\(\int \frac {\sqrt {2-b x}}{x^{7/2}} \, dx\) [519]

Optimal result

Integrand size = 16, antiderivative size = 40 \[ \int \frac {\sqrt {2-b x}}{x^{7/2}} \, dx=-\frac {(2-b x)^{3/2}}{5 x^{5/2}}-\frac {b (2-b x)^{3/2}}{15 x^{3/2}} \]

[Out]

-1/5*(-b*x+2)^(3/2)/x^(5/2)-1/15*b*(-b*x+2)^(3/2)/x^(3/2)

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} \[ \int \frac {\sqrt {2-b x}}{x^{7/2}} \, dx=-\frac {b (2-b x)^{3/2}}{15 x^{3/2}}-\frac {(2-b x)^{3/2}}{5 x^{5/2}} \]

[In]

Int[Sqrt[2 - b*x]/x^(7/2),x]

[Out]

-1/5*(2 - b*x)^(3/2)/x^(5/2) - (b*(2 - b*x)^(3/2))/(15*x^(3/2))

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 {(2-b x)^{3/2}}{5 x^{5/2}}+\frac {1}{5} b \int \frac {\sqrt {2-b x}}{x^{5/2}} \, dx \\ & = -\frac {(2-b x)^{3/2}}{5 x^{5/2}}-\frac {b (2-b x)^{3/2}}{15 x^{3/2}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.06 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.78 \[ \int \frac {\sqrt {2-b x}}{x^{7/2}} \, dx=\frac {\sqrt {2-b x} \left (-6+b x+b^2 x^2\right )}{15 x^{5/2}} \]

[In]

Integrate[Sqrt[2 - b*x]/x^(7/2),x]

[Out]

(Sqrt[2 - b*x]*(-6 + b*x + b^2*x^2))/(15*x^(5/2))

Maple [A] (verified)

Time = 0.08 (sec) , antiderivative size = 19, normalized size of antiderivative = 0.48

method	result	size
gosper	\(-\frac {\left (b x +3\right ) \left (-b x +2\right )^{\frac {3}{2}}}{15 x^{\frac {5}{2}}}\)	\(19\)
meijerg	\(-\frac {2 \sqrt {2}\, \left (-\frac {1}{6} b^{2} x^{2}-\frac {1}{6} b x +1\right ) \sqrt {-\frac {b x}{2}+1}}{5 x^{\frac {5}{2}}}\)	\(31\)
default	\(-\frac {2 \sqrt {-b x +2}}{5 x^{\frac {5}{2}}}-\frac {b \left (-\frac {\sqrt {-b x +2}}{3 x^{\frac {3}{2}}}-\frac {b \sqrt {-b x +2}}{3 \sqrt {x}}\right )}{5}\)	\(46\)
risch	\(-\frac {\sqrt {\left (-b x +2\right ) x}\, \left (b^{3} x^{3}-b^{2} x^{2}-8 b x +12\right )}{15 x^{\frac {5}{2}} \sqrt {-b x +2}\, \sqrt {-x \left (b x -2\right )}}\)	\(55\)

[In]

int((-b*x+2)^(1/2)/x^(7/2),x,method=_RETURNVERBOSE)

[Out]

-1/15/x^(5/2)*(b*x+3)*(-b*x+2)^(3/2)

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.62 \[ \int \frac {\sqrt {2-b x}}{x^{7/2}} \, dx=\frac {{\left (b^{2} x^{2} + b x - 6\right )} \sqrt {-b x + 2}}{15 \, x^{\frac {5}{2}}} \]

[In]

integrate((-b*x+2)^(1/2)/x^(7/2),x, algorithm="fricas")

[Out]

1/15*(b^2*x^2 + b*x - 6)*sqrt(-b*x + 2)/x^(5/2)

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 2.93 (sec) , antiderivative size = 194, normalized size of antiderivative = 4.85 \[ \int \frac {\sqrt {2-b x}}{x^{7/2}} \, dx=\begin {cases} \frac {b^{\frac {9}{2}} x^{2} \sqrt {-1 + \frac {2}{b x}}}{15 b^{2} x^{2} - 30 b x} - \frac {b^{\frac {7}{2}} x \sqrt {-1 + \frac {2}{b x}}}{15 b^{2} x^{2} - 30 b x} - \frac {8 b^{\frac {5}{2}} \sqrt {-1 + \frac {2}{b x}}}{15 b^{2} x^{2} - 30 b x} + \frac {12 b^{\frac {3}{2}} \sqrt {-1 + \frac {2}{b x}}}{x \left (15 b^{2} x^{2} - 30 b x\right )} & \text {for}\: \frac {1}{\left |{b x}\right |} > \frac {1}{2} \\\frac {i b^{\frac {5}{2}} \sqrt {1 - \frac {2}{b x}}}{15} + \frac {i b^{\frac {3}{2}} \sqrt {1 - \frac {2}{b x}}}{15 x} - \frac {2 i \sqrt {b} \sqrt {1 - \frac {2}{b x}}}{5 x^{2}} & \text {otherwise} \end {cases} \]

[In]

integrate((-b*x+2)**(1/2)/x**(7/2),x)

[Out]

Piecewise((b**(9/2)*x**2*sqrt(-1 + 2/(b*x))/(15*b**2*x**2 - 30*b*x) - b**(7/2)*x*sqrt(-1 + 2/(b*x))/(15*b**2*x
**2 - 30*b*x) - 8*b**(5/2)*sqrt(-1 + 2/(b*x))/(15*b**2*x**2 - 30*b*x) + 12*b**(3/2)*sqrt(-1 + 2/(b*x))/(x*(15*
b**2*x**2 - 30*b*x)), 1/Abs(b*x) > 1/2), (I*b**(5/2)*sqrt(1 - 2/(b*x))/15 + I*b**(3/2)*sqrt(1 - 2/(b*x))/(15*x
) - 2*I*sqrt(b)*sqrt(1 - 2/(b*x))/(5*x**2), True))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.70 \[ \int \frac {\sqrt {2-b x}}{x^{7/2}} \, dx=-\frac {{\left (-b x + 2\right )}^{\frac {3}{2}} b}{6 \, x^{\frac {3}{2}}} - \frac {{\left (-b x + 2\right )}^{\frac {5}{2}}}{10 \, x^{\frac {5}{2}}} \]

[In]

integrate((-b*x+2)^(1/2)/x^(7/2),x, algorithm="maxima")

[Out]

-1/6*(-b*x + 2)^(3/2)*b/x^(3/2) - 1/10*(-b*x + 2)^(5/2)/x^(5/2)

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.20 \[ \int \frac {\sqrt {2-b x}}{x^{7/2}} \, dx=\frac {{\left ({\left (b x - 2\right )} b^{5} + 5 \, b^{5}\right )} {\left (b x - 2\right )} \sqrt {-b x + 2} b}{15 \, {\left ({\left (b x - 2\right )} b + 2 \, b\right )}^{\frac {5}{2}} {\left | b \right |}} \]

[In]

integrate((-b*x+2)^(1/2)/x^(7/2),x, algorithm="giac")

[Out]

1/15*((b*x - 2)*b^5 + 5*b^5)*(b*x - 2)*sqrt(-b*x + 2)*b/(((b*x - 2)*b + 2*b)^(5/2)*abs(b))

Mupad [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.65 \[ \int \frac {\sqrt {2-b x}}{x^{7/2}} \, dx=\frac {\sqrt {2-b\,x}\,\left (\frac {b^2\,x^2}{15}+\frac {b\,x}{15}-\frac {2}{5}\right )}{x^{5/2}} \]

[In]

int((2 - b*x)^(1/2)/x^(7/2),x)

[Out]

((2 - b*x)^(1/2)*((b*x)/15 + (b^2*x^2)/15 - 2/5))/x^(5/2)