3.36.100 \(\int (16-2 x+e^{x+2 x^2+x^3} (5+5 x+20 x^2+15 x^3)) \, dx\)

Optimal. Leaf size=27 \[ 5 \left (2+e^{\frac {\left (x+x^2\right )^2}{x}}+\frac {6-x}{5}\right ) x \]

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Rubi [A]  time = 0.04, antiderivative size = 47, normalized size of antiderivative = 1.74, number of steps used = 2, number of rules used = 1, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.030, Rules used = {2288} \begin {gather*} -x^2+\frac {5 e^{x^3+2 x^2+x} \left (3 x^3+4 x^2+x\right )}{3 x^2+4 x+1}+16 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[16 - 2*x + E^(x + 2*x^2 + x^3)*(5 + 5*x + 20*x^2 + 15*x^3),x]

[Out]

16*x - x^2 + (5*E^(x + 2*x^2 + x^3)*(x + 4*x^2 + 3*x^3))/(1 + 4*x + 3*x^2)

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} &=16 x-x^2+\int e^{x+2 x^2+x^3} \left (5+5 x+20 x^2+15 x^3\right ) \, dx\\ &=16 x-x^2+\frac {5 e^{x+2 x^2+x^3} \left (x+4 x^2+3 x^3\right )}{1+4 x+3 x^2}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.02, size = 17, normalized size = 0.63 \begin {gather*} -x \left (-16-5 e^{x (1+x)^2}+x\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[16 - 2*x + E^(x + 2*x^2 + x^3)*(5 + 5*x + 20*x^2 + 15*x^3),x]

[Out]

-(x*(-16 - 5*E^(x*(1 + x)^2) + x))

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fricas [A]  time = 0.83, size = 23, normalized size = 0.85 \begin {gather*} -x^{2} + 5 \, x e^{\left (x^{3} + 2 \, x^{2} + x\right )} + 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^3+20*x^2+5*x+5)*exp(x^3+2*x^2+x)+16-2*x,x, algorithm="fricas")

[Out]

-x^2 + 5*x*e^(x^3 + 2*x^2 + x) + 16*x

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giac [A]  time = 0.20, size = 23, normalized size = 0.85 \begin {gather*} -x^{2} + 5 \, x e^{\left (x^{3} + 2 \, x^{2} + x\right )} + 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^3+20*x^2+5*x+5)*exp(x^3+2*x^2+x)+16-2*x,x, algorithm="giac")

[Out]

-x^2 + 5*x*e^(x^3 + 2*x^2 + x) + 16*x

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maple [A]  time = 0.04, size = 21, normalized size = 0.78

method	result	size
risch	\(16 x +5 \,{\mathrm e}^{x \left (x +1\right )^{2}} x -x^{2}\)	\(21\)
default	\(16 x +5 \,{\mathrm e}^{x^{3}+2 x^{2}+x} x -x^{2}\)	\(24\)
norman	\(16 x +5 \,{\mathrm e}^{x^{3}+2 x^{2}+x} x -x^{2}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((15*x^3+20*x^2+5*x+5)*exp(x^3+2*x^2+x)+16-2*x,x,method=_RETURNVERBOSE)

[Out]

16*x+5*exp(x*(x+1)^2)*x-x^2

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maxima [A]  time = 0.40, size = 23, normalized size = 0.85 \begin {gather*} -x^{2} + 5 \, x e^{\left (x^{3} + 2 \, x^{2} + x\right )} + 16 \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x^3+20*x^2+5*x+5)*exp(x^3+2*x^2+x)+16-2*x,x, algorithm="maxima")

[Out]

-x^2 + 5*x*e^(x^3 + 2*x^2 + x) + 16*x

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mupad [B]  time = 0.06, size = 24, normalized size = 0.89 \begin {gather*} 16\,x-x^2+5\,x\,{\mathrm {e}}^{x^3}\,{\mathrm {e}}^{2\,x^2}\,{\mathrm {e}}^x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(exp(x + 2*x^2 + x^3)*(5*x + 20*x^2 + 15*x^3 + 5) - 2*x + 16,x)

[Out]

16*x - x^2 + 5*x*exp(x^3)*exp(2*x^2)*exp(x)

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sympy [A]  time = 0.11, size = 20, normalized size = 0.74 \begin {gather*} - x^{2} + 5 x e^{x^{3} + 2 x^{2} + x} + 16 x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((15*x**3+20*x**2+5*x+5)*exp(x**3+2*x**2+x)+16-2*x,x)

[Out]

-x**2 + 5*x*exp(x**3 + 2*x**2 + x) + 16*x

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