3.476 \(\int \frac {(25-x^2)^{3/2}}{x^4} \, dx\)

Optimal. Leaf size=40 \[ \frac {\sqrt {25-x^2}}{x}-\frac {\left (25-x^2\right )^{3/2}}{3 x^3}+\sin ^{-1}\left (\frac {x}{5}\right ) \]

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Rubi [A]  time = 0.01, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.133, Rules used = {277, 216} \[ -\frac {\left (25-x^2\right )^{3/2}}{3 x^3}+\frac {\sqrt {25-x^2}}{x}+\sin ^{-1}\left (\frac {x}{5}\right ) \]

Antiderivative was successfully verified.

[In]

Int[(25 - x^2)^(3/2)/x^4,x]

[Out]

Sqrt[25 - x^2]/x - (25 - x^2)^(3/2)/(3*x^3) + ArcSin[x/5]

Rule 216

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

Rule 277

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

Rubi steps

\begin {align*} \int \frac {\left (25-x^2\right )^{3/2}}{x^4} \, dx &=-\frac {\left (25-x^2\right )^{3/2}}{3 x^3}-\int \frac {\sqrt {25-x^2}}{x^2} \, dx\\ &=\frac {\sqrt {25-x^2}}{x}-\frac {\left (25-x^2\right )^{3/2}}{3 x^3}+\int \frac {1}{\sqrt {25-x^2}} \, dx\\ &=\frac {\sqrt {25-x^2}}{x}-\frac {\left (25-x^2\right )^{3/2}}{3 x^3}+\sin ^{-1}\left (\frac {x}{5}\right )\\ \end {align*}

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Mathematica [C]  time = 0.00, size = 24, normalized size = 0.60 \[ -\frac {125 \, _2F_1\left (-\frac {3}{2},-\frac {3}{2};-\frac {1}{2};\frac {x^2}{25}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(25 - x^2)^(3/2)/x^4,x]

[Out]

(-125*Hypergeometric2F1[-3/2, -3/2, -1/2, x^2/25])/(3*x^3)

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IntegrateAlgebraic [A]  time = 0.07, size = 46, normalized size = 1.15 \[ \frac {\sqrt {25-x^2} \left (4 x^2-25\right )}{3 x^3}-2 \tan ^{-1}\left (\frac {\sqrt {25-x^2}}{x+5}\right ) \]

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(25 - x^2)^(3/2)/x^4,x]

[Out]

(Sqrt[25 - x^2]*(-25 + 4*x^2))/(3*x^3) - 2*ArcTan[Sqrt[25 - x^2]/(5 + x)]

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fricas [A]  time = 0.80, size = 45, normalized size = 1.12 \[ -\frac {6 \, x^{3} \arctan \left (\frac {\sqrt {-x^{2} + 25} - 5}{x}\right ) - {\left (4 \, x^{2} - 25\right )} \sqrt {-x^{2} + 25}}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+25)^(3/2)/x^4,x, algorithm="fricas")

[Out]

-1/3*(6*x^3*arctan((sqrt(-x^2 + 25) - 5)/x) - (4*x^2 - 25)*sqrt(-x^2 + 25))/x^3

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giac [B]  time = 0.66, size = 77, normalized size = 1.92 \[ -\frac {x^{3} {\left (\frac {15 \, {\left (\sqrt {-x^{2} + 25} - 5\right )}^{2}}{x^{2}} - 1\right )}}{24 \, {\left (\sqrt {-x^{2} + 25} - 5\right )}^{3}} + \frac {5 \, {\left (\sqrt {-x^{2} + 25} - 5\right )}}{8 \, x} - \frac {{\left (\sqrt {-x^{2} + 25} - 5\right )}^{3}}{24 \, x^{3}} + \arcsin \left (\frac {1}{5} \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+25)^(3/2)/x^4,x, algorithm="giac")

[Out]

-1/24*x^3*(15*(sqrt(-x^2 + 25) - 5)^2/x^2 - 1)/(sqrt(-x^2 + 25) - 5)^3 + 5/8*(sqrt(-x^2 + 25) - 5)/x - 1/24*(s
qrt(-x^2 + 25) - 5)^3/x^3 + arcsin(1/5*x)

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maple [A]  time = 0.33, size = 32, normalized size = 0.80

method	result	size
risch	\(-\frac {4 x^{4}-125 x^{2}+625}{3 x^{3} \sqrt {-x^{2}+25}}+\arcsin \left (\frac {x}{5}\right )\)	\(32\)
meijerg	\(-\frac {3 i \left (-\frac {1000 i \sqrt {\pi }\, \left (-\frac {4 x^{2}}{25}+1\right ) \sqrt {-\frac {x^{2}}{25}+1}}{9 x^{3}}+\frac {8 i \sqrt {\pi }\, \arcsin \left (\frac {x}{5}\right )}{3}\right )}{8 \sqrt {\pi }}\)	\(43\)
trager	\(\frac {\left (4 x^{2}-25\right ) \sqrt {-x^{2}+25}}{3 x^{3}}+\RootOf \left (\textit {\_Z}^{2}+1\right ) \ln \left (-\RootOf \left (\textit {\_Z}^{2}+1\right ) x +\sqrt {-x^{2}+25}\right )\)	\(50\)
default	\(-\frac {\left (-x^{2}+25\right )^{\frac {5}{2}}}{75 x^{3}}+\frac {2 \left (-x^{2}+25\right )^{\frac {5}{2}}}{1875 x}+\frac {2 x \left (-x^{2}+25\right )^{\frac {3}{2}}}{1875}+\frac {\sqrt {-x^{2}+25}\, x}{25}+\arcsin \left (\frac {x}{5}\right )\)	\(58\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-x^2+25)^(3/2)/x^4,x,method=_RETURNVERBOSE)

[Out]

-1/3*(4*x^4-125*x^2+625)/x^3/(-x^2+25)^(1/2)+arcsin(1/5*x)

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maxima [A]  time = 0.97, size = 45, normalized size = 1.12 \[ \frac {1}{25} \, \sqrt {-x^{2} + 25} x + \frac {2 \, {\left (-x^{2} + 25\right )}^{\frac {3}{2}}}{75 \, x} - \frac {{\left (-x^{2} + 25\right )}^{\frac {5}{2}}}{75 \, x^{3}} + \arcsin \left (\frac {1}{5} \, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x^2+25)^(3/2)/x^4,x, algorithm="maxima")

[Out]

1/25*sqrt(-x^2 + 25)*x + 2/75*(-x^2 + 25)^(3/2)/x - 1/75*(-x^2 + 25)^(5/2)/x^3 + arcsin(1/5*x)

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mupad [B]  time = 0.04, size = 33, normalized size = 0.82 \[ \mathrm {asin}\left (\frac {x}{5}\right )+\frac {4\,\sqrt {25-x^2}}{3\,x}-\frac {25\,\sqrt {25-x^2}}{3\,x^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((25 - x^2)^(3/2)/x^4,x)

[Out]

asin(x/5) + (4*(25 - x^2)^(1/2))/(3*x) - (25*(25 - x^2)^(1/2))/(3*x^3)

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sympy [A]  time = 2.23, size = 32, normalized size = 0.80 \[ \operatorname {asin}{\left (\frac {x}{5} \right )} + \frac {4 \sqrt {25 - x^{2}}}{3 x} - \frac {25 \sqrt {25 - x^{2}}}{3 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-x**2+25)**(3/2)/x**4,x)

[Out]

asin(x/5) + 4*sqrt(25 - x**2)/(3*x) - 25*sqrt(25 - x**2)/(3*x**3)

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