3.6.63 \(\int x^{5/2} (2-b x)^{5/2} \, dx\)

Optimal. Leaf size=150 \[ \frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}}-\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{8} x^{7/2} \sqrt {2-b x}-\frac {x^{5/2} \sqrt {2-b x}}{24 b} \]

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Rubi [A]  time = 0.05, antiderivative size = 150, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {50, 54, 216} \begin {gather*} -\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{8} x^{7/2} \sqrt {2-b x}-\frac {x^{5/2} \sqrt {2-b x}}{24 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)*(2 - b*x)^(5/2),x]

[Out]

(-5*Sqrt[x]*Sqrt[2 - b*x])/(16*b^3) - (5*x^(3/2)*Sqrt[2 - b*x])/(48*b^2) - (x^(5/2)*Sqrt[2 - b*x])/(24*b) + (x
^(7/2)*Sqrt[2 - b*x])/8 + (x^(7/2)*(2 - b*x)^(3/2))/6 + (x^(7/2)*(2 - b*x)^(5/2))/6 + (5*ArcSin[(Sqrt[b]*Sqrt[
x])/Sqrt[2]])/(8*b^(7/2))

Rule 50

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 54

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 216

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)^{5/2} \, dx &=\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5}{6} \int x^{5/2} (2-b x)^{3/2} \, dx\\ &=\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {1}{2} \int x^{5/2} \sqrt {2-b x} \, dx\\ &=\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {1}{8} \int \frac {x^{5/2}}{\sqrt {2-b x}} \, dx\\ &=-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx}{24 b}\\ &=-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{16 b^2}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{16 b^3}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{8 b^3}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{16 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{48 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{24 b}+\frac {1}{8} x^{7/2} \sqrt {2-b x}+\frac {1}{6} x^{7/2} (2-b x)^{3/2}+\frac {1}{6} x^{7/2} (2-b x)^{5/2}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.06, size = 87, normalized size = 0.58 \begin {gather*} \frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{8 b^{7/2}}+\frac {\sqrt {x} \sqrt {2-b x} \left (8 b^5 x^5-40 b^4 x^4+54 b^3 x^3-2 b^2 x^2-5 b x-15\right )}{48 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)*(2 - b*x)^(5/2),x]

[Out]

(Sqrt[x]*Sqrt[2 - b*x]*(-15 - 5*b*x - 2*b^2*x^2 + 54*b^3*x^3 - 40*b^4*x^4 + 8*b^5*x^5))/(48*b^3) + (5*ArcSin[(
Sqrt[b]*Sqrt[x])/Sqrt[2]])/(8*b^(7/2))

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IntegrateAlgebraic [A]  time = 0.15, size = 114, normalized size = 0.76 \begin {gather*} \frac {5 \sqrt {-b} \log \left (\sqrt {2-b x}-\sqrt {-b} \sqrt {x}\right )}{8 b^4}+\frac {\sqrt {2-b x} \left (8 b^5 x^{11/2}-40 b^4 x^{9/2}+54 b^3 x^{7/2}-2 b^2 x^{5/2}-5 b x^{3/2}-15 \sqrt {x}\right )}{48 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)*(2 - b*x)^(5/2),x]

[Out]

(Sqrt[2 - b*x]*(-15*Sqrt[x] - 5*b*x^(3/2) - 2*b^2*x^(5/2) + 54*b^3*x^(7/2) - 40*b^4*x^(9/2) + 8*b^5*x^(11/2)))
/(48*b^3) + (5*Sqrt[-b]*Log[-(Sqrt[-b]*Sqrt[x]) + Sqrt[2 - b*x]])/(8*b^4)

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fricas [A]  time = 1.28, size = 173, normalized size = 1.15 \begin {gather*} \left [\frac {{\left (8 \, b^{6} x^{5} - 40 \, b^{5} x^{4} + 54 \, 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 )}{48 \, b^{4}}, \frac {{\left (8 \, b^{6} x^{5} - 40 \, b^{5} x^{4} + 54 \, 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 )}{48 \, b^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*((8*b^6*x^5 - 40*b^5*x^4 + 54*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, 1/48*((8*b^6*x^5 - 40*b^5*x^4 + 54*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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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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maple [A]  time = 0.00, size = 148, normalized size = 0.99 \begin {gather*} -\frac {\left (-b x +2\right )^{\frac {7}{2}} x^{\frac {5}{2}}}{6 b}-\frac {\left (-b x +2\right )^{\frac {7}{2}} x^{\frac {3}{2}}}{6 b^{2}}-\frac {\left (-b x +2\right )^{\frac {7}{2}} \sqrt {x}}{8 b^{3}}+\frac {\left (-b x +2\right )^{\frac {5}{2}} \sqrt {x}}{24 b^{3}}+\frac {5 \left (-b x +2\right )^{\frac {3}{2}} \sqrt {x}}{48 b^{3}}+\frac {5 \sqrt {-b x +2}\, \sqrt {x}}{16 b^{3}}+\frac {5 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\left (x -\frac {1}{b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+2 x}}\right )}{16 \sqrt {-b x +2}\, b^{\frac {7}{2}} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/b*x^(5/2)*(-b*x+2)^(7/2)-1/6/b^2*x^(3/2)*(-b*x+2)^(7/2)-1/8/b^3*x^(1/2)*(-b*x+2)^(7/2)+1/24*(-b*x+2)^(5/2
)/b^3*x^(1/2)+5/48*(-b*x+2)^(3/2)/b^3*x^(1/2)+5/16*(-b*x+2)^(1/2)/b^3*x^(1/2)+5/16*((-b*x+2)*x)^(1/2)/(-b*x+2)
^(1/2)/b^(7/2)/x^(1/2)*arctan((x-1/b)/(-b*x^2+2*x)^(1/2)*b^(1/2))

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maxima [A]  time = 2.96, size = 209, normalized size = 1.39 \begin {gather*} \frac {\frac {15 \, \sqrt {-b x + 2} b^{5}}{\sqrt {x}} + \frac {85 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {198 \, {\left (-b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}} - \frac {198 \, {\left (-b x + 2\right )}^{\frac {7}{2}} b^{2}}{x^{\frac {7}{2}}} - \frac {85 \, {\left (-b x + 2\right )}^{\frac {9}{2}} b}{x^{\frac {9}{2}}} - \frac {15 \, {\left (-b x + 2\right )}^{\frac {11}{2}}}{x^{\frac {11}{2}}}}{24 \, {\left (b^{9} - \frac {6 \, {\left (b x - 2\right )} b^{8}}{x} + \frac {15 \, {\left (b x - 2\right )}^{2} b^{7}}{x^{2}} - \frac {20 \, {\left (b x - 2\right )}^{3} b^{6}}{x^{3}} + \frac {15 \, {\left (b x - 2\right )}^{4} b^{5}}{x^{4}} - \frac {6 \, {\left (b x - 2\right )}^{5} b^{4}}{x^{5}} + \frac {{\left (b x - 2\right )}^{6} b^{3}}{x^{6}}\right )}} - \frac {5 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{8 \, b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(15*sqrt(-b*x + 2)*b^5/sqrt(x) + 85*(-b*x + 2)^(3/2)*b^4/x^(3/2) + 198*(-b*x + 2)^(5/2)*b^3/x^(5/2) - 198
*(-b*x + 2)^(7/2)*b^2/x^(7/2) - 85*(-b*x + 2)^(9/2)*b/x^(9/2) - 15*(-b*x + 2)^(11/2)/x^(11/2))/(b^9 - 6*(b*x -
2)*b^8/x + 15*(b*x - 2)^2*b^7/x^2 - 20*(b*x - 2)^3*b^6/x^3 + 15*(b*x - 2)^4*b^5/x^4 - 6*(b*x - 2)^5*b^4/x^5 +
(b*x - 2)^6*b^3/x^6) - 5/8*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(7/2)

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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 )}^{5/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)*(2 - b*x)^(5/2),x)

[Out]

int(x^(5/2)*(2 - b*x)^(5/2), x)

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sympy [A]  time = 22.84, size = 337, normalized size = 2.25 \begin {gather*} \begin {cases} \frac {i b^{3} x^{\frac {13}{2}}}{6 \sqrt {b x - 2}} - \frac {7 i b^{2} x^{\frac {11}{2}}}{6 \sqrt {b x - 2}} + \frac {67 i b x^{\frac {9}{2}}}{24 \sqrt {b x - 2}} - \frac {55 i x^{\frac {7}{2}}}{24 \sqrt {b x - 2}} - \frac {i x^{\frac {5}{2}}}{48 b \sqrt {b x - 2}} - \frac {5 i x^{\frac {3}{2}}}{48 b^{2} \sqrt {b x - 2}} + \frac {5 i \sqrt {x}}{8 b^{3} \sqrt {b x - 2}} - \frac {5 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{8 b^{\frac {7}{2}}} & \text {for}\: \frac {\left |{b x}\right |}{2} > 1 \\- \frac {b^{3} x^{\frac {13}{2}}}{6 \sqrt {- b x + 2}} + \frac {7 b^{2} x^{\frac {11}{2}}}{6 \sqrt {- b x + 2}} - \frac {67 b x^{\frac {9}{2}}}{24 \sqrt {- b x + 2}} + \frac {55 x^{\frac {7}{2}}}{24 \sqrt {- b x + 2}} + \frac {x^{\frac {5}{2}}}{48 b \sqrt {- b x + 2}} + \frac {5 x^{\frac {3}{2}}}{48 b^{2} \sqrt {- b x + 2}} - \frac {5 \sqrt {x}}{8 b^{3} \sqrt {- b x + 2}} + \frac {5 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{8 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)**(5/2),x)

[Out]

Piecewise((I*b**3*x**(13/2)/(6*sqrt(b*x - 2)) - 7*I*b**2*x**(11/2)/(6*sqrt(b*x - 2)) + 67*I*b*x**(9/2)/(24*sqr
t(b*x - 2)) - 55*I*x**(7/2)/(24*sqrt(b*x - 2)) - I*x**(5/2)/(48*b*sqrt(b*x - 2)) - 5*I*x**(3/2)/(48*b**2*sqrt(
b*x - 2)) + 5*I*sqrt(x)/(8*b**3*sqrt(b*x - 2)) - 5*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(8*b**(7/2)), Abs(b*x)/2
> 1), (-b**3*x**(13/2)/(6*sqrt(-b*x + 2)) + 7*b**2*x**(11/2)/(6*sqrt(-b*x + 2)) - 67*b*x**(9/2)/(24*sqrt(-b*x
+ 2)) + 55*x**(7/2)/(24*sqrt(-b*x + 2)) + x**(5/2)/(48*b*sqrt(-b*x + 2)) + 5*x**(3/2)/(48*b**2*sqrt(-b*x + 2)
) - 5*sqrt(x)/(8*b**3*sqrt(-b*x + 2)) + 5*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/(8*b**(7/2)), True))

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