3.6.67 \(\int \frac {(2-b x)^{5/2}}{x^{3/2}} \, dx\) [567]

Optimal. Leaf size=82 \[ -\frac {15}{2} b \sqrt {x} \sqrt {2-b x}-\frac {5}{2} b \sqrt {x} (2-b x)^{3/2}-\frac {2 (2-b x)^{5/2}}{\sqrt {x}}-15 \sqrt {b} \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 82, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {49, 52, 56, 222} \begin {gather*} -\frac {2 (2-b x)^{5/2}}{\sqrt {x}}-\frac {5}{2} b \sqrt {x} (2-b x)^{3/2}-\frac {15}{2} b \sqrt {x} \sqrt {2-b x}-15 \sqrt {b} \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(2 - b*x)^(5/2)/x^(3/2),x]

[Out]

(-15*b*Sqrt[x]*Sqrt[2 - b*x])/2 - (5*b*Sqrt[x]*(2 - b*x)^(3/2))/2 - (2*(2 - b*x)^(5/2))/Sqrt[x] - 15*Sqrt[b]*A
rcSin[(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*} \int \frac {(2-b x)^{5/2}}{x^{3/2}} \, dx &=-\frac {2 (2-b x)^{5/2}}{\sqrt {x}}-(5 b) \int \frac {(2-b x)^{3/2}}{\sqrt {x}} \, dx\\ &=-\frac {5}{2} b \sqrt {x} (2-b x)^{3/2}-\frac {2 (2-b x)^{5/2}}{\sqrt {x}}-\frac {1}{2} (15 b) \int \frac {\sqrt {2-b x}}{\sqrt {x}} \, dx\\ &=-\frac {15}{2} b \sqrt {x} \sqrt {2-b x}-\frac {5}{2} b \sqrt {x} (2-b x)^{3/2}-\frac {2 (2-b x)^{5/2}}{\sqrt {x}}-\frac {1}{2} (15 b) \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx\\ &=-\frac {15}{2} b \sqrt {x} \sqrt {2-b x}-\frac {5}{2} b \sqrt {x} (2-b x)^{3/2}-\frac {2 (2-b x)^{5/2}}{\sqrt {x}}-(15 b) \text {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )\\ &=-\frac {15}{2} b \sqrt {x} \sqrt {2-b x}-\frac {5}{2} b \sqrt {x} (2-b x)^{3/2}-\frac {2 (2-b x)^{5/2}}{\sqrt {x}}-15 \sqrt {b} \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )\\ \end {align*}

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Mathematica [A]
time = 0.10, size = 68, normalized size = 0.83 \begin {gather*} \frac {\sqrt {2-b x} \left (-16-9 b x+b^2 x^2\right )}{2 \sqrt {x}}-15 \sqrt {-b} \log \left (-\sqrt {-b} \sqrt {x}+\sqrt {2-b x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - b*x)^(5/2)/x^(3/2),x]

[Out]

(Sqrt[2 - b*x]*(-16 - 9*b*x + b^2*x^2))/(2*Sqrt[x]) - 15*Sqrt[-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 = 6.99, size = 156, normalized size = 1.90 \begin {gather*} \text {Piecewise}\left [\left \{\left \{\frac {I \left (30 \sqrt {b} \sqrt {x} \text {ArcCosh}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ] \left (-2+b x\right )+b x \left (2-11 b x+b^2 x^2\right ) \sqrt {-2+b x}+32 \sqrt {-2+b x}\right )}{2 \sqrt {x} \left (-2+b x\right )},\text {Abs}\left [b x\right ]>2\right \}\right \},-15 \sqrt {b} \text {ArcSin}\left [\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2}\right ]-\frac {16}{\sqrt {x} \sqrt {2-b x}}-\frac {b \sqrt {x}}{\sqrt {2-b x}}+\frac {11 b^2 x^{\frac {3}{2}}}{2 \sqrt {2-b x}}-\frac {b^3 x^{\frac {5}{2}}}{2 \sqrt {2-b x}}\right ] \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Piecewise[{{I / 2 (30 Sqrt[b] Sqrt[x] ArcCosh[Sqrt[2] Sqrt[b] Sqrt[x] / 2] (-2 + b x) + b x (2 - 11 b x + b ^
2 x ^ 2) Sqrt[-2 + b x] + 32 Sqrt[-2 + b x]) / (Sqrt[x] (-2 + b x)), Abs[b x] > 2}}, -15 Sqrt[b] ArcSin[Sqrt[2
] Sqrt[b] Sqrt[x] / 2] - 16 / (Sqrt[x] Sqrt[2 - b x]) - b Sqrt[x] / Sqrt[2 - b x] + 11 b ^ 2 x ^ (3 / 2) / (2
Sqrt[2 - b x]) - b ^ 3 x ^ (5 / 2) / (2 Sqrt[2 - b x])]

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Maple [A]
time = 0.12, size = 78, normalized size = 0.95

method	result	size
meijerg	\(\frac {15 \left (-b \right )^{\frac {3}{2}} \left (\frac {16 \sqrt {\pi }\, \sqrt {2}\, \left (-\frac {1}{16} x^{2} b^{2}+\frac {9}{16} b x +1\right ) \sqrt {-\frac {b x}{2}+1}}{15 \sqrt {x}\, \sqrt {-b}}+\frac {2 \sqrt {\pi }\, \sqrt {b}\, \arcsin \left (\frac {\sqrt {b}\, \sqrt {x}\, \sqrt {2}}{2}\right )}{\sqrt {-b}}\right )}{2 \sqrt {\pi }\, b}\)	\(78\)
risch	\(-\frac {\left (b^{3} x^{3}-11 x^{2} b^{2}+2 b x +32\right ) \sqrt {\left (-b x +2\right ) x}}{2 \sqrt {-x \left (b x -2\right )}\, \sqrt {x}\, \sqrt {-b x +2}}-\frac {15 \sqrt {b}\, \arctan \left (\frac {\sqrt {b}\, \left (x -\frac {1}{b}\right )}{\sqrt {-x^{2} b +2 x}}\right ) \sqrt {\left (-b x +2\right ) x}}{2 \sqrt {x}\, \sqrt {-b x +2}}\)	\(106\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15/2*(-b)^(3/2)/Pi^(1/2)/b*(16/15*Pi^(1/2)/x^(1/2)*2^(1/2)/(-b)^(1/2)*(-1/16*x^2*b^2+9/16*b*x+1)*(-1/2*b*x+1)^
(1/2)+2*Pi^(1/2)/(-b)^(1/2)*b^(1/2)*arcsin(1/2*b^(1/2)*x^(1/2)*2^(1/2)))

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Maxima [A]
time = 0.37, size = 96, normalized size = 1.17 \begin {gather*} 15 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) - \frac {\frac {7 \, \sqrt {-b x + 2} b^{2}}{\sqrt {x}} + \frac {9 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b}{x^{\frac {3}{2}}}}{b^{2} - \frac {2 \, {\left (b x - 2\right )} b}{x} + \frac {{\left (b x - 2\right )}^{2}}{x^{2}}} - \frac {8 \, \sqrt {-b x + 2}}{\sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

15*sqrt(b)*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) - (7*sqrt(-b*x + 2)*b^2/sqrt(x) + 9*(-b*x + 2)^(3/2)*b/x^(
3/2))/(b^2 - 2*(b*x - 2)*b/x + (b*x - 2)^2/x^2) - 8*sqrt(-b*x + 2)/sqrt(x)

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Fricas [A]
time = 0.33, size = 117, normalized size = 1.43 \begin {gather*} \left [\frac {15 \, \sqrt {-b} x \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right ) + {\left (b^{2} x^{2} - 9 \, b x - 16\right )} \sqrt {-b x + 2} \sqrt {x}}{2 \, x}, \frac {30 \, \sqrt {b} x \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right ) + {\left (b^{2} x^{2} - 9 \, b x - 16\right )} \sqrt {-b x + 2} \sqrt {x}}{2 \, x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(15*sqrt(-b)*x*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sqrt(x) + 1) + (b^2*x^2 - 9*b*x - 16)*sqrt(-b*x + 2)*sq
rt(x))/x, 1/2*(30*sqrt(b)*x*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x))) + (b^2*x^2 - 9*b*x - 16)*sqrt(-b*x + 2)*s
qrt(x))/x]

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Sympy [A]
time = 3.69, size = 201, normalized size = 2.45 \begin {gather*} \begin {cases} 15 i \sqrt {b} \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )} + \frac {i b^{3} x^{\frac {5}{2}}}{2 \sqrt {b x - 2}} - \frac {11 i b^{2} x^{\frac {3}{2}}}{2 \sqrt {b x - 2}} + \frac {i b \sqrt {x}}{\sqrt {b x - 2}} + \frac {16 i}{\sqrt {x} \sqrt {b x - 2}} & \text {for}\: \left |{b x}\right | > 2 \\- 15 \sqrt {b} \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )} - \frac {b^{3} x^{\frac {5}{2}}}{2 \sqrt {- b x + 2}} + \frac {11 b^{2} x^{\frac {3}{2}}}{2 \sqrt {- b x + 2}} - \frac {b \sqrt {x}}{\sqrt {- b x + 2}} - \frac {16}{\sqrt {x} \sqrt {- b x + 2}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((15*I*sqrt(b)*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2) + I*b**3*x**(5/2)/(2*sqrt(b*x - 2)) - 11*I*b**2*x**(3
/2)/(2*sqrt(b*x - 2)) + I*b*sqrt(x)/sqrt(b*x - 2) + 16*I/(sqrt(x)*sqrt(b*x - 2)), Abs(b*x) > 2), (-15*sqrt(b)*
asin(sqrt(2)*sqrt(b)*sqrt(x)/2) - b**3*x**(5/2)/(2*sqrt(-b*x + 2)) + 11*b**2*x**(3/2)/(2*sqrt(-b*x + 2)) - b*s
qrt(x)/sqrt(-b*x + 2) - 16/(sqrt(x)*sqrt(-b*x + 2)), True))

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Giac [A]
time = 1.26, size = 157, normalized size = 1.91 \begin {gather*} -\frac {b b^{2} \left (\frac {2 \left (\left (-\frac {5}{4}-\frac {1}{4} \sqrt {-b x+2} \sqrt {-b x+2}\right ) \sqrt {-b x+2} \sqrt {-b x+2}+\frac {15}{2}\right ) \sqrt {-b x+2} \sqrt {-b \left (-b x+2\right )+2 b}}{-b \left (-b x+2\right )+2 b}+\frac {15 \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)^(5/2)/x^(3/2),x)

[Out]

1/2*(((b*x - 2)*(b*x - 7) - 30)*sqrt(-b*x + 2)/sqrt((b*x - 2)*b + 2*b) - 30*log(abs(-sqrt(-b*x + 2)*sqrt(-b) +
sqrt((b*x - 2)*b + 2*b)))/sqrt(-b))*b^2/abs(b)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (2-b\,x\right )}^{5/2}}{x^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((2 - b*x)^(5/2)/x^(3/2),x)

[Out]

int((2 - b*x)^(5/2)/x^(3/2), x)

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