3.29.8 \(\int \frac {e^{\frac {16+12 x^2+3 x^3+e^x (4 x^2+x^3)}{4 x^2+x^3}} (-256-96 x+e^x (32 x^3+16 x^4+2 x^5))}{16 x^3+8 x^4+x^5} \, dx\)

Optimal. Leaf size=23 \[ 3+2 \left (-3+e^{3+e^x+\frac {16}{x^2 (4+x)}}\right ) \]

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Rubi [F]  time = 6.86, 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 {\exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{4 x^2+x^3}\right ) \left (-256-96 x+e^x \left (32 x^3+16 x^4+2 x^5\right )\right )}{16 x^3+8 x^4+x^5} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^((16 + 12*x^2 + 3*x^3 + E^x*(4*x^2 + x^3))/(4*x^2 + x^3))*(-256 - 96*x + E^x*(32*x^3 + 16*x^4 + 2*x^5))
)/(16*x^3 + 8*x^4 + x^5),x]

[Out]

2*Defer[Int][E^(x + (16 + 12*x^2 + 3*x^3 + E^x*(4*x^2 + x^3))/(x^2*(4 + x))), x] - 16*Defer[Int][E^((16 + 12*x
^2 + 3*x^3 + E^x*(4*x^2 + x^3))/(x^2*(4 + x)))/x^3, x] + 2*Defer[Int][E^((16 + 12*x^2 + 3*x^3 + E^x*(4*x^2 + x
^3))/(x^2*(4 + x)))/x^2, x] - 2*Defer[Int][E^((16 + 12*x^2 + 3*x^3 + E^x*(4*x^2 + x^3))/(x^2*(4 + x)))/(4 + x)
^2, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {\exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{4 x^2+x^3}\right ) \left (-256-96 x+e^x \left (32 x^3+16 x^4+2 x^5\right )\right )}{x^3 \left (16+8 x+x^2\right )} \, dx\\ &=\int \frac {\exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{4 x^2+x^3}\right ) \left (-256-96 x+e^x \left (32 x^3+16 x^4+2 x^5\right )\right )}{x^3 (4+x)^2} \, dx\\ &=\int \frac {\exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right ) \left (-256-96 x+e^x \left (32 x^3+16 x^4+2 x^5\right )\right )}{x^3 (4+x)^2} \, dx\\ &=\int \left (2 \exp \left (x+\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right )-\frac {32 \exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right ) (8+3 x)}{x^3 (4+x)^2}\right ) \, dx\\ &=2 \int \exp \left (x+\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right ) \, dx-32 \int \frac {\exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right ) (8+3 x)}{x^3 (4+x)^2} \, dx\\ &=2 \int \exp \left (x+\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right ) \, dx-32 \int \left (\frac {\exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right )}{2 x^3}-\frac {\exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right )}{16 x^2}+\frac {\exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right )}{16 (4+x)^2}\right ) \, dx\\ &=2 \int \exp \left (x+\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right ) \, dx+2 \int \frac {\exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right )}{x^2} \, dx-2 \int \frac {\exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right )}{(4+x)^2} \, dx-16 \int \frac {\exp \left (\frac {16+12 x^2+3 x^3+e^x \left (4 x^2+x^3\right )}{x^2 (4+x)}\right )}{x^3} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.25, size = 24, normalized size = 1.04 \begin {gather*} 2 e^{3+e^x+\frac {4}{x^2}-\frac {1}{x}+\frac {1}{4+x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((16 + 12*x^2 + 3*x^3 + E^x*(4*x^2 + x^3))/(4*x^2 + x^3))*(-256 - 96*x + E^x*(32*x^3 + 16*x^4 + 2
*x^5)))/(16*x^3 + 8*x^4 + x^5),x]

[Out]

2*E^(3 + E^x + 4/x^2 - x^(-1) + (4 + x)^(-1))

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fricas [B]  time = 0.59, size = 39, normalized size = 1.70 \begin {gather*} 2 \, e^{\left (\frac {3 \, x^{3} + 12 \, x^{2} + {\left (x^{3} + 4 \, x^{2}\right )} e^{x} + 16}{x^{3} + 4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+16*x^4+32*x^3)*exp(x)-96*x-256)*exp(((x^3+4*x^2)*exp(x)+3*x^3+12*x^2+16)/(x^3+4*x^2))/(x^5+8
*x^4+16*x^3),x, algorithm="fricas")

[Out]

2*e^((3*x^3 + 12*x^2 + (x^3 + 4*x^2)*e^x + 16)/(x^3 + 4*x^2))

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giac [B]  time = 0.36, size = 84, normalized size = 3.65 \begin {gather*} 2 \, e^{\left (\frac {x^{3} e^{x}}{x^{3} + 4 \, x^{2}} + \frac {3 \, x^{3}}{x^{3} + 4 \, x^{2}} + \frac {4 \, x^{2} e^{x}}{x^{3} + 4 \, x^{2}} + \frac {12 \, x^{2}}{x^{3} + 4 \, x^{2}} + \frac {16}{x^{3} + 4 \, x^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+16*x^4+32*x^3)*exp(x)-96*x-256)*exp(((x^3+4*x^2)*exp(x)+3*x^3+12*x^2+16)/(x^3+4*x^2))/(x^5+8
*x^4+16*x^3),x, algorithm="giac")

[Out]

2*e^(x^3*e^x/(x^3 + 4*x^2) + 3*x^3/(x^3 + 4*x^2) + 4*x^2*e^x/(x^3 + 4*x^2) + 12*x^2/(x^3 + 4*x^2) + 16/(x^3 +
4*x^2))

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maple [A]  time = 0.12, size = 38, normalized size = 1.65

method	result	size
risch	\(2 \,{\mathrm e}^{\frac {{\mathrm e}^{x} x^{3}+4 \,{\mathrm e}^{x} x^{2}+3 x^{3}+12 x^{2}+16}{x^{2} \left (4+x \right )}}\)	\(38\)
norman	\(\frac {8 x^{2} {\mathrm e}^{\frac {\left (x^{3}+4 x^{2}\right ) {\mathrm e}^{x}+3 x^{3}+12 x^{2}+16}{x^{3}+4 x^{2}}}+2 x^{3} {\mathrm e}^{\frac {\left (x^{3}+4 x^{2}\right ) {\mathrm e}^{x}+3 x^{3}+12 x^{2}+16}{x^{3}+4 x^{2}}}}{x^{2} \left (4+x \right )}\)	\(95\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((2*x^5+16*x^4+32*x^3)*exp(x)-96*x-256)*exp(((x^3+4*x^2)*exp(x)+3*x^3+12*x^2+16)/(x^3+4*x^2))/(x^5+8*x^4+1
6*x^3),x,method=_RETURNVERBOSE)

[Out]

2*exp((exp(x)*x^3+4*exp(x)*x^2+3*x^3+12*x^2+16)/x^2/(4+x))

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maxima [A]  time = 0.56, size = 22, normalized size = 0.96 \begin {gather*} 2 \, e^{\left (\frac {1}{x + 4} - \frac {1}{x} + \frac {4}{x^{2}} + e^{x} + 3\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x^5+16*x^4+32*x^3)*exp(x)-96*x-256)*exp(((x^3+4*x^2)*exp(x)+3*x^3+12*x^2+16)/(x^3+4*x^2))/(x^5+8
*x^4+16*x^3),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.84, size = 53, normalized size = 2.30 \begin {gather*} 2\,{\mathrm {e}}^{\frac {x\,{\mathrm {e}}^x}{x+4}}\,{\mathrm {e}}^{\frac {3\,x}{x+4}}\,{\mathrm {e}}^{\frac {16}{x^3+4\,x^2}}\,{\mathrm {e}}^{\frac {4\,{\mathrm {e}}^x}{x+4}}\,{\mathrm {e}}^{\frac {12}{x+4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((exp(x)*(4*x^2 + x^3) + 12*x^2 + 3*x^3 + 16)/(4*x^2 + x^3))*(96*x - exp(x)*(32*x^3 + 16*x^4 + 2*x^5)
+ 256))/(16*x^3 + 8*x^4 + x^5),x)

[Out]

2*exp((x*exp(x))/(x + 4))*exp((3*x)/(x + 4))*exp(16/(4*x^2 + x^3))*exp((4*exp(x))/(x + 4))*exp(12/(x + 4))

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sympy [A]  time = 0.33, size = 34, normalized size = 1.48 \begin {gather*} 2 e^{\frac {3 x^{3} + 12 x^{2} + \left (x^{3} + 4 x^{2}\right ) e^{x} + 16}{x^{3} + 4 x^{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((2*x**5+16*x**4+32*x**3)*exp(x)-96*x-256)*exp(((x**3+4*x**2)*exp(x)+3*x**3+12*x**2+16)/(x**3+4*x**2
))/(x**5+8*x**4+16*x**3),x)

[Out]

2*exp((3*x**3 + 12*x**2 + (x**3 + 4*x**2)*exp(x) + 16)/(x**3 + 4*x**2))

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