3.11.81 \(\int \frac {-24 e^{5 x}+e^{2+e^{2-5 x} x^3} (-24 x^2+40 x^3)}{e^{5 x+2 e^{2-5 x} x^3}+e^{5 x+e^{2-5 x} x^3} (6 x+6 \log (2))+e^{5 x} (9 x^2+18 x \log (2)+9 \log ^2(2))} \, dx\)

Optimal. Leaf size=24 \[ \frac {8}{e^{e^{2-5 x} x^3}+3 (x+\log (2))} \]

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Rubi [A]  time = 0.44, antiderivative size = 23, normalized size of antiderivative = 0.96, number of steps used = 2, number of rules used = 2, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6688, 6686} \begin {gather*} \frac {8}{e^{e^{2-5 x} x^3}+3 x+\log (8)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(-24*E^(5*x) + E^(2 + E^(2 - 5*x)*x^3)*(-24*x^2 + 40*x^3))/(E^(5*x + 2*E^(2 - 5*x)*x^3) + E^(5*x + E^(2 -
5*x)*x^3)*(6*x + 6*Log[2]) + E^(5*x)*(9*x^2 + 18*x*Log[2] + 9*Log[2]^2)),x]

[Out]

8/(E^(E^(2 - 5*x)*x^3) + 3*x + Log[8])

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^{-5 x} \left (-24 e^{5 x}+8 e^{2+e^{2-5 x} x^3} x^2 (-3+5 x)\right )}{\left (e^{e^{2-5 x} x^3}+3 x+\log (8)\right )^2} \, dx\\ &=\frac {8}{e^{e^{2-5 x} x^3}+3 x+\log (8)}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.15, size = 23, normalized size = 0.96 \begin {gather*} \frac {8}{e^{e^{2-5 x} x^3}+3 x+\log (8)} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-24*E^(5*x) + E^(2 + E^(2 - 5*x)*x^3)*(-24*x^2 + 40*x^3))/(E^(5*x + 2*E^(2 - 5*x)*x^3) + E^(5*x + E
^(2 - 5*x)*x^3)*(6*x + 6*Log[2]) + E^(5*x)*(9*x^2 + 18*x*Log[2] + 9*Log[2]^2)),x]

[Out]

8/(E^(E^(2 - 5*x)*x^3) + 3*x + Log[8])

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fricas [A]  time = 0.68, size = 39, normalized size = 1.62 \begin {gather*} \frac {8 \, e^{\left (5 \, x\right )}}{3 \, {\left (x + \log \relax (2)\right )} e^{\left (5 \, x\right )} + e^{\left ({\left (x^{3} e^{2} + 5 \, x e^{\left (5 \, x\right )}\right )} e^{\left (-5 \, x\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^3-24*x^2)*exp(1)^2*exp(x^3*exp(1)^2/exp(5*x))-24*exp(5*x))/(exp(5*x)*exp(x^3*exp(1)^2/exp(5*x
))^2+(6*log(2)+6*x)*exp(5*x)*exp(x^3*exp(1)^2/exp(5*x))+(9*log(2)^2+18*x*log(2)+9*x^2)*exp(5*x)),x, algorithm=
"fricas")

[Out]

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

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giac [B]  time = 0.83, size = 621, normalized size = 25.88 \begin {gather*} \frac {8 \, {\left (15 \, x^{5} e^{\left (5 \, x + 2\right )} + 30 \, x^{4} e^{\left (5 \, x + 2\right )} \log \relax (2) + 15 \, x^{3} e^{\left (5 \, x + 2\right )} \log \relax (2)^{2} + 5 \, x^{4} e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} - 9 \, x^{4} e^{\left (5 \, x + 2\right )} + 5 \, x^{3} e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} \log \relax (2) - 18 \, x^{3} e^{\left (5 \, x + 2\right )} \log \relax (2) - 9 \, x^{2} e^{\left (5 \, x + 2\right )} \log \relax (2)^{2} - 3 \, x^{3} e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} - 3 \, x^{2} e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} \log \relax (2) + 3 \, x e^{\left (10 \, x\right )} + 3 \, e^{\left (10 \, x\right )} \log \relax (2) + e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 10 \, x\right )}\right )}}{45 \, x^{6} e^{\left (5 \, x + 2\right )} + 135 \, x^{5} e^{\left (5 \, x + 2\right )} \log \relax (2) + 135 \, x^{4} e^{\left (5 \, x + 2\right )} \log \relax (2)^{2} + 45 \, x^{3} e^{\left (5 \, x + 2\right )} \log \relax (2)^{3} + 30 \, x^{5} e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} - 27 \, x^{5} e^{\left (5 \, x + 2\right )} + 60 \, x^{4} e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} \log \relax (2) - 81 \, x^{4} e^{\left (5 \, x + 2\right )} \log \relax (2) + 30 \, x^{3} e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} \log \relax (2)^{2} - 81 \, x^{3} e^{\left (5 \, x + 2\right )} \log \relax (2)^{2} - 27 \, x^{2} e^{\left (5 \, x + 2\right )} \log \relax (2)^{3} + 5 \, x^{4} e^{\left (2 \, x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} - 18 \, x^{4} e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} + 5 \, x^{3} e^{\left (2 \, x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} \log \relax (2) - 36 \, x^{3} e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} \log \relax (2) - 18 \, x^{2} e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} \log \relax (2)^{2} - 3 \, x^{3} e^{\left (2 \, x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} - 3 \, x^{2} e^{\left (2 \, x^{3} e^{\left (-5 \, x + 2\right )} + 5 \, x + 2\right )} \log \relax (2) + 9 \, x^{2} e^{\left (10 \, x\right )} + 18 \, x e^{\left (10 \, x\right )} \log \relax (2) + 9 \, e^{\left (10 \, x\right )} \log \relax (2)^{2} + 6 \, x e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 10 \, x\right )} + 6 \, e^{\left (x^{3} e^{\left (-5 \, x + 2\right )} + 10 \, x\right )} \log \relax (2) + e^{\left (2 \, x^{3} e^{\left (-5 \, x + 2\right )} + 10 \, x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^3-24*x^2)*exp(1)^2*exp(x^3*exp(1)^2/exp(5*x))-24*exp(5*x))/(exp(5*x)*exp(x^3*exp(1)^2/exp(5*x
))^2+(6*log(2)+6*x)*exp(5*x)*exp(x^3*exp(1)^2/exp(5*x))+(9*log(2)^2+18*x*log(2)+9*x^2)*exp(5*x)),x, algorithm=
"giac")

[Out]

8*(15*x^5*e^(5*x + 2) + 30*x^4*e^(5*x + 2)*log(2) + 15*x^3*e^(5*x + 2)*log(2)^2 + 5*x^4*e^(x^3*e^(-5*x + 2) +
5*x + 2) - 9*x^4*e^(5*x + 2) + 5*x^3*e^(x^3*e^(-5*x + 2) + 5*x + 2)*log(2) - 18*x^3*e^(5*x + 2)*log(2) - 9*x^2
*e^(5*x + 2)*log(2)^2 - 3*x^3*e^(x^3*e^(-5*x + 2) + 5*x + 2) - 3*x^2*e^(x^3*e^(-5*x + 2) + 5*x + 2)*log(2) + 3
*x*e^(10*x) + 3*e^(10*x)*log(2) + e^(x^3*e^(-5*x + 2) + 10*x))/(45*x^6*e^(5*x + 2) + 135*x^5*e^(5*x + 2)*log(2
) + 135*x^4*e^(5*x + 2)*log(2)^2 + 45*x^3*e^(5*x + 2)*log(2)^3 + 30*x^5*e^(x^3*e^(-5*x + 2) + 5*x + 2) - 27*x^
5*e^(5*x + 2) + 60*x^4*e^(x^3*e^(-5*x + 2) + 5*x + 2)*log(2) - 81*x^4*e^(5*x + 2)*log(2) + 30*x^3*e^(x^3*e^(-5
*x + 2) + 5*x + 2)*log(2)^2 - 81*x^3*e^(5*x + 2)*log(2)^2 - 27*x^2*e^(5*x + 2)*log(2)^3 + 5*x^4*e^(2*x^3*e^(-5
*x + 2) + 5*x + 2) - 18*x^4*e^(x^3*e^(-5*x + 2) + 5*x + 2) + 5*x^3*e^(2*x^3*e^(-5*x + 2) + 5*x + 2)*log(2) - 3
6*x^3*e^(x^3*e^(-5*x + 2) + 5*x + 2)*log(2) - 18*x^2*e^(x^3*e^(-5*x + 2) + 5*x + 2)*log(2)^2 - 3*x^3*e^(2*x^3*
e^(-5*x + 2) + 5*x + 2) - 3*x^2*e^(2*x^3*e^(-5*x + 2) + 5*x + 2)*log(2) + 9*x^2*e^(10*x) + 18*x*e^(10*x)*log(2
) + 9*e^(10*x)*log(2)^2 + 6*x*e^(x^3*e^(-5*x + 2) + 10*x) + 6*e^(x^3*e^(-5*x + 2) + 10*x)*log(2) + e^(2*x^3*e^
(-5*x + 2) + 10*x))

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maple [A]  time = 0.24, size = 24, normalized size = 1.00

method	result	size
risch	\(\frac {8}{3 \ln \relax (2)+3 x +{\mathrm e}^{x^{3} {\mathrm e}^{-5 x +2}}}\)	\(24\)
norman	\(\frac {8}{3 \ln \relax (2)+3 x +{\mathrm e}^{x^{3} {\mathrm e}^{2} {\mathrm e}^{-5 x}}}\)	\(28\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((40*x^3-24*x^2)*exp(1)^2*exp(x^3*exp(1)^2/exp(5*x))-24*exp(5*x))/(exp(5*x)*exp(x^3*exp(1)^2/exp(5*x))^2+(
6*ln(2)+6*x)*exp(5*x)*exp(x^3*exp(1)^2/exp(5*x))+(9*ln(2)^2+18*x*ln(2)+9*x^2)*exp(5*x)),x,method=_RETURNVERBOS
E)

[Out]

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

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maxima [A]  time = 0.63, size = 23, normalized size = 0.96 \begin {gather*} \frac {8}{3 \, x + e^{\left (x^{3} e^{\left (-5 \, x + 2\right )}\right )} + 3 \, \log \relax (2)} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x^3-24*x^2)*exp(1)^2*exp(x^3*exp(1)^2/exp(5*x))-24*exp(5*x))/(exp(5*x)*exp(x^3*exp(1)^2/exp(5*x
))^2+(6*log(2)+6*x)*exp(5*x)*exp(x^3*exp(1)^2/exp(5*x))+(9*log(2)^2+18*x*log(2)+9*x^2)*exp(5*x)),x, algorithm=
"maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \begin {gather*} \int -\frac {24\,{\mathrm {e}}^{5\,x}+{\mathrm {e}}^{x^3\,{\mathrm {e}}^{-5\,x}\,{\mathrm {e}}^2}\,{\mathrm {e}}^2\,\left (24\,x^2-40\,x^3\right )}{{\mathrm {e}}^{5\,x}\,\left (9\,x^2+18\,\ln \relax (2)\,x+9\,{\ln \relax (2)}^2\right )+{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{2\,x^3\,{\mathrm {e}}^{-5\,x}\,{\mathrm {e}}^2}+{\mathrm {e}}^{5\,x}\,{\mathrm {e}}^{x^3\,{\mathrm {e}}^{-5\,x}\,{\mathrm {e}}^2}\,\left (6\,x+6\,\ln \relax (2)\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-(24*exp(5*x) + exp(x^3*exp(-5*x)*exp(2))*exp(2)*(24*x^2 - 40*x^3))/(exp(5*x)*(18*x*log(2) + 9*log(2)^2 +
9*x^2) + exp(5*x)*exp(2*x^3*exp(-5*x)*exp(2)) + exp(5*x)*exp(x^3*exp(-5*x)*exp(2))*(6*x + 6*log(2))),x)

[Out]

int(-(24*exp(5*x) + exp(x^3*exp(-5*x)*exp(2))*exp(2)*(24*x^2 - 40*x^3))/(exp(5*x)*(18*x*log(2) + 9*log(2)^2 +
9*x^2) + exp(5*x)*exp(2*x^3*exp(-5*x)*exp(2)) + exp(5*x)*exp(x^3*exp(-5*x)*exp(2))*(6*x + 6*log(2))), x)

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sympy [A]  time = 0.20, size = 22, normalized size = 0.92 \begin {gather*} \frac {8}{3 x + e^{x^{3} e^{2} e^{- 5 x}} + 3 \log {\relax (2 )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((40*x**3-24*x**2)*exp(1)**2*exp(x**3*exp(1)**2/exp(5*x))-24*exp(5*x))/(exp(5*x)*exp(x**3*exp(1)**2/
exp(5*x))**2+(6*ln(2)+6*x)*exp(5*x)*exp(x**3*exp(1)**2/exp(5*x))+(9*ln(2)**2+18*x*ln(2)+9*x**2)*exp(5*x)),x)

[Out]

8/(3*x + exp(x**3*exp(2)*exp(-5*x)) + 3*log(2))

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