3.442 \(\int x^4 \sqrt {9+4 x^2} \, dx\)

Optimal. Leaf size=63 \[ -\frac {81}{256} \sqrt {4 x^2+9} x+\frac {1}{6} \sqrt {4 x^2+9} x^5+\frac {3}{32} \sqrt {4 x^2+9} x^3+\frac {729}{512} \sinh ^{-1}\left (\frac {2 x}{3}\right ) \]

[Out]

729/512*arcsinh(2/3*x)-81/256*x*(4*x^2+9)^(1/2)+3/32*x^3*(4*x^2+9)^(1/2)+1/6*x^5*(4*x^2+9)^(1/2)

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Rubi [A]  time = 0.02, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {279, 321, 215} \[ \frac {1}{6} \sqrt {4 x^2+9} x^5+\frac {3}{32} \sqrt {4 x^2+9} x^3-\frac {81}{256} \sqrt {4 x^2+9} x+\frac {729}{512} \sinh ^{-1}\left (\frac {2 x}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^4*Sqrt[9 + 4*x^2],x]

[Out]

(-81*x*Sqrt[9 + 4*x^2])/256 + (3*x^3*Sqrt[9 + 4*x^2])/32 + (x^5*Sqrt[9 + 4*x^2])/6 + (729*ArcSinh[(2*x)/3])/51
2

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]

Rule 279

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 321

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^4 \sqrt {9+4 x^2} \, dx &=\frac {1}{6} x^5 \sqrt {9+4 x^2}+\frac {3}{2} \int \frac {x^4}{\sqrt {9+4 x^2}} \, dx\\ &=\frac {3}{32} x^3 \sqrt {9+4 x^2}+\frac {1}{6} x^5 \sqrt {9+4 x^2}-\frac {81}{32} \int \frac {x^2}{\sqrt {9+4 x^2}} \, dx\\ &=-\frac {81}{256} x \sqrt {9+4 x^2}+\frac {3}{32} x^3 \sqrt {9+4 x^2}+\frac {1}{6} x^5 \sqrt {9+4 x^2}+\frac {729}{256} \int \frac {1}{\sqrt {9+4 x^2}} \, dx\\ &=-\frac {81}{256} x \sqrt {9+4 x^2}+\frac {3}{32} x^3 \sqrt {9+4 x^2}+\frac {1}{6} x^5 \sqrt {9+4 x^2}+\frac {729}{512} \sinh ^{-1}\left (\frac {2 x}{3}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 39, normalized size = 0.62 \[ \frac {1}{768} x \sqrt {4 x^2+9} \left (128 x^4+72 x^2-243\right )+\frac {729}{512} \sinh ^{-1}\left (\frac {2 x}{3}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*Sqrt[9 + 4*x^2],x]

[Out]

(x*Sqrt[9 + 4*x^2]*(-243 + 72*x^2 + 128*x^4))/768 + (729*ArcSinh[(2*x)/3])/512

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fricas [A]  time = 0.97, size = 42, normalized size = 0.67 \[ \frac {1}{768} \, {\left (128 \, x^{5} + 72 \, x^{3} - 243 \, x\right )} \sqrt {4 \, x^{2} + 9} - \frac {729}{512} \, \log \left (-2 \, x + \sqrt {4 \, x^{2} + 9}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(128*x^5 + 72*x^3 - 243*x)*sqrt(4*x^2 + 9) - 729/512*log(-2*x + sqrt(4*x^2 + 9))

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giac [A]  time = 1.17, size = 43, normalized size = 0.68 \[ \frac {1}{768} \, {\left (8 \, {\left (16 \, x^{2} + 9\right )} x^{2} - 243\right )} \sqrt {4 \, x^{2} + 9} x - \frac {729}{512} \, \log \left (-2 \, x + \sqrt {4 \, x^{2} + 9}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/768*(8*(16*x^2 + 9)*x^2 - 243)*sqrt(4*x^2 + 9)*x - 729/512*log(-2*x + sqrt(4*x^2 + 9))

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maple [A]  time = 0.01, size = 46, normalized size = 0.73 \[ \frac {\left (4 x^{2}+9\right )^{\frac {3}{2}} x^{3}}{24}-\frac {9 \left (4 x^{2}+9\right )^{\frac {3}{2}} x}{128}+\frac {81 \sqrt {4 x^{2}+9}\, x}{256}+\frac {729 \arcsinh \left (\frac {2 x}{3}\right )}{512} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*x^3*(4*x^2+9)^(3/2)-9/128*x*(4*x^2+9)^(3/2)+81/256*x*(4*x^2+9)^(1/2)+729/512*arcsinh(2/3*x)

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maxima [A]  time = 2.94, size = 45, normalized size = 0.71 \[ \frac {1}{24} \, {\left (4 \, x^{2} + 9\right )}^{\frac {3}{2}} x^{3} - \frac {9}{128} \, {\left (4 \, x^{2} + 9\right )}^{\frac {3}{2}} x + \frac {81}{256} \, \sqrt {4 \, x^{2} + 9} x + \frac {729}{512} \, \operatorname {arsinh}\left (\frac {2}{3} \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(4*x^2 + 9)^(3/2)*x^3 - 9/128*(4*x^2 + 9)^(3/2)*x + 81/256*sqrt(4*x^2 + 9)*x + 729/512*arcsinh(2/3*x)

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mupad [B]  time = 0.03, size = 30, normalized size = 0.48 \[ \frac {729\,\mathrm {asinh}\left (\frac {2\,x}{3}\right )}{512}+\frac {\sqrt {x^2+\frac {9}{4}}\,\left (\frac {2\,x^5}{3}+\frac {3\,x^3}{8}-\frac {81\,x}{64}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*(4*x^2 + 9)^(1/2),x)

[Out]

(729*asinh((2*x)/3))/512 + ((x^2 + 9/4)^(1/2)*((3*x^3)/8 - (81*x)/64 + (2*x^5)/3))/2

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sympy [A]  time = 4.56, size = 75, normalized size = 1.19 \[ \frac {2 x^{7}}{3 \sqrt {4 x^{2} + 9}} + \frac {15 x^{5}}{8 \sqrt {4 x^{2} + 9}} - \frac {27 x^{3}}{64 \sqrt {4 x^{2} + 9}} - \frac {729 x}{256 \sqrt {4 x^{2} + 9}} + \frac {729 \operatorname {asinh}{\left (\frac {2 x}{3} \right )}}{512} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*x**7/(3*sqrt(4*x**2 + 9)) + 15*x**5/(8*sqrt(4*x**2 + 9)) - 27*x**3/(64*sqrt(4*x**2 + 9)) - 729*x/(256*sqrt(4
*x**2 + 9)) + 729*asinh(2*x/3)/512

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