int \frac {1}{x} \, dx\right )-\frac {1}{2} e^2 \int \left (-\frac {5 x}{\left (-5+e^x+x^2\right )^2}-\frac {2 x^2}{\left (-5+e^x+x^2\right )^2}+\frac {x^3}{\left (-5+e^x+x^2\right )^2}\right ) \, dx+\frac {1}{2} e^2 \int \left (-\frac {1}{-5+e^x+x^2}+\frac {x}{-5+e^x+x^2}\right ) \, dx+\int \frac {1}{-5+3 x} \, dx\\ &=\frac {1}{3} \log (5-3 x)-\frac {\log (x)}{3}-\frac {1}{2} e^2 \int \frac {x^3}{\left (-5+e^x+x^2\right )^2} \, dx-\frac {1}{2} e^2 \int \frac {1}{-5+e^x+x^2} \, dx+\frac {1}{2} e^2 \int \frac {x}{-5+e^x+x^2} \, dx+e^2 \int \frac {x^2}{\left (-5+e^x+x^2\right )^2} \, dx+\frac {1}{2} \left (5 e^2\right ) \int \frac {x}{\left (-5+e^x+x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.06, size = 33, normalized size = 0.94 \begin {gather*} \frac {1}{6} \left (-\frac {3 e^2 x}{-5+e^x+x^2}+2 \log (5-3 x)-2 \log (x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(250 + 10*E^(2*x) - 100*x^2 + 10*x^4 + E^2*(-75*x + 45*x^2 - 15*x^3 + 9*x^4) + E^x*(-100 + 20*x^2 +
E^2*(15*x - 24*x^2 + 9*x^3)))/(-750*x + 450*x^2 + 300*x^3 - 180*x^4 - 30*x^5 + 18*x^6 + E^(2*x)*(-30*x + 18*x^
2) + E^x*(300*x - 180*x^2 - 60*x^3 + 36*x^4)),x]

[Out]

((-3*E^2*x)/(-5 + E^x + x^2) + 2*Log[5 - 3*x] - 2*Log[x])/6

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fricas [A]  time = 0.63, size = 43, normalized size = 1.23 \begin {gather*} -\frac {3 \, x e^{2} - 2 \, {\left (x^{2} + e^{x} - 5\right )} \log \left (3 \, x - 5\right ) + 2 \, {\left (x^{2} + e^{x} - 5\right )} \log \relax (x)}{6 \, {\left (x^{2} + e^{x} - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(x)^2+((9*x^3-24*x^2+15*x)*exp(2)+20*x^2-100)*exp(x)+(9*x^4-15*x^3+45*x^2-75*x)*exp(2)+10*x^4
-100*x^2+250)/((18*x^2-30*x)*exp(x)^2+(36*x^4-60*x^3-180*x^2+300*x)*exp(x)+18*x^6-30*x^5-180*x^4+300*x^3+450*x
^2-750*x),x, algorithm="fricas")

[Out]

-1/6*(3*x*e^2 - 2*(x^2 + e^x - 5)*log(3*x - 5) + 2*(x^2 + e^x - 5)*log(x))/(x^2 + e^x - 5)

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giac [B]  time = 0.54, size = 63, normalized size = 1.80 \begin {gather*} \frac {2 \, x^{2} \log \left (3 \, x - 5\right ) - 2 \, x^{2} \log \relax (x) - 3 \, x e^{2} + 2 \, e^{x} \log \left (3 \, x - 5\right ) - 2 \, e^{x} \log \relax (x) - 10 \, \log \left (3 \, x - 5\right ) + 10 \, \log \relax (x)}{6 \, {\left (x^{2} + e^{x} - 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(x)^2+((9*x^3-24*x^2+15*x)*exp(2)+20*x^2-100)*exp(x)+(9*x^4-15*x^3+45*x^2-75*x)*exp(2)+10*x^4
-100*x^2+250)/((18*x^2-30*x)*exp(x)^2+(36*x^4-60*x^3-180*x^2+300*x)*exp(x)+18*x^6-30*x^5-180*x^4+300*x^3+450*x
^2-750*x),x, algorithm="giac")

[Out]

1/6*(2*x^2*log(3*x - 5) - 2*x^2*log(x) - 3*x*e^2 + 2*e^x*log(3*x - 5) - 2*e^x*log(x) - 10*log(3*x - 5) + 10*lo
g(x))/(x^2 + e^x - 5)

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maple [A]  time = 0.19, size = 28, normalized size = 0.80

method	result	size
norman	\(-\frac {{\mathrm e}^{2} x}{2 \left (x^{2}+{\mathrm e}^{x}-5\right )}-\frac {\ln \relax (x )}{3}+\frac {\ln \left (3 x -5\right )}{3}\)	\(28\)
risch	\(-\frac {{\mathrm e}^{2} x}{2 \left (x^{2}+{\mathrm e}^{x}-5\right )}-\frac {\ln \relax (x )}{3}+\frac {\ln \left (3 x -5\right )}{3}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((10*exp(x)^2+((9*x^3-24*x^2+15*x)*exp(2)+20*x^2-100)*exp(x)+(9*x^4-15*x^3+45*x^2-75*x)*exp(2)+10*x^4-100*x
^2+250)/((18*x^2-30*x)*exp(x)^2+(36*x^4-60*x^3-180*x^2+300*x)*exp(x)+18*x^6-30*x^5-180*x^4+300*x^3+450*x^2-750
*x),x,method=_RETURNVERBOSE)

[Out]

-1/2*exp(2)*x/(x^2+exp(x)-5)-1/3*ln(x)+1/3*ln(3*x-5)

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maxima [A]  time = 0.71, size = 27, normalized size = 0.77 \begin {gather*} -\frac {x e^{2}}{2 \, {\left (x^{2} + e^{x} - 5\right )}} + \frac {1}{3} \, \log \left (3 \, x - 5\right ) - \frac {1}{3} \, \log \relax (x) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(x)^2+((9*x^3-24*x^2+15*x)*exp(2)+20*x^2-100)*exp(x)+(9*x^4-15*x^3+45*x^2-75*x)*exp(2)+10*x^4
-100*x^2+250)/((18*x^2-30*x)*exp(x)^2+(36*x^4-60*x^3-180*x^2+300*x)*exp(x)+18*x^6-30*x^5-180*x^4+300*x^3+450*x
^2-750*x),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 0.37, size = 52, normalized size = 1.49 \begin {gather*} -\frac {2\,\mathrm {atanh}\left (\frac {6\,x}{5}-1\right )}{3}-\frac {-{\mathrm {e}}^2\,x^3+2\,{\mathrm {e}}^2\,x^2+5\,{\mathrm {e}}^2\,x}{2\,\left ({\mathrm {e}}^x+x^2-5\right )\,\left (-x^2+2\,x+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*exp(2*x) + exp(x)*(exp(2)*(15*x - 24*x^2 + 9*x^3) + 20*x^2 - 100) - exp(2)*(75*x - 45*x^2 + 15*x^3 -
9*x^4) - 100*x^2 + 10*x^4 + 250)/(750*x + exp(2*x)*(30*x - 18*x^2) - exp(x)*(300*x - 180*x^2 - 60*x^3 + 36*x^4
) - 450*x^2 - 300*x^3 + 180*x^4 + 30*x^5 - 18*x^6),x)

[Out]

- (2*atanh((6*x)/5 - 1))/3 - (5*x*exp(2) + 2*x^2*exp(2) - x^3*exp(2))/(2*(exp(x) + x^2 - 5)*(2*x - x^2 + 5))

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sympy [A]  time = 0.20, size = 29, normalized size = 0.83 \begin {gather*} - \frac {x e^{2}}{2 x^{2} + 2 e^{x} - 10} - \frac {\log {\relax (x )}}{3} + \frac {\log {\left (x - \frac {5}{3} \right )}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((10*exp(x)**2+((9*x**3-24*x**2+15*x)*exp(2)+20*x**2-100)*exp(x)+(9*x**4-15*x**3+45*x**2-75*x)*exp(2)
+10*x**4-100*x**2+250)/((18*x**2-30*x)*exp(x)**2+(36*x**4-60*x**3-180*x**2+300*x)*exp(x)+18*x**6-30*x**5-180*x
**4+300*x**3+450*x**2-750*x),x)

[Out]

-x*exp(2)/(2*x**2 + 2*exp(x) - 10) - log(x)/3 + log(x - 5/3)/3

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