3.37.90 \(\int \frac {e^{\frac {1-2 x^3+2 x^3 \log (16)}{-2 x^2+2 x^2 \log (16)}} (-1-x^2-x^3+(x^2+x^3) \log (16))}{-x^2+x^2 \log (16)} \, dx\)

Optimal. Leaf size=19 \[ e^{x+\frac {1}{2 x^2 (-1+\log (16))}} x \]

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Rubi [B]  time = 0.60, antiderivative size = 83, normalized size of antiderivative = 4.37, number of steps used = 4, number of rules used = 4, integrand size = 69, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.058, Rules used = {6, 12, 6741, 2288} \begin {gather*} \frac {\left (x^3 (1-\log (16))+1\right ) \exp \left (-\frac {1-2 x^3 (1-\log (16))}{2 x^2 (1-\log (16))}\right )}{x^2 (1-\log (16)) \left (\frac {1-2 x^3 (1-\log (16))}{x^3 (1-\log (16))}+3\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((1 - 2*x^3 + 2*x^3*Log[16])/(-2*x^2 + 2*x^2*Log[16]))*(-1 - x^2 - x^3 + (x^2 + x^3)*Log[16]))/(-x^2 +
x^2*Log[16]),x]

[Out]

(1 + x^3*(1 - Log[16]))/(E^((1 - 2*x^3*(1 - Log[16]))/(2*x^2*(1 - Log[16])))*x^2*(3 + (1 - 2*x^3*(1 - Log[16])
)/(x^3*(1 - Log[16])))*(1 - Log[16]))

Rule 6

Int[(u_.)*((w_.) + (a_.)*(v_) + (b_.)*(v_))^(p_.), x_Symbol] :> Int[u*((a + b)*v + w)^p, x] /; FreeQ[{a, b}, x
] &&  !FreeQ[v, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2288

Int[(y_.)*(F_)^(u_)*((v_) + (w_)), x_Symbol] :> With[{z = (v*y)/(Log[F]*D[u, x])}, Simp[F^u*z, x] /; EqQ[D[z,
x], w*y]] /; FreeQ[F, x]

Rule 6741

Int[u_, x_Symbol] :> With[{v = NormalizeIntegrand[u, x]}, Int[v, x] /; v =!= u]

Rubi steps

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

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Mathematica [A]  time = 0.46, size = 34, normalized size = 1.79 \begin {gather*} 16^{\frac {x}{-1+\log (16)}} e^{\frac {1-2 x^3}{2 x^2 (-1+\log (16))}} x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((1 - 2*x^3 + 2*x^3*Log[16])/(-2*x^2 + 2*x^2*Log[16]))*(-1 - x^2 - x^3 + (x^2 + x^3)*Log[16]))/(-
x^2 + x^2*Log[16]),x]

[Out]

16^(x/(-1 + Log[16]))*E^((1 - 2*x^3)/(2*x^2*(-1 + Log[16])))*x

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fricas [A]  time = 0.63, size = 34, normalized size = 1.79 \begin {gather*} x e^{\left (\frac {8 \, x^{3} \log \relax (2) - 2 \, x^{3} + 1}{2 \, {\left (4 \, x^{2} \log \relax (2) - x^{2}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(x^3+x^2)*log(2)-x^3-x^2-1)*exp((8*x^3*log(2)-2*x^3+1)/(8*x^2*log(2)-2*x^2))/(4*x^2*log(2)-x^2),x
, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.21, size = 34, normalized size = 1.79 \begin {gather*} x e^{\left (\frac {8 \, x^{3} \log \relax (2) - 2 \, x^{3} + 1}{2 \, {\left (4 \, x^{2} \log \relax (2) - x^{2}\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(x^3+x^2)*log(2)-x^3-x^2-1)*exp((8*x^3*log(2)-2*x^3+1)/(8*x^2*log(2)-2*x^2))/(4*x^2*log(2)-x^2),x
, algorithm="giac")

[Out]

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

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maple [A]  time = 0.14, size = 31, normalized size = 1.63

method	result	size
gosper	\({\mathrm e}^{\frac {8 x^{3} \ln \relax (2)-2 x^{3}+1}{2 x^{2} \left (4 \ln \relax (2)-1\right )}} x\)	\(31\)
risch	\({\mathrm e}^{\frac {8 x^{3} \ln \relax (2)-2 x^{3}+1}{2 x^{2} \left (4 \ln \relax (2)-1\right )}} x\)	\(31\)
norman	\(x \,{\mathrm e}^{\frac {8 x^{3} \ln \relax (2)-2 x^{3}+1}{8 x^{2} \ln \relax (2)-2 x^{2}}}\)	\(34\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((4*(x^3+x^2)*ln(2)-x^3-x^2-1)*exp((8*x^3*ln(2)-2*x^3+1)/(8*x^2*ln(2)-2*x^2))/(4*x^2*ln(2)-x^2),x,method=_R
ETURNVERBOSE)

[Out]

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

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maxima [B]  time = 0.56, size = 41, normalized size = 2.16 \begin {gather*} x e^{\left (\frac {4 \, x \log \relax (2)}{4 \, \log \relax (2) - 1} - \frac {x}{4 \, \log \relax (2) - 1} + \frac {1}{2 \, x^{2} {\left (4 \, \log \relax (2) - 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(x^3+x^2)*log(2)-x^3-x^2-1)*exp((8*x^3*log(2)-2*x^3+1)/(8*x^2*log(2)-2*x^2))/(4*x^2*log(2)-x^2),x
, algorithm="maxima")

[Out]

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

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mupad [B]  time = 2.38, size = 61, normalized size = 3.21 \begin {gather*} 2^{\frac {8\,x^3}{8\,x^2\,\ln \relax (2)-2\,x^2}}\,x\,{\mathrm {e}}^{-\frac {2\,x^3}{8\,x^2\,\ln \relax (2)-2\,x^2}}\,{\mathrm {e}}^{\frac {1}{8\,x^2\,\ln \relax (2)-2\,x^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp((8*x^3*log(2) - 2*x^3 + 1)/(8*x^2*log(2) - 2*x^2))*(x^2 + x^3 - 4*log(2)*(x^2 + x^3) + 1))/(4*x^2*lo
g(2) - x^2),x)

[Out]

2^((8*x^3)/(8*x^2*log(2) - 2*x^2))*x*exp(-(2*x^3)/(8*x^2*log(2) - 2*x^2))*exp(1/(8*x^2*log(2) - 2*x^2))

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sympy [A]  time = 0.23, size = 31, normalized size = 1.63 \begin {gather*} x e^{\frac {- 2 x^{3} + 8 x^{3} \log {\relax (2 )} + 1}{- 2 x^{2} + 8 x^{2} \log {\relax (2 )}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((4*(x**3+x**2)*ln(2)-x**3-x**2-1)*exp((8*x**3*ln(2)-2*x**3+1)/(8*x**2*ln(2)-2*x**2))/(4*x**2*ln(2)-x
**2),x)

[Out]

x*exp((-2*x**3 + 8*x**3*log(2) + 1)/(-2*x**2 + 8*x**2*log(2)))

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