3.1.5 \(\int \sqrt {3 i x+4 x^2} \, dx\) [5]

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

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {626, 633, 221} \begin {gather*} \frac {9}{64} i \text {ArcSin}\left (1-\frac {8 i x}{3}\right )+\frac {1}{16} \sqrt {4 x^2+3 i x} (8 x+3 i) \end {gather*}

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 221

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 626

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 633

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 \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} \text {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*}

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Mathematica [A]
time = 0.05, size = 62, normalized size = 1.44 \begin {gather*} \frac {1}{32} \sqrt {x (3 i+4 x)} \left (6 i+16 x-\frac {9 \log \left (-2 \sqrt {x}+\sqrt {3 i+4 x}\right )}{\sqrt {x} \sqrt {3 i+4 x}}\right ) \end {gather*}

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*Log[-2*Sqrt[x] + Sqrt[3*I + 4*x]])/(Sqrt[x]*Sqrt[3*I + 4*x])))/32

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Maple [A]
time = 0.46, size = 31, normalized size = 0.72

method	result	size
default	\(\frac {\left (3 i+8 x \right ) \sqrt {4 x^{2}+3 i x}}{16}+\frac {9 \arcsinh \left (i+\frac {8 x}{3}\right )}{64}\)	\(31\)
risch	\(\frac {\left (3 i+8 x \right ) x \left (3 i+4 x \right )}{16 \sqrt {x \left (3 i+4 x \right )}}+\frac {9 \arcsinh \left (i+\frac {8 x}{3}\right )}{64}\)	\(36\)
trager	\(\left (\frac {3 i}{16}+\frac {x}{2}\right ) \sqrt {4 x^{2}+3 i x}+\frac {9 \ln \left (440 x +144+165 i-192 i \sqrt {4 x^{2}+3 i x}-384 i x +220 \sqrt {4 x^{2}+3 i x}\right )}{64}\)	\(64\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.52, size = 49, normalized size = 1.14 \begin {gather*} \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 {gather*}

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 = 1.44, size = 39, normalized size = 0.91 \begin {gather*} \frac {1}{16} \, \sqrt {4 \, x^{2} + 3 i \, x} {\left (8 \, x + 3 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 {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*sqrt(4*x^2 + 3*I*x)*(8*x + 3*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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {4 x^{2} + 3 i x}\, dx \end {gather*}

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 [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (27) = 54\).
time = 2.61, size = 110, normalized size = 2.56 \begin {gather*} \frac {1}{32} \, \sqrt {8 \, x^{2} + 2 \, \sqrt {16 \, x^{2} + 9} x} {\left (8 \, x + 3 i\right )} {\left (\frac {3 i \, x}{4 \, x^{2} + \sqrt {16 \, x^{4} + 9 \, x^{2}}} + 1\right )} - \frac {9}{64} \, \log \left (2 \, \sqrt {8 \, x^{2} + 2 \, \sqrt {16 \, x^{2} + 9} x} {\left (\frac {3 i \, x}{4 \, x^{2} + \sqrt {16 \, x^{4} + 9 \, x^{2}}} + 1\right )} - 8 \, x - 3 i\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*sqrt(8*x^2 + 2*sqrt(16*x^2 + 9)*x)*(8*x + 3*I)*(3*I*x/(4*x^2 + sqrt(16*x^4 + 9*x^2)) + 1) - 9/64*log(2*sq
rt(8*x^2 + 2*sqrt(16*x^2 + 9)*x)*(3*I*x/(4*x^2 + sqrt(16*x^4 + 9*x^2)) + 1) - 8*x - 3*I)

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Mupad [B]
time = 0.09, size = 39, normalized size = 0.91 \begin {gather*} \frac {9\,\ln \left (x+\frac {\sqrt {x\,\left (4\,x+3{}\mathrm {i}\right )}}{2}+\frac {3}{8}{}\mathrm {i}\right )}{64}+\left (\frac {x}{2}+\frac {3}{16}{}\mathrm {i}\right )\,\sqrt {4\,x^2+x\,3{}\mathrm {i}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x*3i + 4*x^2)^(1/2),x)

[Out]

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

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