3.17 \(\int \frac{1}{\sqrt{3 i x+4 x^2}} \, dx\)

Optimal. Leaf size=16 \[ \frac{1}{2} i \sin ^{-1}\left (1-\frac{8 i x}{3}\right ) \]

[Out]

(I/2)*ArcSin[1 - ((8*I)/3)*x]

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Rubi [A]  time = 0.0064543, antiderivative size = 16, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.133, Rules used = {619, 215} \[ \frac{1}{2} i \sin ^{-1}\left (1-\frac{8 i x}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[(3*I)*x + 4*x^2],x]

[Out]

(I/2)*ArcSin[1 - ((8*I)/3)*x]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rubi steps

\begin{align*} \int \frac{1}{\sqrt{3 i x+4 x^2}} \, dx &=\frac{1}{6} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{9}}} \, dx,x,3 i+8 x\right )\\ &=\frac{1}{2} i \sin ^{-1}\left (1-\frac{8 i x}{3}\right )\\ \end{align*}

Mathematica [B]  time = 0.0173975, size = 53, normalized size = 3.31 \[ -\frac{(-1)^{3/4} \sqrt{3-4 i x} \sqrt{x} \sin ^{-1}\left ((1+i) \sqrt{\frac{2}{3}} \sqrt{x}\right )}{\sqrt{x (4 x+3 i)}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[(3*I)*x + 4*x^2],x]

[Out]

-(((-1)^(3/4)*Sqrt[3 - (4*I)*x]*Sqrt[x]*ArcSin[(1 + I)*Sqrt[2/3]*Sqrt[x]])/Sqrt[x*(3*I + 4*x)])

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Maple [A]  time = 0.105, size = 10, normalized size = 0.6 \begin{align*}{\frac{1}{2}{\it Arcsinh} \left ({\frac{8\,x}{3}}+i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*arcsinh(8/3*x+I)

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Maxima [B]  time = 1.79627, size = 28, normalized size = 1.75 \begin{align*} \frac{1}{2} \, \log \left (8 \, x + 4 \, \sqrt{4 \, x^{2} + 3 i \, x} + 3 i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*log(8*x + 4*sqrt(4*x^2 + 3*I*x) + 3*I)

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Fricas [B]  time = 2.4758, size = 62, normalized size = 3.88 \begin{align*} -\frac{1}{2} \, \log \left (-2 \, x + \sqrt{4 \, x^{2} + 3 i \, x} - \frac{3}{4} i\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*log(-2*x + sqrt(4*x^2 + 3*I*x) - 3/4*I)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{4 x^{2} + 3 i x}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(1/sqrt(4*x**2 + 3*I*x), x)

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: TypeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError