3.88.50 \(\int (2 e^{2 x}+2 x+69 x^2+529 x^3+e^x (-1-48 x-23 x^2)) \, dx\)

Optimal. Leaf size=21 \[ -4+e^x+\left (e^x-x-\frac {23 x^2}{2}\right )^2 \]

________________________________________________________________________________________

Rubi [A]  time = 0.04, antiderivative size = 38, normalized size of antiderivative = 1.81, number of steps used = 10, number of rules used = 3, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.086, Rules used = {2194, 2196, 2176} \begin {gather*} \frac {529 x^4}{4}+23 x^3-23 e^x x^2+x^2-2 e^x x+e^x+e^{2 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[2*E^(2*x) + 2*x + 69*x^2 + 529*x^3 + E^x*(-1 - 48*x - 23*x^2),x]

[Out]

E^x + E^(2*x) - 2*E^x*x + x^2 - 23*E^x*x^2 + 23*x^3 + (529*x^4)/4

Rule 2176

Int[((b_.)*(F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[((c + d*x)^m
*(b*F^(g*(e + f*x)))^n)/(f*g*n*Log[F]), x] - Dist[(d*m)/(f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*(b*F^(g*(e + f*x
)))^n, x], x] /; FreeQ[{F, b, c, d, e, f, g, n}, x] && GtQ[m, 0] && IntegerQ[2*m] &&  !$UseGamma === True

Rule 2194

Int[((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.), x_Symbol] :> Simp[(F^(c*(a + b*x)))^n/(b*c*n*Log[F]), x] /; Fre
eQ[{F, a, b, c, n}, x]

Rule 2196

Int[(F_)^((c_.)*(v_))*(u_), x_Symbol] :> Int[ExpandIntegrand[F^(c*ExpandToSum[v, x]), u, x], x] /; FreeQ[{F, c
}, x] && PolynomialQ[u, x] && LinearQ[v, x] &&  !$UseGamma === True

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=x^2+23 x^3+\frac {529 x^4}{4}+2 \int e^{2 x} \, dx+\int e^x \left (-1-48 x-23 x^2\right ) \, dx\\ &=e^{2 x}+x^2+23 x^3+\frac {529 x^4}{4}+\int \left (-e^x-48 e^x x-23 e^x x^2\right ) \, dx\\ &=e^{2 x}+x^2+23 x^3+\frac {529 x^4}{4}-23 \int e^x x^2 \, dx-48 \int e^x x \, dx-\int e^x \, dx\\ &=-e^x+e^{2 x}-48 e^x x+x^2-23 e^x x^2+23 x^3+\frac {529 x^4}{4}+46 \int e^x x \, dx+48 \int e^x \, dx\\ &=47 e^x+e^{2 x}-2 e^x x+x^2-23 e^x x^2+23 x^3+\frac {529 x^4}{4}-46 \int e^x \, dx\\ &=e^x+e^{2 x}-2 e^x x+x^2-23 e^x x^2+23 x^3+\frac {529 x^4}{4}\\ \end {aligned} \end {gather*}

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 36, normalized size = 1.71 \begin {gather*} e^{2 x}+x^2+23 x^3+\frac {529 x^4}{4}-e^x \left (-1+2 x+23 x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[2*E^(2*x) + 2*x + 69*x^2 + 529*x^3 + E^x*(-1 - 48*x - 23*x^2),x]

[Out]

E^(2*x) + x^2 + 23*x^3 + (529*x^4)/4 - E^x*(-1 + 2*x + 23*x^2)

________________________________________________________________________________________

fricas [A]  time = 0.94, size = 32, normalized size = 1.52 \begin {gather*} \frac {529}{4} \, x^{4} + 23 \, x^{3} + x^{2} - {\left (23 \, x^{2} + 2 \, x - 1\right )} e^{x} + e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x)^2+(-23*x^2-48*x-1)*exp(x)+529*x^3+69*x^2+2*x,x, algorithm="fricas")

[Out]

529/4*x^4 + 23*x^3 + x^2 - (23*x^2 + 2*x - 1)*e^x + e^(2*x)

________________________________________________________________________________________

giac [A]  time = 0.25, size = 32, normalized size = 1.52 \begin {gather*} \frac {529}{4} \, x^{4} + 23 \, x^{3} + x^{2} - {\left (23 \, x^{2} + 2 \, x - 1\right )} e^{x} + e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x)^2+(-23*x^2-48*x-1)*exp(x)+529*x^3+69*x^2+2*x,x, algorithm="giac")

[Out]

529/4*x^4 + 23*x^3 + x^2 - (23*x^2 + 2*x - 1)*e^x + e^(2*x)

________________________________________________________________________________________

maple [A]  time = 0.03, size = 32, normalized size = 1.52

method	result	size
risch	\({\mathrm e}^{2 x}+\left (-23 x^{2}-2 x +1\right ) {\mathrm e}^{x}+\frac {529 x^{4}}{4}+23 x^{3}+x^{2}\)	\(32\)
default	\(-2 \,{\mathrm e}^{x} x +{\mathrm e}^{x}-23 \,{\mathrm e}^{x} x^{2}+x^{2}+23 x^{3}+\frac {529 x^{4}}{4}+{\mathrm e}^{2 x}\)	\(33\)
norman	\(-2 \,{\mathrm e}^{x} x +{\mathrm e}^{x}-23 \,{\mathrm e}^{x} x^{2}+x^{2}+23 x^{3}+\frac {529 x^{4}}{4}+{\mathrm e}^{2 x}\)	\(33\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*exp(x)^2+(-23*x^2-48*x-1)*exp(x)+529*x^3+69*x^2+2*x,x,method=_RETURNVERBOSE)

[Out]

exp(2*x)+(-23*x^2-2*x+1)*exp(x)+529/4*x^4+23*x^3+x^2

________________________________________________________________________________________

maxima [A]  time = 0.41, size = 32, normalized size = 1.52 \begin {gather*} \frac {529}{4} \, x^{4} + 23 \, x^{3} + x^{2} - {\left (23 \, x^{2} + 2 \, x - 1\right )} e^{x} + e^{\left (2 \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x)^2+(-23*x^2-48*x-1)*exp(x)+529*x^3+69*x^2+2*x,x, algorithm="maxima")

[Out]

529/4*x^4 + 23*x^3 + x^2 - (23*x^2 + 2*x - 1)*e^x + e^(2*x)

________________________________________________________________________________________

mupad [B]  time = 0.05, size = 32, normalized size = 1.52 \begin {gather*} {\mathrm {e}}^{2\,x}+{\mathrm {e}}^x-23\,x^2\,{\mathrm {e}}^x-2\,x\,{\mathrm {e}}^x+x^2+23\,x^3+\frac {529\,x^4}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x + 2*exp(2*x) - exp(x)*(48*x + 23*x^2 + 1) + 69*x^2 + 529*x^3,x)

[Out]

exp(2*x) + exp(x) - 23*x^2*exp(x) - 2*x*exp(x) + x^2 + 23*x^3 + (529*x^4)/4

________________________________________________________________________________________

sympy [A]  time = 0.10, size = 32, normalized size = 1.52 \begin {gather*} \frac {529 x^{4}}{4} + 23 x^{3} + x^{2} + \left (- 23 x^{2} - 2 x + 1\right ) e^{x} + e^{2 x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*exp(x)**2+(-23*x**2-48*x-1)*exp(x)+529*x**3+69*x**2+2*x,x)

[Out]

529*x**4/4 + 23*x**3 + x**2 + (-23*x**2 - 2*x + 1)*exp(x) + exp(2*x)

________________________________________________________________________________________