\(\int \frac {(-10+x^2)^{5/2}}{x} \, dx\) [462]

Optimal result

Integrand size = 13, antiderivative size = 61 \[ \int \frac {\left (-10+x^2\right )^{5/2}}{x} \, dx=100 \sqrt {-10+x^2}-\frac {10}{3} \left (-10+x^2\right )^{3/2}+\frac {1}{5} \left (-10+x^2\right )^{5/2}-100 \sqrt {10} \arctan \left (\frac {\sqrt {-10+x^2}}{\sqrt {10}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {272, 52, 65, 209} \[ \int \frac {\left (-10+x^2\right )^{5/2}}{x} \, dx=-100 \sqrt {10} \arctan \left (\frac {\sqrt {x^2-10}}{\sqrt {10}}\right )+\frac {1}{5} \left (x^2-10\right )^{5/2}-\frac {10}{3} \left (x^2-10\right )^{3/2}+100 \sqrt {x^2-10} \]

[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 52

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 65

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 209

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

Rule 272

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]]

Rubi steps \begin{align*} \text {integral}& = \frac {1}{2} \text {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 \text {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 \text {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 \text {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 \text {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} \arctan \left (\frac {\sqrt {-10+x^2}}{\sqrt {10}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.80 \[ \int \frac {\left (-10+x^2\right )^{5/2}}{x} \, dx=\frac {1}{15} \sqrt {-10+x^2} \left (2300-110 x^2+3 x^4\right )-100 \sqrt {10} \arctan \left (\frac {\sqrt {-10+x^2}}{\sqrt {10}}\right ) \]

[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[-10 + x^2]/Sqrt[10]]

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.67

method	result	size
pseudoelliptic	\(-100 \arctan \left (\frac {\sqrt {x^{2}-10}\, \sqrt {10}}{10}\right ) \sqrt {10}+\frac {\sqrt {x^{2}-10}\, \left (3 x^{4}-110 x^{2}+2300\right )}{15}\)	\(41\)
default	\(\frac {\left (x^{2}-10\right )^{\frac {5}{2}}}{5}-\frac {10 \left (x^{2}-10\right )^{\frac {3}{2}}}{3}+100 \sqrt {x^{2}-10}+100 \sqrt {10}\, \arctan \left (\frac {\sqrt {10}}{\sqrt {x^{2}-10}}\right )\)	\(46\)
trager	\(\left (\frac {1}{5} x^{4}-\frac {22}{3} x^{2}+\frac {460}{3}\right ) \sqrt {x^{2}-10}-100 \operatorname {RootOf}\left (\textit {\_Z}^{2}+10\right ) \ln \left (\frac {\sqrt {x^{2}-10}+\operatorname {RootOf}\left (\textit {\_Z}^{2}+10\right )}{x}\right )\)	\(49\)
meijerg	\(-\frac {375 \sqrt {2}\, \sqrt {5}\, \operatorname {signum}\left (-1+\frac {x^{2}}{10}\right )^{\frac {5}{2}} \left (-\frac {8 \left (\frac {46}{15}-3 \ln \left (2\right )+2 \ln \left (x \right )-\ln \left (5\right )+i \pi \right ) \sqrt {\pi }}{15}+\frac {368 \sqrt {\pi }}{225}-\frac {4 \sqrt {\pi }\, \left (\frac {3}{25} x^{4}-\frac {22}{5} x^{2}+92\right ) \sqrt {1-\frac {x^{2}}{10}}}{225}+\frac {16 \sqrt {\pi }\, \ln \left (\frac {1}{2}+\frac {\sqrt {1-\frac {x^{2}}{10}}}{2}\right )}{15}\right )}{4 \sqrt {\pi }\, {\left (-\operatorname {signum}\left (-1+\frac {x^{2}}{10}\right )\right )}^{\frac {5}{2}}}\)	\(108\)

[In]

int((x^2-10)^(5/2)/x,x,method=_RETURNVERBOSE)

[Out]

-100*arctan(1/10*(x^2-10)^(1/2)*10^(1/2))*10^(1/2)+1/15*(x^2-10)^(1/2)*(3*x^4-110*x^2+2300)

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.77 \[ \int \frac {\left (-10+x^2\right )^{5/2}}{x} \, dx=\frac {1}{15} \, {\left (3 \, x^{4} - 110 \, x^{2} + 2300\right )} \sqrt {x^{2} - 10} - 200 \, \sqrt {10} \arctan \left (-\frac {1}{10} \, \sqrt {10} x + \frac {1}{10} \, \sqrt {10} \sqrt {x^{2} - 10}\right ) \]

[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))

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 3.43 (sec) , antiderivative size = 165, normalized size of antiderivative = 2.70 \[ \int \frac {\left (-10+x^2\right )^{5/2}}{x} \, dx=\begin {cases} \frac {x^{4} \sqrt {x^{2} - 10}}{5} - \frac {22 x^{2} \sqrt {x^{2} - 10}}{3} + \frac {460 \sqrt {x^{2} - 10}}{3} - 100 \sqrt {10} i \log {\left (x \right )} + 50 \sqrt {10} i \log {\left (x^{2} \right )} + 100 \sqrt {10} \operatorname {asin}{\left (\frac {\sqrt {10}}{x} \right )} & \text {for}\: \left |{x^{2}}\right | > 10 \\\frac {i x^{4} \sqrt {10 - x^{2}}}{5} - \frac {22 i x^{2} \sqrt {10 - x^{2}}}{3} + \frac {460 i \sqrt {10 - x^{2}}}{3} + 50 \sqrt {10} i \log {\left (x^{2} \right )} - 100 \sqrt {10} i \log {\left (\sqrt {1 - \frac {x^{2}}{10}} + 1 \right )} & \text {otherwise} \end {cases} \]

[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), (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**2/1
0) + 1), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 42, normalized size of antiderivative = 0.69 \[ \int \frac {\left (-10+x^2\right )^{5/2}}{x} \, dx=\frac {1}{5} \, {\left (x^{2} - 10\right )}^{\frac {5}{2}} - \frac {10}{3} \, {\left (x^{2} - 10\right )}^{\frac {3}{2}} + 100 \, \sqrt {10} \arcsin \left (\frac {\sqrt {10}}{{\left | x \right |}}\right ) + 100 \, \sqrt {x^{2} - 10} \]

[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)

Giac [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-10+x^2\right )^{5/2}}{x} \, dx=\frac {1}{5} \, {\left (x^{2} - 10\right )}^{\frac {5}{2}} - \frac {10}{3} \, {\left (x^{2} - 10\right )}^{\frac {3}{2}} - 100 \, \sqrt {10} \arctan \left (\frac {1}{10} \, \sqrt {10} \sqrt {x^{2} - 10}\right ) + 100 \, \sqrt {x^{2} - 10} \]

[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)

Mupad [B] (verification not implemented)

Time = 0.49 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.75 \[ \int \frac {\left (-10+x^2\right )^{5/2}}{x} \, dx=100\,\sqrt {x^2-10}-100\,\sqrt {10}\,\mathrm {atan}\left (\frac {\sqrt {10}\,\sqrt {x^2-10}}{10}\right )-\frac {10\,{\left (x^2-10\right )}^{3/2}}{3}+\frac {{\left (x^2-10\right )}^{5/2}}{5} \]

[In]

int((x^2 - 10)^(5/2)/x,x)

[Out]

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