3.100.71 \(\int (131+44 x+2 e^{\frac {1}{3} (1+3 x^2)} x+3 x^2) \, dx\)

Optimal. Leaf size=20 \[ e^{\frac {1}{3}+x^2}+x+x \left (9+(11+x)^2\right ) \]

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Rubi [A]  time = 0.03, antiderivative size = 21, normalized size of antiderivative = 1.05, number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2225, 2209} \begin {gather*} x^3+22 x^2+e^{x^2+\frac {1}{3}}+131 x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[131 + 44*x + 2*E^((1 + 3*x^2)/3)*x + 3*x^2,x]

[Out]

E^(1/3 + x^2) + 131*x + 22*x^2 + x^3

Rule 2209

Int[(F_)^((a_.) + (b_.)*((c_.) + (d_.)*(x_))^(n_))*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Simp[((e + f*x)^n*
F^(a + b*(c + d*x)^n))/(b*f*n*(c + d*x)^n*Log[F]), x] /; FreeQ[{F, a, b, c, d, e, f, n}, x] && EqQ[m, n - 1] &
& EqQ[d*e - c*f, 0]

Rule 2225

Int[(F_)^(v_)*(u_)^(m_.), x_Symbol] :> Int[ExpandToSum[u, x]^m*F^ExpandToSum[v, x], x] /; FreeQ[{F, m}, x] &&
LinearQ[u, x] && BinomialQ[v, x] &&  !(LinearMatchQ[u, x] && BinomialMatchQ[v, x])

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=131 x+22 x^2+x^3+2 \int e^{\frac {1}{3} \left (1+3 x^2\right )} x \, dx\\ &=131 x+22 x^2+x^3+2 \int e^{\frac {1}{3}+x^2} x \, dx\\ &=e^{\frac {1}{3}+x^2}+131 x+22 x^2+x^3\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.01, size = 21, normalized size = 1.05 \begin {gather*} e^{\frac {1}{3}+x^2}+131 x+22 x^2+x^3 \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[131 + 44*x + 2*E^((1 + 3*x^2)/3)*x + 3*x^2,x]

[Out]

E^(1/3 + x^2) + 131*x + 22*x^2 + x^3

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fricas [A]  time = 0.48, size = 18, normalized size = 0.90 \begin {gather*} x^{3} + 22 \, x^{2} + 131 \, x + e^{\left (x^{2} + \frac {1}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(x^2+1/3)+3*x^2+44*x+131,x, algorithm="fricas")

[Out]

x^3 + 22*x^2 + 131*x + e^(x^2 + 1/3)

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giac [A]  time = 0.15, size = 18, normalized size = 0.90 \begin {gather*} x^{3} + 22 \, x^{2} + 131 \, x + e^{\left (x^{2} + \frac {1}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(x^2+1/3)+3*x^2+44*x+131,x, algorithm="giac")

[Out]

x^3 + 22*x^2 + 131*x + e^(x^2 + 1/3)

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maple [A]  time = 0.02, size = 19, normalized size = 0.95

method	result	size
default	\(x^{3}+22 x^{2}+131 x +{\mathrm e}^{x^{2}+\frac {1}{3}}\)	\(19\)
norman	\(x^{3}+22 x^{2}+131 x +{\mathrm e}^{x^{2}+\frac {1}{3}}\)	\(19\)
risch	\(x^{3}+22 x^{2}+131 x +{\mathrm e}^{x^{2}+\frac {1}{3}}\)	\(19\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(2*x*exp(x^2+1/3)+3*x^2+44*x+131,x,method=_RETURNVERBOSE)

[Out]

x^3+22*x^2+131*x+exp(x^2+1/3)

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maxima [A]  time = 0.35, size = 18, normalized size = 0.90 \begin {gather*} x^{3} + 22 \, x^{2} + 131 \, x + e^{\left (x^{2} + \frac {1}{3}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(x^2+1/3)+3*x^2+44*x+131,x, algorithm="maxima")

[Out]

x^3 + 22*x^2 + 131*x + e^(x^2 + 1/3)

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mupad [B]  time = 7.40, size = 18, normalized size = 0.90 \begin {gather*} 131\,x+{\mathrm {e}}^{x^2+\frac {1}{3}}+22\,x^2+x^3 \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(44*x + 2*x*exp(x^2 + 1/3) + 3*x^2 + 131,x)

[Out]

131*x + exp(x^2 + 1/3) + 22*x^2 + x^3

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sympy [A]  time = 0.08, size = 19, normalized size = 0.95 \begin {gather*} x^{3} + 22 x^{2} + 131 x + e^{x^{2} + \frac {1}{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(2*x*exp(x**2+1/3)+3*x**2+44*x+131,x)

[Out]

x**3 + 22*x**2 + 131*x + exp(x**2 + 1/3)

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