3.16.68 \(\int \frac {-4-10 x-4 e^3 x+e^{2 e+2 x^2} (-1-3 x-e^3 x)+e^{e+x^2} (-4-11 x-4 e^3 x-10 x^2-2 x^3)}{12 x+4 e^3 x+e^{2 e+2 x^2} (3 x+e^3 x)+e^{e+x^2} (12 x+4 e^3 x)} \, dx\)

Optimal. Leaf size=32 \[ -x+\frac {\frac {5+x}{2+e^{e+x^2}}-\log (x)}{3+e^3} \]

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Rubi [F]  time = 1.48, 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 {-4-10 x-4 e^3 x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{12 x+4 e^3 x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-4 - 10*x - 4*E^3*x + E^(2*E + 2*x^2)*(-1 - 3*x - E^3*x) + E^(E + x^2)*(-4 - 11*x - 4*E^3*x - 10*x^2 - 2*
x^3))/(12*x + 4*E^3*x + E^(2*E + 2*x^2)*(3*x + E^3*x) + E^(E + x^2)*(12*x + 4*E^3*x)),x]

[Out]

5/((3 + E^3)*(2 + E^(E + x^2))) - x - Log[x]/(3 + E^3) + Defer[Int][(2 + E^(E + x^2))^(-1), x]/(3 + E^3) + (4*
Defer[Int][x^2/(2 + E^(E + x^2))^2, x])/(3 + E^3) - (2*Defer[Int][x^2/(2 + E^(E + x^2)), x])/(3 + E^3)

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-4-10 x-4 e^3 x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{\left (12+4 e^3\right ) x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx\\ &=\int \frac {-4+\left (-10-4 e^3\right ) x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{\left (12+4 e^3\right ) x+e^{2 e+2 x^2} \left (3 x+e^3 x\right )+e^{e+x^2} \left (12 x+4 e^3 x\right )} \, dx\\ &=\int \frac {-4+\left (-10-4 e^3\right ) x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{\left (3+e^3\right ) \left (2+e^{e+x^2}\right )^2 x} \, dx\\ &=\frac {\int \frac {-4+\left (-10-4 e^3\right ) x+e^{2 e+2 x^2} \left (-1-3 x-e^3 x\right )+e^{e+x^2} \left (-4-11 x-4 e^3 x-10 x^2-2 x^3\right )}{\left (2+e^{e+x^2}\right )^2 x} \, dx}{3+e^3}\\ &=\frac {\int \left (\frac {4 x (5+x)}{\left (2+e^{e+x^2}\right )^2}+\frac {-1-\left (3+e^3\right ) x}{x}-\frac {-1+10 x+2 x^2}{2+e^{e+x^2}}\right ) \, dx}{3+e^3}\\ &=\frac {\int \frac {-1-\left (3+e^3\right ) x}{x} \, dx}{3+e^3}-\frac {\int \frac {-1+10 x+2 x^2}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {4 \int \frac {x (5+x)}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}\\ &=\frac {\int \left (-3-e^3-\frac {1}{x}\right ) \, dx}{3+e^3}-\frac {\int \left (-\frac {1}{2+e^{e+x^2}}+\frac {10 x}{2+e^{e+x^2}}+\frac {2 x^2}{2+e^{e+x^2}}\right ) \, dx}{3+e^3}+\frac {4 \int \left (\frac {5 x}{\left (2+e^{e+x^2}\right )^2}+\frac {x^2}{\left (2+e^{e+x^2}\right )^2}\right ) \, dx}{3+e^3}\\ &=-x-\frac {\log (x)}{3+e^3}+\frac {\int \frac {1}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {2 \int \frac {x^2}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {4 \int \frac {x^2}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}-\frac {10 \int \frac {x}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {20 \int \frac {x}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}\\ &=-x-\frac {\log (x)}{3+e^3}+\frac {\int \frac {1}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {2 \int \frac {x^2}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {4 \int \frac {x^2}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{2+e^{e+x}} \, dx,x,x^2\right )}{3+e^3}+\frac {10 \operatorname {Subst}\left (\int \frac {1}{\left (2+e^{e+x}\right )^2} \, dx,x,x^2\right )}{3+e^3}\\ &=-x-\frac {\log (x)}{3+e^3}+\frac {\int \frac {1}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {2 \int \frac {x^2}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {4 \int \frac {x^2}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{x (2+x)} \, dx,x,e^{e+x^2}\right )}{3+e^3}+\frac {10 \operatorname {Subst}\left (\int \frac {1}{x (2+x)^2} \, dx,x,e^{e+x^2}\right )}{3+e^3}\\ &=-x-\frac {\log (x)}{3+e^3}+\frac {\int \frac {1}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {2 \int \frac {x^2}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {5 \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^{e+x^2}\right )}{2 \left (3+e^3\right )}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{2+x} \, dx,x,e^{e+x^2}\right )}{2 \left (3+e^3\right )}+\frac {4 \int \frac {x^2}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}+\frac {10 \operatorname {Subst}\left (\int \left (\frac {1}{4 x}-\frac {1}{2 (2+x)^2}-\frac {1}{4 (2+x)}\right ) \, dx,x,e^{e+x^2}\right )}{3+e^3}\\ &=\frac {5}{\left (3+e^3\right ) \left (2+e^{e+x^2}\right )}-x-\frac {\log (x)}{3+e^3}+\frac {\int \frac {1}{2+e^{e+x^2}} \, dx}{3+e^3}-\frac {2 \int \frac {x^2}{2+e^{e+x^2}} \, dx}{3+e^3}+\frac {4 \int \frac {x^2}{\left (2+e^{e+x^2}\right )^2} \, dx}{3+e^3}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.11, size = 36, normalized size = 1.12 \begin {gather*} -\frac {\frac {-5-x}{2+e^{e+x^2}}+\left (3+e^3\right ) x+\log (x)}{3+e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-4 - 10*x - 4*E^3*x + E^(2*E + 2*x^2)*(-1 - 3*x - E^3*x) + E^(E + x^2)*(-4 - 11*x - 4*E^3*x - 10*x^
2 - 2*x^3))/(12*x + 4*E^3*x + E^(2*E + 2*x^2)*(3*x + E^3*x) + E^(E + x^2)*(12*x + 4*E^3*x)),x]

[Out]

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

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fricas [A]  time = 0.87, size = 60, normalized size = 1.88 \begin {gather*} -\frac {2 \, x e^{3} + {\left (x e^{3} + 3 \, x\right )} e^{\left (x^{2} + e\right )} + {\left (e^{\left (x^{2} + e\right )} + 2\right )} \log \relax (x) + 5 \, x - 5}{{\left (e^{3} + 3\right )} e^{\left (x^{2} + e\right )} + 2 \, e^{3} + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(3)-3*x-1)*exp(exp(1)+x^2)^2+(-4*x*exp(3)-2*x^3-10*x^2-11*x-4)*exp(exp(1)+x^2)-4*x*exp(3)-10
*x-4)/((x*exp(3)+3*x)*exp(exp(1)+x^2)^2+(4*x*exp(3)+12*x)*exp(exp(1)+x^2)+4*x*exp(3)+12*x),x, algorithm="frica
s")

[Out]

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

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giac [B]  time = 0.22, size = 71, normalized size = 2.22 \begin {gather*} -\frac {2 \, x e^{3} + x e^{\left (x^{2} + e + 3\right )} + 3 \, x e^{\left (x^{2} + e\right )} + e^{\left (x^{2} + e\right )} \log \relax (x) + 5 \, x + 2 \, \log \relax (x) - 5}{2 \, e^{3} + e^{\left (x^{2} + e + 3\right )} + 3 \, e^{\left (x^{2} + e\right )} + 6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(3)-3*x-1)*exp(exp(1)+x^2)^2+(-4*x*exp(3)-2*x^3-10*x^2-11*x-4)*exp(exp(1)+x^2)-4*x*exp(3)-10
*x-4)/((x*exp(3)+3*x)*exp(exp(1)+x^2)^2+(4*x*exp(3)+12*x)*exp(exp(1)+x^2)+4*x*exp(3)+12*x),x, algorithm="giac"
)

[Out]

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

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maple [A]  time = 0.13, size = 36, normalized size = 1.12

method	result	size
risch	\(-x -\frac {\ln \relax (x )}{{\mathrm e}^{3}+3}+\frac {5+x}{\left ({\mathrm e}^{3}+3\right ) \left (2+{\mathrm e}^{{\mathrm e}+x^{2}}\right )}\)	\(36\)
norman	\(\frac {-x \,{\mathrm e}^{{\mathrm e}+x^{2}}-\frac {\left (5+2 \,{\mathrm e}^{3}\right ) x}{{\mathrm e}^{3}+3}+\frac {5}{{\mathrm e}^{3}+3}}{2+{\mathrm e}^{{\mathrm e}+x^{2}}}-\frac {\ln \relax (x )}{{\mathrm e}^{3}+3}\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-x*exp(3)-3*x-1)*exp(exp(1)+x^2)^2+(-4*x*exp(3)-2*x^3-10*x^2-11*x-4)*exp(exp(1)+x^2)-4*x*exp(3)-10*x-4)/
((x*exp(3)+3*x)*exp(exp(1)+x^2)^2+(4*x*exp(3)+12*x)*exp(exp(1)+x^2)+4*x*exp(3)+12*x),x,method=_RETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.52, size = 64, normalized size = 2.00 \begin {gather*} -\frac {x {\left (e^{\left (e + 3\right )} + 3 \, e^{e}\right )} e^{\left (x^{2}\right )} + x {\left (2 \, e^{3} + 5\right )} - 5}{{\left (e^{\left (e + 3\right )} + 3 \, e^{e}\right )} e^{\left (x^{2}\right )} + 2 \, e^{3} + 6} - \frac {\log \relax (x)}{e^{3} + 3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(3)-3*x-1)*exp(exp(1)+x^2)^2+(-4*x*exp(3)-2*x^3-10*x^2-11*x-4)*exp(exp(1)+x^2)-4*x*exp(3)-10
*x-4)/((x*exp(3)+3*x)*exp(exp(1)+x^2)^2+(4*x*exp(3)+12*x)*exp(exp(1)+x^2)+4*x*exp(3)+12*x),x, algorithm="maxim
a")

[Out]

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

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mupad [B]  time = 1.76, size = 76, normalized size = 2.38 \begin {gather*} -\frac {3\,x+\ln \relax (x)+x\,{\mathrm {e}}^3}{{\mathrm {e}}^3+3}-\frac {5\,x+2\,\ln \relax (x)+2\,x\,{\mathrm {e}}^3-\frac {\left (2\,{\mathrm {e}}^3+6\right )\,\left (3\,x+\ln \relax (x)+x\,{\mathrm {e}}^3+\frac {5}{2}\right )}{{\mathrm {e}}^3+3}}{\left ({\mathrm {e}}^{x^2+\mathrm {e}}+2\right )\,\left ({\mathrm {e}}^3+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(10*x + 4*x*exp(3) + exp(exp(1) + x^2)*(11*x + 4*x*exp(3) + 10*x^2 + 2*x^3 + 4) + exp(2*exp(1) + 2*x^2)*(
3*x + x*exp(3) + 1) + 4)/(12*x + 4*x*exp(3) + exp(2*exp(1) + 2*x^2)*(3*x + x*exp(3)) + exp(exp(1) + x^2)*(12*x
+ 4*x*exp(3))),x)

[Out]

- (3*x + log(x) + x*exp(3))/(exp(3) + 3) - (5*x + 2*log(x) + 2*x*exp(3) - ((2*exp(3) + 6)*(3*x + log(x) + x*ex
p(3) + 5/2))/(exp(3) + 3))/((exp(exp(1) + x^2) + 2)*(exp(3) + 3))

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sympy [A]  time = 0.19, size = 39, normalized size = 1.22 \begin {gather*} \frac {x + 5}{\left (3 + e^{3}\right ) e^{x^{2} + e} + 6 + 2 e^{3}} + \frac {- x \left (3 + e^{3}\right ) - \log {\relax (x )}}{3 + e^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-x*exp(3)-3*x-1)*exp(exp(1)+x**2)**2+(-4*x*exp(3)-2*x**3-10*x**2-11*x-4)*exp(exp(1)+x**2)-4*x*exp(
3)-10*x-4)/((x*exp(3)+3*x)*exp(exp(1)+x**2)**2+(4*x*exp(3)+12*x)*exp(exp(1)+x**2)+4*x*exp(3)+12*x),x)

[Out]

(x + 5)/((3 + exp(3))*exp(x**2 + E) + 6 + 2*exp(3)) + (-x*(3 + exp(3)) - log(x))/(3 + exp(3))

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