3.4 \(\int (3 i x+4 x^2)^{3/2} \, dx\)

Optimal. Leaf size=69 \[ \frac{1}{32} (8 x+3 i) \left (4 x^2+3 i x\right )^{3/2}+\frac{27 (8 x+3 i) \sqrt{4 x^2+3 i x}}{1024}+\frac{243 i \sin ^{-1}\left (1-\frac{8 i x}{3}\right )}{4096} \]

[Out]

(27*(3*I + 8*x)*Sqrt[(3*I)*x + 4*x^2])/1024 + ((3*I + 8*x)*((3*I)*x + 4*x^2)^(3/2))/32 + ((243*I)/4096)*ArcSin
[1 - ((8*I)/3)*x]

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Rubi [A]  time = 0.0148143, antiderivative size = 69, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {612, 619, 215} \[ \frac{1}{32} (8 x+3 i) \left (4 x^2+3 i x\right )^{3/2}+\frac{27 (8 x+3 i) \sqrt{4 x^2+3 i x}}{1024}+\frac{243 i \sin ^{-1}\left (1-\frac{8 i x}{3}\right )}{4096} \]

Antiderivative was successfully verified.

[In]

Int[((3*I)*x + 4*x^2)^(3/2),x]

[Out]

(27*(3*I + 8*x)*Sqrt[(3*I)*x + 4*x^2])/1024 + ((3*I + 8*x)*((3*I)*x + 4*x^2)^(3/2))/32 + ((243*I)/4096)*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 \left (3 i x+4 x^2\right )^{3/2} \, dx &=\frac{1}{32} (3 i+8 x) \left (3 i x+4 x^2\right )^{3/2}+\frac{27}{64} \int \sqrt{3 i x+4 x^2} \, dx\\ &=\frac{27 (3 i+8 x) \sqrt{3 i x+4 x^2}}{1024}+\frac{1}{32} (3 i+8 x) \left (3 i x+4 x^2\right )^{3/2}+\frac{243 \int \frac{1}{\sqrt{3 i x+4 x^2}} \, dx}{2048}\\ &=\frac{27 (3 i+8 x) \sqrt{3 i x+4 x^2}}{1024}+\frac{1}{32} (3 i+8 x) \left (3 i x+4 x^2\right )^{3/2}+\frac{81 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{9}}} \, dx,x,3 i+8 x\right )}{4096}\\ &=\frac{27 (3 i+8 x) \sqrt{3 i x+4 x^2}}{1024}+\frac{1}{32} (3 i+8 x) \left (3 i x+4 x^2\right )^{3/2}+\frac{243 i \sin ^{-1}\left (1-\frac{8 i x}{3}\right )}{4096}\\ \end{align*}

Mathematica [A]  time = 0.0703425, size = 76, normalized size = 1.1 \[ \frac{\sqrt{x (4 x+3 i)} \left (2048 x^3+2304 i x^2-144 x-\frac{243 \sqrt [4]{-1} \sin ^{-1}\left ((1+i) \sqrt{\frac{2}{3}} \sqrt{x}\right )}{\sqrt{3-4 i x} \sqrt{x}}+162 i\right )}{2048} \]

Antiderivative was successfully verified.

[In]

Integrate[((3*I)*x + 4*x^2)^(3/2),x]

[Out]

(Sqrt[x*(3*I + 4*x)]*(162*I - 144*x + (2304*I)*x^2 + 2048*x^3 - (243*(-1)^(1/4)*ArcSin[(1 + I)*Sqrt[2/3]*Sqrt[
x]])/(Sqrt[3 - (4*I)*x]*Sqrt[x])))/2048

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Maple [A]  time = 0.095, size = 51, normalized size = 0.7 \begin{align*}{\frac{3\,i+8\,x}{32} \left ( 3\,ix+4\,{x}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{81\,i+216\,x}{1024}\sqrt{3\,ix+4\,{x}^{2}}}+{\frac{243}{4096}{\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)^(3/2),x)

[Out]

1/32*(3*I+8*x)*(3*I*x+4*x^2)^(3/2)+27/1024*(3*I+8*x)*(3*I*x+4*x^2)^(1/2)+243/4096*arcsinh(8/3*x+I)

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Maxima [A]  time = 1.71097, size = 103, normalized size = 1.49 \begin{align*} \frac{1}{4} \,{\left (4 \, x^{2} + 3 i \, x\right )}^{\frac{3}{2}} x + \frac{3}{32} i \,{\left (4 \, x^{2} + 3 i \, x\right )}^{\frac{3}{2}} + \frac{27}{128} \, \sqrt{4 \, x^{2} + 3 i \, x} x + \frac{81}{1024} i \, \sqrt{4 \, x^{2} + 3 i \, x} + \frac{243}{4096} \, \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)^(3/2),x, algorithm="maxima")

[Out]

1/4*(4*x^2 + 3*I*x)^(3/2)*x + 3/32*I*(4*x^2 + 3*I*x)^(3/2) + 27/128*sqrt(4*x^2 + 3*I*x)*x + 81/1024*I*sqrt(4*x
^2 + 3*I*x) + 243/4096*log(8*x + 4*sqrt(4*x^2 + 3*I*x) + 3*I)

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Fricas [A]  time = 2.28958, size = 184, normalized size = 2.67 \begin{align*} \frac{1}{32768} \,{\left (32768 \, x^{3} + 36864 i \, x^{2} - 2304 \, x + 2592 i\right )} \sqrt{4 \, x^{2} + 3 i \, x} - \frac{243}{4096} \, \log \left (-2 \, x + \sqrt{4 \, x^{2} + 3 i \, x} - \frac{3}{4} i\right ) - \frac{567}{32768} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32768*(32768*x^3 + 36864*I*x^2 - 2304*x + 2592*I)*sqrt(4*x^2 + 3*I*x) - 243/4096*log(-2*x + sqrt(4*x^2 + 3*I
*x) - 3/4*I) - 567/32768

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((4*x**2 + 3*I*x)**(3/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((4*x^2 + 3*I*x)^(3/2), x)