3.6.65 \(\int \sqrt {x} (2-b x)^{5/2} \, dx\) [565]

Optimal. Leaf size=106 \[ -\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b}+\frac {5}{8} x^{3/2} \sqrt {2-b x}+\frac {5}{12} x^{3/2} (2-b x)^{3/2}+\frac {1}{4} x^{3/2} (2-b x)^{5/2}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 106, 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^{3/2}}+\frac {1}{4} x^{3/2} (2-b x)^{5/2}+\frac {5}{12} x^{3/2} (2-b x)^{3/2}+\frac {5}{8} x^{3/2} \sqrt {2-b x}-\frac {5 \sqrt {x} \sqrt {2-b x}}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(2 - b*x)^(5/2),x]

[Out]

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

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Mathematica [A]
time = 0.12, size = 82, normalized size = 0.77 \begin {gather*} \frac {\sqrt {x} \sqrt {2-b x} \left (-15+59 b x-34 b^2 x^2+6 b^3 x^3\right )}{24 b}+\frac {5 \log \left (-\sqrt {-b} \sqrt {x}+\sqrt {2-b x}\right )}{4 (-b)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(2 - b*x)^(5/2),x]

[Out]

(Sqrt[x]*Sqrt[2 - b*x]*(-15 + 59*b*x - 34*b^2*x^2 + 6*b^3*x^3))/(24*b) + (5*Log[-(Sqrt[-b]*Sqrt[x]) + Sqrt[2 -
b*x]])/(4*(-b)^(3/2))

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Maple [A]
time = 0.13, size = 107, normalized size = 1.01

method	result	size
meijerg	\(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {3}{2}} \left (-6 b^{3} x^{3}+34 x^{2} b^{2}-59 b x +15\right ) \sqrt {-\frac {b x}{2}+1}}{24 b}-\frac {5 \sqrt {\pi }\, \left (-b \right )^{\frac {3}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{4 b^{\frac {3}{2}}}}{\sqrt {-b}\, \sqrt {\pi }\, b}\)	\(89\)
default	\(\frac {x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {5}{2}}}{4}+\frac {5 x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {3}{2}}}{12}+\frac {5 x^{\frac {3}{2}} \sqrt {-b x +2}}{8}-\frac {5 \sqrt {x}\, \sqrt {-b x +2}}{8 b}+\frac {5 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right )}{8 b^{\frac {3}{2}} \sqrt {-b x +2}\, \sqrt {x}}\)	\(107\)
risch	\(-\frac {\left (6 b^{3} x^{3}-34 x^{2} b^{2}+59 b x -15\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{24 b \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {5 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right )}{8 b^{\frac {3}{2}} \sqrt {-b x +2}\, \sqrt {x}}\)	\(115\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.50, size = 145, normalized size = 1.37 \begin {gather*} \frac {\frac {15 \, \sqrt {-b x + 2} b^{3}}{\sqrt {x}} + \frac {55 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}} + \frac {73 \, {\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^{5} - \frac {4 \, {\left (b x - 2\right )} b^{4}}{x} + \frac {6 \, {\left (b x - 2\right )}^{2} b^{3}}{x^{2}} - \frac {4 \, {\left (b x - 2\right )}^{3} b^{2}}{x^{3}} + \frac {{\left (b x - 2\right )}^{4} b}{x^{4}}\right )}} - \frac {5 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{4 \, b^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.73, size = 141, normalized size = 1.33 \begin {gather*} \left [\frac {{\left (6 \, b^{4} x^{3} - 34 \, b^{3} x^{2} + 59 \, 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^{2}}, \frac {{\left (6 \, b^{4} x^{3} - 34 \, b^{3} x^{2} + 59 \, 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^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/24*((6*b^4*x^3 - 34*b^3*x^2 + 59*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^2, 1/24*((6*b^4*x^3 - 34*b^3*x^2 + 59*b^2*x - 15*b)*sqrt(-b*x + 2)*sqrt(x) - 30*sq
rt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))))/b^2]

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Sympy [C] Result contains complex when optimal does not.
time = 8.50, size = 253, normalized size = 2.39 \begin {gather*} \begin {cases} \frac {i b^{3} x^{\frac {9}{2}}}{4 \sqrt {b x - 2}} - \frac {23 i b^{2} x^{\frac {7}{2}}}{12 \sqrt {b x - 2}} + \frac {127 i b x^{\frac {5}{2}}}{24 \sqrt {b x - 2}} - \frac {133 i x^{\frac {3}{2}}}{24 \sqrt {b x - 2}} + \frac {5 i \sqrt {x}}{4 b \sqrt {b x - 2}} - \frac {5 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {3}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {b^{3} x^{\frac {9}{2}}}{4 \sqrt {- b x + 2}} + \frac {23 b^{2} x^{\frac {7}{2}}}{12 \sqrt {- b x + 2}} - \frac {127 b x^{\frac {5}{2}}}{24 \sqrt {- b x + 2}} + \frac {133 x^{\frac {3}{2}}}{24 \sqrt {- b x + 2}} - \frac {5 \sqrt {x}}{4 b \sqrt {- b x + 2}} + \frac {5 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*b**3*x**(9/2)/(4*sqrt(b*x - 2)) - 23*I*b**2*x**(7/2)/(12*sqrt(b*x - 2)) + 127*I*b*x**(5/2)/(24*sq
rt(b*x - 2)) - 133*I*x**(3/2)/(24*sqrt(b*x - 2)) + 5*I*sqrt(x)/(4*b*sqrt(b*x - 2)) - 5*I*acosh(sqrt(2)*sqrt(b)
*sqrt(x)/2)/(4*b**(3/2)), Abs(b*x) > 2), (-b**3*x**(9/2)/(4*sqrt(-b*x + 2)) + 23*b**2*x**(7/2)/(12*sqrt(-b*x +
2)) - 127*b*x**(5/2)/(24*sqrt(-b*x + 2)) + 133*x**(3/2)/(24*sqrt(-b*x + 2)) - 5*sqrt(x)/(4*b*sqrt(-b*x + 2))
+ 5*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/(4*b**(3/2)), 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((-b*x+2)^(5/2)*x^(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 \sqrt {x}\,{\left (2-b\,x\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(2 - b*x)^(5/2),x)

[Out]

int(x^(1/2)*(2 - b*x)^(5/2), x)

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