3.28.93 \(\int \frac {-24 x-8 e x-80 x^2+e^{\frac {4+x}{x}} (12+23 x+10 x^2+e (4+x))}{192 x+64 e x+e^{\frac {4+x}{x}} (-48 x-16 e x)+e^{\frac {2 (4+x)}{x}} (3 x+e x)} \, dx\)

Optimal. Leaf size=28 \[ 3+\frac {x+\frac {5 x^2}{3+e}}{-8+e^{\frac {4+x}{x}}} \]

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Rubi [F]  time = 1.38, 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 {-24 x-8 e x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{192 x+64 e x+e^{\frac {4+x}{x}} (-48 x-16 e x)+e^{\frac {2 (4+x)}{x}} (3 x+e x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-24*x - 8*E*x - 80*x^2 + E^((4 + x)/x)*(12 + 23*x + 10*x^2 + E*(4 + x)))/(192*x + 64*E*x + E^((4 + x)/x)*
(-48*x - 16*E*x) + E^((2*(4 + x))/x)*(3*x + E*x)),x]

[Out]

(160*Defer[Int][(-8 + E^(1 + 4/x))^(-2), x])/(3 + E) + ((23 + E)*Defer[Int][(-8 + E^(1 + 4/x))^(-1), x])/(3 +
E) + (10*Defer[Int][x/(-8 + E^(1 + 4/x)), x])/(3 + E) - 32*Defer[Subst][Defer[Int][1/((-8 + E^(1 + 4*x))^2*x),
x], x, x^(-1)] - 4*Defer[Subst][Defer[Int][1/((-8 + E^(1 + 4*x))*x), x], x, x^(-1)]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {-24 x-8 e x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{(192+64 e) x+e^{\frac {4+x}{x}} (-48 x-16 e x)+e^{\frac {2 (4+x)}{x}} (3 x+e x)} \, dx\\ &=\int \frac {(-24-8 e) x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{(192+64 e) x+e^{\frac {4+x}{x}} (-48 x-16 e x)+e^{\frac {2 (4+x)}{x}} (3 x+e x)} \, dx\\ &=\int \frac {(-24-8 e) x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{(3+e) \left (8-e^{1+\frac {4}{x}}\right )^2 x} \, dx\\ &=\frac {\int \frac {(-24-8 e) x-80 x^2+e^{\frac {4+x}{x}} \left (12+23 x+10 x^2+e (4+x)\right )}{\left (8-e^{1+\frac {4}{x}}\right )^2 x} \, dx}{3+e}\\ &=\frac {\int \left (\frac {32 (3+e+5 x)}{\left (-8+e^{1+\frac {4}{x}}\right )^2 x}+\frac {-4 (3+e)-(23+e) x-10 x^2}{\left (8-e^{1+\frac {4}{x}}\right ) x}\right ) \, dx}{3+e}\\ &=\frac {\int \frac {-4 (3+e)-(23+e) x-10 x^2}{\left (8-e^{1+\frac {4}{x}}\right ) x} \, dx}{3+e}+\frac {32 \int \frac {3+e+5 x}{\left (-8+e^{1+\frac {4}{x}}\right )^2 x} \, dx}{3+e}\\ &=\frac {\int \left (\frac {23 \left (1+\frac {e}{23}\right )}{-8+e^{1+\frac {4}{x}}}+\frac {4 (3+e)}{\left (-8+e^{1+\frac {4}{x}}\right ) x}+\frac {10 x}{-8+e^{1+\frac {4}{x}}}\right ) \, dx}{3+e}+\frac {32 \int \left (\frac {5}{\left (-8+e^{1+\frac {4}{x}}\right )^2}+\frac {3+e}{\left (-8+e^{1+\frac {4}{x}}\right )^2 x}\right ) \, dx}{3+e}\\ &=4 \int \frac {1}{\left (-8+e^{1+\frac {4}{x}}\right ) x} \, dx+32 \int \frac {1}{\left (-8+e^{1+\frac {4}{x}}\right )^2 x} \, dx+\frac {10 \int \frac {x}{-8+e^{1+\frac {4}{x}}} \, dx}{3+e}+\frac {160 \int \frac {1}{\left (-8+e^{1+\frac {4}{x}}\right )^2} \, dx}{3+e}+\frac {(23+e) \int \frac {1}{-8+e^{1+\frac {4}{x}}} \, dx}{3+e}\\ &=-\left (4 \operatorname {Subst}\left (\int \frac {1}{\left (-8+e^{1+4 x}\right ) x} \, dx,x,\frac {1}{x}\right )\right )-32 \operatorname {Subst}\left (\int \frac {1}{\left (-8+e^{1+4 x}\right )^2 x} \, dx,x,\frac {1}{x}\right )+\frac {10 \int \frac {x}{-8+e^{1+\frac {4}{x}}} \, dx}{3+e}+\frac {160 \int \frac {1}{\left (-8+e^{1+\frac {4}{x}}\right )^2} \, dx}{3+e}+\frac {(23+e) \int \frac {1}{-8+e^{1+\frac {4}{x}}} \, dx}{3+e}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.33, size = 26, normalized size = 0.93 \begin {gather*} \frac {x (3+e+5 x)}{(3+e) \left (-8+e^{1+\frac {4}{x}}\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24*x - 8*E*x - 80*x^2 + E^((4 + x)/x)*(12 + 23*x + 10*x^2 + E*(4 + x)))/(192*x + 64*E*x + E^((4 +
x)/x)*(-48*x - 16*E*x) + E^((2*(4 + x))/x)*(3*x + E*x)),x]

[Out]

(x*(3 + E + 5*x))/((3 + E)*(-8 + E^(1 + 4/x)))

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fricas [A]  time = 1.06, size = 35, normalized size = 1.25 \begin {gather*} \frac {5 \, x^{2} + x e + 3 \, x}{{\left (e + 3\right )} e^{\left (\frac {x + 4}{x}\right )} - 8 \, e - 24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4+x)*exp(1)+10*x^2+23*x+12)*exp((4+x)/x)-8*x*exp(1)-80*x^2-24*x)/((x*exp(1)+3*x)*exp((4+x)/x)^2+(
-16*x*exp(1)-48*x)*exp((4+x)/x)+64*x*exp(1)+192*x),x, algorithm="fricas")

[Out]

(5*x^2 + x*e + 3*x)/((e + 3)*e^((x + 4)/x) - 8*e - 24)

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giac [A]  time = 0.46, size = 56, normalized size = 2.00 \begin {gather*} -\frac {\frac {e}{x} + \frac {3}{x} + 5}{\frac {8 \, e}{x^{2}} - \frac {e^{\left (\frac {4}{x} + 2\right )}}{x^{2}} - \frac {3 \, e^{\left (\frac {4}{x} + 1\right )}}{x^{2}} + \frac {24}{x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4+x)*exp(1)+10*x^2+23*x+12)*exp((4+x)/x)-8*x*exp(1)-80*x^2-24*x)/((x*exp(1)+3*x)*exp((4+x)/x)^2+(
-16*x*exp(1)-48*x)*exp((4+x)/x)+64*x*exp(1)+192*x),x, algorithm="giac")

[Out]

-(e/x + 3/x + 5)/(8*e/x^2 - e^(4/x + 2)/x^2 - 3*e^(4/x + 1)/x^2 + 24/x^2)

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maple [A]  time = 0.47, size = 27, normalized size = 0.96

method	result	size
norman	\(\frac {\frac {5 x^{2}}{3+{\mathrm e}}+x}{{\mathrm e}^{\frac {4+x}{x}}-8}\)	\(27\)
risch	\(\frac {x \left ({\mathrm e}+5 x +3\right )}{\left (3+{\mathrm e}\right ) \left ({\mathrm e}^{\frac {4+x}{x}}-8\right )}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((((4+x)*exp(1)+10*x^2+23*x+12)*exp((4+x)/x)-8*x*exp(1)-80*x^2-24*x)/((x*exp(1)+3*x)*exp((4+x)/x)^2+(-16*x*
exp(1)-48*x)*exp((4+x)/x)+64*x*exp(1)+192*x),x,method=_RETURNVERBOSE)

[Out]

(5*x^2/(3+exp(1))+x)/(exp((4+x)/x)-8)

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maxima [A]  time = 0.46, size = 35, normalized size = 1.25 \begin {gather*} \frac {5 \, x^{2} + x {\left (e + 3\right )}}{{\left (e^{2} + 3 \, e\right )} e^{\frac {4}{x}} - 8 \, e - 24} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4+x)*exp(1)+10*x^2+23*x+12)*exp((4+x)/x)-8*x*exp(1)-80*x^2-24*x)/((x*exp(1)+3*x)*exp((4+x)/x)^2+(
-16*x*exp(1)-48*x)*exp((4+x)/x)+64*x*exp(1)+192*x),x, algorithm="maxima")

[Out]

(5*x^2 + x*(e + 3))/((e^2 + 3*e)*e^(4/x) - 8*e - 24)

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mupad [B]  time = 1.84, size = 27, normalized size = 0.96 \begin {gather*} \frac {x\,\left (5\,x+\mathrm {e}+3\right )}{\left ({\mathrm {e}}^{\frac {4}{x}+1}-8\right )\,\left (\mathrm {e}+3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(24*x + 8*x*exp(1) + 80*x^2 - exp((x + 4)/x)*(23*x + exp(1)*(x + 4) + 10*x^2 + 12))/(192*x + 64*x*exp(1)
+ exp((2*(x + 4))/x)*(3*x + x*exp(1)) - exp((x + 4)/x)*(48*x + 16*x*exp(1))),x)

[Out]

(x*(5*x + exp(1) + 3))/((exp(4/x + 1) - 8)*(exp(1) + 3))

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sympy [A]  time = 0.16, size = 31, normalized size = 1.11 \begin {gather*} \frac {5 x^{2} + e x + 3 x}{\left (e + 3\right ) e^{\frac {x + 4}{x}} - 24 - 8 e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((((4+x)*exp(1)+10*x**2+23*x+12)*exp((4+x)/x)-8*x*exp(1)-80*x**2-24*x)/((x*exp(1)+3*x)*exp((4+x)/x)**
2+(-16*x*exp(1)-48*x)*exp((4+x)/x)+64*x*exp(1)+192*x),x)

[Out]

(5*x**2 + E*x + 3*x)/((E + 3)*exp((x + 4)/x) - 24 - 8*E)

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