\(\int \frac {1}{x^{5/2} (2-b x)^{5/2}} \, dx\) [646]

Optimal result

Integrand size = 16, antiderivative size = 75 \[ \int \frac {1}{x^{5/2} (2-b x)^{5/2}} \, dx=\frac {1}{3 x^{3/2} (2-b x)^{3/2}}+\frac {1}{x^{3/2} \sqrt {2-b x}}-\frac {2 \sqrt {2-b x}}{3 x^{3/2}}-\frac {2 b \sqrt {2-b x}}{3 \sqrt {x}} \]

[Out]

1/3/x^(3/2)/(-b*x+2)^(3/2)+1/x^(3/2)/(-b*x+2)^(1/2)-2/3*(-b*x+2)^(1/2)/x^(3/2)-2/3*b*(-b*x+2)^(1/2)/x^(1/2)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {47, 37} \[ \int \frac {1}{x^{5/2} (2-b x)^{5/2}} \, dx=-\frac {2 \sqrt {2-b x}}{3 x^{3/2}}+\frac {1}{x^{3/2} \sqrt {2-b x}}+\frac {1}{3 x^{3/2} (2-b x)^{3/2}}-\frac {2 b \sqrt {2-b x}}{3 \sqrt {x}} \]

[In]

Int[1/(x^(5/2)*(2 - b*x)^(5/2)),x]

[Out]

1/(3*x^(3/2)*(2 - b*x)^(3/2)) + 1/(x^(3/2)*Sqrt[2 - b*x]) - (2*Sqrt[2 - b*x])/(3*x^(3/2)) - (2*b*Sqrt[2 - b*x]
)/(3*Sqrt[x])

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 47

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

Rubi steps \begin{align*} \text {integral}& = \frac {1}{3 x^{3/2} (2-b x)^{3/2}}+\int \frac {1}{x^{5/2} (2-b x)^{3/2}} \, dx \\ & = \frac {1}{3 x^{3/2} (2-b x)^{3/2}}+\frac {1}{x^{3/2} \sqrt {2-b x}}+2 \int \frac {1}{x^{5/2} \sqrt {2-b x}} \, dx \\ & = \frac {1}{3 x^{3/2} (2-b x)^{3/2}}+\frac {1}{x^{3/2} \sqrt {2-b x}}-\frac {2 \sqrt {2-b x}}{3 x^{3/2}}+\frac {1}{3} (2 b) \int \frac {1}{x^{3/2} \sqrt {2-b x}} \, dx \\ & = \frac {1}{3 x^{3/2} (2-b x)^{3/2}}+\frac {1}{x^{3/2} \sqrt {2-b x}}-\frac {2 \sqrt {2-b x}}{3 x^{3/2}}-\frac {2 b \sqrt {2-b x}}{3 \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 41, normalized size of antiderivative = 0.55 \[ \int \frac {1}{x^{5/2} (2-b x)^{5/2}} \, dx=-\frac {1+3 b x-6 b^2 x^2+2 b^3 x^3}{3 x^{3/2} (2-b x)^{3/2}} \]

[In]

Integrate[1/(x^(5/2)*(2 - b*x)^(5/2)),x]

[Out]

-1/3*(1 + 3*b*x - 6*b^2*x^2 + 2*b^3*x^3)/(x^(3/2)*(2 - b*x)^(3/2))

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 36, normalized size of antiderivative = 0.48

method	result	size
gosper	\(-\frac {2 b^{3} x^{3}-6 b^{2} x^{2}+3 b x +1}{3 x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {3}{2}}}\)	\(36\)
meijerg	\(-\frac {\sqrt {2}\, \left (2 b^{3} x^{3}-6 b^{2} x^{2}+3 b x +1\right )}{12 x^{\frac {3}{2}} \left (-\frac {b x}{2}+1\right )^{\frac {3}{2}}}\)	\(39\)
default	\(-\frac {1}{3 x^{\frac {3}{2}} \left (-b x +2\right )^{\frac {3}{2}}}+b \left (-\frac {1}{\left (-b x +2\right )^{\frac {3}{2}} \sqrt {x}}+2 b \left (\frac {\sqrt {x}}{3 \left (-b x +2\right )^{\frac {3}{2}}}+\frac {\sqrt {x}}{3 \sqrt {-b x +2}}\right )\right )\)	\(61\)
risch	\(\frac {\left (4 b^{2} x^{2}-7 b x -2\right ) \sqrt {\left (-b x +2\right ) x}}{12 x^{\frac {3}{2}} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {b^{2} \left (4 b x -9\right ) \sqrt {x}\, \sqrt {\left (-b x +2\right ) x}}{12 \sqrt {-x \left (b x -2\right )}\, \left (b x -2\right ) \sqrt {-b x +2}}\)	\(98\)

[In]

int(1/x^(5/2)/(-b*x+2)^(5/2),x,method=_RETURNVERBOSE)

[Out]

-1/3/x^(3/2)/(-b*x+2)^(3/2)*(2*b^3*x^3-6*b^2*x^2+3*b*x+1)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.75 \[ \int \frac {1}{x^{5/2} (2-b x)^{5/2}} \, dx=-\frac {{\left (2 \, b^{3} x^{3} - 6 \, b^{2} x^{2} + 3 \, b x + 1\right )} \sqrt {-b x + 2} \sqrt {x}}{3 \, {\left (b^{2} x^{4} - 4 \, b x^{3} + 4 \, x^{2}\right )}} \]

[In]

integrate(1/x^(5/2)/(-b*x+2)^(5/2),x, algorithm="fricas")

[Out]

-1/3*(2*b^3*x^3 - 6*b^2*x^2 + 3*b*x + 1)*sqrt(-b*x + 2)*sqrt(x)/(b^2*x^4 - 4*b*x^3 + 4*x^2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 5.56 (sec) , antiderivative size = 530, normalized size of antiderivative = 7.07 \[ \int \frac {1}{x^{5/2} (2-b x)^{5/2}} \, dx=\begin {cases} - \frac {2 b^{\frac {27}{2}} x^{4} \sqrt {-1 + \frac {2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac {10 b^{\frac {25}{2}} x^{3} \sqrt {-1 + \frac {2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} - \frac {15 b^{\frac {23}{2}} x^{2} \sqrt {-1 + \frac {2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac {5 b^{\frac {21}{2}} x \sqrt {-1 + \frac {2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac {2 b^{\frac {19}{2}} \sqrt {-1 + \frac {2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} & \text {for}\: \frac {1}{\left |{b x}\right |} > \frac {1}{2} \\- \frac {2 i b^{\frac {27}{2}} x^{4} \sqrt {1 - \frac {2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac {10 i b^{\frac {25}{2}} x^{3} \sqrt {1 - \frac {2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} - \frac {15 i b^{\frac {23}{2}} x^{2} \sqrt {1 - \frac {2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac {5 i b^{\frac {21}{2}} x \sqrt {1 - \frac {2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} + \frac {2 i b^{\frac {19}{2}} \sqrt {1 - \frac {2}{b x}}}{3 b^{12} x^{4} - 18 b^{11} x^{3} + 36 b^{10} x^{2} - 24 b^{9} x} & \text {otherwise} \end {cases} \]

[In]

integrate(1/x**(5/2)/(-b*x+2)**(5/2),x)

[Out]

Piecewise((-2*b**(27/2)*x**4*sqrt(-1 + 2/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) + 1
0*b**(25/2)*x**3*sqrt(-1 + 2/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) - 15*b**(23/2)*
x**2*sqrt(-1 + 2/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) + 5*b**(21/2)*x*sqrt(-1 + 2
/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) + 2*b**(19/2)*sqrt(-1 + 2/(b*x))/(3*b**12*x
**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x), 1/Abs(b*x) > 1/2), (-2*I*b**(27/2)*x**4*sqrt(1 - 2/(b*x))/(3
*b**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) + 10*I*b**(25/2)*x**3*sqrt(1 - 2/(b*x))/(3*b**12*x**4
- 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*x) - 15*I*b**(23/2)*x**2*sqrt(1 - 2/(b*x))/(3*b**12*x**4 - 18*b**11
*x**3 + 36*b**10*x**2 - 24*b**9*x) + 5*I*b**(21/2)*x*sqrt(1 - 2/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**1
0*x**2 - 24*b**9*x) + 2*I*b**(19/2)*sqrt(1 - 2/(b*x))/(3*b**12*x**4 - 18*b**11*x**3 + 36*b**10*x**2 - 24*b**9*
x), True))

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.77 \[ \int \frac {1}{x^{5/2} (2-b x)^{5/2}} \, dx=-\frac {3 \, \sqrt {-b x + 2} b}{8 \, \sqrt {x}} + \frac {{\left (b^{3} - \frac {9 \, {\left (b x - 2\right )} b^{2}}{x}\right )} x^{\frac {3}{2}}}{24 \, {\left (-b x + 2\right )}^{\frac {3}{2}}} - \frac {{\left (-b x + 2\right )}^{\frac {3}{2}}}{24 \, x^{\frac {3}{2}}} \]

[In]

integrate(1/x^(5/2)/(-b*x+2)^(5/2),x, algorithm="maxima")

[Out]

-3/8*sqrt(-b*x + 2)*b/sqrt(x) + 1/24*(b^3 - 9*(b*x - 2)*b^2/x)*x^(3/2)/(-b*x + 2)^(3/2) - 1/24*(-b*x + 2)^(3/2
)/x^(3/2)

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 183 vs. \(2 (53) = 106\).

Time = 0.45 (sec) , antiderivative size = 183, normalized size of antiderivative = 2.44 \[ \int \frac {1}{x^{5/2} (2-b x)^{5/2}} \, dx=-\frac {{\left (4 \, {\left (b x - 2\right )} b^{2} {\left | b \right |} + 9 \, b^{2} {\left | b \right |}\right )} \sqrt {-b x + 2}}{12 \, {\left ({\left (b x - 2\right )} b + 2 \, b\right )}^{\frac {3}{2}}} - \frac {3 \, {\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{4} \sqrt {-b} b^{3} - 18 \, {\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} \sqrt {-b} b^{4} + 16 \, \sqrt {-b} b^{5}}{3 \, {\left ({\left (\sqrt {-b x + 2} \sqrt {-b} - \sqrt {{\left (b x - 2\right )} b + 2 \, b}\right )}^{2} - 2 \, b\right )}^{3} {\left | b \right |}} \]

[In]

integrate(1/x^(5/2)/(-b*x+2)^(5/2),x, algorithm="giac")

[Out]

-1/12*(4*(b*x - 2)*b^2*abs(b) + 9*b^2*abs(b))*sqrt(-b*x + 2)/((b*x - 2)*b + 2*b)^(3/2) - 1/3*(3*(sqrt(-b*x + 2
)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^4*sqrt(-b)*b^3 - 18*(sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^
2*sqrt(-b)*b^4 + 16*sqrt(-b)*b^5)/(((sqrt(-b*x + 2)*sqrt(-b) - sqrt((b*x - 2)*b + 2*b))^2 - 2*b)^3*abs(b))

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.97 \[ \int \frac {1}{x^{5/2} (2-b x)^{5/2}} \, dx=\frac {\sqrt {2-b\,x}+3\,b\,x\,\sqrt {2-b\,x}-6\,b^2\,x^2\,\sqrt {2-b\,x}+2\,b^3\,x^3\,\sqrt {2-b\,x}}{x^{3/2}\,\left (x\,\left (12\,b-3\,b^2\,x\right )-12\right )} \]

[In]

int(1/(x^(5/2)*(2 - b*x)^(5/2)),x)

[Out]

((2 - b*x)^(1/2) + 3*b*x*(2 - b*x)^(1/2) - 6*b^2*x^2*(2 - b*x)^(1/2) + 2*b^3*x^3*(2 - b*x)^(1/2))/(x^(3/2)*(x*
(12*b - 3*b^2*x) - 12))