3.5.44 \(\int x^2 \sqrt {9+4 x^2} \, dx\) [444]

Optimal. Leaf size=45 \[ \frac {9}{32} x \sqrt {9+4 x^2}+\frac {1}{4} x^3 \sqrt {9+4 x^2}-\frac {81}{64} \sinh ^{-1}\left (\frac {2 x}{3}\right ) \]

[Out]

-81/64*arcsinh(2/3*x)+9/32*x*(4*x^2+9)^(1/2)+1/4*x^3*(4*x^2+9)^(1/2)

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Rubi [A]
time = 0.01, antiderivative size = 45, 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 = {285, 327, 221} \begin {gather*} \frac {9}{32} \sqrt {4 x^2+9} x+\frac {1}{4} \sqrt {4 x^2+9} x^3-\frac {81}{64} \sinh ^{-1}\left (\frac {2 x}{3}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2*Sqrt[9 + 4*x^2],x]

[Out]

(9*x*Sqrt[9 + 4*x^2])/32 + (x^3*Sqrt[9 + 4*x^2])/4 - (81*ArcSinh[(2*x)/3])/64

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 285

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

Rule 327

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

Rubi steps

\begin {align*} \int x^2 \sqrt {9+4 x^2} \, dx &=\frac {1}{4} x^3 \sqrt {9+4 x^2}+\frac {9}{4} \int \frac {x^2}{\sqrt {9+4 x^2}} \, dx\\ &=\frac {9}{32} x \sqrt {9+4 x^2}+\frac {1}{4} x^3 \sqrt {9+4 x^2}-\frac {81}{32} \int \frac {1}{\sqrt {9+4 x^2}} \, dx\\ &=\frac {9}{32} x \sqrt {9+4 x^2}+\frac {1}{4} x^3 \sqrt {9+4 x^2}-\frac {81}{64} \sinh ^{-1}\left (\frac {2 x}{3}\right )\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 44, normalized size = 0.98 \begin {gather*} \frac {1}{32} x \sqrt {9+4 x^2} \left (9+8 x^2\right )+\frac {81}{64} \log \left (-2 x+\sqrt {9+4 x^2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2*Sqrt[9 + 4*x^2],x]

[Out]

(x*Sqrt[9 + 4*x^2]*(9 + 8*x^2))/32 + (81*Log[-2*x + Sqrt[9 + 4*x^2]])/64

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Maple [A]
time = 0.08, size = 32, normalized size = 0.71

method	result	size
risch	\(\frac {x \left (8 x^{2}+9\right ) \sqrt {4 x^{2}+9}}{32}-\frac {81 \arcsinh \left (\frac {2 x}{3}\right )}{64}\)	\(27\)
default	\(\frac {x \left (4 x^{2}+9\right )^{\frac {3}{2}}}{16}-\frac {81 \arcsinh \left (\frac {2 x}{3}\right )}{64}-\frac {9 x \sqrt {4 x^{2}+9}}{32}\)	\(32\)
meijerg	\(-\frac {81 \left (-\frac {\sqrt {\pi }\, x \left (\frac {8 x^{2}}{3}+3\right ) \sqrt {1+\frac {4 x^{2}}{9}}}{9}+\frac {\sqrt {\pi }\, \arcsinh \left (\frac {2 x}{3}\right )}{2}\right )}{32 \sqrt {\pi }}\)	\(38\)
trager	\(\frac {x \left (8 x^{2}+9\right ) \sqrt {4 x^{2}+9}}{32}+\frac {81 \ln \left (2 x -\sqrt {4 x^{2}+9}\right )}{64}\)	\(39\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(4*x^2+9)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/16*x*(4*x^2+9)^(3/2)-81/64*arcsinh(2/3*x)-9/32*x*(4*x^2+9)^(1/2)

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Maxima [A]
time = 0.61, size = 31, normalized size = 0.69 \begin {gather*} \frac {1}{16} \, {\left (4 \, x^{2} + 9\right )}^{\frac {3}{2}} x - \frac {9}{32} \, \sqrt {4 \, x^{2} + 9} x - \frac {81}{64} \, \operatorname {arsinh}\left (\frac {2}{3} \, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(4*x^2+9)^(1/2),x, algorithm="maxima")

[Out]

1/16*(4*x^2 + 9)^(3/2)*x - 9/32*sqrt(4*x^2 + 9)*x - 81/64*arcsinh(2/3*x)

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Fricas [A]
time = 1.11, size = 37, normalized size = 0.82 \begin {gather*} \frac {1}{32} \, {\left (8 \, x^{3} + 9 \, x\right )} \sqrt {4 \, x^{2} + 9} + \frac {81}{64} \, \log \left (-2 \, x + \sqrt {4 \, x^{2} + 9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(4*x^2+9)^(1/2),x, algorithm="fricas")

[Out]

1/32*(8*x^3 + 9*x)*sqrt(4*x^2 + 9) + 81/64*log(-2*x + sqrt(4*x^2 + 9))

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Sympy [A]
time = 1.64, size = 54, normalized size = 1.20 \begin {gather*} \frac {x^{5}}{\sqrt {4 x^{2} + 9}} + \frac {27 x^{3}}{8 \sqrt {4 x^{2} + 9}} + \frac {81 x}{32 \sqrt {4 x^{2} + 9}} - \frac {81 \operatorname {asinh}{\left (\frac {2 x}{3} \right )}}{64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(4*x**2+9)**(1/2),x)

[Out]

x**5/sqrt(4*x**2 + 9) + 27*x**3/(8*sqrt(4*x**2 + 9)) + 81*x/(32*sqrt(4*x**2 + 9)) - 81*asinh(2*x/3)/64

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Giac [A]
time = 0.62, size = 36, normalized size = 0.80 \begin {gather*} \frac {1}{32} \, {\left (8 \, x^{2} + 9\right )} \sqrt {4 \, x^{2} + 9} x + \frac {81}{64} \, \log \left (-2 \, x + \sqrt {4 \, x^{2} + 9}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(4*x^2+9)^(1/2),x, algorithm="giac")

[Out]

1/32*(8*x^2 + 9)*sqrt(4*x^2 + 9)*x + 81/64*log(-2*x + sqrt(4*x^2 + 9))

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Mupad [B]
time = 0.03, size = 23, normalized size = 0.51 \begin {gather*} \frac {\left (x^3+\frac {9\,x}{8}\right )\,\sqrt {x^2+\frac {9}{4}}}{2}-\frac {81\,\mathrm {asinh}\left (\frac {2\,x}{3}\right )}{64} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(4*x^2 + 9)^(1/2),x)

[Out]

(((9*x)/8 + x^3)*(x^2 + 9/4)^(1/2))/2 - (81*asinh((2*x)/3))/64

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