3.74.65 \(\int (e^3+2 x+e^{25-10 e^x+e^{2 x}+2 x} (-1-2 e^3+e^{2 x} (-2 e^3-2 x)-2 x+e^x (10 e^3+10 x))-\log (4+e^4)) \, dx\)

Optimal. Leaf size=33 \[ \left (e^3+x\right ) \left (-e^{\left (5-e^x\right )^2+2 x}+x-\log \left (4+e^4\right )\right ) \]

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Rubi [B]  time = 0.39, antiderivative size = 77, normalized size of antiderivative = 2.33, number of steps used = 2, number of rules used = 1, integrand size = 71, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.014, Rules used = {2288} \begin {gather*} x^2-\frac {e^{2 x-10 e^x+e^{2 x}+25} \left (x-5 e^x \left (x+e^3\right )+e^{2 x} \left (x+e^3\right )+e^3\right )}{-5 e^x+e^{2 x}+1}+x \left (e^3-\log \left (4+e^4\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[E^3 + 2*x + E^(25 - 10*E^x + E^(2*x) + 2*x)*(-1 - 2*E^3 + E^(2*x)*(-2*E^3 - 2*x) - 2*x + E^x*(10*E^3 + 10*
x)) - Log[4 + E^4],x]

[Out]

x^2 - (E^(25 - 10*E^x + E^(2*x) + 2*x)*(E^3 + x - 5*E^x*(E^3 + x) + E^(2*x)*(E^3 + x)))/(1 - 5*E^x + E^(2*x))
+ x*(E^3 - Log[4 + E^4])

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} &=x^2+x \left (e^3-\log \left (4+e^4\right )\right )+\int e^{25-10 e^x+e^{2 x}+2 x} \left (-1-2 e^3+e^{2 x} \left (-2 e^3-2 x\right )-2 x+e^x \left (10 e^3+10 x\right )\right ) \, dx\\ &=x^2-\frac {e^{25-10 e^x+e^{2 x}+2 x} \left (e^3+x-5 e^x \left (e^3+x\right )+e^{2 x} \left (e^3+x\right )\right )}{1-5 e^x+e^{2 x}}+x \left (e^3-\log \left (4+e^4\right )\right )\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.42, size = 57, normalized size = 1.73 \begin {gather*} -e^{28-10 e^x+e^{2 x}+2 x}+e^3 x-e^{25-10 e^x+e^{2 x}+2 x} x+x \left (x-\log \left (4+e^4\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[E^3 + 2*x + E^(25 - 10*E^x + E^(2*x) + 2*x)*(-1 - 2*E^3 + E^(2*x)*(-2*E^3 - 2*x) - 2*x + E^x*(10*E^3
+ 10*x)) - Log[4 + E^4],x]

[Out]

-E^(28 - 10*E^x + E^(2*x) + 2*x) + E^3*x - E^(25 - 10*E^x + E^(2*x) + 2*x)*x + x*(x - Log[4 + E^4])

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fricas [A]  time = 0.66, size = 36, normalized size = 1.09 \begin {gather*} x^{2} + x e^{3} - {\left (x + e^{3}\right )} e^{\left (2 \, x + e^{\left (2 \, x\right )} - 10 \, e^{x} + 25\right )} - x \log \left (e^{4} + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(3)-2*x)*exp(x)^2+(10*exp(3)+10*x)*exp(x)-2*exp(3)-2*x-1)*exp(exp(x)^2-10*exp(x)+2*x+25)-log
(4+exp(4))+2*x+exp(3),x, algorithm="fricas")

[Out]

x^2 + x*e^3 - (x + e^3)*e^(2*x + e^(2*x) - 10*e^x + 25) - x*log(e^4 + 4)

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giac [A]  time = 0.17, size = 49, normalized size = 1.48 \begin {gather*} x^{2} + x e^{3} - x e^{\left (2 \, x + e^{\left (2 \, x\right )} - 10 \, e^{x} + 25\right )} - x \log \left (e^{4} + 4\right ) - e^{\left (2 \, x + e^{\left (2 \, x\right )} - 10 \, e^{x} + 28\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(3)-2*x)*exp(x)^2+(10*exp(3)+10*x)*exp(x)-2*exp(3)-2*x-1)*exp(exp(x)^2-10*exp(x)+2*x+25)-log
(4+exp(4))+2*x+exp(3),x, algorithm="giac")

[Out]

x^2 + x*e^3 - x*e^(2*x + e^(2*x) - 10*e^x + 25) - x*log(e^4 + 4) - e^(2*x + e^(2*x) - 10*e^x + 28)

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maple [A]  time = 0.08, size = 40, normalized size = 1.21

method	result	size
risch	\(\left (-{\mathrm e}^{3}-x \right ) {\mathrm e}^{{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+2 x +25}-\ln \left (4+{\mathrm e}^{4}\right ) x +x^{2}+x \,{\mathrm e}^{3}\)	\(40\)
default	\(-{\mathrm e}^{{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+2 x +25} x -{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+2 x +25}+x^{2}-\ln \left (4+{\mathrm e}^{4}\right ) x +x \,{\mathrm e}^{3}\)	\(52\)
norman	\(x^{2}+\left ({\mathrm e}^{3}-\ln \left (4+{\mathrm e}^{4}\right )\right ) x -{\mathrm e}^{3} {\mathrm e}^{{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+2 x +25}-{\mathrm e}^{{\mathrm e}^{2 x}-10 \,{\mathrm e}^{x}+2 x +25} x\)	\(52\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((-2*exp(3)-2*x)*exp(x)^2+(10*exp(3)+10*x)*exp(x)-2*exp(3)-2*x-1)*exp(exp(x)^2-10*exp(x)+2*x+25)-ln(4+exp(
4))+2*x+exp(3),x,method=_RETURNVERBOSE)

[Out]

(-exp(3)-x)*exp(exp(2*x)-10*exp(x)+2*x+25)-ln(4+exp(4))*x+x^2+x*exp(3)

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maxima [A]  time = 0.37, size = 38, normalized size = 1.15 \begin {gather*} x^{2} + x e^{3} - {\left (x e^{25} + e^{28}\right )} e^{\left (2 \, x + e^{\left (2 \, x\right )} - 10 \, e^{x}\right )} - x \log \left (e^{4} + 4\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(3)-2*x)*exp(x)^2+(10*exp(3)+10*x)*exp(x)-2*exp(3)-2*x-1)*exp(exp(x)^2-10*exp(x)+2*x+25)-log
(4+exp(4))+2*x+exp(3),x, algorithm="maxima")

[Out]

x^2 + x*e^3 - (x*e^25 + e^28)*e^(2*x + e^(2*x) - 10*e^x) - x*log(e^4 + 4)

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mupad [B]  time = 5.18, size = 49, normalized size = 1.48 \begin {gather*} x\,{\mathrm {e}}^3-x\,\ln \left ({\mathrm {e}}^4+4\right )-{\mathrm {e}}^{2\,x+{\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^x+28}+x^2-x\,{\mathrm {e}}^{2\,x+{\mathrm {e}}^{2\,x}-10\,{\mathrm {e}}^x+25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x - log(exp(4) + 4) + exp(3) - exp(2*x + exp(2*x) - 10*exp(x) + 25)*(2*x + 2*exp(3) - exp(x)*(10*x + 10*
exp(3)) + exp(2*x)*(2*x + 2*exp(3)) + 1),x)

[Out]

x*exp(3) - x*log(exp(4) + 4) - exp(2*x + exp(2*x) - 10*exp(x) + 28) + x^2 - x*exp(2*x + exp(2*x) - 10*exp(x) +
25)

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sympy [A]  time = 0.25, size = 37, normalized size = 1.12 \begin {gather*} x^{2} + x \left (- \log {\left (4 + e^{4} \right )} + e^{3}\right ) + \left (- x - e^{3}\right ) e^{2 x + e^{2 x} - 10 e^{x} + 25} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(((-2*exp(3)-2*x)*exp(x)**2+(10*exp(3)+10*x)*exp(x)-2*exp(3)-2*x-1)*exp(exp(x)**2-10*exp(x)+2*x+25)-l
n(4+exp(4))+2*x+exp(3),x)

[Out]

x**2 + x*(-log(4 + exp(4)) + exp(3)) + (-x - exp(3))*exp(2*x + exp(2*x) - 10*exp(x) + 25)

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