3.15.35 \(\int \frac {1}{4} e^{\frac {1}{4} (x+4 x^3)} (1+12 x^2) \, dx\)

Optimal. Leaf size=13 \[ 3+e^{x \left (\frac {1}{4}+x^2\right )} \]

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Rubi [A]  time = 0.04, antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {12, 6706} \begin {gather*} e^{\frac {1}{4} \left (4 x^3+x\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(E^((x + 4*x^3)/4)*(1 + 12*x^2))/4,x]

[Out]

E^((x + 4*x^3)/4)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 6706

Int[(F_)^(v_)*(u_), x_Symbol] :> With[{q = DerivativeDivides[v, u, x]}, Simp[(q*F^v)/Log[F], x] /;  !FalseQ[q]
] /; FreeQ[F, x]

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\frac {1}{4} \int e^{\frac {1}{4} \left (x+4 x^3\right )} \left (1+12 x^2\right ) \, dx\\ &=e^{\frac {1}{4} \left (x+4 x^3\right )}\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.03, size = 11, normalized size = 0.85 \begin {gather*} e^{\frac {x}{4}+x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(E^((x + 4*x^3)/4)*(1 + 12*x^2))/4,x]

[Out]

E^(x/4 + x^3)

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fricas [A]  time = 0.91, size = 8, normalized size = 0.62 \begin {gather*} e^{\left (x^{3} + \frac {1}{4} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(12*x^2+1)*exp(x^3+1/4*x),x, algorithm="fricas")

[Out]

e^(x^3 + 1/4*x)

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giac [A]  time = 0.66, size = 8, normalized size = 0.62 \begin {gather*} e^{\left (x^{3} + \frac {1}{4} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(12*x^2+1)*exp(x^3+1/4*x),x, algorithm="giac")

[Out]

e^(x^3 + 1/4*x)

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maple [A]  time = 0.02, size = 9, normalized size = 0.69

method	result	size
gosper	\({\mathrm e}^{x^{3}+\frac {1}{4} x}\)	\(9\)
norman	\({\mathrm e}^{x^{3}+\frac {1}{4} x}\)	\(9\)
risch	\({\mathrm e}^{\frac {x \left (4 x^{2}+1\right )}{4}}\)	\(12\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/4*(12*x^2+1)*exp(x^3+1/4*x),x,method=_RETURNVERBOSE)

[Out]

exp(x^3+1/4*x)

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maxima [A]  time = 0.49, size = 8, normalized size = 0.62 \begin {gather*} e^{\left (x^{3} + \frac {1}{4} \, x\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(12*x^2+1)*exp(x^3+1/4*x),x, algorithm="maxima")

[Out]

e^(x^3 + 1/4*x)

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mupad [B]  time = 0.04, size = 8, normalized size = 0.62 \begin {gather*} {\mathrm {e}}^{x^3+\frac {x}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((exp(x/4 + x^3)*(12*x^2 + 1))/4,x)

[Out]

exp(x/4 + x^3)

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sympy [A]  time = 0.09, size = 7, normalized size = 0.54 \begin {gather*} e^{x^{3} + \frac {x}{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/4*(12*x**2+1)*exp(x**3+1/4*x),x)

[Out]

exp(x**3 + x/4)

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