\(\int \frac {(2-b x)^{5/2}}{x^{5/2}} \, dx\) [568]

Optimal result

Integrand size = 16, antiderivative size = 84 \[ \int \frac {(2-b x)^{5/2}}{x^{5/2}} \, dx=5 b^2 \sqrt {x} \sqrt {2-b x}+\frac {10 b (2-b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (2-b x)^{5/2}}{3 x^{3/2}}+10 b^{3/2} \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {49, 52, 56, 222} \[ \int \frac {(2-b x)^{5/2}}{x^{5/2}} \, dx=10 b^{3/2} \arcsin \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )+5 b^2 \sqrt {x} \sqrt {2-b x}-\frac {2 (2-b x)^{5/2}}{3 x^{3/2}}+\frac {10 b (2-b x)^{3/2}}{3 \sqrt {x}} \]

[In]

Int[(2 - b*x)^(5/2)/x^(5/2),x]

[Out]

5*b^2*Sqrt[x]*Sqrt[2 - b*x] + (10*b*(2 - b*x)^(3/2))/(3*Sqrt[x]) - (2*(2 - b*x)^(5/2))/(3*x^(3/2)) + 10*b^(3/2
)*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[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*} \text {integral}& = -\frac {2 (2-b x)^{5/2}}{3 x^{3/2}}-\frac {1}{3} (5 b) \int \frac {(2-b x)^{3/2}}{x^{3/2}} \, dx \\ & = \frac {10 b (2-b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (2-b x)^{5/2}}{3 x^{3/2}}+\left (5 b^2\right ) \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx \\ & = 5 b^2 \sqrt {x} \sqrt {2-b x}+\frac {10 b (2-b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (2-b x)^{5/2}}{3 x^{3/2}}+\left (5 b^2\right ) \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx \\ & = 5 b^2 \sqrt {x} \sqrt {2-b x}+\frac {10 b (2-b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (2-b x)^{5/2}}{3 x^{3/2}}+\left (10 b^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right ) \\ & = 5 b^2 \sqrt {x} \sqrt {2-b x}+\frac {10 b (2-b x)^{3/2}}{3 \sqrt {x}}-\frac {2 (2-b x)^{5/2}}{3 x^{3/2}}+10 b^{3/2} \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.15 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.87 \[ \int \frac {(2-b x)^{5/2}}{x^{5/2}} \, dx=\frac {\sqrt {2-b x} \left (-8+28 b x+3 b^2 x^2\right )}{3 x^{3/2}}-20 b^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}-\sqrt {2-b x}}\right ) \]

[In]

Integrate[(2 - b*x)^(5/2)/x^(5/2),x]

[Out]

(Sqrt[2 - b*x]*(-8 + 28*b*x + 3*b^2*x^2))/(3*x^(3/2)) - 20*b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[x])/(Sqrt[2] - Sqrt[2
- b*x])]

Maple [A] (verified)

Time = 0.11 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.93

method	result	size
meijerg	\(\frac {15 \left (-b \right )^{\frac {5}{2}} \left (\frac {32 \sqrt {\pi }\, \sqrt {2}\, \left (-\frac {3}{8} b^{2} x^{2}-\frac {7}{2} b x +1\right ) \sqrt {-\frac {b x}{2}+1}}{45 x^{\frac {3}{2}} \left (-b \right )^{\frac {3}{2}}}-\frac {8 \sqrt {\pi }\, b^{\frac {3}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{3 \left (-b \right )^{\frac {3}{2}}}\right )}{4 \sqrt {\pi }\, b}\)	\(78\)
risch	\(-\frac {\left (3 b^{3} x^{3}+22 b^{2} x^{2}-64 b x +16\right ) \sqrt {\left (-b x +2\right ) x}}{3 x^{\frac {3}{2}} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {5 b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-b \,x^{2}+2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{\sqrt {x}\, \sqrt {-b x +2}}\)	\(107\)

[In]

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

[Out]

15/4*(-b)^(5/2)/Pi^(1/2)/b*(32/45*Pi^(1/2)/x^(3/2)*2^(1/2)/(-b)^(3/2)*(-3/8*b^2*x^2-7/2*b*x+1)*(-1/2*b*x+1)^(1
/2)-8/3*Pi^(1/2)/(-b)^(3/2)*b^(3/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.50 \[ \int \frac {(2-b x)^{5/2}}{x^{5/2}} \, dx=\left [\frac {15 \, \sqrt {-b} b x^{2} \log \left (-b x - \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right ) + {\left (3 \, b^{2} x^{2} + 28 \, b x - 8\right )} \sqrt {-b x + 2} \sqrt {x}}{3 \, x^{2}}, -\frac {30 \, b^{\frac {3}{2}} x^{2} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) - {\left (3 \, b^{2} x^{2} + 28 \, b x - 8\right )} \sqrt {-b x + 2} \sqrt {x}}{3 \, x^{2}}\right ] \]

[In]

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

[Out]

[1/3*(15*sqrt(-b)*b*x^2*log(-b*x - sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) + (3*b^2*x^2 + 28*b*x - 8)*sqrt(-b*x +
2)*sqrt(x))/x^2, -1/3*(30*b^(3/2)*x^2*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) - (3*b^2*x^2 + 28*b*x - 8)*sqr
t(-b*x + 2)*sqrt(x))/x^2]

Sympy [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 4.50 (sec) , antiderivative size = 223, normalized size of antiderivative = 2.65 \[ \int \frac {(2-b x)^{5/2}}{x^{5/2}} \, dx=\begin {cases} b^{\frac {5}{2}} x \sqrt {-1 + \frac {2}{b x}} + \frac {28 b^{\frac {3}{2}} \sqrt {-1 + \frac {2}{b x}}}{3} + 5 i b^{\frac {3}{2}} \log {\left (\frac {1}{b x} \right )} - 10 i b^{\frac {3}{2}} \log {\left (\frac {1}{\sqrt {b} \sqrt {x}} \right )} + 10 b^{\frac {3}{2}} \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )} - \frac {8 \sqrt {b} \sqrt {-1 + \frac {2}{b x}}}{3 x} & \text {for}\: \frac {1}{\left |{b x}\right |} > \frac {1}{2} \\i b^{\frac {5}{2}} x \sqrt {1 - \frac {2}{b x}} + \frac {28 i b^{\frac {3}{2}} \sqrt {1 - \frac {2}{b x}}}{3} + 5 i b^{\frac {3}{2}} \log {\left (\frac {1}{b x} \right )} - 10 i b^{\frac {3}{2}} \log {\left (\sqrt {1 - \frac {2}{b x}} + 1 \right )} - \frac {8 i \sqrt {b} \sqrt {1 - \frac {2}{b x}}}{3 x} & \text {otherwise} \end {cases} \]

[In]

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

[Out]

Piecewise((b**(5/2)*x*sqrt(-1 + 2/(b*x)) + 28*b**(3/2)*sqrt(-1 + 2/(b*x))/3 + 5*I*b**(3/2)*log(1/(b*x)) - 10*I
*b**(3/2)*log(1/(sqrt(b)*sqrt(x))) + 10*b**(3/2)*asin(sqrt(2)*sqrt(b)*sqrt(x)/2) - 8*sqrt(b)*sqrt(-1 + 2/(b*x)
)/(3*x), 1/Abs(b*x) > 1/2), (I*b**(5/2)*x*sqrt(1 - 2/(b*x)) + 28*I*b**(3/2)*sqrt(1 - 2/(b*x))/3 + 5*I*b**(3/2)
*log(1/(b*x)) - 10*I*b**(3/2)*log(sqrt(1 - 2/(b*x)) + 1) - 8*I*sqrt(b)*sqrt(1 - 2/(b*x))/(3*x), True))

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.94 \[ \int \frac {(2-b x)^{5/2}}{x^{5/2}} \, dx=-10 \, b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) + \frac {8 \, \sqrt {-b x + 2} b}{\sqrt {x}} + \frac {2 \, \sqrt {-b x + 2} b^{2}}{{\left (b - \frac {b x - 2}{x}\right )} \sqrt {x}} - \frac {4 \, {\left (-b x + 2\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \]

[In]

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

[Out]

-10*b^(3/2)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) + 8*sqrt(-b*x + 2)*b/sqrt(x) + 2*sqrt(-b*x + 2)*b^2/((b -
(b*x - 2)/x)*sqrt(x)) - 4/3*(-b*x + 2)^(3/2)/x^(3/2)

Giac [A] (verification not implemented)

none

Time = 5.85 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.18 \[ \int \frac {(2-b x)^{5/2}}{x^{5/2}} \, dx=\frac {{\left (\frac {30 \, b^{2} \log \left ({\left | -\sqrt {-b x + 2} \sqrt {-b} + \sqrt {{\left (b x - 2\right )} b + 2 \, b} \right |}\right )}{\sqrt {-b}} + \frac {{\left (60 \, b^{3} + {\left (3 \, {\left (b x - 2\right )} b^{3} + 40 \, b^{3}\right )} {\left (b x - 2\right )}\right )} \sqrt {-b x + 2}}{{\left ({\left (b x - 2\right )} b + 2 \, b\right )}^{\frac {3}{2}}}\right )} b}{3 \, {\left | b \right |}} \]

[In]

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

[Out]

1/3*(30*b^2*log(abs(-sqrt(-b*x + 2)*sqrt(-b) + sqrt((b*x - 2)*b + 2*b)))/sqrt(-b) + (60*b^3 + (3*(b*x - 2)*b^3
+ 40*b^3)*(b*x - 2))*sqrt(-b*x + 2)/((b*x - 2)*b + 2*b)^(3/2))*b/abs(b)

Mupad [F(-1)]

Timed out. \[ \int \frac {(2-b x)^{5/2}}{x^{5/2}} \, dx=\int \frac {{\left (2-b\,x\right )}^{5/2}}{x^{5/2}} \,d x \]

[In]

int((2 - b*x)^(5/2)/x^(5/2),x)

[Out]

int((2 - b*x)^(5/2)/x^(5/2), x)