3.2812 \(\int \frac{1}{\sqrt{-(2+3 x)^2}} \, dx\)

Optimal. Leaf size=28 \[ \frac{(3 x+2) \log (3 x+2)}{3 \sqrt{-(3 x+2)^2}} \]

[Out]

((2 + 3*x)*Log[2 + 3*x])/(3*Sqrt[-(2 + 3*x)^2])

________________________________________________________________________________________

Rubi [A]  time = 0.008147, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.231, Rules used = {247, 15, 29} \[ \frac{(3 x+2) \log (3 x+2)}{3 \sqrt{-(3 x+2)^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[-(2 + 3*x)^2],x]

[Out]

((2 + 3*x)*Log[2 + 3*x])/(3*Sqrt[-(2 + 3*x)^2])

Rule 247

Int[((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[1/Coefficient[v, x, 1], Subst[Int[(a + b*x^n)^p, x], x,
v], x] /; FreeQ[{a, b, n, p}, x] && LinearQ[v, x] && NeQ[v, x]

Rule 15

Int[(u_.)*((a_.)*(x_)^(n_))^(m_), x_Symbol] :> Dist[(a^IntPart[m]*(a*x^n)^FracPart[m])/x^(n*FracPart[m]), Int[
u*x^(m*n), x], x] /; FreeQ[{a, m, n}, x] &&  !IntegerQ[m]

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{-(2+3 x)^2}} \, dx &=\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{\sqrt{-x^2}} \, dx,x,2+3 x\right )\\ &=\frac{(2+3 x) \operatorname{Subst}\left (\int \frac{1}{x} \, dx,x,2+3 x\right )}{3 \sqrt{-(2+3 x)^2}}\\ &=\frac{(2+3 x) \log (2+3 x)}{3 \sqrt{-(2+3 x)^2}}\\ \end{align*}

Mathematica [A]  time = 0.0057903, size = 28, normalized size = 1. \[ \frac{(3 x+2) \log (3 x+2)}{3 \sqrt{-(3 x+2)^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[-(2 + 3*x)^2],x]

[Out]

((2 + 3*x)*Log[2 + 3*x])/(3*Sqrt[-(2 + 3*x)^2])

________________________________________________________________________________________

Maple [A]  time = 0.005, size = 25, normalized size = 0.9 \begin{align*}{\frac{ \left ( 2+3\,x \right ) \ln \left ( 2+3\,x \right ) }{3}{\frac{1}{\sqrt{- \left ( 2+3\,x \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-(2+3*x)^2)^(1/2),x)

[Out]

1/3*(2+3*x)*ln(2+3*x)/(-(2+3*x)^2)^(1/2)

________________________________________________________________________________________

Maxima [C]  time = 1.46836, size = 8, normalized size = 0.29 \begin{align*} \frac{1}{3} i \, \log \left (x + \frac{2}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*I*log(x + 2/3)

________________________________________________________________________________________

Fricas [C]  time = 1.18553, size = 28, normalized size = 1. \begin{align*} -\frac{1}{3} i \, \log \left (x + \frac{2}{3}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*I*log(x + 2/3)

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{- \left (3 x + 2\right )^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(-(2+3*x)**2)**(1/2),x)

[Out]

Integral(1/sqrt(-(3*x + 2)**2), x)

________________________________________________________________________________________

Giac [C]  time = 1.16224, size = 31, normalized size = 1.11 \begin{align*} \frac{i \, \log \left ({\left (-3 i \, x - 2 i\right )} \mathrm{sgn}\left (-3 \, x - 2\right )\right )}{3 \, \mathrm{sgn}\left (-3 \, x - 2\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*I*log((-3*I*x - 2*I)*sgn(-3*x - 2))/sgn(-3*x - 2)