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]

-1/2*I*arcsin(-1+8/3*I*x)

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Rubi [A]  time = 0.01, antiderivative size = 16, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [B]  time = 0.01, 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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fricas [B]  time = 0.94, size = 19, normalized size = 1.19 \[ -\frac {1}{2} \, \log \left (-2 \, x + \sqrt {4 \, x^{2} + 3 i \, x} - \frac {3}{4} i\right ) \]

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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

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 >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Warn
ing, replacing 0 by ` u`, a substitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a sub
stitution variable should perhaps be purged.Warning, replacing 0 by ` u`, a substitution variable should perha
ps be purged.

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maple [A]  time = 0.10, size = 10, normalized size = 0.62 \[ \frac {\arcsinh \left (\frac {8 x}{3}+i\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 2.93, size = 21, normalized size = 1.31 \[ \frac {1}{2} \, \log \left (8 \, x + 4 \, \sqrt {4 \, x^{2} + 3 i \, x} + 3 i\right ) \]

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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mupad [B]  time = 0.28, size = 19, normalized size = 1.19 \[ \frac {\ln \left (x+\frac {\sqrt {x\,\left (4\,x+3{}\mathrm {i}\right )}}{2}+\frac {3}{8}{}\mathrm {i}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x*3i + 4*x^2)^(1/2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {4 x^{2} + 3 i x}}\, dx \]

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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