3.7.41 \(\int \frac {x^{5/2}}{(2-b x)^{5/2}} \, dx\) [641]

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

[Out]

2/3*x^(5/2)/b/(-b*x+2)^(3/2)+10*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2))/b^(7/2)-10/3*x^(3/2)/b^2/(-b*x+2)^(1/2)-5*
x^(1/2)*(-b*x+2)^(1/2)/b^3

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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 = {49, 52, 56, 222} \begin {gather*} \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}} \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 49

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 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 \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 \text {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 [A]
time = 0.14, size = 72, normalized size = 0.81 \begin {gather*} -\frac {\sqrt {x} \left (60-40 b x+3 b^2 x^2\right )}{3 b^3 (2-b x)^{3/2}}+\frac {10 \log \left (-\sqrt {-b} \sqrt {x}+\sqrt {2-b x}\right )}{(-b)^{7/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/(2 - b*x)^(5/2),x]

[Out]

-1/3*(Sqrt[x]*(60 - 40*b*x + 3*b^2*x^2))/(b^3*(2 - b*x)^(3/2)) + (10*Log[-(Sqrt[-b]*Sqrt[x]) + Sqrt[2 - b*x]])
/(-b)^(7/2)

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Mathics [F(-1)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Timed out

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Maple [A]
time = 0.12, size = 81, normalized size = 0.91

method	result	size
meijerg	\(-\frac {8 \left (-\frac {\sqrt {\pi }\, \sqrt {x}\, \sqrt {2}\, \left (-b \right )^{\frac {7}{2}} \left (\frac {21}{4} x^{2} b^{2}-70 b x +105\right )}{56 b^{3} \left (-\frac {b x}{2}+1\right )^{\frac {3}{2}}}+\frac {15 \sqrt {\pi }\, \left (-b \right )^{\frac {7}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{4 b^{\frac {7}{2}}}\right )}{3 \left (-b \right )^{\frac {5}{2}} \sqrt {\pi }\, b}\)	\(81\)
risch	\(\frac {\sqrt {x}\, \left (b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{b^{3} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {\left (\frac {5 \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right )}{b^{\frac {7}{2}}}+\frac {28 \sqrt {-\left (x -\frac {2}{b}\right )^{2} b -2 x +\frac {4}{b}}}{3 b^{4} \left (x -\frac {2}{b}\right )}+\frac {8 \sqrt {-\left (x -\frac {2}{b}\right )^{2} b -2 x +\frac {4}{b}}}{3 b^{5} \left (x -\frac {2}{b}\right )^{2}}\right ) \sqrt {\left (-b x +2\right ) x}}{\sqrt {x}\, \sqrt {-b x +2}}\)	\(168\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-8/3/(-b)^(5/2)/Pi^(1/2)/b*(-1/56*Pi^(1/2)*x^(1/2)*2^(1/2)*(-b)^(7/2)*(21/4*x^2*b^2-70*b*x+105)/b^3/(-1/2*b*x+
1)^(3/2)+15/4*Pi^(1/2)*(-b)^(7/2)/b^(7/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

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Maxima [A]
time = 0.36, 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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Fricas [A]
time = 0.31, 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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Sympy [C] Result contains complex when optimal does not.
time = 4.25, size = 751, normalized size = 8.44 \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}\: \left |{b x}\right | > 2 \\\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), (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)*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)), Tru
e))

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Giac [A]
time = 0.01, size = 129, normalized size = 1.45 \begin {gather*} 2 \left (\frac {2 \left (\left (-\frac {\frac {1}{36}\cdot 9 b^{4} \sqrt {x} \sqrt {x}}{b^{5}}+\frac {\frac {1}{36}\cdot 120 b^{3}}{b^{5}}\right ) \sqrt {x} \sqrt {x}-\frac {\frac {1}{36}\cdot 180 b^{2}}{b^{5}}\right ) \sqrt {x} \sqrt {-b x+2}}{\left (-b x+2\right )^{2}}-\frac {5 \ln \left (\sqrt {-b x+2}-\sqrt {-b} \sqrt {x}\right )}{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)^(5/2),x)

[Out]

-1/3*sqrt(-b*x + 2)*(x*(3*x/b - 40/b^2) + 60/b^3)*sqrt(x)/(b*x - 2)^2 - 10*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 \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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