3.6.44 \(\int \frac {(2-b x)^{3/2}}{x^{5/2}} \, dx\) [544]

Optimal. Leaf size=62 \[ \frac {2 b \sqrt {2-b x}}{\sqrt {x}}-\frac {2 (2-b x)^{3/2}}{3 x^{3/2}}+2 b^{3/2} \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {49, 56, 222} \begin {gather*} 2 b^{3/2} \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )-\frac {2 (2-b x)^{3/2}}{3 x^{3/2}}+\frac {2 b \sqrt {2-b x}}{\sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - b*x)^(3/2)/x^(5/2),x]

[Out]

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

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Mathematica [A]
time = 0.09, size = 62, normalized size = 1.00 \begin {gather*} \frac {4 \sqrt {2-b x} (-1+2 b x)}{3 x^{3/2}}+2 \sqrt {-b} b \log \left (-\sqrt {-b} \sqrt {x}+\sqrt {2-b x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - b*x)^(3/2)/x^(5/2),x]

[Out]

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

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Mathics [C] Result contains higher order function than in optimal. Order 9 vs. order 3 in optimal.
time = 3.71, size = 163, normalized size = 2.63 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {\sqrt {b} \left (b x \left (-6 I \text {Log}\left [\frac {1}{\sqrt {b} \sqrt {x}}\right ]+3 I \text {Log}\left [\frac {1}{b x}\right ]+6 \text {ArcSin}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ]+8 \sqrt {\frac {2-b x}{b x}}\right )-4 \sqrt {\frac {2-b x}{b x}}\right )}{3 x},\frac {1}{\text {Abs}\left [b x\right ]}>\frac {1}{2}\right \}\right \},-2 I b^{\frac {3}{2}} \text {Log}\left [1+\sqrt {1-\frac {2}{b x}}\right ]+\frac {I 8 b^{\frac {3}{2}} \sqrt {1-\frac {2}{b x}}}{3}+I b^{\frac {3}{2}} \text {Log}\left [\frac {1}{b x}\right ]-\frac {4 I \sqrt {b} \sqrt {1-\frac {2}{b x}}}{3 x}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

mathics('Integrate[(2 - b*x)^(3/2)/x^(5/2),x]')

[Out]

Piecewise[{{Sqrt[b] (b x (-6 I Log[1 / (Sqrt[b] Sqrt[x])] + 3 I Log[1 / (b x)] + 6 ArcSin[Sqrt[2] Sqrt[b] Sqrt
[x] / 2] + 8 Sqrt[(2 - b x) / (b x)]) - 4 Sqrt[(2 - b x) / (b x)]) / (3 x), 1 / Abs[b x] > 1 / 2}}, -2 I b ^ (
3 / 2) Log[1 + Sqrt[1 - 2 / (b x)]] + I 8 b ^ (3 / 2) Sqrt[1 - 2 / (b x)] / 3 + I b ^ (3 / 2) Log[1 / (b x)] -
4 I Sqrt[b] Sqrt[1 - 2 / (b x)] / (3 x)]

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Maple [A]
time = 0.12, size = 70, normalized size = 1.13

method	result	size
meijerg	\(-\frac {3 \left (-b \right )^{\frac {5}{2}} \left (-\frac {16 \sqrt {\pi }\, \sqrt {2}\, \left (-2 b x +1\right ) \sqrt {-\frac {b x}{2}+1}}{9 x^{\frac {3}{2}} \left (-b \right )^{\frac {3}{2}}}+\frac {8 \sqrt {\pi }\, b^{\frac {3}{2}} \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{3 \left (-b \right )^{\frac {3}{2}}}\right )}{4 \sqrt {\pi }\, b}\)	\(70\)
risch	\(-\frac {4 \left (2 x^{2} b^{2}-5 b x +2\right ) \sqrt {\left (-b x +2\right ) x}}{3 x^{\frac {3}{2}} \sqrt {-x \left (b x -2\right )}\, \sqrt {-b x +2}}+\frac {b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{\sqrt {x}\, \sqrt {-b x +2}}\)	\(98\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-b*x+2)^(3/2)/x^(5/2),x,method=_RETURNVERBOSE)

[Out]

-3/4*(-b)^(5/2)/Pi^(1/2)/b*(-16/9*Pi^(1/2)/x^(3/2)*2^(1/2)/(-b)^(3/2)*(-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)))

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Maxima [A]
time = 0.35, size = 49, normalized size = 0.79 \begin {gather*} -2 \, b^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) + \frac {2 \, \sqrt {-b x + 2} b}{\sqrt {x}} - \frac {2 \, {\left (-b x + 2\right )}^{\frac {3}{2}}}{3 \, x^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+2)^(3/2)/x^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.32, size = 111, normalized size = 1.79 \begin {gather*} \left [\frac {3 \, \sqrt {-b} b x^{2} \log \left (-b x - \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right ) + 4 \, {\left (2 \, b x - 1\right )} \sqrt {-b x + 2} \sqrt {x}}{3 \, x^{2}}, -\frac {2 \, {\left (3 \, b^{\frac {3}{2}} x^{2} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) - 2 \, {\left (2 \, b x - 1\right )} \sqrt {-b x + 2} \sqrt {x}\right )}}{3 \, x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+2)^(3/2)/x^(5/2),x, algorithm="fricas")

[Out]

[1/3*(3*sqrt(-b)*b*x^2*log(-b*x - sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) + 4*(2*b*x - 1)*sqrt(-b*x + 2)*sqrt(x))
/x^2, -2/3*(3*b^(3/2)*x^2*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) - 2*(2*b*x - 1)*sqrt(-b*x + 2)*sqrt(x))/x^2
]

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Sympy [C] Result contains complex when optimal does not.
time = 1.81, size = 184, normalized size = 2.97 \begin {gather*} \begin {cases} \frac {8 b^{\frac {3}{2}} \sqrt {-1 + \frac {2}{b x}}}{3} + i b^{\frac {3}{2}} \log {\left (\frac {1}{b x} \right )} - 2 i b^{\frac {3}{2}} \log {\left (\frac {1}{\sqrt {b} \sqrt {x}} \right )} + 2 b^{\frac {3}{2}} \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )} - \frac {4 \sqrt {b} \sqrt {-1 + \frac {2}{b x}}}{3 x} & \text {for}\: \frac {1}{\left |{b x}\right |} > \frac {1}{2} \\\frac {8 i b^{\frac {3}{2}} \sqrt {1 - \frac {2}{b x}}}{3} + i b^{\frac {3}{2}} \log {\left (\frac {1}{b x} \right )} - 2 i b^{\frac {3}{2}} \log {\left (\sqrt {1 - \frac {2}{b x}} + 1 \right )} - \frac {4 i \sqrt {b} \sqrt {1 - \frac {2}{b x}}}{3 x} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+2)**(3/2)/x**(5/2),x)

[Out]

Piecewise((8*b**(3/2)*sqrt(-1 + 2/(b*x))/3 + I*b**(3/2)*log(1/(b*x)) - 2*I*b**(3/2)*log(1/(sqrt(b)*sqrt(x))) +
2*b**(3/2)*asin(sqrt(2)*sqrt(b)*sqrt(x)/2) - 4*sqrt(b)*sqrt(-1 + 2/(b*x))/(3*x), 1/Abs(b*x) > 1/2), (8*I*b**(
3/2)*sqrt(1 - 2/(b*x))/3 + I*b**(3/2)*log(1/(b*x)) - 2*I*b**(3/2)*log(sqrt(1 - 2/(b*x)) + 1) - 4*I*sqrt(b)*sqr
t(1 - 2/(b*x))/(3*x), True))

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Giac [A]
time = 1.13, size = 141, normalized size = 2.27 \begin {gather*} \frac {b^{2} \left (\frac {2 \left (-\frac {12}{9} b^{3} \sqrt {-b x+2} \sqrt {-b x+2}+\frac {18}{9} b^{3}\right ) \sqrt {-b x+2} \sqrt {-b \left (-b x+2\right )+2 b}}{\left (-b \left (-b x+2\right )+2 b\right )^{2}}+\frac {2 b^{2} \ln \left |\sqrt {-b \left (-b x+2\right )+2 b}-\sqrt {-b} \sqrt {-b x+2}\right |}{\sqrt {-b}}\right )}{\left |b\right | b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+2)^(3/2)/x^(5/2),x)

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (2-b\,x\right )}^{3/2}}{x^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2 - b*x)^(3/2)/x^(5/2),x)

[Out]

int((2 - b*x)^(3/2)/x^(5/2), x)

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