3.29.12 \(\int \frac {1}{\sqrt {-(2+3 x)^2}} \, dx\) [2812]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 28, normalized size of antiderivative = 1.00, 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 = {253, 15, 29} \begin {gather*} \frac {(3 x+2) \log (3 x+2)}{3 \sqrt {-(3 x+2)^2}} \end {gather*}

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 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]

Rule 253

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]

Rubi steps

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

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Mathematica [A]
time = 0.00, size = 28, normalized size = 1.00 \begin {gather*} \frac {(2+3 x) \log (2+3 x)}{3 \sqrt {-(2+3 x)^2}} \end {gather*}

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])

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Maple [A]
time = 0.24, size = 25, normalized size = 0.89

method	result	size
meijerg	\(-\frac {i \ln \left (1+\frac {3 x}{2}\right )}{3}\)	\(10\)
default	\(\frac {\left (2+3 x \right ) \ln \left (2+3 x \right )}{3 \sqrt {-\left (2+3 x \right )^{2}}}\)	\(25\)
risch	\(\frac {\left (2+3 x \right ) \ln \left (2+3 x \right )}{3 \sqrt {-\left (2+3 x \right )^{2}}}\)	\(25\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [C] Result contains complex when optimal does not.
time = 0.48, size = 6, normalized size = 0.21 \begin {gather*} \frac {1}{3} i \, \log \left (x + \frac {2}{3}\right ) \end {gather*}

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)

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Fricas [C] Result contains complex when optimal does not.
time = 0.36, size = 6, normalized size = 0.21 \begin {gather*} -\frac {1}{3} i \, \log \left (x + \frac {2}{3}\right ) \end {gather*}

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)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {- \left (3 x + 2\right )^{2}}}\, dx \end {gather*}

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)

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Giac [C] Result contains complex when optimal does not.
time = 1.16, size = 23, normalized size = 0.82 \begin {gather*} \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 {gather*}

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)

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Mupad [B]
time = 1.36, size = 15, normalized size = 0.54 \begin {gather*} -\frac {\ln \left (-3\,x-2\right )\,\mathrm {sign}\left (3\,x+2\right )\,1{}\mathrm {i}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(-(3*x + 2)^2)^(1/2),x)

[Out]

-(log(- 3*x - 2)*sign(3*x + 2)*1i)/3

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