3.16.60 \(\int (6 x-2 e^6 x+(3-e^6) \log (5)+e^{-4+10 e^{4 x}} (-2 x-\log (5)+e^{4 x} (-40 x^2-40 x \log (5)))+e^{-2+5 e^{4 x}} (-4 e^3 x-2 e^3 \log (5)+e^{4 x} (-40 e^3 x^2-40 e^3 x \log (5)))) \, dx\)

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

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Rubi [B]  time = 0.10, antiderivative size = 72, normalized size of antiderivative = 2.67, number of steps used = 4, number of rules used = 2, integrand size = 106, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.019, Rules used = {6, 2288} \begin {gather*} \left (3-e^6\right ) x^2-e^{10 e^{4 x}-4} \left (x^2+x \log (5)\right )-2 e^{5 e^{4 x}-2} \left (e^3 x^2+e^3 x \log (5)\right )+\left (3-e^6\right ) x \log (5) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[6*x - 2*E^6*x + (3 - E^6)*Log[5] + E^(-4 + 10*E^(4*x))*(-2*x - Log[5] + E^(4*x)*(-40*x^2 - 40*x*Log[5])) +
E^(-2 + 5*E^(4*x))*(-4*E^3*x - 2*E^3*Log[5] + E^(4*x)*(-40*E^3*x^2 - 40*E^3*x*Log[5])),x]

[Out]

(3 - E^6)*x^2 + (3 - E^6)*x*Log[5] - E^(-4 + 10*E^(4*x))*(x^2 + x*Log[5]) - 2*E^(-2 + 5*E^(4*x))*(E^3*x^2 + E^
3*x*Log[5])

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 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]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \left (\left (6-2 e^6\right ) x+\left (3-e^6\right ) \log (5)+e^{-4+10 e^{4 x}} \left (-2 x-\log (5)+e^{4 x} \left (-40 x^2-40 x \log (5)\right )\right )+e^{-2+5 e^{4 x}} \left (-4 e^3 x-2 e^3 \log (5)+e^{4 x} \left (-40 e^3 x^2-40 e^3 x \log (5)\right )\right )\right ) \, dx\\ &=\left (3-e^6\right ) x^2+\left (3-e^6\right ) x \log (5)+\int e^{-4+10 e^{4 x}} \left (-2 x-\log (5)+e^{4 x} \left (-40 x^2-40 x \log (5)\right )\right ) \, dx+\int e^{-2+5 e^{4 x}} \left (-4 e^3 x-2 e^3 \log (5)+e^{4 x} \left (-40 e^3 x^2-40 e^3 x \log (5)\right )\right ) \, dx\\ &=\left (3-e^6\right ) x^2+\left (3-e^6\right ) x \log (5)-e^{-4+10 e^{4 x}} \left (x^2+x \log (5)\right )-2 e^{-2+5 e^{4 x}} \left (e^3 x^2+e^3 x \log (5)\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.29, size = 51, normalized size = 1.89 \begin {gather*} x \left (3 x-e^6 (x+\log (5))-2 e^{1+5 e^{4 x}} (x+\log (5))-e^{-4+10 e^{4 x}} (x+\log (5))+\log (125)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[6*x - 2*E^6*x + (3 - E^6)*Log[5] + E^(-4 + 10*E^(4*x))*(-2*x - Log[5] + E^(4*x)*(-40*x^2 - 40*x*Log[
5])) + E^(-2 + 5*E^(4*x))*(-4*E^3*x - 2*E^3*Log[5] + E^(4*x)*(-40*E^3*x^2 - 40*E^3*x*Log[5])),x]

[Out]

x*(3*x - E^6*(x + Log[5]) - 2*E^(1 + 5*E^(4*x))*(x + Log[5]) - E^(-4 + 10*E^(4*x))*(x + Log[5]) + Log[125])

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fricas [B]  time = 0.70, size = 68, normalized size = 2.52 \begin {gather*} -x^{2} e^{6} + 3 \, x^{2} - {\left (x^{2} + x \log \relax (5)\right )} e^{\left (10 \, e^{\left (4 \, x\right )} - 4\right )} - 2 \, {\left (x^{2} e^{3} + x e^{3} \log \relax (5)\right )} e^{\left (5 \, e^{\left (4 \, x\right )} - 2\right )} - {\left (x e^{6} - 3 \, x\right )} \log \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*log(5)-40*x^2)*exp(4*x)-log(5)-2*x)*exp(5*exp(4*x)-2)^2+((-40*x*exp(3)*log(5)-40*x^2*exp(3))
*exp(4*x)-2*exp(3)*log(5)-4*x*exp(3))*exp(5*exp(4*x)-2)+(-exp(3)^2+3)*log(5)-2*x*exp(3)^2+6*x,x, algorithm="fr
icas")

[Out]

-x^2*e^6 + 3*x^2 - (x^2 + x*log(5))*e^(10*e^(4*x) - 4) - 2*(x^2*e^3 + x*e^3*log(5))*e^(5*e^(4*x) - 2) - (x*e^6
- 3*x)*log(5)

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giac [B]  time = 1.05, size = 77, normalized size = 2.85 \begin {gather*} -x^{2} e^{6} - 2 \, x^{2} e^{\left (5 \, e^{\left (4 \, x\right )} + 1\right )} - x {\left (e^{6} - 3\right )} \log \relax (5) - 2 \, x e^{\left (5 \, e^{\left (4 \, x\right )} + 1\right )} \log \relax (5) + 3 \, x^{2} - {\left (x^{2} e^{\left (10 \, e^{\left (4 \, x\right )}\right )} + x e^{\left (10 \, e^{\left (4 \, x\right )}\right )} \log \relax (5)\right )} e^{\left (-4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*log(5)-40*x^2)*exp(4*x)-log(5)-2*x)*exp(5*exp(4*x)-2)^2+((-40*x*exp(3)*log(5)-40*x^2*exp(3))
*exp(4*x)-2*exp(3)*log(5)-4*x*exp(3))*exp(5*exp(4*x)-2)+(-exp(3)^2+3)*log(5)-2*x*exp(3)^2+6*x,x, algorithm="gi
ac")

[Out]

-x^2*e^6 - 2*x^2*e^(5*e^(4*x) + 1) - x*(e^6 - 3)*log(5) - 2*x*e^(5*e^(4*x) + 1)*log(5) + 3*x^2 - (x^2*e^(10*e^
(4*x)) + x*e^(10*e^(4*x))*log(5))*e^(-4)

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maple [B]  time = 0.19, size = 63, normalized size = 2.33

method	result	size
risch	\(\left (-x \ln \relax (5)-x^{2}\right ) {\mathrm e}^{10 \,{\mathrm e}^{4 x}-4}-2 x \left (\ln \relax (5)+x \right ) {\mathrm e}^{1+5 \,{\mathrm e}^{4 x}}-x \ln \relax (5) {\mathrm e}^{6}+3 x \ln \relax (5)-x^{2} {\mathrm e}^{6}+3 x^{2}\)	\(63\)
norman	\(\left (-{\mathrm e}^{6}+3\right ) x^{2}+\left (-{\mathrm e}^{6} \ln \relax (5)+3 \ln \relax (5)\right ) x -x^{2} {\mathrm e}^{10 \,{\mathrm e}^{4 x}-4}-x \ln \relax (5) {\mathrm e}^{10 \,{\mathrm e}^{4 x}-4}-2 x^{2} {\mathrm e}^{3} {\mathrm e}^{5 \,{\mathrm e}^{4 x}-2}-2 x \,{\mathrm e}^{3} \ln \relax (5) {\mathrm e}^{5 \,{\mathrm e}^{4 x}-2}\)	\(93\)
default	\(-x \ln \relax (5) {\mathrm e}^{6}+3 x \ln \relax (5)-x^{2} {\mathrm e}^{10 \,{\mathrm e}^{4 x}-4}-x \ln \relax (5) {\mathrm e}^{10 \,{\mathrm e}^{4 x}-4}-2 x^{2} {\mathrm e}^{3} {\mathrm e}^{5 \,{\mathrm e}^{4 x}-2}-2 x \,{\mathrm e}^{3} \ln \relax (5) {\mathrm e}^{5 \,{\mathrm e}^{4 x}-2}+3 x^{2}-x^{2} {\mathrm e}^{6}\)	\(94\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-40*x*ln(5)-40*x^2)*exp(4*x)-ln(5)-2*x)*exp(5*exp(4*x)-2)^2+((-40*x*exp(3)*ln(5)-40*x^2*exp(3))*exp(4*x)
-2*exp(3)*ln(5)-4*x*exp(3))*exp(5*exp(4*x)-2)+(-exp(3)^2+3)*ln(5)-2*x*exp(3)^2+6*x,x,method=_RETURNVERBOSE)

[Out]

(-x*ln(5)-x^2)*exp(10*exp(4*x)-4)-2*x*(ln(5)+x)*exp(1+5*exp(4*x))-x*ln(5)*exp(6)+3*x*ln(5)-x^2*exp(6)+3*x^2

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maxima [B]  time = 0.50, size = 63, normalized size = 2.33 \begin {gather*} -x^{2} e^{6} - x {\left (e^{6} - 3\right )} \log \relax (5) + 3 \, x^{2} - 2 \, {\left (x^{2} e + x e \log \relax (5)\right )} e^{\left (5 \, e^{\left (4 \, x\right )}\right )} - {\left (x^{2} + x \log \relax (5)\right )} e^{\left (10 \, e^{\left (4 \, x\right )} - 4\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*log(5)-40*x^2)*exp(4*x)-log(5)-2*x)*exp(5*exp(4*x)-2)^2+((-40*x*exp(3)*log(5)-40*x^2*exp(3))
*exp(4*x)-2*exp(3)*log(5)-4*x*exp(3))*exp(5*exp(4*x)-2)+(-exp(3)^2+3)*log(5)-2*x*exp(3)^2+6*x,x, algorithm="ma
xima")

[Out]

-x^2*e^6 - x*(e^6 - 3)*log(5) + 3*x^2 - 2*(x^2*e + x*e*log(5))*e^(5*e^(4*x)) - (x^2 + x*log(5))*e^(10*e^(4*x)
- 4)

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mupad [B]  time = 1.07, size = 79, normalized size = 2.93 \begin {gather*} x\,\left (3\,\ln \relax (5)-{\mathrm {e}}^6\,\ln \relax (5)\right )-2\,x^2\,{\mathrm {e}}^{5\,{\mathrm {e}}^{4\,x}+1}-x^2\,{\mathrm {e}}^{10\,{\mathrm {e}}^{4\,x}-4}-x^2\,\left ({\mathrm {e}}^6-3\right )-2\,x\,{\mathrm {e}}^{5\,{\mathrm {e}}^{4\,x}+1}\,\ln \relax (5)-x\,{\mathrm {e}}^{10\,{\mathrm {e}}^{4\,x}-4}\,\ln \relax (5) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(6*x - 2*x*exp(6) - log(5)*(exp(6) - 3) - exp(5*exp(4*x) - 2)*(2*exp(3)*log(5) + 4*x*exp(3) + exp(4*x)*(40*
x^2*exp(3) + 40*x*exp(3)*log(5))) - exp(10*exp(4*x) - 4)*(2*x + log(5) + exp(4*x)*(40*x*log(5) + 40*x^2)),x)

[Out]

x*(3*log(5) - exp(6)*log(5)) - 2*x^2*exp(5*exp(4*x) + 1) - x^2*exp(10*exp(4*x) - 4) - x^2*(exp(6) - 3) - 2*x*e
xp(5*exp(4*x) + 1)*log(5) - x*exp(10*exp(4*x) - 4)*log(5)

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sympy [B]  time = 0.29, size = 71, normalized size = 2.63 \begin {gather*} x^{2} \left (3 - e^{6}\right ) + x \left (- e^{6} \log {\relax (5 )} + 3 \log {\relax (5 )}\right ) + \left (- x^{2} - x \log {\relax (5 )}\right ) e^{10 e^{4 x} - 4} + \left (- 2 x^{2} e^{3} - 2 x e^{3} \log {\relax (5 )}\right ) e^{5 e^{4 x} - 2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-40*x*ln(5)-40*x**2)*exp(4*x)-ln(5)-2*x)*exp(5*exp(4*x)-2)**2+((-40*x*exp(3)*ln(5)-40*x**2*exp(3))
*exp(4*x)-2*exp(3)*ln(5)-4*x*exp(3))*exp(5*exp(4*x)-2)+(-exp(3)**2+3)*ln(5)-2*x*exp(3)**2+6*x,x)

[Out]

x**2*(3 - exp(6)) + x*(-exp(6)*log(5) + 3*log(5)) + (-x**2 - x*log(5))*exp(10*exp(4*x) - 4) + (-2*x**2*exp(3)
- 2*x*exp(3)*log(5))*exp(5*exp(4*x) - 2)

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