3.15.35 \(\int \frac {1}{4} e^{\frac {1}{4} (x+4 x^3)} (1+12 x^2) \, dx\) [1435]

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

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, 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, 6838} \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 6838

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*}

________________________________________________________________________________________

Mathematica [A]
time = 0.01, 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)

________________________________________________________________________________________

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)

________________________________________________________________________________________

Maxima [A]
time = 0.25, 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)

________________________________________________________________________________________

Fricas [A]
time = 0.37, 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)

________________________________________________________________________________________

Sympy [A]
time = 0.03, 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)

________________________________________________________________________________________

Giac [A]
time = 0.41, 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)

________________________________________________________________________________________

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)

________________________________________________________________________________________