3.462 \(\int \frac{(-10+x^2)^{5/2}}{x} \, dx\)

Optimal. Leaf size=61 \[ \frac{1}{5} \left (x^2-10\right )^{5/2}-\frac{10}{3} \left (x^2-10\right )^{3/2}+100 \sqrt{x^2-10}-100 \sqrt{10} \tan ^{-1}\left (\frac{\sqrt{x^2-10}}{\sqrt{10}}\right ) \]

[Out]

100*Sqrt[-10 + x^2] - (10*(-10 + x^2)^(3/2))/3 + (-10 + x^2)^(5/2)/5 - 100*Sqrt[10]*ArcTan[Sqrt[-10 + x^2]/Sqr
t[10]]

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Rubi [A]  time = 0.0318791, antiderivative size = 61, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 13, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.308, Rules used = {266, 50, 63, 203} \[ \frac{1}{5} \left (x^2-10\right )^{5/2}-\frac{10}{3} \left (x^2-10\right )^{3/2}+100 \sqrt{x^2-10}-100 \sqrt{10} \tan ^{-1}\left (\frac{\sqrt{x^2-10}}{\sqrt{10}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(-10 + x^2)^(5/2)/x,x]

[Out]

100*Sqrt[-10 + x^2] - (10*(-10 + x^2)^(3/2))/3 + (-10 + x^2)^(5/2)/5 - 100*Sqrt[10]*ArcTan[Sqrt[-10 + x^2]/Sqr
t[10]]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\left (-10+x^2\right )^{5/2}}{x} \, dx &=\frac{1}{2} \operatorname{Subst}\left (\int \frac{(-10+x)^{5/2}}{x} \, dx,x,x^2\right )\\ &=\frac{1}{5} \left (-10+x^2\right )^{5/2}-5 \operatorname{Subst}\left (\int \frac{(-10+x)^{3/2}}{x} \, dx,x,x^2\right )\\ &=-\frac{10}{3} \left (-10+x^2\right )^{3/2}+\frac{1}{5} \left (-10+x^2\right )^{5/2}+50 \operatorname{Subst}\left (\int \frac{\sqrt{-10+x}}{x} \, dx,x,x^2\right )\\ &=100 \sqrt{-10+x^2}-\frac{10}{3} \left (-10+x^2\right )^{3/2}+\frac{1}{5} \left (-10+x^2\right )^{5/2}-500 \operatorname{Subst}\left (\int \frac{1}{\sqrt{-10+x} x} \, dx,x,x^2\right )\\ &=100 \sqrt{-10+x^2}-\frac{10}{3} \left (-10+x^2\right )^{3/2}+\frac{1}{5} \left (-10+x^2\right )^{5/2}-1000 \operatorname{Subst}\left (\int \frac{1}{10+x^2} \, dx,x,\sqrt{-10+x^2}\right )\\ &=100 \sqrt{-10+x^2}-\frac{10}{3} \left (-10+x^2\right )^{3/2}+\frac{1}{5} \left (-10+x^2\right )^{5/2}-100 \sqrt{10} \tan ^{-1}\left (\frac{\sqrt{-10+x^2}}{\sqrt{10}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0177892, size = 47, normalized size = 0.77 \[ \frac{1}{15} \sqrt{x^2-10} \left (3 x^4-110 x^2+2300\right )-100 \sqrt{10} \tan ^{-1}\left (\sqrt{\frac{x^2}{10}-1}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(-10 + x^2)^(5/2)/x,x]

[Out]

(Sqrt[-10 + x^2]*(2300 - 110*x^2 + 3*x^4))/15 - 100*Sqrt[10]*ArcTan[Sqrt[-1 + x^2/10]]

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Maple [A]  time = 0.005, size = 46, normalized size = 0.8 \begin{align*}{\frac{1}{5} \left ({x}^{2}-10 \right ) ^{{\frac{5}{2}}}}-{\frac{10}{3} \left ({x}^{2}-10 \right ) ^{{\frac{3}{2}}}}+100\,\sqrt{{x}^{2}-10}+100\,\sqrt{10}\arctan \left ({\frac{\sqrt{10}}{\sqrt{{x}^{2}-10}}} \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^2-10)^(5/2)/x,x)

[Out]

1/5*(x^2-10)^(5/2)-10/3*(x^2-10)^(3/2)+100*(x^2-10)^(1/2)+100*10^(1/2)*arctan(10^(1/2)/(x^2-10)^(1/2))

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Maxima [A]  time = 1.45614, size = 57, normalized size = 0.93 \begin{align*} \frac{1}{5} \,{\left (x^{2} - 10\right )}^{\frac{5}{2}} - \frac{10}{3} \,{\left (x^{2} - 10\right )}^{\frac{3}{2}} + 100 \, \sqrt{10} \arcsin \left (\frac{\sqrt{10}}{{\left | x \right |}}\right ) + 100 \, \sqrt{x^{2} - 10} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-10)^(5/2)/x,x, algorithm="maxima")

[Out]

1/5*(x^2 - 10)^(5/2) - 10/3*(x^2 - 10)^(3/2) + 100*sqrt(10)*arcsin(sqrt(10)/abs(x)) + 100*sqrt(x^2 - 10)

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Fricas [A]  time = 2.05554, size = 158, normalized size = 2.59 \begin{align*} \frac{1}{15} \,{\left (3 \, x^{4} - 110 \, x^{2} + 2300\right )} \sqrt{x^{2} - 10} - 200 \, \sqrt{10} \arctan \left (-\frac{1}{10} \, \sqrt{10} x + \frac{1}{10} \, \sqrt{10} \sqrt{x^{2} - 10}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-10)^(5/2)/x,x, algorithm="fricas")

[Out]

1/15*(3*x^4 - 110*x^2 + 2300)*sqrt(x^2 - 10) - 200*sqrt(10)*arctan(-1/10*sqrt(10)*x + 1/10*sqrt(10)*sqrt(x^2 -
10))

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Sympy [C]  time = 9.04472, size = 167, normalized size = 2.74 \begin{align*} \begin{cases} \frac{x^{4} \sqrt{x^{2} - 10}}{5} - \frac{22 x^{2} \sqrt{x^{2} - 10}}{3} + \frac{460 \sqrt{x^{2} - 10}}{3} - 100 \sqrt{10} i \log{\left (x \right )} + 50 \sqrt{10} i \log{\left (x^{2} \right )} + 100 \sqrt{10} \operatorname{asin}{\left (\frac{\sqrt{10}}{x} \right )} & \text{for}\: \frac{\left |{x^{2}}\right |}{10} > 1 \\\frac{i x^{4} \sqrt{10 - x^{2}}}{5} - \frac{22 i x^{2} \sqrt{10 - x^{2}}}{3} + \frac{460 i \sqrt{10 - x^{2}}}{3} + 50 \sqrt{10} i \log{\left (x^{2} \right )} - 100 \sqrt{10} i \log{\left (\sqrt{1 - \frac{x^{2}}{10}} + 1 \right )} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x**2-10)**(5/2)/x,x)

[Out]

Piecewise((x**4*sqrt(x**2 - 10)/5 - 22*x**2*sqrt(x**2 - 10)/3 + 460*sqrt(x**2 - 10)/3 - 100*sqrt(10)*I*log(x)
+ 50*sqrt(10)*I*log(x**2) + 100*sqrt(10)*asin(sqrt(10)/x), Abs(x**2)/10 > 1), (I*x**4*sqrt(10 - x**2)/5 - 22*I
*x**2*sqrt(10 - x**2)/3 + 460*I*sqrt(10 - x**2)/3 + 50*sqrt(10)*I*log(x**2) - 100*sqrt(10)*I*log(sqrt(1 - x**2
/10) + 1), True))

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Giac [A]  time = 1.06717, size = 62, normalized size = 1.02 \begin{align*} \frac{1}{5} \,{\left (x^{2} - 10\right )}^{\frac{5}{2}} - \frac{10}{3} \,{\left (x^{2} - 10\right )}^{\frac{3}{2}} - 100 \, \sqrt{10} \arctan \left (\frac{1}{10} \, \sqrt{10} \sqrt{x^{2} - 10}\right ) + 100 \, \sqrt{x^{2} - 10} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((x^2-10)^(5/2)/x,x, algorithm="giac")

[Out]

1/5*(x^2 - 10)^(5/2) - 10/3*(x^2 - 10)^(3/2) - 100*sqrt(10)*arctan(1/10*sqrt(10)*sqrt(x^2 - 10)) + 100*sqrt(x^
2 - 10)