3.3 \(\int (3 i x+4 x^2)^{5/2} \, dx\)

Optimal. Leaf size=95 \[ \frac{1}{48} (8 x+3 i) \left (4 x^2+3 i x\right )^{5/2}+\frac{15 (8 x+3 i) \left (4 x^2+3 i x\right )^{3/2}}{1024}+\frac{405 (8 x+3 i) \sqrt{4 x^2+3 i x}}{32768}+\frac{3645 i \sin ^{-1}\left (1-\frac{8 i x}{3}\right )}{131072} \]

[Out]

(405*(3*I + 8*x)*Sqrt[(3*I)*x + 4*x^2])/32768 + (15*(3*I + 8*x)*((3*I)*x + 4*x^2)^(3/2))/1024 + ((3*I + 8*x)*(
(3*I)*x + 4*x^2)^(5/2))/48 + ((3645*I)/131072)*ArcSin[1 - ((8*I)/3)*x]

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Rubi [A]  time = 0.02172, antiderivative size = 95, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {612, 619, 215} \[ \frac{1}{48} (8 x+3 i) \left (4 x^2+3 i x\right )^{5/2}+\frac{15 (8 x+3 i) \left (4 x^2+3 i x\right )^{3/2}}{1024}+\frac{405 (8 x+3 i) \sqrt{4 x^2+3 i x}}{32768}+\frac{3645 i \sin ^{-1}\left (1-\frac{8 i x}{3}\right )}{131072} \]

Antiderivative was successfully verified.

[In]

Int[((3*I)*x + 4*x^2)^(5/2),x]

[Out]

(405*(3*I + 8*x)*Sqrt[(3*I)*x + 4*x^2])/32768 + (15*(3*I + 8*x)*((3*I)*x + 4*x^2)^(3/2))/1024 + ((3*I + 8*x)*(
(3*I)*x + 4*x^2)^(5/2))/48 + ((3645*I)/131072)*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 )^{5/2} \, dx &=\frac{1}{48} (3 i+8 x) \left (3 i x+4 x^2\right )^{5/2}+\frac{15}{32} \int \left (3 i x+4 x^2\right )^{3/2} \, dx\\ &=\frac{15 (3 i+8 x) \left (3 i x+4 x^2\right )^{3/2}}{1024}+\frac{1}{48} (3 i+8 x) \left (3 i x+4 x^2\right )^{5/2}+\frac{405 \int \sqrt{3 i x+4 x^2} \, dx}{2048}\\ &=\frac{405 (3 i+8 x) \sqrt{3 i x+4 x^2}}{32768}+\frac{15 (3 i+8 x) \left (3 i x+4 x^2\right )^{3/2}}{1024}+\frac{1}{48} (3 i+8 x) \left (3 i x+4 x^2\right )^{5/2}+\frac{3645 \int \frac{1}{\sqrt{3 i x+4 x^2}} \, dx}{65536}\\ &=\frac{405 (3 i+8 x) \sqrt{3 i x+4 x^2}}{32768}+\frac{15 (3 i+8 x) \left (3 i x+4 x^2\right )^{3/2}}{1024}+\frac{1}{48} (3 i+8 x) \left (3 i x+4 x^2\right )^{5/2}+\frac{1215 \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{9}}} \, dx,x,3 i+8 x\right )}{131072}\\ &=\frac{405 (3 i+8 x) \sqrt{3 i x+4 x^2}}{32768}+\frac{15 (3 i+8 x) \left (3 i x+4 x^2\right )^{3/2}}{1024}+\frac{1}{48} (3 i+8 x) \left (3 i x+4 x^2\right )^{5/2}+\frac{3645 i \sin ^{-1}\left (1-\frac{8 i x}{3}\right )}{131072}\\ \end{align*}

Mathematica [A]  time = 0.0870257, size = 88, normalized size = 0.93 \[ \frac{\sqrt{x (4 x+3 i)} \left (524288 x^5+983040 i x^4-497664 x^3-6912 i x^2-6480 x-\frac{10935 \sqrt [4]{-1} \sin ^{-1}\left ((1+i) \sqrt{\frac{2}{3}} \sqrt{x}\right )}{\sqrt{3-4 i x} \sqrt{x}}+7290 i\right )}{196608} \]

Antiderivative was successfully verified.

[In]

Integrate[((3*I)*x + 4*x^2)^(5/2),x]

[Out]

(Sqrt[x*(3*I + 4*x)]*(7290*I - 6480*x - (6912*I)*x^2 - 497664*x^3 + (983040*I)*x^4 + 524288*x^5 - (10935*(-1)^
(1/4)*ArcSin[(1 + I)*Sqrt[2/3]*Sqrt[x]])/(Sqrt[3 - (4*I)*x]*Sqrt[x])))/196608

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Maple [A]  time = 0.094, size = 71, normalized size = 0.8 \begin{align*}{\frac{3\,i+8\,x}{48} \left ( 3\,ix+4\,{x}^{2} \right ) ^{{\frac{5}{2}}}}+{\frac{45\,i+120\,x}{1024} \left ( 3\,ix+4\,{x}^{2} \right ) ^{{\frac{3}{2}}}}+{\frac{1215\,i+3240\,x}{32768}\sqrt{3\,ix+4\,{x}^{2}}}+{\frac{3645}{131072}{\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)^(5/2),x)

[Out]

1/48*(3*I+8*x)*(3*I*x+4*x^2)^(5/2)+15/1024*(3*I+8*x)*(3*I*x+4*x^2)^(3/2)+405/32768*(3*I+8*x)*(3*I*x+4*x^2)^(1/
2)+3645/131072*arcsinh(8/3*x+I)

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Maxima [A]  time = 1.78468, size = 139, normalized size = 1.46 \begin{align*} \frac{1}{6} \,{\left (4 \, x^{2} + 3 i \, x\right )}^{\frac{5}{2}} x + \frac{1}{16} i \,{\left (4 \, x^{2} + 3 i \, x\right )}^{\frac{5}{2}} + \frac{15}{128} \,{\left (4 \, x^{2} + 3 i \, x\right )}^{\frac{3}{2}} x + \frac{45}{1024} i \,{\left (4 \, x^{2} + 3 i \, x\right )}^{\frac{3}{2}} + \frac{405}{4096} \, \sqrt{4 \, x^{2} + 3 i \, x} x + \frac{1215}{32768} i \, \sqrt{4 \, x^{2} + 3 i \, x} + \frac{3645}{131072} \, \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)^(5/2),x, algorithm="maxima")

[Out]

1/6*(4*x^2 + 3*I*x)^(5/2)*x + 1/16*I*(4*x^2 + 3*I*x)^(5/2) + 15/128*(4*x^2 + 3*I*x)^(3/2)*x + 45/1024*I*(4*x^2
+ 3*I*x)^(3/2) + 405/4096*sqrt(4*x^2 + 3*I*x)*x + 1215/32768*I*sqrt(4*x^2 + 3*I*x) + 3645/131072*log(8*x + 4*
sqrt(4*x^2 + 3*I*x) + 3*I)

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Fricas [A]  time = 2.2787, size = 246, normalized size = 2.59 \begin{align*} \frac{1}{3145728} \,{\left (8388608 \, x^{5} + 15728640 i \, x^{4} - 7962624 \, x^{3} - 110592 i \, x^{2} - 103680 \, x + 116640 i\right )} \sqrt{4 \, x^{2} + 3 i \, x} - \frac{3645}{131072} \, \log \left (-2 \, x + \sqrt{4 \, x^{2} + 3 i \, x} - \frac{3}{4} i\right ) - \frac{8991}{1048576} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*I*x+4*x^2)^(5/2),x, algorithm="fricas")

[Out]

1/3145728*(8388608*x^5 + 15728640*I*x^4 - 7962624*x^3 - 110592*I*x^2 - 103680*x + 116640*I)*sqrt(4*x^2 + 3*I*x
) - 3645/131072*log(-2*x + sqrt(4*x^2 + 3*I*x) - 3/4*I) - 8991/1048576

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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{5}{2}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*I*x+4*x**2)**(5/2),x)

[Out]

Integral((4*x**2 + 3*I*x)**(5/2), 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{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((3*I*x+4*x^2)^(5/2),x, algorithm="giac")

[Out]

integrate((4*x^2 + 3*I*x)^(5/2), x)