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

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

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Rubi [A]  time = 0.02, 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 = {47, 50, 54, 216} \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 47

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 50

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 54

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 216

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) \operatorname {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 [C]  time = 0.01, size = 28, normalized size = 0.34 \begin {gather*} -\frac {8 \sqrt {2} \, _2F_1\left (-\frac {5}{2},-\frac {1}{2};\frac {1}{2};\frac {b x}{2}\right )}{\sqrt {x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(2 - b*x)^(5/2)/x^(3/2),x]

[Out]

(-8*Sqrt[2]*Hypergeometric2F1[-5/2, -1/2, 1/2, (b*x)/2])/Sqrt[x]

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IntegrateAlgebraic [A]  time = 0.15, size = 68, normalized size = 0.83 \begin {gather*} \frac {\sqrt {2-b x} \left (b^2 x^2-9 b x-16\right )}{2 \sqrt {x}}-15 \sqrt {-b} \log \left (\sqrt {2-b x}-\sqrt {-b} \sqrt {x}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(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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fricas [A]  time = 1.32, 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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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: NotImplementedError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Warning, choosing root of [1,0,%%%{4,[1,
1]%%%}+%%%{4,[1,0]%%%}+%%%{-4,[0,1]%%%}+%%%{-8,[0,0]%%%},0,%%%{6,[2,2]%%%}+%%%{4,[2,1]%%%}+%%%{6,[2,0]%%%}+%%%
{-4,[1,2]%%%}+%%%{-28,[1,1]%%%}+%%%{-8,[1,0]%%%}+%%%{6,[0,2]%%%}+%%%{8,[0,1]%%%}+%%%{24,[0,0]%%%},0,%%%{4,[3,3
]%%%}+%%%{-4,[3,2]%%%}+%%%{-4,[3,1]%%%}+%%%{4,[3,0]%%%}+%%%{4,[2,3]%%%}+%%%{-64,[2,2]%%%}+%%%{20,[2,1]%%%}+%%%
{8,[2,0]%%%}+%%%{-4,[1,3]%%%}+%%%{-20,[1,2]%%%}+%%%{128,[1,1]%%%}+%%%{-16,[1,0]%%%}+%%%{-4,[0,3]%%%}+%%%{8,[0,
2]%%%}+%%%{16,[0,1]%%%}+%%%{-32,[0,0]%%%},0,%%%{1,[4,4]%%%}+%%%{-4,[4,3]%%%}+%%%{6,[4,2]%%%}+%%%{-4,[4,1]%%%}+
%%%{1,[4,0]%%%}+%%%{4,[3,4]%%%}+%%%{-12,[3,3]%%%}+%%%{20,[3,2]%%%}+%%%{-20,[3,1]%%%}+%%%{8,[3,0]%%%}+%%%{6,[2,
4]%%%}+%%%{-20,[2,3]%%%}+%%%{46,[2,2]%%%}+%%%{-40,[2,1]%%%}+%%%{24,[2,0]%%%}+%%%{4,[1,4]%%%}+%%%{-20,[1,3]%%%}
+%%%{40,[1,2]%%%}+%%%{-48,[1,1]%%%}+%%%{32,[1,0]%%%}+%%%{1,[0,4]%%%}+%%%{-8,[0,3]%%%}+%%%{24,[0,2]%%%}+%%%{-32
,[0,1]%%%}+%%%{16,[0,0]%%%}] at parameters values [-15.6438432182,61.7937478349]Warning, choosing root of [1,0
,%%%{4,[1,1]%%%}+%%%{4,[1,0]%%%}+%%%{-4,[0,1]%%%}+%%%{-8,[0,0]%%%},0,%%%{6,[2,2]%%%}+%%%{4,[2,1]%%%}+%%%{6,[2,
0]%%%}+%%%{-4,[1,2]%%%}+%%%{-28,[1,1]%%%}+%%%{-8,[1,0]%%%}+%%%{6,[0,2]%%%}+%%%{8,[0,1]%%%}+%%%{24,[0,0]%%%},0,
%%%{4,[3,3]%%%}+%%%{-4,[3,2]%%%}+%%%{-4,[3,1]%%%}+%%%{4,[3,0]%%%}+%%%{4,[2,3]%%%}+%%%{-64,[2,2]%%%}+%%%{20,[2,
1]%%%}+%%%{8,[2,0]%%%}+%%%{-4,[1,3]%%%}+%%%{-20,[1,2]%%%}+%%%{128,[1,1]%%%}+%%%{-16,[1,0]%%%}+%%%{-4,[0,3]%%%}
+%%%{8,[0,2]%%%}+%%%{16,[0,1]%%%}+%%%{-32,[0,0]%%%},0,%%%{1,[4,4]%%%}+%%%{-4,[4,3]%%%}+%%%{6,[4,2]%%%}+%%%{-4,
[4,1]%%%}+%%%{1,[4,0]%%%}+%%%{4,[3,4]%%%}+%%%{-12,[3,3]%%%}+%%%{20,[3,2]%%%}+%%%{-20,[3,1]%%%}+%%%{8,[3,0]%%%}
+%%%{6,[2,4]%%%}+%%%{-20,[2,3]%%%}+%%%{46,[2,2]%%%}+%%%{-40,[2,1]%%%}+%%%{24,[2,0]%%%}+%%%{4,[1,4]%%%}+%%%{-20
,[1,3]%%%}+%%%{40,[1,2]%%%}+%%%{-48,[1,1]%%%}+%%%{32,[1,0]%%%}+%%%{1,[0,4]%%%}+%%%{-8,[0,3]%%%}+%%%{24,[0,2]%%
%}+%%%{-32,[0,1]%%%}+%%%{16,[0,0]%%%}] at parameters values [-29.292030761,78.6493344628]-b/abs(b)*b^2/b*(2*((
-5/4-1/4*sqrt(-b*x+2)*sqrt(-b*x+2))*sqrt(-b*x+2)*sqrt(-b*x+2)+15/2)*sqrt(-b*x+2)*sqrt(-b*(-b*x+2)+2*b)/(-b*(-b
*x+2)+2*b)+15/sqrt(-b)*ln(abs(sqrt(-b*(-b*x+2)+2*b)-sqrt(-b)*sqrt(-b*x+2))))

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maple [A]  time = 0.02, size = 106, normalized size = 1.29 \begin {gather*} -\frac {15 \sqrt {\left (-b x +2\right ) x}\, \sqrt {b}\, \arctan \left (\frac {\left (x -\frac {1}{b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+2 x}}\right )}{2 \sqrt {-b x +2}\, \sqrt {x}}-\frac {\left (b^{3} x^{3}-11 b^{2} x^{2}+2 b x +32\right ) \sqrt {\left (-b x +2\right ) x}}{2 \sqrt {-\left (b x -2\right ) x}\, \sqrt {-b x +2}\, \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(b^3*x^3-11*b^2*x^2+2*b*x+32)/(-(b*x-2)*x)^(1/2)*((-b*x+2)*x)^(1/2)/(-b*x+2)^(1/2)/x^(1/2)-15/2*((-b*x+2)
*x)^(1/2)/(-b*x+2)^(1/2)*b^(1/2)/x^(1/2)*arctan((x-1/b)/(-b*x^2+2*x)^(1/2)*b^(1/2))

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maxima [A]  time = 2.92, 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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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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sympy [A]  time = 5.62, size = 202, normalized size = 2.46 \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}\: \frac {\left |{b x}\right |}{2} > 1 \\- 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 > 1), (-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
*sqrt(x)/sqrt(-b*x + 2) - 16/(sqrt(x)*sqrt(-b*x + 2)), True))

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