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

Optimal. Leaf size=43 \[ \frac{1}{16} \sqrt{4 x^2+3 i x} (8 x+3 i)+\frac{9}{64} i \sin ^{-1}\left (1-\frac{8 i x}{3}\right ) \]

[Out]

((3*I + 8*x)*Sqrt[(3*I)*x + 4*x^2])/16 + ((9*I)/64)*ArcSin[1 - ((8*I)/3)*x]

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Rubi [A]  time = 0.0101643, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {612, 619, 215} \[ \frac{1}{16} \sqrt{4 x^2+3 i x} (8 x+3 i)+\frac{9}{64} i \sin ^{-1}\left (1-\frac{8 i x}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[(3*I)*x + 4*x^2],x]

[Out]

((3*I + 8*x)*Sqrt[(3*I)*x + 4*x^2])/16 + ((9*I)/64)*ArcSin[1 - ((8*I)/3)*x]

Rule 612

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

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 \sqrt{3 i x+4 x^2} \, dx &=\frac{1}{16} (3 i+8 x) \sqrt{3 i x+4 x^2}+\frac{9}{32} \int \frac{1}{\sqrt{3 i x+4 x^2}} \, dx\\ &=\frac{1}{16} (3 i+8 x) \sqrt{3 i x+4 x^2}+\frac{3}{64} \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{9}}} \, dx,x,3 i+8 x\right )\\ &=\frac{1}{16} (3 i+8 x) \sqrt{3 i x+4 x^2}+\frac{9}{64} i \sin ^{-1}\left (1-\frac{8 i x}{3}\right )\\ \end{align*}

Mathematica [A]  time = 0.0520036, size = 64, normalized size = 1.49 \[ \frac{1}{32} \sqrt{x (4 x+3 i)} \left (16 x-\frac{9 \sqrt [4]{-1} \sin ^{-1}\left ((1+i) \sqrt{\frac{2}{3}} \sqrt{x}\right )}{\sqrt{3-4 i x} \sqrt{x}}+6 i\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[(3*I)*x + 4*x^2],x]

[Out]

(Sqrt[x*(3*I + 4*x)]*(6*I + 16*x - (9*(-1)^(1/4)*ArcSin[(1 + I)*Sqrt[2/3]*Sqrt[x]])/(Sqrt[3 - (4*I)*x]*Sqrt[x]
)))/32

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Maple [A]  time = 0.094, size = 31, normalized size = 0.7 \begin{align*}{\frac{3\,i+8\,x}{16}\sqrt{3\,ix+4\,{x}^{2}}}+{\frac{9}{64}{\it Arcsinh} \left ({\frac{8\,x}{3}}+i \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(3*I+8*x)*(3*I*x+4*x^2)^(1/2)+9/64*arcsinh(8/3*x+I)

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Maxima [A]  time = 1.79029, size = 66, normalized size = 1.53 \begin{align*} \frac{1}{2} \, \sqrt{4 \, x^{2} + 3 i \, x} x + \frac{3}{16} i \, \sqrt{4 \, x^{2} + 3 i \, x} + \frac{9}{64} \, \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((3*I*x+4*x^2)^(1/2),x, algorithm="maxima")

[Out]

1/2*sqrt(4*x^2 + 3*I*x)*x + 3/16*I*sqrt(4*x^2 + 3*I*x) + 9/64*log(8*x + 4*sqrt(4*x^2 + 3*I*x) + 3*I)

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Fricas [A]  time = 2.11607, size = 131, normalized size = 3.05 \begin{align*} \frac{1}{256} \, \sqrt{4 \, x^{2} + 3 i \, x}{\left (128 \, x + 48 i\right )} - \frac{9}{64} \, \log \left (-2 \, x + \sqrt{4 \, x^{2} + 3 i \, x} - \frac{3}{4} i\right ) - \frac{9}{256} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/256*sqrt(4*x^2 + 3*I*x)*(128*x + 48*I) - 9/64*log(-2*x + sqrt(4*x^2 + 3*I*x) - 3/4*I) - 9/256

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(4*x**2 + 3*I*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(4*x^2 + 3*I*x), x)