3.6.39 \(\int x^{5/2} (2-b x)^{3/2} \, dx\) [539]

Optimal. Leaf size=131 \[ -\frac {3 \sqrt {x} \sqrt {2-b x}}{8 b^3}-\frac {x^{3/2} \sqrt {2-b x}}{8 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2-b x}+\frac {1}{5} x^{7/2} (2-b x)^{3/2}+\frac {3 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}} \]

[Out]

1/5*x^(7/2)*(-b*x+2)^(3/2)+3/4*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)-1/8*x^(3/2)*(-b*x+2)^(1/2)/b^2-1/20
*x^(5/2)*(-b*x+2)^(1/2)/b+3/20*x^(7/2)*(-b*x+2)^(1/2)-3/8*x^(1/2)*(-b*x+2)^(1/2)/b^3

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Rubi [A]
time = 0.03, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 7, 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 {3 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}}-\frac {3 \sqrt {x} \sqrt {2-b x}}{8 b^3}-\frac {x^{3/2} \sqrt {2-b x}}{8 b^2}+\frac {1}{5} x^{7/2} (2-b x)^{3/2}+\frac {3}{20} x^{7/2} \sqrt {2-b x}-\frac {x^{5/2} \sqrt {2-b x}}{20 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(2 - b*x)^(3/2),x]

[Out]

(-3*Sqrt[x]*Sqrt[2 - b*x])/(8*b^3) - (x^(3/2)*Sqrt[2 - b*x])/(8*b^2) - (x^(5/2)*Sqrt[2 - b*x])/(20*b) + (3*x^(
7/2)*Sqrt[2 - b*x])/20 + (x^(7/2)*(2 - b*x)^(3/2))/5 + (3*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} (2-b x)^{3/2} \, dx &=\frac {1}{5} x^{7/2} (2-b x)^{3/2}+\frac {3}{5} \int x^{5/2} \sqrt {2-b x} \, dx\\ &=\frac {3}{20} x^{7/2} \sqrt {2-b x}+\frac {1}{5} x^{7/2} (2-b x)^{3/2}+\frac {3}{20} \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx\\ &=-\frac {x^{5/2} \sqrt {2-b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2-b x}+\frac {1}{5} x^{7/2} (2-b x)^{3/2}+\frac {\int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx}{4 b}\\ &=-\frac {x^{3/2} \sqrt {2-b x}}{8 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2-b x}+\frac {1}{5} x^{7/2} (2-b x)^{3/2}+\frac {3 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{8 b^2}\\ &=-\frac {3 \sqrt {x} \sqrt {2-b x}}{8 b^3}-\frac {x^{3/2} \sqrt {2-b x}}{8 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2-b x}+\frac {1}{5} x^{7/2} (2-b x)^{3/2}+\frac {3 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{8 b^3}\\ &=-\frac {3 \sqrt {x} \sqrt {2-b x}}{8 b^3}-\frac {x^{3/2} \sqrt {2-b x}}{8 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2-b x}+\frac {1}{5} x^{7/2} (2-b x)^{3/2}+\frac {3 \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{4 b^3}\\ &=-\frac {3 \sqrt {x} \sqrt {2-b x}}{8 b^3}-\frac {x^{3/2} \sqrt {2-b x}}{8 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{20 b}+\frac {3}{20} x^{7/2} \sqrt {2-b x}+\frac {1}{5} x^{7/2} (2-b x)^{3/2}+\frac {3 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{7/2}}\\ \end {align*}

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Mathematica [A]
time = 0.13, size = 90, normalized size = 0.69 \begin {gather*} -\frac {\sqrt {x} \sqrt {2-b x} \left (15+5 b x+2 b^2 x^2-22 b^3 x^3+8 b^4 x^4\right )}{40 b^3}+\frac {3 \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)*(2 - b*x)^(3/2),x]

[Out]

-1/40*(Sqrt[x]*Sqrt[2 - b*x]*(15 + 5*b*x + 2*b^2*x^2 - 22*b^3*x^3 + 8*b^4*x^4))/b^3 + (3*Log[-(Sqrt[-b]*Sqrt[x
]) + Sqrt[2 - b*x]])/(4*(-b)^(7/2))

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 57.77, size = 229, normalized size = 1.75 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {I \left (-30 b^6 \text {ArcCosh}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ] \left (-2+b x\right )^2+30 b^{\frac {13}{2}} \sqrt {x} \left (-2+b x\right )^{\frac {3}{2}}-5 b^{\frac {15}{2}} x^{\frac {3}{2}} \left (-2+b x\right )^{\frac {3}{2}}-b^{\frac {17}{2}} x^{\frac {5}{2}} \left (-2+b x\right )^{\frac {3}{2}}+2 b^{\frac {19}{2}} x^{\frac {7}{2}} \left (-23+19 b x-4 b^2 x^2\right ) \left (-2+b x\right )^{\frac {3}{2}}\right )}{40 b^{\frac {19}{2}} \left (-2+b x\right )^2},\text {Abs}\left [b x\right ]>2\right \}\right \},\frac {3 \text {ArcSin}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ]}{4 b^{\frac {7}{2}}}-\frac {3 \sqrt {x}}{4 b^3 \sqrt {2-b x}}+\frac {x^{\frac {3}{2}}}{8 b^2 \sqrt {2-b x}}+\frac {x^{\frac {5}{2}}}{40 b \sqrt {2-b x}}+\frac {23 x^{\frac {7}{2}}}{20 \sqrt {2-b x}}-\frac {19 b x^{\frac {9}{2}}}{20 \sqrt {2-b x}}+\frac {b^2 x^{\frac {11}{2}}}{5 \sqrt {2-b x}}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[x^(5/2)*(2 - b*x)^(3/2),x]')

[Out]

Piecewise[{{I / 40 (-30 b ^ 6 ArcCosh[Sqrt[2] Sqrt[b] Sqrt[x] / 2] (-2 + b x) ^ 2 + 30 b ^ (13 / 2) Sqrt[x] (-
2 + b x) ^ (3 / 2) - 5 b ^ (15 / 2) x ^ (3 / 2) (-2 + b x) ^ (3 / 2) - b ^ (17 / 2) x ^ (5 / 2) (-2 + b x) ^ (
3 / 2) + 2 b ^ (19 / 2) x ^ (7 / 2) (-23 + 19 b x - 4 b ^ 2 x ^ 2) (-2 + b x) ^ (3 / 2)) / (b ^ (19 / 2) (-2 +
b x) ^ 2), Abs[b x] > 2}}, 3 ArcSin[Sqrt[2] Sqrt[b] Sqrt[x] / 2] / (4 b ^ (7 / 2)) - 3 Sqrt[x] / (4 b ^ 3 Sqr
t[2 - b x]) + x ^ (3 / 2) / (8 b ^ 2 Sqrt[2 - b x]) + x ^ (5 / 2) / (40 b Sqrt[2 - b x]) + 23 x ^ (7 / 2) / (2
0 Sqrt[2 - b x]) - 19 b x ^ (9 / 2) / (20 Sqrt[2 - b x]) + b ^ 2 x ^ (11 / 2) / (5 Sqrt[2 - b x])]

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Maple [A]
time = 0.11, size = 143, normalized size = 1.09

method	result	size
meijerg	\(-\frac {24 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {7}{2}} \left (56 b^{4} x^{4}-154 b^{3} x^{3}+14 x^{2} b^{2}+35 b x +105\right ) \sqrt {-\frac {b x}{2}+1}}{6720 b^{3}}+\frac {\sqrt {\pi }\, \left (-b \right )^{\frac {7}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{32 b^{\frac {7}{2}}}\right )}{\left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, b}\)	\(97\)
risch	\(\frac {\left (8 b^{4} x^{4}-22 b^{3} x^{3}+2 x^{2} b^{2}+5 b x +15\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{40 b^{3} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {3 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{8 b^{\frac {7}{2}} \sqrt {x}\, \sqrt {-b x +2}}\)	\(123\)
default	\(-\frac {x^{\frac {5}{2}} \left (-b x +2\right )^{\frac {5}{2}}}{5 b}+\frac {-\frac {x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {5}{2}}}{4 b}+\frac {-\frac {\sqrt {x}\, \left (-b x +2\right )^{\frac {5}{2}}}{4 b}+\frac {\frac {\left (-b x +2\right )^{\frac {3}{2}} \sqrt {x}}{2}+\frac {3 \sqrt {x}\, \sqrt {-b x +2}}{2}+\frac {3 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right )}{2 \sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}}{4 b}}{b}}{b}\)	\(143\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(-b*x+2)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/5/b*x^(5/2)*(-b*x+2)^(5/2)+1/b*(-1/4/b*x^(3/2)*(-b*x+2)^(5/2)+3/4/b*(-1/3/b*x^(1/2)*(-b*x+2)^(5/2)+1/3/b*(1
/2*(-b*x+2)^(3/2)*x^(1/2)+3/2*x^(1/2)*(-b*x+2)^(1/2)+3/2*((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)/x^(1/2)/b^(1/2)*arc
tan(b^(1/2)*(x-1/b)/(-b*x^2+2*x)^(1/2)))))

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Maxima [A]
time = 0.34, size = 179, normalized size = 1.37 \begin {gather*} \frac {\frac {15 \, \sqrt {-b x + 2} b^{4}}{\sqrt {x}} + \frac {70 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{3}}{x^{\frac {3}{2}}} - \frac {128 \, {\left (-b x + 2\right )}^{\frac {5}{2}} b^{2}}{x^{\frac {5}{2}}} - \frac {70 \, {\left (-b x + 2\right )}^{\frac {7}{2}} b}{x^{\frac {7}{2}}} - \frac {15 \, {\left (-b x + 2\right )}^{\frac {9}{2}}}{x^{\frac {9}{2}}}}{20 \, {\left (b^{8} - \frac {5 \, {\left (b x - 2\right )} b^{7}}{x} + \frac {10 \, {\left (b x - 2\right )}^{2} b^{6}}{x^{2}} - \frac {10 \, {\left (b x - 2\right )}^{3} b^{5}}{x^{3}} + \frac {5 \, {\left (b x - 2\right )}^{4} b^{4}}{x^{4}} - \frac {{\left (b x - 2\right )}^{5} b^{3}}{x^{5}}\right )}} - \frac {3 \, \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)^(3/2),x, algorithm="maxima")

[Out]

1/20*(15*sqrt(-b*x + 2)*b^4/sqrt(x) + 70*(-b*x + 2)^(3/2)*b^3/x^(3/2) - 128*(-b*x + 2)^(5/2)*b^2/x^(5/2) - 70*
(-b*x + 2)^(7/2)*b/x^(7/2) - 15*(-b*x + 2)^(9/2)/x^(9/2))/(b^8 - 5*(b*x - 2)*b^7/x + 10*(b*x - 2)^2*b^6/x^2 -
10*(b*x - 2)^3*b^5/x^3 + 5*(b*x - 2)^4*b^4/x^4 - (b*x - 2)^5*b^3/x^5) - 3/4*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqr
t(x)))/b^(7/2)

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Fricas [A]
time = 0.32, size = 157, normalized size = 1.20 \begin {gather*} \left [-\frac {{\left (8 \, b^{5} x^{4} - 22 \, 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 )}{40 \, b^{4}}, -\frac {{\left (8 \, b^{5} x^{4} - 22 \, 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 )}{40 \, b^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(-b*x+2)^(3/2),x, algorithm="fricas")

[Out]

[-1/40*((8*b^5*x^4 - 22*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, -1/40*((8*b^5*x^4 - 22*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 [A]
time = 58.58, size = 289, normalized size = 2.21 \begin {gather*} \begin {cases} - \frac {i b^{2} x^{\frac {11}{2}}}{5 \sqrt {b x - 2}} + \frac {19 i b x^{\frac {9}{2}}}{20 \sqrt {b x - 2}} - \frac {23 i x^{\frac {7}{2}}}{20 \sqrt {b x - 2}} - \frac {i x^{\frac {5}{2}}}{40 b \sqrt {b x - 2}} - \frac {i x^{\frac {3}{2}}}{8 b^{2} \sqrt {b x - 2}} + \frac {3 i \sqrt {x}}{4 b^{3} \sqrt {b x - 2}} - \frac {3 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^{2} x^{\frac {11}{2}}}{5 \sqrt {- b x + 2}} - \frac {19 b x^{\frac {9}{2}}}{20 \sqrt {- b x + 2}} + \frac {23 x^{\frac {7}{2}}}{20 \sqrt {- b x + 2}} + \frac {x^{\frac {5}{2}}}{40 b \sqrt {- b x + 2}} + \frac {x^{\frac {3}{2}}}{8 b^{2} \sqrt {- b x + 2}} - \frac {3 \sqrt {x}}{4 b^{3} \sqrt {- b x + 2}} + \frac {3 \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)**(3/2),x)

[Out]

Piecewise((-I*b**2*x**(11/2)/(5*sqrt(b*x - 2)) + 19*I*b*x**(9/2)/(20*sqrt(b*x - 2)) - 23*I*x**(7/2)/(20*sqrt(b
*x - 2)) - I*x**(5/2)/(40*b*sqrt(b*x - 2)) - I*x**(3/2)/(8*b**2*sqrt(b*x - 2)) + 3*I*sqrt(x)/(4*b**3*sqrt(b*x
- 2)) - 3*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(4*b**(7/2)), Abs(b*x) > 2), (b**2*x**(11/2)/(5*sqrt(-b*x + 2)) -
19*b*x**(9/2)/(20*sqrt(-b*x + 2)) + 23*x**(7/2)/(20*sqrt(-b*x + 2)) + x**(5/2)/(40*b*sqrt(-b*x + 2)) + x**(3/
2)/(8*b**2*sqrt(-b*x + 2)) - 3*sqrt(x)/(4*b**3*sqrt(-b*x + 2)) + 3*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/(4*b**(7/2)
), True))

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Giac [A]
time = 0.01, size = 322, normalized size = 2.46 \begin {gather*} -2 b \left (2 \left (\left (\left (\left (\frac {\frac {1}{100800}\cdot 5040 b^{8} \sqrt {x} \sqrt {x}}{b^{8}}-\frac {\frac {1}{100800}\cdot 1260 b^{7}}{b^{8}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{100800}\cdot 2940 b^{6}}{b^{8}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{100800}\cdot 7350 b^{5}}{b^{8}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{100800}\cdot 22050 b^{4}}{b^{8}}\right ) \sqrt {x} \sqrt {-b x+2}-\frac {7 \ln \left (\sqrt {-b x+2}-\sqrt {-b} \sqrt {x}\right )}{8 b^{4} \sqrt {-b}}\right )+4 \left (2 \left (\left (\left (\frac {\frac {1}{2880}\cdot 180 b^{6} \sqrt {x} \sqrt {x}}{b^{6}}-\frac {\frac {1}{2880}\cdot 60 b^{5}}{b^{6}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{2880}\cdot 150 b^{4}}{b^{6}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{2880}\cdot 450 b^{3}}{b^{6}}\right ) \sqrt {x} \sqrt {-b x+2}-\frac {5 \ln \left (\sqrt {-b x+2}-\sqrt {-b} \sqrt {x}\right )}{8 b^{3} \sqrt {-b}}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)*(-b*x+2)^(3/2),x)

[Out]

-1/120*(((2*(3*(4*x - 1/b)*x - 7/b^2)*x - 35/b^3)*x - 105/b^4)*sqrt(-b*x + 2)*sqrt(x) - 210*log(-sqrt(-b)*sqrt
(x) + sqrt(-b*x + 2))/(sqrt(-b)*b^4))*b + 1/12*((2*(3*x - 1/b)*x - 5/b^2)*x - 15/b^3)*sqrt(-b*x + 2)*sqrt(x) -
5/2*log(-sqrt(-b)*sqrt(x) + sqrt(-b*x + 2))/(sqrt(-b)*b^3)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{5/2}\,{\left (2-b\,x\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(2 - b*x)^(3/2),x)

[Out]

int(x^(5/2)*(2 - b*x)^(3/2), x)

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