3.6.15 \(\int \sqrt {x} \sqrt {2-b x} \, dx\) [515]

Optimal. Leaf size=65 \[ -\frac {\sqrt {x} \sqrt {2-b x}}{2 b}+\frac {1}{2} x^{3/2} \sqrt {2-b x}+\frac {\sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {52, 56, 222} \begin {gather*} \frac {\sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{3/2}}+\frac {1}{2} x^{3/2} \sqrt {2-b x}-\frac {\sqrt {x} \sqrt {2-b x}}{2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*Sqrt[2 - b*x],x]

[Out]

-1/2*(Sqrt[x]*Sqrt[2 - b*x])/b + (x^(3/2)*Sqrt[2 - b*x])/2 + ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]]/b^(3/2)

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

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Mathematica [A]
time = 0.07, size = 62, normalized size = 0.95 \begin {gather*} \frac {\sqrt {x} \sqrt {2-b x} (-1+b x)}{2 b}+\frac {\log \left (-\sqrt {-b} \sqrt {x}+\sqrt {2-b x}\right )}{(-b)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*Sqrt[2 - b*x],x]

[Out]

(Sqrt[x]*Sqrt[2 - b*x]*(-1 + b*x))/(2*b) + Log[-(Sqrt[-b]*Sqrt[x]) + Sqrt[2 - b*x]]/(-b)^(3/2)

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Mathics [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[Sqrt[x]*Sqrt[2 - b*x],x]')

[Out]

Timed out

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Maple [A]
time = 0.12, size = 85, normalized size = 1.31

method	result	size
meijerg	\(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {3}{2}} \left (-3 b x +3\right ) \sqrt {-\frac {b x}{2}+1}}{6 b}-\frac {\sqrt {\pi }\, \left (-b \right )^{\frac {3}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{b^{\frac {3}{2}}}}{\sqrt {-b}\, \sqrt {\pi }\, b}\)	\(73\)
default	\(-\frac {\sqrt {x}\, \left (-b x +2\right )^{\frac {3}{2}}}{2 b}+\frac {\sqrt {x}\, \sqrt {-b x +2}+\frac {\sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right )}{\sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}}{2 b}\)	\(85\)
risch	\(-\frac {\left (b x -1\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{2 b \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {\arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{2 b^{\frac {3}{2}} \sqrt {x}\, \sqrt {-b x +2}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/b*x^(1/2)*(-b*x+2)^(3/2)+1/2/b*(x^(1/2)*(-b*x+2)^(1/2)+((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)/x^(1/2)/b^(1/2)*
arctan(b^(1/2)*(x-1/b)/(-b*x^2+2*x)^(1/2)))

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Maxima [A]
time = 0.36, size = 81, normalized size = 1.25 \begin {gather*} \frac {\frac {\sqrt {-b x + 2} b}{\sqrt {x}} - \frac {{\left (-b x + 2\right )}^{\frac {3}{2}}}{x^{\frac {3}{2}}}}{b^{3} - \frac {2 \, {\left (b x - 2\right )} b^{2}}{x} + \frac {{\left (b x - 2\right )}^{2} b}{x^{2}}} - \frac {\arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(sqrt(-b*x + 2)*b/sqrt(x) - (-b*x + 2)^(3/2)/x^(3/2))/(b^3 - 2*(b*x - 2)*b^2/x + (b*x - 2)^2*b/x^2) - arctan(s
qrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(3/2)

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Fricas [A]
time = 0.31, size = 107, normalized size = 1.65 \begin {gather*} \left [\frac {{\left (b^{2} x - b\right )} \sqrt {-b x + 2} \sqrt {x} - \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{2 \, b^{2}}, \frac {{\left (b^{2} x - b\right )} \sqrt {-b x + 2} \sqrt {x} - 2 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{2 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*((b^2*x - b)*sqrt(-b*x + 2)*sqrt(x) - sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1))/b^2, 1/2*
((b^2*x - b)*sqrt(-b*x + 2)*sqrt(x) - 2*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))))/b^2]

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Sympy [A]
time = 1.79, size = 155, normalized size = 2.38 \begin {gather*} \begin {cases} \frac {i b x^{\frac {5}{2}}}{2 \sqrt {b x - 2}} - \frac {3 i x^{\frac {3}{2}}}{2 \sqrt {b x - 2}} + \frac {i \sqrt {x}}{b \sqrt {b x - 2}} - \frac {i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {b x^{\frac {5}{2}}}{2 \sqrt {- b x + 2}} + \frac {3 x^{\frac {3}{2}}}{2 \sqrt {- b x + 2}} - \frac {\sqrt {x}}{b \sqrt {- b x + 2}} + \frac {\operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*b*x**(5/2)/(2*sqrt(b*x - 2)) - 3*I*x**(3/2)/(2*sqrt(b*x - 2)) + I*sqrt(x)/(b*sqrt(b*x - 2)) - I*a
cosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(3/2), Abs(b*x) > 2), (-b*x**(5/2)/(2*sqrt(-b*x + 2)) + 3*x**(3/2)/(2*sqrt(
-b*x + 2)) - sqrt(x)/(b*sqrt(-b*x + 2)) + asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(3/2), True))

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Giac [A]
time = 0.00, size = 92, normalized size = 1.42 \begin {gather*} 2 \left (2 \left (\frac {\frac {1}{16}\cdot 2 b^{2} \sqrt {x} \sqrt {x}}{b^{2}}-\frac {\frac {1}{16}\cdot 2 b}{b^{2}}\right ) \sqrt {x} \sqrt {-b x+2}-\frac {\ln \left (\sqrt {-b x+2}-\sqrt {-b} \sqrt {x}\right )}{2 b \sqrt {-b}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*sqrt(-b*x + 2)*(x - 1/b)*sqrt(x) - log(-sqrt(-b)*sqrt(x) + sqrt(-b*x + 2))/(sqrt(-b)*b)

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Mupad [B]
time = 0.10, size = 53, normalized size = 0.82 \begin {gather*} \sqrt {x}\,\left (\frac {x}{2}-\frac {1}{2\,b}\right )\,\sqrt {2-b\,x}-\frac {\ln \left (\sqrt {-b}\,\sqrt {x}\,\sqrt {2-b\,x}-b\,x+1\right )}{2\,{\left (-b\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(2 - b*x)^(1/2),x)

[Out]

x^(1/2)*(x/2 - 1/(2*b))*(2 - b*x)^(1/2) - log((-b)^(1/2)*x^(1/2)*(2 - b*x)^(1/2) - b*x + 1)/(2*(-b)^(3/2))

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