3.52.81 \(\int \frac {-e^{12+x} x+e^3 (-3-8 x^2)}{100 x+4 e^{18+2 x} x+160 x^3+64 x^5+e^{9+x} (40 x+32 x^3)+(40 x+8 e^{9+x} x+32 x^3) \log (2 x^3)+4 x \log ^2(2 x^3)} \, dx\)

Optimal. Leaf size=27 \[ \frac {e^3}{4 \left (5+e^{9+x}+4 x^2+\log \left (2 x^3\right )\right )} \]

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Rubi [A]  time = 0.33, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 97, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.031, Rules used = {6688, 12, 6686} \begin {gather*} \frac {e^3}{4 \left (\log \left (2 x^3\right )+4 x^2+e^{x+9}+5\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-(E^(12 + x)*x) + E^3*(-3 - 8*x^2))/(100*x + 4*E^(18 + 2*x)*x + 160*x^3 + 64*x^5 + E^(9 + x)*(40*x + 32*x
^3) + (40*x + 8*E^(9 + x)*x + 32*x^3)*Log[2*x^3] + 4*x*Log[2*x^3]^2),x]

[Out]

E^3/(4*(5 + E^(9 + x) + 4*x^2 + Log[2*x^3]))

Rule 12

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

Rule 6686

Int[(u_)*(y_)^(m_.), x_Symbol] :> With[{q = DerivativeDivides[y, u, x]}, Simp[(q*y^(m + 1))/(m + 1), x] /;  !F
alseQ[q]] /; FreeQ[m, x] && NeQ[m, -1]

Rule 6688

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {e^3 \left (-3-e^{9+x} x-8 x^2\right )}{4 x \left (5+e^{9+x}+4 x^2+\log \left (2 x^3\right )\right )^2} \, dx\\ &=\frac {1}{4} e^3 \int \frac {-3-e^{9+x} x-8 x^2}{x \left (5+e^{9+x}+4 x^2+\log \left (2 x^3\right )\right )^2} \, dx\\ &=\frac {e^3}{4 \left (5+e^{9+x}+4 x^2+\log \left (2 x^3\right )\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.05, size = 27, normalized size = 1.00 \begin {gather*} \frac {e^3}{4 \left (5+e^{9+x}+4 x^2+\log \left (2 x^3\right )\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-(E^(12 + x)*x) + E^3*(-3 - 8*x^2))/(100*x + 4*E^(18 + 2*x)*x + 160*x^3 + 64*x^5 + E^(9 + x)*(40*x
+ 32*x^3) + (40*x + 8*E^(9 + x)*x + 32*x^3)*Log[2*x^3] + 4*x*Log[2*x^3]^2),x]

[Out]

E^3/(4*(5 + E^(9 + x) + 4*x^2 + Log[2*x^3]))

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fricas [A]  time = 0.49, size = 30, normalized size = 1.11 \begin {gather*} \frac {e^{6}}{4 \, {\left ({\left (4 \, x^{2} + 5\right )} e^{3} + e^{3} \log \left (2 \, x^{3}\right ) + e^{\left (x + 12\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(3)*exp(x+9)+(-8*x^2-3)*exp(3))/(4*x*log(2*x^3)^2+(8*x*exp(x+9)+32*x^3+40*x)*log(2*x^3)+4*x*e
xp(x+9)^2+(32*x^3+40*x)*exp(x+9)+64*x^5+160*x^3+100*x),x, algorithm="fricas")

[Out]

1/4*e^6/((4*x^2 + 5)*e^3 + e^3*log(2*x^3) + e^(x + 12))

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giac [A]  time = 0.17, size = 23, normalized size = 0.85 \begin {gather*} \frac {e^{3}}{4 \, {\left (4 \, x^{2} + e^{\left (x + 9\right )} + \log \relax (2) + 3 \, \log \relax (x) + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(3)*exp(x+9)+(-8*x^2-3)*exp(3))/(4*x*log(2*x^3)^2+(8*x*exp(x+9)+32*x^3+40*x)*log(2*x^3)+4*x*e
xp(x+9)^2+(32*x^3+40*x)*exp(x+9)+64*x^5+160*x^3+100*x),x, algorithm="giac")

[Out]

1/4*e^3/(4*x^2 + e^(x + 9) + log(2) + 3*log(x) + 5)

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maple [C]  time = 0.19, size = 146, normalized size = 5.41

method	result	size
risch	\(\frac {i {\mathrm e}^{3}}{2 \pi \mathrm {csgn}\left (i x \right )^{2} \mathrm {csgn}\left (i x^{2}\right )-4 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right )^{2}+2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )-2 \pi \,\mathrm {csgn}\left (i x \right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+2 \pi \mathrm {csgn}\left (i x^{2}\right )^{3}-2 \pi \,\mathrm {csgn}\left (i x^{2}\right ) \mathrm {csgn}\left (i x^{3}\right )^{2}+2 \pi \mathrm {csgn}\left (i x^{3}\right )^{3}+12 i \ln \relax (x )+16 i x^{2}+4 i \ln \relax (2)+4 i {\mathrm e}^{x +9}+20 i}\)	\(146\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x*exp(3)*exp(x+9)+(-8*x^2-3)*exp(3))/(4*x*ln(2*x^3)^2+(8*x*exp(x+9)+32*x^3+40*x)*ln(2*x^3)+4*x*exp(x+9)^
2+(32*x^3+40*x)*exp(x+9)+64*x^5+160*x^3+100*x),x,method=_RETURNVERBOSE)

[Out]

1/2*I*exp(3)/(Pi*csgn(I*x)^2*csgn(I*x^2)-2*Pi*csgn(I*x)*csgn(I*x^2)^2+Pi*csgn(I*x)*csgn(I*x^2)*csgn(I*x^3)-Pi*
csgn(I*x)*csgn(I*x^3)^2+Pi*csgn(I*x^2)^3-Pi*csgn(I*x^2)*csgn(I*x^3)^2+Pi*csgn(I*x^3)^3+6*I*ln(x)+8*I*x^2+2*I*l
n(2)+2*I*exp(x+9)+10*I)

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maxima [A]  time = 0.48, size = 23, normalized size = 0.85 \begin {gather*} \frac {e^{3}}{4 \, {\left (4 \, x^{2} + e^{\left (x + 9\right )} + \log \relax (2) + 3 \, \log \relax (x) + 5\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(3)*exp(x+9)+(-8*x^2-3)*exp(3))/(4*x*log(2*x^3)^2+(8*x*exp(x+9)+32*x^3+40*x)*log(2*x^3)+4*x*e
xp(x+9)^2+(32*x^3+40*x)*exp(x+9)+64*x^5+160*x^3+100*x),x, algorithm="maxima")

[Out]

1/4*e^3/(4*x^2 + e^(x + 9) + log(2) + 3*log(x) + 5)

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mupad [B]  time = 3.52, size = 26, normalized size = 0.96 \begin {gather*} \frac {{\mathrm {e}}^3}{4\,\left ({\mathrm {e}}^{x+9}+\ln \left (2\,x^3\right )+4\,x^2+5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(exp(3)*(8*x^2 + 3) + x*exp(x + 9)*exp(3))/(100*x + exp(x + 9)*(40*x + 32*x^3) + 4*x*exp(2*x + 18) + log(
2*x^3)*(40*x + 8*x*exp(x + 9) + 32*x^3) + 160*x^3 + 64*x^5 + 4*x*log(2*x^3)^2),x)

[Out]

exp(3)/(4*(exp(x + 9) + log(2*x^3) + 4*x^2 + 5))

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sympy [A]  time = 0.30, size = 24, normalized size = 0.89 \begin {gather*} \frac {e^{3}}{16 x^{2} + 4 e^{x + 9} + 4 \log {\left (2 x^{3} \right )} + 20} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x*exp(3)*exp(x+9)+(-8*x**2-3)*exp(3))/(4*x*ln(2*x**3)**2+(8*x*exp(x+9)+32*x**3+40*x)*ln(2*x**3)+4*
x*exp(x+9)**2+(32*x**3+40*x)*exp(x+9)+64*x**5+160*x**3+100*x),x)

[Out]

exp(3)/(16*x**2 + 4*exp(x + 9) + 4*log(2*x**3) + 20)

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