3.12.26 \(\int \frac {2+6 x+3 e^3 (1+3 x)}{3+9 x} \, dx\)

Optimal. Leaf size=15 \[ \frac {2 x}{3}+e^3 \left (\frac {1}{3}+x\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 12, normalized size of antiderivative = 0.80, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {186, 21, 8} \begin {gather*} \frac {1}{3} \left (2+3 e^3\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 + 6*x + 3*E^3*(1 + 3*x))/(3 + 9*x),x]

[Out]

((2 + 3*E^3)*x)/3

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 186

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

Rubi steps

\begin {gather*} \begin {aligned} \text {integral} &=\int \frac {2+3 e^3+3 \left (2+3 e^3\right ) x}{3+9 x} \, dx\\ &=\frac {1}{3} \left (2+3 e^3\right ) \int 1 \, dx\\ &=\frac {1}{3} \left (2+3 e^3\right ) x\\ \end {aligned} \end {gather*}

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Mathematica [A]  time = 0.00, size = 9, normalized size = 0.60 \begin {gather*} \left (\frac {2}{3}+e^3\right ) x \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 + 6*x + 3*E^3*(1 + 3*x))/(3 + 9*x),x]

[Out]

(2/3 + E^3)*x

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fricas [A]  time = 0.54, size = 8, normalized size = 0.53 \begin {gather*} x e^{3} + \frac {2}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(log(x+1/3)+3)+6*x+2)/(9*x+3),x, algorithm="fricas")

[Out]

x*e^3 + 2/3*x

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giac [A]  time = 0.23, size = 8, normalized size = 0.53 \begin {gather*} x e^{3} + \frac {2}{3} \, x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(log(x+1/3)+3)+6*x+2)/(9*x+3),x, algorithm="giac")

[Out]

x*e^3 + 2/3*x

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maple [A]  time = 0.19, size = 7, normalized size = 0.47

method	result	size
norman	\(\left ({\mathrm e}^{3}+\frac {2}{3}\right ) x\)	\(7\)
default	\(\frac {2 x}{3}+x \,{\mathrm e}^{3}\)	\(9\)
risch	\(\frac {2 x}{3}+x \,{\mathrm e}^{3}\)	\(9\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((9*exp(ln(x+1/3)+3)+6*x+2)/(9*x+3),x,method=_RETURNVERBOSE)

[Out]

(exp(3)+2/3)*x

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maxima [A]  time = 0.44, size = 9, normalized size = 0.60 \begin {gather*} \frac {1}{3} \, x {\left (3 \, e^{3} + 2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(log(x+1/3)+3)+6*x+2)/(9*x+3),x, algorithm="maxima")

[Out]

1/3*x*(3*e^3 + 2)

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mupad [B]  time = 0.05, size = 6, normalized size = 0.40 \begin {gather*} x\,\left ({\mathrm {e}}^3+\frac {2}{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((6*x + 9*exp(log(x + 1/3) + 3) + 2)/(9*x + 3),x)

[Out]

x*(exp(3) + 2/3)

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sympy [A]  time = 0.06, size = 7, normalized size = 0.47 \begin {gather*} x \left (\frac {2}{3} + e^{3}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((9*exp(ln(x+1/3)+3)+6*x+2)/(9*x+3),x)

[Out]

x*(2/3 + exp(3))

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