\(\int \frac {1}{x^{5/2} \sqrt {2-b x}} \, dx\) [634]

Optimal result

Integrand size = 16, antiderivative size = 40 \[ \int \frac {1}{x^{5/2} \sqrt {2-b x}} \, dx=-\frac {\sqrt {2-b x}}{3 x^{3/2}}-\frac {b \sqrt {2-b x}}{3 \sqrt {x}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {47, 37} \[ \int \frac {1}{x^{5/2} \sqrt {2-b x}} \, dx=-\frac {\sqrt {2-b x}}{3 x^{3/2}}-\frac {b \sqrt {2-b x}}{3 \sqrt {x}} \]

[In]

Int[1/(x^(5/2)*Sqrt[2 - b*x]),x]

[Out]

-1/3*Sqrt[2 - b*x]/x^(3/2) - (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 {\sqrt {2-b x}}{3 x^{3/2}}+\frac {1}{3} b \int \frac {1}{x^{3/2} \sqrt {2-b x}} \, dx \\ & = -\frac {\sqrt {2-b x}}{3 x^{3/2}}-\frac {b \sqrt {2-b x}}{3 \sqrt {x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.62 \[ \int \frac {1}{x^{5/2} \sqrt {2-b x}} \, dx=\frac {(-1-b x) \sqrt {2-b x}}{3 x^{3/2}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.10 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.48

method	result	size
gosper	\(-\frac {\left (b x +1\right ) \sqrt {-b x +2}}{3 x^{\frac {3}{2}}}\)	\(19\)
meijerg	\(-\frac {\sqrt {2}\, \left (b x +1\right ) \sqrt {-\frac {b x}{2}+1}}{3 x^{\frac {3}{2}}}\)	\(22\)
default	\(-\frac {\sqrt {-b x +2}}{3 x^{\frac {3}{2}}}-\frac {b \sqrt {-b x +2}}{3 \sqrt {x}}\)	\(29\)
risch	\(\frac {\sqrt {\left (-b x +2\right ) x}\, \left (b^{2} x^{2}-b x -2\right )}{3 x^{\frac {3}{2}} \sqrt {-b x +2}\, \sqrt {-x \left (b x -2\right )}}\)	\(47\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.45 \[ \int \frac {1}{x^{5/2} \sqrt {2-b x}} \, dx=-\frac {{\left (b x + 1\right )} \sqrt {-b x + 2}}{3 \, x^{\frac {3}{2}}} \]

[In]

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

[Out]

-1/3*(b*x + 1)*sqrt(-b*x + 2)/x^(3/2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 1.03 (sec) , antiderivative size = 139, normalized size of antiderivative = 3.48 \[ \int \frac {1}{x^{5/2} \sqrt {2-b x}} \, dx=\begin {cases} - \frac {b^{\frac {7}{2}} x^{2} \sqrt {-1 + \frac {2}{b x}}}{3 b^{2} x^{2} - 6 b x} + \frac {b^{\frac {5}{2}} x \sqrt {-1 + \frac {2}{b x}}}{3 b^{2} x^{2} - 6 b x} + \frac {2 b^{\frac {3}{2}} \sqrt {-1 + \frac {2}{b x}}}{3 b^{2} x^{2} - 6 b x} & \text {for}\: \frac {1}{\left |{b x}\right |} > \frac {1}{2} \\- \frac {i b^{\frac {3}{2}} \sqrt {1 - \frac {2}{b x}}}{3} - \frac {i \sqrt {b} \sqrt {1 - \frac {2}{b x}}}{3 x} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((-b**(7/2)*x**2*sqrt(-1 + 2/(b*x))/(3*b**2*x**2 - 6*b*x) + b**(5/2)*x*sqrt(-1 + 2/(b*x))/(3*b**2*x**
2 - 6*b*x) + 2*b**(3/2)*sqrt(-1 + 2/(b*x))/(3*b**2*x**2 - 6*b*x), 1/Abs(b*x) > 1/2), (-I*b**(3/2)*sqrt(1 - 2/(
b*x))/3 - I*sqrt(b)*sqrt(1 - 2/(b*x))/(3*x), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70 \[ \int \frac {1}{x^{5/2} \sqrt {2-b x}} \, dx=-\frac {\sqrt {-b x + 2} b}{2 \, \sqrt {x}} - \frac {{\left (-b x + 2\right )}^{\frac {3}{2}}}{6 \, x^{\frac {3}{2}}} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.08 \[ \int \frac {1}{x^{5/2} \sqrt {2-b x}} \, dx=-\frac {{\left ({\left (b x - 2\right )} b^{3} + 3 \, b^{3}\right )} \sqrt {-b x + 2} b}{3 \, {\left ({\left (b x - 2\right )} b + 2 \, b\right )}^{\frac {3}{2}} {\left | b \right |}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.45 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.48 \[ \int \frac {1}{x^{5/2} \sqrt {2-b x}} \, dx=-\frac {\sqrt {2-b\,x}\,\left (\frac {b\,x}{3}+\frac {1}{3}\right )}{x^{3/2}} \]

[In]

int(1/(x^(5/2)*(2 - b*x)^(1/2)),x)

[Out]

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