3.58.46 \(\int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} (16+16 x+28 x^2+19 x^3+4 x^4+(8+4 x+6 x^2+3 x^3) \log (2+x)+\log ^2(2+x))}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx\)

Optimal. Leaf size=21 \[ e^{\frac {x^3}{4+2 x+\log (2+x)}} (2+x) \]

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Rubi [F]  time = 3.56, 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 {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{16+16 x+4 x^2+(8+4 x) \log (2+x)+\log ^2(2+x)} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(E^(x^3/(4 + 2*x + Log[2 + x]))*(16 + 16*x + 28*x^2 + 19*x^3 + 4*x^4 + (8 + 4*x + 6*x^2 + 3*x^3)*Log[2 + x
] + Log[2 + x]^2))/(16 + 16*x + 4*x^2 + (8 + 4*x)*Log[2 + x] + Log[2 + x]^2),x]

[Out]

Defer[Int][E^(x^3/(4 + 2*x + Log[2 + x])), x] - 5*Defer[Int][(E^(x^3/(4 + 2*x + Log[2 + x]))*x^3)/(4 + 2*x + L
og[2 + x])^2, x] - 2*Defer[Int][(E^(x^3/(4 + 2*x + Log[2 + x]))*x^4)/(4 + 2*x + Log[2 + x])^2, x] + 6*Defer[In
t][(E^(x^3/(4 + 2*x + Log[2 + x]))*x^2)/(4 + 2*x + Log[2 + x]), x] + 3*Defer[Int][(E^(x^3/(4 + 2*x + Log[2 + x
]))*x^3)/(4 + 2*x + Log[2 + x]), x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} \left (16+16 x+28 x^2+19 x^3+4 x^4+\left (8+4 x+6 x^2+3 x^3\right ) \log (2+x)+\log ^2(2+x)\right )}{(4+2 x+\log (2+x))^2} \, dx\\ &=\int \left (e^{\frac {x^3}{4+2 x+\log (2+x)}}-\frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3 (5+2 x)}{(4+2 x+\log (2+x))^2}+\frac {3 e^{\frac {x^3}{4+2 x+\log (2+x)}} x^2 (2+x)}{4+2 x+\log (2+x)}\right ) \, dx\\ &=3 \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^2 (2+x)}{4+2 x+\log (2+x)} \, dx+\int e^{\frac {x^3}{4+2 x+\log (2+x)}} \, dx-\int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3 (5+2 x)}{(4+2 x+\log (2+x))^2} \, dx\\ &=3 \int \left (\frac {2 e^{\frac {x^3}{4+2 x+\log (2+x)}} x^2}{4+2 x+\log (2+x)}+\frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3}{4+2 x+\log (2+x)}\right ) \, dx+\int e^{\frac {x^3}{4+2 x+\log (2+x)}} \, dx-\int \left (\frac {5 e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3}{(4+2 x+\log (2+x))^2}+\frac {2 e^{\frac {x^3}{4+2 x+\log (2+x)}} x^4}{(4+2 x+\log (2+x))^2}\right ) \, dx\\ &=-\left (2 \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^4}{(4+2 x+\log (2+x))^2} \, dx\right )+3 \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3}{4+2 x+\log (2+x)} \, dx-5 \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^3}{(4+2 x+\log (2+x))^2} \, dx+6 \int \frac {e^{\frac {x^3}{4+2 x+\log (2+x)}} x^2}{4+2 x+\log (2+x)} \, dx+\int e^{\frac {x^3}{4+2 x+\log (2+x)}} \, dx\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 1.20, size = 21, normalized size = 1.00 \begin {gather*} e^{\frac {x^3}{4+2 x+\log (2+x)}} (2+x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^(x^3/(4 + 2*x + Log[2 + x]))*(16 + 16*x + 28*x^2 + 19*x^3 + 4*x^4 + (8 + 4*x + 6*x^2 + 3*x^3)*Log
[2 + x] + Log[2 + x]^2))/(16 + 16*x + 4*x^2 + (8 + 4*x)*Log[2 + x] + Log[2 + x]^2),x]

[Out]

E^(x^3/(4 + 2*x + Log[2 + x]))*(2 + x)

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fricas [A]  time = 0.58, size = 20, normalized size = 0.95 \begin {gather*} {\left (x + 2\right )} e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2+x)^2+(3*x^3+6*x^2+4*x+8)*log(2+x)+4*x^4+19*x^3+28*x^2+16*x+16)*exp(x^3/(log(2+x)+2*x+4))/(log
(2+x)^2+(4*x+8)*log(2+x)+4*x^2+16*x+16),x, algorithm="fricas")

[Out]

(x + 2)*e^(x^3/(2*x + log(x + 2) + 4))

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giac [A]  time = 0.27, size = 37, normalized size = 1.76 \begin {gather*} x e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )} + 2 \, e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2+x)^2+(3*x^3+6*x^2+4*x+8)*log(2+x)+4*x^4+19*x^3+28*x^2+16*x+16)*exp(x^3/(log(2+x)+2*x+4))/(log
(2+x)^2+(4*x+8)*log(2+x)+4*x^2+16*x+16),x, algorithm="giac")

[Out]

x*e^(x^3/(2*x + log(x + 2) + 4)) + 2*e^(x^3/(2*x + log(x + 2) + 4))

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maple [A]  time = 0.03, size = 21, normalized size = 1.00

method	result	size
risch	\(\left (2+x \right ) {\mathrm e}^{\frac {x^{3}}{\ln \left (2+x \right )+2 x +4}}\)	\(21\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((ln(2+x)^2+(3*x^3+6*x^2+4*x+8)*ln(2+x)+4*x^4+19*x^3+28*x^2+16*x+16)*exp(x^3/(ln(2+x)+2*x+4))/(ln(2+x)^2+(4
*x+8)*ln(2+x)+4*x^2+16*x+16),x,method=_RETURNVERBOSE)

[Out]

(2+x)*exp(x^3/(ln(2+x)+2*x+4))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {{\left (4 \, x^{4} + 19 \, x^{3} + 28 \, x^{2} + {\left (3 \, x^{3} + 6 \, x^{2} + 4 \, x + 8\right )} \log \left (x + 2\right ) + \log \left (x + 2\right )^{2} + 16 \, x + 16\right )} e^{\left (\frac {x^{3}}{2 \, x + \log \left (x + 2\right ) + 4}\right )}}{4 \, x^{2} + 4 \, {\left (x + 2\right )} \log \left (x + 2\right ) + \log \left (x + 2\right )^{2} + 16 \, x + 16}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((log(2+x)^2+(3*x^3+6*x^2+4*x+8)*log(2+x)+4*x^4+19*x^3+28*x^2+16*x+16)*exp(x^3/(log(2+x)+2*x+4))/(log
(2+x)^2+(4*x+8)*log(2+x)+4*x^2+16*x+16),x, algorithm="maxima")

[Out]

integrate((4*x^4 + 19*x^3 + 28*x^2 + (3*x^3 + 6*x^2 + 4*x + 8)*log(x + 2) + log(x + 2)^2 + 16*x + 16)*e^(x^3/(
2*x + log(x + 2) + 4))/(4*x^2 + 4*(x + 2)*log(x + 2) + log(x + 2)^2 + 16*x + 16), x)

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mupad [B]  time = 3.89, size = 20, normalized size = 0.95 \begin {gather*} {\mathrm {e}}^{\frac {x^3}{2\,x+\ln \left (x+2\right )+4}}\,\left (x+2\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x^3/(2*x + log(x + 2) + 4))*(16*x + log(x + 2)*(4*x + 6*x^2 + 3*x^3 + 8) + log(x + 2)^2 + 28*x^2 + 19
*x^3 + 4*x^4 + 16))/(16*x + log(x + 2)^2 + 4*x^2 + log(x + 2)*(4*x + 8) + 16),x)

[Out]

exp(x^3/(2*x + log(x + 2) + 4))*(x + 2)

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sympy [A]  time = 2.85, size = 17, normalized size = 0.81 \begin {gather*} \left (x + 2\right ) e^{\frac {x^{3}}{2 x + \log {\left (x + 2 \right )} + 4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((ln(2+x)**2+(3*x**3+6*x**2+4*x+8)*ln(2+x)+4*x**4+19*x**3+28*x**2+16*x+16)*exp(x**3/(ln(2+x)+2*x+4))/
(ln(2+x)**2+(4*x+8)*ln(2+x)+4*x**2+16*x+16),x)

[Out]

(x + 2)*exp(x**3/(2*x + log(x + 2) + 4))

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