3.54.49 \(\int \frac {-3 x^4-x^6+e^{2 x} (-630-225 x^2-675 \log (2))-3 x^4 \log (2)+e^x (81 x^2+3 x^3+30 x^4+90 x^2 \log (2))}{225 e^{2 x} x^2-30 e^x x^4+x^6} \, dx\)

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

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Rubi [F]  time = 1.39, 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 {-3 x^4-x^6+e^{2 x} \left (-630-225 x^2-675 \log (2)\right )-3 x^4 \log (2)+e^x \left (81 x^2+3 x^3+30 x^4+90 x^2 \log (2)\right )}{225 e^{2 x} x^2-30 e^x x^4+x^6} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3*x^4 - x^6 + E^(2*x)*(-630 - 225*x^2 - 675*Log[2]) - 3*x^4*Log[2] + E^x*(81*x^2 + 3*x^3 + 30*x^4 + 90*x
^2*Log[2]))/(225*E^(2*x)*x^2 - 30*E^x*x^4 + x^6),x]

[Out]

-x + (14 + 15*Log[2])/(5*x) - Defer[Int][(15*E^x - x^2)^(-1), x]/5 - (2*Defer[Int][x^2/(-15*E^x + x^2)^2, x])/
5 + Defer[Int][x^3/(-15*E^x + x^2)^2, x]/5 - Defer[Int][x/(-15*E^x + x^2), x]/5

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-x^6+e^{2 x} \left (-630-225 x^2-675 \log (2)\right )+x^4 (-3-3 \log (2))+e^x \left (81 x^2+3 x^3+30 x^4+90 x^2 \log (2)\right )}{225 e^{2 x} x^2-30 e^x x^4+x^6} \, dx\\ &=\int \frac {-x^6+e^{2 x} \left (-630-225 x^2-675 \log (2)\right )+x^4 (-3-3 \log (2))+e^x \left (81 x^2+3 x^3+30 x^4+90 x^2 \log (2)\right )}{x^2 \left (15 e^x-x^2\right )^2} \, dx\\ &=\int \left (\frac {(-2+x) x^2}{5 \left (-15 e^x+x^2\right )^2}-\frac {-1+x}{5 \left (-15 e^x+x^2\right )}+\frac {-14-5 x^2-15 \log (2)}{5 x^2}\right ) \, dx\\ &=\frac {1}{5} \int \frac {(-2+x) x^2}{\left (-15 e^x+x^2\right )^2} \, dx-\frac {1}{5} \int \frac {-1+x}{-15 e^x+x^2} \, dx+\frac {1}{5} \int \frac {-14-5 x^2-15 \log (2)}{x^2} \, dx\\ &=\frac {1}{5} \int \left (-\frac {2 x^2}{\left (-15 e^x+x^2\right )^2}+\frac {x^3}{\left (-15 e^x+x^2\right )^2}\right ) \, dx-\frac {1}{5} \int \left (\frac {1}{15 e^x-x^2}+\frac {x}{-15 e^x+x^2}\right ) \, dx+\frac {1}{5} \int \left (-5+\frac {-14-15 \log (2)}{x^2}\right ) \, dx\\ &=-x+\frac {14+15 \log (2)}{5 x}-\frac {1}{5} \int \frac {1}{15 e^x-x^2} \, dx+\frac {1}{5} \int \frac {x^3}{\left (-15 e^x+x^2\right )^2} \, dx-\frac {1}{5} \int \frac {x}{-15 e^x+x^2} \, dx-\frac {2}{5} \int \frac {x^2}{\left (-15 e^x+x^2\right )^2} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.32, size = 43, normalized size = 1.23 \begin {gather*} \frac {3 e^x \left (-14+5 x^2-15 \log (2)\right )+x^2 \left (3-x^2+\log (8)\right )}{-15 e^x x+x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3*x^4 - x^6 + E^(2*x)*(-630 - 225*x^2 - 675*Log[2]) - 3*x^4*Log[2] + E^x*(81*x^2 + 3*x^3 + 30*x^4
+ 90*x^2*Log[2]))/(225*E^(2*x)*x^2 - 30*E^x*x^4 + x^6),x]

[Out]

(3*E^x*(-14 + 5*x^2 - 15*Log[2]) + x^2*(3 - x^2 + Log[8]))/(-15*E^x*x + x^3)

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fricas [A]  time = 0.65, size = 44, normalized size = 1.26 \begin {gather*} -\frac {x^{4} - 3 \, x^{2} \log \relax (2) - 3 \, x^{2} - 3 \, {\left (5 \, x^{2} - 15 \, \log \relax (2) - 14\right )} e^{x}}{x^{3} - 15 \, x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-675*log(2)-225*x^2-630)*exp(x)^2+(90*x^2*log(2)+30*x^4+3*x^3+81*x^2)*exp(x)-3*x^4*log(2)-x^6-3*x^
4)/(225*exp(x)^2*x^2-30*exp(x)*x^4+x^6),x, algorithm="fricas")

[Out]

-(x^4 - 3*x^2*log(2) - 3*x^2 - 3*(5*x^2 - 15*log(2) - 14)*e^x)/(x^3 - 15*x*e^x)

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giac [A]  time = 0.14, size = 46, normalized size = 1.31 \begin {gather*} -\frac {x^{4} - 15 \, x^{2} e^{x} - 3 \, x^{2} \log \relax (2) - 3 \, x^{2} + 45 \, e^{x} \log \relax (2) + 42 \, e^{x}}{x^{3} - 15 \, x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-675*log(2)-225*x^2-630)*exp(x)^2+(90*x^2*log(2)+30*x^4+3*x^3+81*x^2)*exp(x)-3*x^4*log(2)-x^6-3*x^
4)/(225*exp(x)^2*x^2-30*exp(x)*x^4+x^6),x, algorithm="giac")

[Out]

-(x^4 - 15*x^2*e^x - 3*x^2*log(2) - 3*x^2 + 45*e^x*log(2) + 42*e^x)/(x^3 - 15*x*e^x)

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maple [A]  time = 0.08, size = 30, normalized size = 0.86

method	result	size
risch	\(-x +\frac {3 \ln \relax (2)}{x}+\frac {14}{5 x}+\frac {x}{5 x^{2}-75 \,{\mathrm e}^{x}}\)	\(30\)
norman	\(\frac {\left (-42-45 \ln \relax (2)\right ) {\mathrm e}^{x}+\left (3 \ln \relax (2)+3\right ) x^{2}-x^{4}+15 \,{\mathrm e}^{x} x^{2}}{x \left (x^{2}-15 \,{\mathrm e}^{x}\right )}\)	\(47\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-675*ln(2)-225*x^2-630)*exp(x)^2+(90*x^2*ln(2)+30*x^4+3*x^3+81*x^2)*exp(x)-3*x^4*ln(2)-x^6-3*x^4)/(225*e
xp(x)^2*x^2-30*exp(x)*x^4+x^6),x,method=_RETURNVERBOSE)

[Out]

-x+3*ln(2)/x+14/5/x+1/5*x/(x^2-15*exp(x))

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maxima [A]  time = 0.51, size = 41, normalized size = 1.17 \begin {gather*} -\frac {x^{4} - 3 \, x^{2} {\left (\log \relax (2) + 1\right )} - 3 \, {\left (5 \, x^{2} - 15 \, \log \relax (2) - 14\right )} e^{x}}{x^{3} - 15 \, x e^{x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-675*log(2)-225*x^2-630)*exp(x)^2+(90*x^2*log(2)+30*x^4+3*x^3+81*x^2)*exp(x)-3*x^4*log(2)-x^6-3*x^
4)/(225*exp(x)^2*x^2-30*exp(x)*x^4+x^6),x, algorithm="maxima")

[Out]

-(x^4 - 3*x^2*(log(2) + 1) - 3*(5*x^2 - 15*log(2) - 14)*e^x)/(x^3 - 15*x*e^x)

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mupad [B]  time = 3.59, size = 38, normalized size = 1.09 \begin {gather*} -x-\frac {x^2\,\left (\ln \relax (8)+3\right )-{\mathrm {e}}^x\,\left (45\,\ln \relax (2)+42\right )}{15\,x\,{\mathrm {e}}^x-x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(3*x^4*log(2) + 3*x^4 + x^6 - exp(x)*(90*x^2*log(2) + 81*x^2 + 3*x^3 + 30*x^4) + exp(2*x)*(675*log(2) + 2
25*x^2 + 630))/(225*x^2*exp(2*x) - 30*x^4*exp(x) + x^6),x)

[Out]

- x - (x^2*(log(8) + 3) - exp(x)*(45*log(2) + 42))/(15*x*exp(x) - x^3)

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sympy [A]  time = 0.16, size = 26, normalized size = 0.74 \begin {gather*} - x - \frac {x}{- 5 x^{2} + 75 e^{x}} - \frac {-14 - 15 \log {\relax (2 )}}{5 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-675*ln(2)-225*x**2-630)*exp(x)**2+(90*x**2*ln(2)+30*x**4+3*x**3+81*x**2)*exp(x)-3*x**4*ln(2)-x**6
-3*x**4)/(225*exp(x)**2*x**2-30*exp(x)*x**4+x**6),x)

[Out]

-x - x/(-5*x**2 + 75*exp(x)) - (-14 - 15*log(2))/(5*x)

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