3.39.49 \(\int \frac {1}{3} (-3+e^{\frac {1}{3} (e^{2 x}-10 x+e^x (-5+2 x))} (-10+2 e^{2 x}+e^x (-3+2 x))) \, dx\)

Optimal. Leaf size=22 \[ e^{\frac {1}{3} \left (-5+e^x\right ) \left (e^x+2 x\right )}-x \]

________________________________________________________________________________________

Rubi [F]  time = 0.60, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int \frac {1}{3} \left (-3+e^{\frac {1}{3} \left (e^{2 x}-10 x+e^x (-5+2 x)\right )} \left (-10+2 e^{2 x}+e^x (-3+2 x)\right )\right ) \, dx \end {gather*}

Verification is not applicable to the result.

[In]

Int[(-3 + E^((E^(2*x) - 10*x + E^x*(-5 + 2*x))/3)*(-10 + 2*E^(2*x) + E^x*(-3 + 2*x)))/3,x]

[Out]

-x - (10*Defer[Int][E^(((-5 + E^x)*(E^x + 2*x))/3), x])/3 - Defer[Int][E^(x + ((-5 + E^x)*(E^x + 2*x))/3), x]
+ (2*Defer[Int][E^(2*x + ((-5 + E^x)*(E^x + 2*x))/3), x])/3 + (2*Defer[Int][E^(x + ((-5 + E^x)*(E^x + 2*x))/3)
*x, x])/3

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.21, size = 27, normalized size = 1.23 \begin {gather*} e^{-\frac {10 x}{3}+\frac {1}{3} e^x \left (-5+e^x+2 x\right )}-x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(-3 + E^((E^(2*x) - 10*x + E^x*(-5 + 2*x))/3)*(-10 + 2*E^(2*x) + E^x*(-3 + 2*x)))/3,x]

[Out]

E^((-10*x)/3 + (E^x*(-5 + E^x + 2*x))/3) - x

________________________________________________________________________________________

fricas [A]  time = 0.71, size = 24, normalized size = 1.09 \begin {gather*} -x + e^{\left (\frac {1}{3} \, {\left (2 \, x - 5\right )} e^{x} - \frac {10}{3} \, x + \frac {1}{3} \, e^{\left (2 \, x\right )}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(2*exp(x)^2+(2*x-3)*exp(x)-10)*exp(1/3*exp(x)^2+1/3*(2*x-5)*exp(x)-10/3*x)-1,x, algorithm="frica
s")

[Out]

-x + e^(1/3*(2*x - 5)*e^x - 10/3*x + 1/3*e^(2*x))

________________________________________________________________________________________

giac [A]  time = 0.19, size = 24, normalized size = 1.09 \begin {gather*} -x + e^{\left (\frac {2}{3} \, x e^{x} - \frac {10}{3} \, x + \frac {1}{3} \, e^{\left (2 \, x\right )} - \frac {5}{3} \, e^{x}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(2*exp(x)^2+(2*x-3)*exp(x)-10)*exp(1/3*exp(x)^2+1/3*(2*x-5)*exp(x)-10/3*x)-1,x, algorithm="giac"
)

[Out]

-x + e^(2/3*x*e^x - 10/3*x + 1/3*e^(2*x) - 5/3*e^x)

________________________________________________________________________________________

maple [A]  time = 0.06, size = 25, normalized size = 1.14

method	result	size
default	\(-x +{\mathrm e}^{\frac {{\mathrm e}^{2 x}}{3}+\frac {\left (2 x -5\right ) {\mathrm e}^{x}}{3}-\frac {10 x}{3}}\)	\(25\)
norman	\(-x +{\mathrm e}^{\frac {{\mathrm e}^{2 x}}{3}+\frac {\left (2 x -5\right ) {\mathrm e}^{x}}{3}-\frac {10 x}{3}}\)	\(25\)
risch	\(-x +{\mathrm e}^{\frac {{\mathrm e}^{2 x}}{3}+\frac {2 \,{\mathrm e}^{x} x}{3}-\frac {5 \,{\mathrm e}^{x}}{3}-\frac {10 x}{3}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/3*(2*exp(x)^2+(2*x-3)*exp(x)-10)*exp(1/3*exp(x)^2+1/3*(2*x-5)*exp(x)-10/3*x)-1,x,method=_RETURNVERBOSE)

[Out]

-x+exp(1/3*exp(x)^2+1/3*(2*x-5)*exp(x)-10/3*x)

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -x + \frac {1}{3} \, \int {\left ({\left (2 \, x - 3\right )} e^{x} + 2 \, e^{\left (2 \, x\right )} - 10\right )} e^{\left (\frac {2}{3} \, x e^{x} - \frac {10}{3} \, x + \frac {1}{3} \, e^{\left (2 \, x\right )} - \frac {5}{3} \, e^{x}\right )}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(2*exp(x)^2+(2*x-3)*exp(x)-10)*exp(1/3*exp(x)^2+1/3*(2*x-5)*exp(x)-10/3*x)-1,x, algorithm="maxim
a")

[Out]

-x + 1/3*integrate(((2*x - 3)*e^x + 2*e^(2*x) - 10)*e^(2/3*x*e^x - 10/3*x + 1/3*e^(2*x) - 5/3*e^x), x)

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 24, normalized size = 1.09 \begin {gather*} {\mathrm {e}}^{\frac {{\mathrm {e}}^{2\,x}}{3}-\frac {10\,x}{3}-\frac {5\,{\mathrm {e}}^x}{3}+\frac {2\,x\,{\mathrm {e}}^x}{3}}-x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(exp(2*x)/3 - (10*x)/3 + (exp(x)*(2*x - 5))/3)*(2*exp(2*x) + exp(x)*(2*x - 3) - 10))/3 - 1,x)

[Out]

exp(exp(2*x)/3 - (10*x)/3 - (5*exp(x))/3 + (2*x*exp(x))/3) - x

________________________________________________________________________________________

sympy [A]  time = 0.25, size = 26, normalized size = 1.18 \begin {gather*} - x + e^{- \frac {10 x}{3} + \left (\frac {2 x}{3} - \frac {5}{3}\right ) e^{x} + \frac {e^{2 x}}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/3*(2*exp(x)**2+(2*x-3)*exp(x)-10)*exp(1/3*exp(x)**2+1/3*(2*x-5)*exp(x)-10/3*x)-1,x)

[Out]

-x + exp(-10*x/3 + (2*x/3 - 5/3)*exp(x) + exp(2*x)/3)

________________________________________________________________________________________