3.6.64 \(\int x^{3/2} (2-b x)^{5/2} \, dx\) [564]

Optimal. Leaf size=128 \[ -\frac {3 \sqrt {x} \sqrt {2-b x}}{8 b^2}-\frac {x^{3/2} \sqrt {2-b x}}{8 b}+\frac {1}{4} x^{5/2} \sqrt {2-b x}+\frac {1}{4} x^{5/2} (2-b x)^{3/2}+\frac {1}{5} x^{5/2} (2-b x)^{5/2}+\frac {3 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{5/2}} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 128, 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 \text {ArcSin}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{4 b^{5/2}}-\frac {3 \sqrt {x} \sqrt {2-b x}}{8 b^2}+\frac {1}{5} x^{5/2} (2-b x)^{5/2}+\frac {1}{4} x^{5/2} (2-b x)^{3/2}+\frac {1}{4} x^{5/2} \sqrt {2-b x}-\frac {x^{3/2} \sqrt {2-b x}}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*(2 - b*x)^(5/2),x]

[Out]

(-3*Sqrt[x]*Sqrt[2 - b*x])/(8*b^2) - (x^(3/2)*Sqrt[2 - b*x])/(8*b) + (x^(5/2)*Sqrt[2 - b*x])/4 + (x^(5/2)*(2 -
b*x)^(3/2))/4 + (x^(5/2)*(2 - b*x)^(5/2))/5 + (3*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/(4*b^(5/2))

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

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Mathematica [A]
time = 0.14, size = 90, normalized size = 0.70 \begin {gather*} \frac {\sqrt {x} \sqrt {2-b x} \left (-15-5 b x+62 b^2 x^2-42 b^3 x^3+8 b^4 x^4\right )}{40 b^2}-\frac {3 \log \left (-\sqrt {-b} \sqrt {x}+\sqrt {2-b x}\right )}{4 (-b)^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*(2 - b*x)^(5/2),x]

[Out]

(Sqrt[x]*Sqrt[2 - b*x]*(-15 - 5*b*x + 62*b^2*x^2 - 42*b^3*x^3 + 8*b^4*x^4))/(40*b^2) - (3*Log[-(Sqrt[-b]*Sqrt[
x]) + Sqrt[2 - b*x]])/(4*(-b)^(5/2))

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Maple [A]
time = 0.12, size = 135, normalized size = 1.05

method	result	size
meijerg	\(\frac {\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {5}{2}} \left (-8 b^{4} x^{4}+42 b^{3} x^{3}-62 x^{2} b^{2}+5 b x +15\right ) \sqrt {-\frac {b x}{2}+1}}{40 b^{2}}-\frac {3 \sqrt {\pi }\, \left (-b \right )^{\frac {5}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{4 b^{\frac {5}{2}}}}{\left (-b \right )^{\frac {3}{2}} \sqrt {\pi }\, b}\)	\(97\)
risch	\(-\frac {\left (8 b^{4} x^{4}-42 b^{3} x^{3}+62 x^{2} b^{2}-5 b x -15\right ) \sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{40 b^{2} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {3 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{8 b^{\frac {5}{2}} \sqrt {x}\, \sqrt {-b x +2}}\)	\(123\)
default	\(-\frac {x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {7}{2}}}{5 b}+\frac {-\frac {3 \sqrt {x}\, \left (-b x +2\right )^{\frac {7}{2}}}{20 b}+\frac {3 \left (\frac {\left (-b x +2\right )^{\frac {5}{2}} \sqrt {x}}{3}+\frac {5 \left (-b x +2\right )^{\frac {3}{2}} \sqrt {x}}{6}+\frac {5 \sqrt {x}\, \sqrt {-b x +2}}{2}+\frac {5 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right )}{2 \sqrt {-b x +2}\, \sqrt {x}\, \sqrt {b}}\right )}{20 b}}{b}\)	\(135\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/5/b*x^(3/2)*(-b*x+2)^(7/2)+3/5/b*(-1/4/b*x^(1/2)*(-b*x+2)^(7/2)+1/4/b*(1/3*(-b*x+2)^(5/2)*x^(1/2)+5/6*(-b*x
+2)^(3/2)*x^(1/2)+5/2*x^(1/2)*(-b*x+2)^(1/2)+5/2*((-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 [B] Leaf count of result is larger than twice the leaf count of optimal. 179 vs. \(2 (89) = 178\).
time = 0.54, size = 179, normalized size = 1.40 \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^{7} - \frac {5 \, {\left (b x - 2\right )} b^{6}}{x} + \frac {10 \, {\left (b x - 2\right )}^{2} b^{5}}{x^{2}} - \frac {10 \, {\left (b x - 2\right )}^{3} b^{4}}{x^{3}} + \frac {5 \, {\left (b x - 2\right )}^{4} b^{3}}{x^{4}} - \frac {{\left (b x - 2\right )}^{5} b^{2}}{x^{5}}\right )}} - \frac {3 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{4 \, b^{\frac {5}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(-b*x+2)^(5/2),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^7 - 5*(b*x - 2)*b^6/x + 10*(b*x - 2)^2*b^5/x^2 -
10*(b*x - 2)^3*b^4/x^3 + 5*(b*x - 2)^4*b^3/x^4 - (b*x - 2)^5*b^2/x^5) - 3/4*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqr
t(x)))/b^(5/2)

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Fricas [A]
time = 0.71, size = 157, normalized size = 1.23 \begin {gather*} \left [\frac {{\left (8 \, b^{5} x^{4} - 42 \, b^{4} x^{3} + 62 \, 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^{3}}, \frac {{\left (8 \, b^{5} x^{4} - 42 \, b^{4} x^{3} + 62 \, 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^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/40*((8*b^5*x^4 - 42*b^4*x^3 + 62*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^3, 1/40*((8*b^5*x^4 - 42*b^4*x^3 + 62*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^3]

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Sympy [C] Result contains complex when optimal does not.
time = 33.12, size = 292, normalized size = 2.28 \begin {gather*} \begin {cases} \frac {i b^{3} x^{\frac {11}{2}}}{5 \sqrt {b x - 2}} - \frac {29 i b^{2} x^{\frac {9}{2}}}{20 \sqrt {b x - 2}} + \frac {73 i b x^{\frac {7}{2}}}{20 \sqrt {b x - 2}} - \frac {129 i x^{\frac {5}{2}}}{40 \sqrt {b x - 2}} - \frac {i x^{\frac {3}{2}}}{8 b \sqrt {b x - 2}} + \frac {3 i \sqrt {x}}{4 b^{2} \sqrt {b x - 2}} - \frac {3 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {5}{2}}} & \text {for}\: \left |{b x}\right | > 2 \\- \frac {b^{3} x^{\frac {11}{2}}}{5 \sqrt {- b x + 2}} + \frac {29 b^{2} x^{\frac {9}{2}}}{20 \sqrt {- b x + 2}} - \frac {73 b x^{\frac {7}{2}}}{20 \sqrt {- b x + 2}} + \frac {129 x^{\frac {5}{2}}}{40 \sqrt {- b x + 2}} + \frac {x^{\frac {3}{2}}}{8 b \sqrt {- b x + 2}} - \frac {3 \sqrt {x}}{4 b^{2} \sqrt {- b x + 2}} + \frac {3 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{4 b^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((I*b**3*x**(11/2)/(5*sqrt(b*x - 2)) - 29*I*b**2*x**(9/2)/(20*sqrt(b*x - 2)) + 73*I*b*x**(7/2)/(20*sq
rt(b*x - 2)) - 129*I*x**(5/2)/(40*sqrt(b*x - 2)) - I*x**(3/2)/(8*b*sqrt(b*x - 2)) + 3*I*sqrt(x)/(4*b**2*sqrt(b
*x - 2)) - 3*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(4*b**(5/2)), Abs(b*x) > 2), (-b**3*x**(11/2)/(5*sqrt(-b*x + 2
)) + 29*b**2*x**(9/2)/(20*sqrt(-b*x + 2)) - 73*b*x**(7/2)/(20*sqrt(-b*x + 2)) + 129*x**(5/2)/(40*sqrt(-b*x + 2
)) + x**(3/2)/(8*b*sqrt(-b*x + 2)) - 3*sqrt(x)/(4*b**2*sqrt(-b*x + 2)) + 3*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/(4*
b**(5/2)), 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^(3/2)*(-b*x+2)^(5/2),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^{3/2}\,{\left (2-b\,x\right )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(2 - b*x)^(5/2),x)

[Out]

int(x^(3/2)*(2 - b*x)^(5/2), x)

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