\(\int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx\) [629]

Optimal result

Integrand size = 16, antiderivative size = 91 \[ \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx=-\frac {5 \sqrt {x} \sqrt {2-b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{3 b}+\frac {5 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \]

[Out]

5*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)-5/6*x^(3/2)*(-b*x+2)^(1/2)/b^2-1/3*x^(5/2)*(-b*x+2)^(1/2)/b-5/2*
x^(1/2)*(-b*x+2)^(1/2)/b^3

Rubi [A] (verified)

Time = 0.02 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {52, 56, 222} \[ \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx=\frac {5 \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}-\frac {5 \sqrt {x} \sqrt {2-b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{3 b} \]

[In]

Int[x^(5/2)/Sqrt[2 - b*x],x]

[Out]

(-5*Sqrt[x]*Sqrt[2 - b*x])/(2*b^3) - (5*x^(3/2)*Sqrt[2 - b*x])/(6*b^2) - (x^(5/2)*Sqrt[2 - b*x])/(3*b) + (5*Ar
cSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

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 56

Int[1/(Sqrt[(a_.) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Dist[2/Sqrt[b], Subst[Int[1/Sqrt[b*c -
a*d + d*x^2], x], x, Sqrt[a + b*x]], x] /; FreeQ[{a, b, c, d}, x] && GtQ[b*c - a*d, 0] && GtQ[b, 0]

Rule 222

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]

Rubi steps \begin{align*} \text {integral}& = -\frac {x^{5/2} \sqrt {2-b x}}{3 b}+\frac {5 \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx}{3 b} \\ & = -\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{3 b}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{2 b^2} \\ & = -\frac {5 \sqrt {x} \sqrt {2-b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{3 b}+\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{2 b^3} \\ & = -\frac {5 \sqrt {x} \sqrt {2-b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{3 b}+\frac {5 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{b^3} \\ & = -\frac {5 \sqrt {x} \sqrt {2-b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{3 b}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.84 \[ \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx=-\frac {\sqrt {x} \sqrt {2-b x} \left (15+5 b x+2 b^2 x^2\right )}{6 b^3}-\frac {10 \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right )}{b^{7/2}} \]

[In]

Integrate[x^(5/2)/Sqrt[2 - b*x],x]

[Out]

-1/6*(Sqrt[x]*Sqrt[2 - b*x]*(15 + 5*b*x + 2*b^2*x^2))/b^3 - (10*ArcTan[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2 - b
*x])])/b^(7/2)

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.89

method	result	size
meijerg	\(-\frac {8 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {7}{2}} \left (14 b^{2} x^{2}+35 b x +105\right ) \sqrt {-\frac {b x}{2}+1}}{336 b^{3}}+\frac {5 \sqrt {\pi }\, \left (-b \right )^{\frac {7}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{8 b^{\frac {7}{2}}}\right )}{\left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, b}\)	\(81\)
risch	\(\frac {\left (2 b^{2} x^{2}+5 b x +15\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{6 b^{3} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {5 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{2 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {-b x +2}}\)	\(107\)
default	\(-\frac {x^{\frac {5}{2}} \sqrt {-b x +2}}{3 b}+\frac {-\frac {5 x^{\frac {3}{2}} \sqrt {-b x +2}}{6 b}+\frac {5 \left (-\frac {3 \sqrt {x}\, \sqrt {-b x +2}}{2 b}+\frac {3 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{2 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +2}}\right )}{3 b}}{b}\)	\(111\)

[In]

int(x^(5/2)/(-b*x+2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-8/(-b)^(5/2)/Pi^(1/2)/b*(-1/336*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(7/2)*(14*b^2*x^2+35*b*x+105)/b^3*(-1/2*b*x+1)^
(1/2)+5/8*Pi^(1/2)*(-b)^(7/2)/b^(7/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.37 \[ \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx=\left [-\frac {{\left (2 \, b^{3} x^{2} + 5 \, b^{2} x + 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} + 15 \, \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{6 \, b^{4}}, -\frac {{\left (2 \, b^{3} x^{2} + 5 \, b^{2} x + 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} + 30 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{6 \, b^{4}}\right ] \]

[In]

integrate(x^(5/2)/(-b*x+2)^(1/2),x, algorithm="fricas")

[Out]

[-1/6*((2*b^3*x^2 + 5*b^2*x + 15*b)*sqrt(-b*x + 2)*sqrt(x) + 15*sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sq
rt(x) + 1))/b^4, -1/6*((2*b^3*x^2 + 5*b^2*x + 15*b)*sqrt(-b*x + 2)*sqrt(x) + 30*sqrt(b)*arctan(sqrt(-b*x + 2)/
(sqrt(b)*sqrt(x))))/b^4]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 10.83 (sec) , antiderivative size = 204, normalized size of antiderivative = 2.24 \[ \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx=\begin {cases} - \frac {i x^{\frac {7}{2}}}{3 \sqrt {b x - 2}} - \frac {i x^{\frac {5}{2}}}{6 b \sqrt {b x - 2}} - \frac {5 i x^{\frac {3}{2}}}{6 b^{2} \sqrt {b x - 2}} + \frac {5 i \sqrt {x}}{b^{3} \sqrt {b x - 2}} - \frac {5 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\\frac {x^{\frac {7}{2}}}{3 \sqrt {- b x + 2}} + \frac {x^{\frac {5}{2}}}{6 b \sqrt {- b x + 2}} + \frac {5 x^{\frac {3}{2}}}{6 b^{2} \sqrt {- b x + 2}} - \frac {5 \sqrt {x}}{b^{3} \sqrt {- b x + 2}} + \frac {5 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} & \text {otherwise} \end {cases} \]

[In]

integrate(x**(5/2)/(-b*x+2)**(1/2),x)

[Out]

Piecewise((-I*x**(7/2)/(3*sqrt(b*x - 2)) - I*x**(5/2)/(6*b*sqrt(b*x - 2)) - 5*I*x**(3/2)/(6*b**2*sqrt(b*x - 2)
) + 5*I*sqrt(x)/(b**3*sqrt(b*x - 2)) - 5*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/2), Abs(b*x) > 2), (x**(7/2)
/(3*sqrt(-b*x + 2)) + x**(5/2)/(6*b*sqrt(-b*x + 2)) + 5*x**(3/2)/(6*b**2*sqrt(-b*x + 2)) - 5*sqrt(x)/(b**3*sqr
t(-b*x + 2)) + 5*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/2), True))

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.29 \[ \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx=-\frac {\frac {33 \, \sqrt {-b x + 2} b^{2}}{\sqrt {x}} + \frac {40 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b}{x^{\frac {3}{2}}} + \frac {15 \, {\left (-b x + 2\right )}^{\frac {5}{2}}}{x^{\frac {5}{2}}}}{3 \, {\left (b^{6} - \frac {3 \, {\left (b x - 2\right )} b^{5}}{x} + \frac {3 \, {\left (b x - 2\right )}^{2} b^{4}}{x^{2}} - \frac {{\left (b x - 2\right )}^{3} b^{3}}{x^{3}}\right )}} - \frac {5 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {7}{2}}} \]

[In]

integrate(x^(5/2)/(-b*x+2)^(1/2),x, algorithm="maxima")

[Out]

-1/3*(33*sqrt(-b*x + 2)*b^2/sqrt(x) + 40*(-b*x + 2)^(3/2)*b/x^(3/2) + 15*(-b*x + 2)^(5/2)/x^(5/2))/(b^6 - 3*(b
*x - 2)*b^5/x + 3*(b*x - 2)^2*b^4/x^2 - (b*x - 2)^3*b^3/x^3) - 5*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(7
/2)

Giac [A] (verification not implemented)

none

Time = 5.87 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.09 \[ \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx=-\frac {{\left (\sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2} {\left ({\left (b x - 2\right )} {\left (\frac {2 \, {\left (b x - 2\right )}}{b^{2}} + \frac {13}{b^{2}}\right )} + \frac {33}{b^{2}}\right )} - \frac {30 \, \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b} b}\right )} {\left | b \right |}}{6 \, b^{3}} \]

[In]

integrate(x^(5/2)/(-b*x+2)^(1/2),x, algorithm="giac")

[Out]

-1/6*(sqrt((b*x - 2)*b + 2*b)*sqrt(-b*x + 2)*((b*x - 2)*(2*(b*x - 2)/b^2 + 13/b^2) + 33/b^2) - 30*log(abs(-sqr
t(-b*x + 2)*sqrt(-b) + sqrt((b*x - 2)*b + 2*b)))/(sqrt(-b)*b))*abs(b)/b^3

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx=\int \frac {x^{5/2}}{\sqrt {2-b\,x}} \,d x \]

[In]

int(x^(5/2)/(2 - b*x)^(1/2),x)

[Out]

int(x^(5/2)/(2 - b*x)^(1/2), x)