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

Optimal. Leaf size=89 \[ \frac {10 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}-\frac {5 \sqrt {x} \sqrt {2-b x}}{b^3}-\frac {10 x^{3/2}}{3 b^2 \sqrt {2-b x}}+\frac {2 x^{5/2}}{3 b (2-b x)^{3/2}} \]

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Rubi [A]  time = 0.02, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {47, 50, 54, 216} \begin {gather*} -\frac {10 x^{3/2}}{3 b^2 \sqrt {2-b x}}-\frac {5 \sqrt {x} \sqrt {2-b x}}{b^3}+\frac {10 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}+\frac {2 x^{5/2}}{3 b (2-b x)^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/(2 - b*x)^(5/2),x]

[Out]

(2*x^(5/2))/(3*b*(2 - b*x)^(3/2)) - (10*x^(3/2))/(3*b^2*Sqrt[2 - b*x]) - (5*Sqrt[x]*Sqrt[2 - b*x])/b^3 + (10*A
rcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

Rule 47

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + 1)), x] - Dist[(d*n)/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, c, d},
x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && LtQ[m, -1] &&  !(IntegerQ[n] &&  !IntegerQ[m]) &&  !(ILeQ[m + n + 2, 0
] && (FractionQ[m] || GeQ[2*n + m + 1, 0])) && IntLinearQ[a, b, c, d, m, n, x]

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 \frac {x^{5/2}}{(2-b x)^{5/2}} \, dx &=\frac {2 x^{5/2}}{3 b (2-b x)^{3/2}}-\frac {5 \int \frac {x^{3/2}}{(2-b x)^{3/2}} \, dx}{3 b}\\ &=\frac {2 x^{5/2}}{3 b (2-b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {2-b x}}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{b^2}\\ &=\frac {2 x^{5/2}}{3 b (2-b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {2-b x}}-\frac {5 \sqrt {x} \sqrt {2-b x}}{b^3}+\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{b^3}\\ &=\frac {2 x^{5/2}}{3 b (2-b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {2-b x}}-\frac {5 \sqrt {x} \sqrt {2-b x}}{b^3}+\frac {10 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=\frac {2 x^{5/2}}{3 b (2-b x)^{3/2}}-\frac {10 x^{3/2}}{3 b^2 \sqrt {2-b x}}-\frac {5 \sqrt {x} \sqrt {2-b x}}{b^3}+\frac {10 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 30, normalized size = 0.34 \begin {gather*} \frac {x^{7/2} \, _2F_1\left (\frac {5}{2},\frac {7}{2};\frac {9}{2};\frac {b x}{2}\right )}{14 \sqrt {2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(2 - b*x)^(5/2),x]

[Out]

(x^(7/2)*Hypergeometric2F1[5/2, 7/2, 9/2, (b*x)/2])/(14*Sqrt[2])

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IntegrateAlgebraic [A]  time = 0.19, size = 89, normalized size = 1.00 \begin {gather*} \frac {10 \sqrt {-b} \log \left (\sqrt {2-b x}-\sqrt {-b} \sqrt {x}\right )}{b^4}+\frac {\sqrt {2-b x} \left (-3 b^2 x^{5/2}+40 b x^{3/2}-60 \sqrt {x}\right )}{3 b^3 (b x-2)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[2 - b*x]*(-60*Sqrt[x] + 40*b*x^(3/2) - 3*b^2*x^(5/2)))/(3*b^3*(-2 + b*x)^2) + (10*Sqrt[-b]*Log[-(Sqrt[-b
]*Sqrt[x]) + Sqrt[2 - b*x]])/b^4

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fricas [A]  time = 1.14, size = 187, normalized size = 2.10 \begin {gather*} \left [-\frac {15 \, {\left (b^{2} x^{2} - 4 \, b x + 4\right )} \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right ) + {\left (3 \, b^{3} x^{2} - 40 \, b^{2} x + 60 \, b\right )} \sqrt {-b x + 2} \sqrt {x}}{3 \, {\left (b^{6} x^{2} - 4 \, b^{5} x + 4 \, b^{4}\right )}}, -\frac {30 \, {\left (b^{2} x^{2} - 4 \, b x + 4\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) + {\left (3 \, b^{3} x^{2} - 40 \, b^{2} x + 60 \, b\right )} \sqrt {-b x + 2} \sqrt {x}}{3 \, {\left (b^{6} x^{2} - 4 \, b^{5} x + 4 \, b^{4}\right )}}\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]

[-1/3*(15*(b^2*x^2 - 4*b*x + 4)*sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) + (3*b^3*x^2 - 40*b^2
*x + 60*b)*sqrt(-b*x + 2)*sqrt(x))/(b^6*x^2 - 4*b^5*x + 4*b^4), -1/3*(30*(b^2*x^2 - 4*b*x + 4)*sqrt(b)*arctan(
sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) + (3*b^3*x^2 - 40*b^2*x + 60*b)*sqrt(-b*x + 2)*sqrt(x))/(b^6*x^2 - 4*b^5*x +
4*b^4)]

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giac [B]  time = 10.72, size = 200, normalized size = 2.25 \begin {gather*} \frac {{\left (\frac {15 \, \log \left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2}\right )}{\sqrt {-b} b^{2}} - \frac {3 \, \sqrt {{\left (b x - 2\right )} b + 2 \, b} \sqrt {-b x + 2}}{b^{3}} - \frac {16 \, {\left (9 \, {\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{4} - 24 \, {\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} b + 28 \, b^{2}\right )}}{{\left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b\right )}^{3} \sqrt {-b} b}\right )} {\left | b \right |}}{3 \, b^{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="giac")

[Out]

1/3*(15*log((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2)/(sqrt(-b)*b^2) - 3*sqrt((b*x - 2)*b + 2*b)*
sqrt(-b*x + 2)/b^3 - 16*(9*(sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^4 - 24*(sqrt(-b*x + 2)*sqrt(-b)
- sqrt((b*x - 2)*b + 2*b))^2*b + 28*b^2)/(((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2 - 2*b)^3*sqr
t(-b)*b))*abs(b)/b^2

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maple [B]  time = 0.04, size = 168, normalized size = 1.89 \begin {gather*} \frac {\left (\frac {5 \arctan \left (\frac {\left (x -\frac {1}{b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+2 x}}\right )}{b^{\frac {7}{2}}}+\frac {28 \sqrt {-\left (x -\frac {2}{b}\right )^{2} b -2 x +\frac {4}{b}}}{3 \left (x -\frac {2}{b}\right ) b^{4}}+\frac {8 \sqrt {-\left (x -\frac {2}{b}\right )^{2} b -2 x +\frac {4}{b}}}{3 \left (x -\frac {2}{b}\right )^{2} b^{5}}\right ) \sqrt {\left (-b x +2\right ) x}}{\sqrt {-b x +2}\, \sqrt {x}}+\frac {\left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}\, \sqrt {x}}{\sqrt {-\left (b x -2\right ) x}\, \sqrt {-b x +2}\, b^{3}} \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/b^3*(b*x-2)*x^(1/2)/(-(b*x-2)*x)^(1/2)*((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)+(5/b^(7/2)*arctan((x-1/b)/(-b*x^2+2
*x)^(1/2)*b^(1/2))+8/3/b^5/(x-2/b)^2*(-(x-2/b)^2*b-2*x+4/b)^(1/2)+28/3/(x-2/b)*(-(x-2/b)^2*b-2*x+4/b)^(1/2)/b^
4)*((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)/x^(1/2)

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maxima [A]  time = 2.99, size = 86, normalized size = 0.97 \begin {gather*} \frac {2 \, {\left (2 \, b^{2} + \frac {10 \, {\left (b x - 2\right )} b}{x} - \frac {15 \, {\left (b x - 2\right )}^{2}}{x^{2}}\right )}}{3 \, {\left (\frac {{\left (-b x + 2\right )}^{\frac {3}{2}} b^{4}}{x^{\frac {3}{2}}} + \frac {{\left (-b x + 2\right )}^{\frac {5}{2}} b^{3}}{x^{\frac {5}{2}}}\right )}} - \frac {10 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{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]

2/3*(2*b^2 + 10*(b*x - 2)*b/x - 15*(b*x - 2)^2/x^2)/((-b*x + 2)^(3/2)*b^4/x^(3/2) + (-b*x + 2)^(5/2)*b^3/x^(5/
2)) - 10*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 \frac {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 [B]  time = 6.80, size = 753, normalized size = 8.46 \begin {gather*} \begin {cases} - \frac {3 i b^{\frac {23}{2}} x^{15}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x - 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x - 2}} + \frac {40 i b^{\frac {21}{2}} x^{14}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x - 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x - 2}} - \frac {60 i b^{\frac {19}{2}} x^{13}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x - 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x - 2}} - \frac {30 i b^{10} x^{\frac {27}{2}} \sqrt {b x - 2} \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x - 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x - 2}} + \frac {15 \pi b^{10} x^{\frac {27}{2}} \sqrt {b x - 2}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x - 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x - 2}} + \frac {60 i b^{9} x^{\frac {25}{2}} \sqrt {b x - 2} \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x - 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x - 2}} - \frac {30 \pi b^{9} x^{\frac {25}{2}} \sqrt {b x - 2}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {b x - 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {b x - 2}} & \text {for}\: \frac {\left |{b x}\right |}{2} > 1 \\\frac {3 b^{\frac {23}{2}} x^{15}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {- b x + 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {- b x + 2}} - \frac {40 b^{\frac {21}{2}} x^{14}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {- b x + 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {- b x + 2}} + \frac {60 b^{\frac {19}{2}} x^{13}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {- b x + 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {- b x + 2}} + \frac {30 b^{10} x^{\frac {27}{2}} \sqrt {- b x + 2} \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {- b x + 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {- b x + 2}} - \frac {60 b^{9} x^{\frac {25}{2}} \sqrt {- b x + 2} \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{3 b^{\frac {27}{2}} x^{\frac {27}{2}} \sqrt {- b x + 2} - 6 b^{\frac {25}{2}} x^{\frac {25}{2}} \sqrt {- b x + 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((-3*I*b**(23/2)*x**15/(3*b**(27/2)*x**(27/2)*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqrt(b*x - 2)) +
40*I*b**(21/2)*x**14/(3*b**(27/2)*x**(27/2)*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqrt(b*x - 2)) - 60*I*b**(19
/2)*x**13/(3*b**(27/2)*x**(27/2)*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqrt(b*x - 2)) - 30*I*b**10*x**(27/2)*s
qrt(b*x - 2)*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(27/2)*x**(27/2)*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqr
t(b*x - 2)) + 15*pi*b**10*x**(27/2)*sqrt(b*x - 2)/(3*b**(27/2)*x**(27/2)*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)
*sqrt(b*x - 2)) + 60*I*b**9*x**(25/2)*sqrt(b*x - 2)*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(27/2)*x**(27/2)*sq
rt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqrt(b*x - 2)) - 30*pi*b**9*x**(25/2)*sqrt(b*x - 2)/(3*b**(27/2)*x**(27/2)
*sqrt(b*x - 2) - 6*b**(25/2)*x**(25/2)*sqrt(b*x - 2)), Abs(b*x)/2 > 1), (3*b**(23/2)*x**15/(3*b**(27/2)*x**(27
/2)*sqrt(-b*x + 2) - 6*b**(25/2)*x**(25/2)*sqrt(-b*x + 2)) - 40*b**(21/2)*x**14/(3*b**(27/2)*x**(27/2)*sqrt(-b
*x + 2) - 6*b**(25/2)*x**(25/2)*sqrt(-b*x + 2)) + 60*b**(19/2)*x**13/(3*b**(27/2)*x**(27/2)*sqrt(-b*x + 2) - 6
*b**(25/2)*x**(25/2)*sqrt(-b*x + 2)) + 30*b**10*x**(27/2)*sqrt(-b*x + 2)*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b*
*(27/2)*x**(27/2)*sqrt(-b*x + 2) - 6*b**(25/2)*x**(25/2)*sqrt(-b*x + 2)) - 60*b**9*x**(25/2)*sqrt(-b*x + 2)*as
in(sqrt(2)*sqrt(b)*sqrt(x)/2)/(3*b**(27/2)*x**(27/2)*sqrt(-b*x + 2) - 6*b**(25/2)*x**(25/2)*sqrt(-b*x + 2)), T
rue))

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