3.6.13 \(\int x^{5/2} \sqrt {2-b x} \, dx\) [513]

Optimal. Leaf size=112 \[ -\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{24 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2-b x}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}} \]

[Out]

5/4*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)-5/24*x^(3/2)*(-b*x+2)^(1/2)/b^2-1/12*x^(5/2)*(-b*x+2)^(1/2)/b+
1/4*x^(7/2)*(-b*x+2)^(1/2)-5/8*x^(1/2)*(-b*x+2)^(1/2)/b^3

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Rubi [A]
time = 0.02, antiderivative size = 112, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {5 \text {ArcSin}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}}-\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{24 b^2}+\frac {1}{4} x^{7/2} \sqrt {2-b x}-\frac {x^{5/2} \sqrt {2-b x}}{12 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*Sqrt[2 - b*x],x]

[Out]

(-5*Sqrt[x]*Sqrt[2 - b*x])/(8*b^3) - (5*x^(3/2)*Sqrt[2 - b*x])/(24*b^2) - (x^(5/2)*Sqrt[2 - b*x])/(12*b) + (x^
(7/2)*Sqrt[2 - b*x])/4 + (5*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(4*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*} \int x^{5/2} \sqrt {2-b x} \, dx &=\frac {1}{4} x^{7/2} \sqrt {2-b x}+\frac {1}{4} \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx\\ &=-\frac {x^{5/2} \sqrt {2-b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2-b x}+\frac {5 \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx}{12 b}\\ &=-\frac {5 x^{3/2} \sqrt {2-b x}}{24 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2-b x}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{8 b^2}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{24 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2-b x}+\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{8 b^3}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{24 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2-b x}+\frac {5 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^3}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{24 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{12 b}+\frac {1}{4} x^{7/2} \sqrt {2-b x}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 82, normalized size = 0.73 \begin {gather*} \frac {\sqrt {x} \sqrt {2-b x} \left (-15-5 b x-2 b^2 x^2+6 b^3 x^3\right )}{24 b^3}+\frac {5 \log \left (-\sqrt {-b} \sqrt {x}+\sqrt {2-b x}\right )}{4 (-b)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*Sqrt[2 - b*x],x]

[Out]

(Sqrt[x]*Sqrt[2 - b*x]*(-15 - 5*b*x - 2*b^2*x^2 + 6*b^3*x^3))/(24*b^3) + (5*Log[-(Sqrt[-b]*Sqrt[x]) + Sqrt[2 -
b*x]])/(4*(-b)^(7/2))

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Maple [A]
time = 0.12, size = 128, normalized size = 1.14

method	result	size
meijerg	\(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {7}{2}} \left (-42 b^{3} x^{3}+14 x^{2} b^{2}+35 b x +105\right ) \sqrt {-\frac {b x}{2}+1}}{168 b^{3}}-\frac {5 \sqrt {\pi }\, \left (-b \right )^{\frac {7}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{4 b^{\frac {7}{2}}}}{\left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, b}\)	\(89\)
risch	\(-\frac {\left (6 b^{3} x^{3}-2 x^{2} b^{2}-5 b x -15\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{24 b^{3} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {5 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{8 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {-b x +2}}\)	\(115\)
default	\(-\frac {x^{\frac {5}{2}} \left (-b x +2\right )^{\frac {3}{2}}}{4 b}+\frac {-\frac {5 x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {3}{2}}}{12 b}+\frac {5 \left (-\frac {\sqrt {x}\, \left (-b x +2\right )^{\frac {3}{2}}}{2 b}+\frac {\sqrt {x}\, \sqrt {-b x +2}+\frac {\sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right )}{\sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}}{2 b}\right )}{4 b}}{b}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4/b*x^(5/2)*(-b*x+2)^(3/2)+5/4/b*(-1/3/b*x^(3/2)*(-b*x+2)^(3/2)+1/b*(-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 = 0.52, size = 147, normalized size = 1.31 \begin {gather*} \frac {\frac {15 \, \sqrt {-b x + 2} b^{3}}{\sqrt {x}} - \frac {73 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}} - \frac {55 \, {\left (-b x + 2\right )}^{\frac {5}{2}} b}{x^{\frac {5}{2}}} - \frac {15 \, {\left (-b x + 2\right )}^{\frac {7}{2}}}{x^{\frac {7}{2}}}}{12 \, {\left (b^{7} - \frac {4 \, {\left (b x - 2\right )} b^{6}}{x} + \frac {6 \, {\left (b x - 2\right )}^{2} b^{5}}{x^{2}} - \frac {4 \, {\left (b x - 2\right )}^{3} b^{4}}{x^{3}} + \frac {{\left (b x - 2\right )}^{4} b^{3}}{x^{4}}\right )}} - \frac {5 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{4 \, b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(15*sqrt(-b*x + 2)*b^3/sqrt(x) - 73*(-b*x + 2)^(3/2)*b^2/x^(3/2) - 55*(-b*x + 2)^(5/2)*b/x^(5/2) - 15*(-b
*x + 2)^(7/2)/x^(7/2))/(b^7 - 4*(b*x - 2)*b^6/x + 6*(b*x - 2)^2*b^5/x^2 - 4*(b*x - 2)^3*b^4/x^3 + (b*x - 2)^4*
b^3/x^4) - 5/4*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(7/2)

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Fricas [A]
time = 0.50, size = 141, normalized size = 1.26 \begin {gather*} \left [\frac {{\left (6 \, b^{4} x^{3} - 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 )}{24 \, b^{4}}, \frac {{\left (6 \, b^{4} x^{3} - 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 )}{24 \, b^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*((6*b^4*x^3 - 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)*sqrt(x) + 1))/b^4, 1/24*((6*b^4*x^3 - 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]

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Sympy [C] Result contains complex when optimal does not.
time = 22.59, size = 250, normalized size = 2.23 \begin {gather*} \begin {cases} \frac {i b x^{\frac {9}{2}}}{4 \sqrt {b x - 2}} - \frac {7 i x^{\frac {7}{2}}}{12 \sqrt {b x - 2}} - \frac {i x^{\frac {5}{2}}}{24 b \sqrt {b x - 2}} - \frac {5 i x^{\frac {3}{2}}}{24 b^{2} \sqrt {b x - 2}} + \frac {5 i \sqrt {x}}{4 b^{3} \sqrt {b x - 2}} - \frac {5 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {7}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {b x^{\frac {9}{2}}}{4 \sqrt {- b x + 2}} + \frac {7 x^{\frac {7}{2}}}{12 \sqrt {- b x + 2}} + \frac {x^{\frac {5}{2}}}{24 b \sqrt {- b x + 2}} + \frac {5 x^{\frac {3}{2}}}{24 b^{2} \sqrt {- b x + 2}} - \frac {5 \sqrt {x}}{4 b^{3} \sqrt {- b x + 2}} + \frac {5 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {7}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*b*x**(9/2)/(4*sqrt(b*x - 2)) - 7*I*x**(7/2)/(12*sqrt(b*x - 2)) - I*x**(5/2)/(24*b*sqrt(b*x - 2))
- 5*I*x**(3/2)/(24*b**2*sqrt(b*x - 2)) + 5*I*sqrt(x)/(4*b**3*sqrt(b*x - 2)) - 5*I*acosh(sqrt(2)*sqrt(b)*sqrt(x
)/2)/(4*b**(7/2)), Abs(b*x) > 2), (-b*x**(9/2)/(4*sqrt(-b*x + 2)) + 7*x**(7/2)/(12*sqrt(-b*x + 2)) + x**(5/2)/
(24*b*sqrt(-b*x + 2)) + 5*x**(3/2)/(24*b**2*sqrt(-b*x + 2)) - 5*sqrt(x)/(4*b**3*sqrt(-b*x + 2)) + 5*asin(sqrt(
2)*sqrt(b)*sqrt(x)/2)/(4*b**(7/2)), True))

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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{4,[1,
1]%%%}+%%%{4,[1,0]%%%}+%%%{-4,[0,1]%%%}+%%%{-8,[0,0]%%%},0,%%%{6,[2,2]%%%}+%%%{4,[2,1]%%%}+%%%{6,[2,0]%%%}+%%%
{-4,[1,2]%%%}+%%%{-28

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{5/2}\,\sqrt {2-b\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(2 - b*x)^(1/2),x)

[Out]

int(x^(5/2)*(2 - b*x)^(1/2), x)

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