3.40.78 \(\int \frac {12+e^x (3+3 x+x^2)}{16 x^2+8 e^x x^2+e^{2 x} x^2} \, dx\)

Optimal. Leaf size=21 \[ \frac {x-\frac {4 (3+x)}{4+e^x}}{4 x} \]

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Rubi [F]  time = 0.66, 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 {12+e^x \left (3+3 x+x^2\right )}{16 x^2+8 e^x x^2+e^{2 x} x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(12 + E^x*(3 + 3*x + x^2))/(16*x^2 + 8*E^x*x^2 + E^(2*x)*x^2),x]

[Out]

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {12+e^x \left (3+3 x+x^2\right )}{\left (4+e^x\right )^2 x^2} \, dx\\ &=\int \left (-\frac {4 (3+x)}{\left (4+e^x\right )^2 x}+\frac {3+3 x+x^2}{\left (4+e^x\right ) x^2}\right ) \, dx\\ &=-\left (4 \int \frac {3+x}{\left (4+e^x\right )^2 x} \, dx\right )+\int \frac {3+3 x+x^2}{\left (4+e^x\right ) x^2} \, dx\\ &=-\left (4 \int \left (\frac {1}{\left (4+e^x\right )^2}+\frac {3}{\left (4+e^x\right )^2 x}\right ) \, dx\right )+\int \left (\frac {1}{4+e^x}+\frac {3}{\left (4+e^x\right ) x^2}+\frac {3}{\left (4+e^x\right ) x}\right ) \, dx\\ &=3 \int \frac {1}{\left (4+e^x\right ) x^2} \, dx+3 \int \frac {1}{\left (4+e^x\right ) x} \, dx-4 \int \frac {1}{\left (4+e^x\right )^2} \, dx-12 \int \frac {1}{\left (4+e^x\right )^2 x} \, dx+\int \frac {1}{4+e^x} \, dx\\ &=3 \int \frac {1}{\left (4+e^x\right ) x^2} \, dx+3 \int \frac {1}{\left (4+e^x\right ) x} \, dx-4 \operatorname {Subst}\left (\int \frac {1}{x (4+x)^2} \, dx,x,e^x\right )-12 \int \frac {1}{\left (4+e^x\right )^2 x} \, dx+\operatorname {Subst}\left (\int \frac {1}{x (4+x)} \, dx,x,e^x\right )\\ &=\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{x} \, dx,x,e^x\right )-\frac {1}{4} \operatorname {Subst}\left (\int \frac {1}{4+x} \, dx,x,e^x\right )+3 \int \frac {1}{\left (4+e^x\right ) x^2} \, dx+3 \int \frac {1}{\left (4+e^x\right ) x} \, dx-4 \operatorname {Subst}\left (\int \left (\frac {1}{16 x}-\frac {1}{4 (4+x)^2}-\frac {1}{16 (4+x)}\right ) \, dx,x,e^x\right )-12 \int \frac {1}{\left (4+e^x\right )^2 x} \, dx\\ &=-\frac {1}{4+e^x}+3 \int \frac {1}{\left (4+e^x\right ) x^2} \, dx+3 \int \frac {1}{\left (4+e^x\right ) x} \, dx-12 \int \frac {1}{\left (4+e^x\right )^2 x} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.12, size = 16, normalized size = 0.76 \begin {gather*} \frac {-3-x}{\left (4+e^x\right ) x} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(12 + E^x*(3 + 3*x + x^2))/(16*x^2 + 8*E^x*x^2 + E^(2*x)*x^2),x]

[Out]

(-3 - x)/((4 + E^x)*x)

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fricas [A]  time = 0.62, size = 15, normalized size = 0.71 \begin {gather*} -\frac {x + 3}{x e^{x} + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+3*x+3)*exp(x)+12)/(exp(x)^2*x^2+8*exp(x)*x^2+16*x^2),x, algorithm="fricas")

[Out]

-(x + 3)/(x*e^x + 4*x)

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giac [A]  time = 0.16, size = 15, normalized size = 0.71 \begin {gather*} -\frac {x + 3}{x e^{x} + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+3*x+3)*exp(x)+12)/(exp(x)^2*x^2+8*exp(x)*x^2+16*x^2),x, algorithm="giac")

[Out]

-(x + 3)/(x*e^x + 4*x)

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maple [A]  time = 0.06, size = 15, normalized size = 0.71

method	result	size
risch	\(-\frac {3+x}{x \left ({\mathrm e}^{x}+4\right )}\)	\(15\)
norman	\(\frac {-3-x}{\left ({\mathrm e}^{x}+4\right ) x}\)	\(16\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((x^2+3*x+3)*exp(x)+12)/(exp(x)^2*x^2+8*exp(x)*x^2+16*x^2),x,method=_RETURNVERBOSE)

[Out]

-(3+x)/x/(exp(x)+4)

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maxima [A]  time = 0.40, size = 15, normalized size = 0.71 \begin {gather*} -\frac {x + 3}{x e^{x} + 4 \, x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x^2+3*x+3)*exp(x)+12)/(exp(x)^2*x^2+8*exp(x)*x^2+16*x^2),x, algorithm="maxima")

[Out]

-(x + 3)/(x*e^x + 4*x)

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mupad [B]  time = 0.08, size = 14, normalized size = 0.67 \begin {gather*} -\frac {x+3}{x\,\left ({\mathrm {e}}^x+4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x)*(3*x + x^2 + 3) + 12)/(8*x^2*exp(x) + x^2*exp(2*x) + 16*x^2),x)

[Out]

-(x + 3)/(x*(exp(x) + 4))

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sympy [A]  time = 0.10, size = 12, normalized size = 0.57 \begin {gather*} \frac {- x - 3}{x e^{x} + 4 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((x**2+3*x+3)*exp(x)+12)/(exp(x)**2*x**2+8*exp(x)*x**2+16*x**2),x)

[Out]

(-x - 3)/(x*exp(x) + 4*x)

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