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

Optimal. Leaf size=91 \[ \frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}-\frac {5 \sqrt {x} \sqrt {2-b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{3 b} \]

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Rubi [A]  time = 0.02, antiderivative size = 91, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.188, Rules used = {50, 54, 216} \begin {gather*} -\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {5 \sqrt {x} \sqrt {2-b x}}{2 b^3}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}-\frac {x^{5/2} \sqrt {2-b x}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(5/2)/Sqrt[2 - b*x],x]

[Out]

(-5*Sqrt[x]*Sqrt[2 - b*x])/(2*b^3) - (5*x^(3/2)*Sqrt[2 - b*x])/(6*b^2) - (x^(5/2)*Sqrt[2 - b*x])/(3*b) + (5*Ar
cSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

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 {x^{5/2}}{\sqrt {2-b x}} \, dx &=-\frac {x^{5/2} \sqrt {2-b x}}{3 b}+\frac {5 \int \frac {x^{3/2}}{\sqrt {2-b x}} \, dx}{3 b}\\ &=-\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{3 b}+\frac {5 \int \frac {\sqrt {x}}{\sqrt {2-b x}} \, dx}{2 b^2}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{3 b}+\frac {5 \int \frac {1}{\sqrt {x} \sqrt {2-b x}} \, dx}{2 b^3}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{3 b}+\frac {5 \operatorname {Subst}\left (\int \frac {1}{\sqrt {2-b x^2}} \, dx,x,\sqrt {x}\right )}{b^3}\\ &=-\frac {5 \sqrt {x} \sqrt {2-b x}}{2 b^3}-\frac {5 x^{3/2} \sqrt {2-b x}}{6 b^2}-\frac {x^{5/2} \sqrt {2-b x}}{3 b}+\frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}\\ \end {align*}

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Mathematica [A]  time = 0.04, size = 61, normalized size = 0.67 \begin {gather*} \frac {5 \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {2}}\right )}{b^{7/2}}-\frac {\sqrt {x} \sqrt {2-b x} \left (2 b^2 x^2+5 b x+15\right )}{6 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(5/2)/Sqrt[2 - b*x],x]

[Out]

-1/6*(Sqrt[x]*Sqrt[2 - b*x]*(15 + 5*b*x + 2*b^2*x^2))/b^3 + (5*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[2]])/b^(7/2)

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IntegrateAlgebraic [A]  time = 0.11, size = 82, normalized size = 0.90 \begin {gather*} \frac {5 \sqrt {-b} \log \left (\sqrt {2-b x}-\sqrt {-b} \sqrt {x}\right )}{b^4}+\frac {\sqrt {2-b x} \left (-2 b^2 x^{5/2}-5 b x^{3/2}-15 \sqrt {x}\right )}{6 b^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(5/2)/Sqrt[2 - b*x],x]

[Out]

(Sqrt[2 - b*x]*(-15*Sqrt[x] - 5*b*x^(3/2) - 2*b^2*x^(5/2)))/(6*b^3) + (5*Sqrt[-b]*Log[-(Sqrt[-b]*Sqrt[x]) + Sq
rt[2 - b*x]])/b^4

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fricas [A]  time = 1.44, size = 125, normalized size = 1.37 \begin {gather*} \left [-\frac {{\left (2 \, b^{3} x^{2} + 5 \, b^{2} x + 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} + 15 \, \sqrt {-b} \log \left (-b x + \sqrt {-b x + 2} \sqrt {-b} \sqrt {x} + 1\right )}{6 \, b^{4}}, -\frac {{\left (2 \, b^{3} x^{2} + 5 \, b^{2} x + 15 \, b\right )} \sqrt {-b x + 2} \sqrt {x} + 30 \, \sqrt {b} \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{6 \, b^{4}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(-b*x+2)^(1/2),x, algorithm="fricas")

[Out]

[-1/6*((2*b^3*x^2 + 5*b^2*x + 15*b)*sqrt(-b*x + 2)*sqrt(x) + 15*sqrt(-b)*log(-b*x + sqrt(-b*x + 2)*sqrt(-b)*sq
rt(x) + 1))/b^4, -1/6*((2*b^3*x^2 + 5*b^2*x + 15*b)*sqrt(-b*x + 2)*sqrt(x) + 30*sqrt(b)*arctan(sqrt(-b*x + 2)/
(sqrt(b)*sqrt(x))))/b^4]

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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(x^(5/2)/(-b*x+2)^(1/2),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]-2*abs(b)/b^2/b*(2*((
12*b^5/144/b^7*sqrt(-b*x+2)*sqrt(-b*x+2)-78*b^5/144/b^7)*sqrt(-b*x+2)*sqrt(-b*x+2)+198*b^5/144/b^7)*sqrt(-b*x+
2)*sqrt(-b*(-b*x+2)+2*b)-5/2/b/sqrt(-b)*ln(abs(sqrt(-b*(-b*x+2)+2*b)-sqrt(-b)*sqrt(-b*x+2))))

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maple [A]  time = 0.00, size = 100, normalized size = 1.10 \begin {gather*} -\frac {\sqrt {-b x +2}\, x^{\frac {5}{2}}}{3 b}-\frac {5 \sqrt {-b x +2}\, x^{\frac {3}{2}}}{6 b^{2}}-\frac {5 \sqrt {-b x +2}\, \sqrt {x}}{2 b^{3}}+\frac {5 \sqrt {\left (-b x +2\right ) x}\, \arctan \left (\frac {\left (x -\frac {1}{b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+2 x}}\right )}{2 \sqrt {-b x +2}\, b^{\frac {7}{2}} \sqrt {x}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*x^(5/2)*(-b*x+2)^(1/2)/b-5/6*x^(3/2)*(-b*x+2)^(1/2)/b^2-5/2*(-b*x+2)^(1/2)/b^3*x^(1/2)+5/2*((-b*x+2)*x)^(
1/2)/(-b*x+2)^(1/2)/b^(7/2)/x^(1/2)*arctan((x-1/b)/(-b*x^2+2*x)^(1/2)*b^(1/2))

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maxima [A]  time = 3.03, size = 117, normalized size = 1.29 \begin {gather*} -\frac {\frac {33 \, \sqrt {-b x + 2} b^{2}}{\sqrt {x}} + \frac {40 \, {\left (-b x + 2\right )}^{\frac {3}{2}} b}{x^{\frac {3}{2}}} + \frac {15 \, {\left (-b x + 2\right )}^{\frac {5}{2}}}{x^{\frac {5}{2}}}}{3 \, {\left (b^{6} - \frac {3 \, {\left (b x - 2\right )} b^{5}}{x} + \frac {3 \, {\left (b x - 2\right )}^{2} b^{4}}{x^{2}} - \frac {{\left (b x - 2\right )}^{3} b^{3}}{x^{3}}\right )}} - \frac {5 \, \arctan \left (\frac {\sqrt {-b x + 2}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {7}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(5/2)/(-b*x+2)^(1/2),x, algorithm="maxima")

[Out]

-1/3*(33*sqrt(-b*x + 2)*b^2/sqrt(x) + 40*(-b*x + 2)^(3/2)*b/x^(3/2) + 15*(-b*x + 2)^(5/2)/x^(5/2))/(b^6 - 3*(b
*x - 2)*b^5/x + 3*(b*x - 2)^2*b^4/x^2 - (b*x - 2)^3*b^3/x^3) - 5*arctan(sqrt(-b*x + 2)/(sqrt(b)*sqrt(x)))/b^(7
/2)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^{5/2}}{\sqrt {2-b\,x}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(5/2)/(2 - b*x)^(1/2),x)

[Out]

int(x^(5/2)/(2 - b*x)^(1/2), x)

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sympy [A]  time = 7.51, size = 206, normalized size = 2.26 \begin {gather*} \begin {cases} - \frac {i x^{\frac {7}{2}}}{3 \sqrt {b x - 2}} - \frac {i x^{\frac {5}{2}}}{6 b \sqrt {b x - 2}} - \frac {5 i x^{\frac {3}{2}}}{6 b^{2} \sqrt {b x - 2}} + \frac {5 i \sqrt {x}}{b^{3} \sqrt {b x - 2}} - \frac {5 i \operatorname {acosh}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} & \text {for}\: \frac {\left |{b x}\right |}{2} > 1 \\\frac {x^{\frac {7}{2}}}{3 \sqrt {- b x + 2}} + \frac {x^{\frac {5}{2}}}{6 b \sqrt {- b x + 2}} + \frac {5 x^{\frac {3}{2}}}{6 b^{2} \sqrt {- b x + 2}} - \frac {5 \sqrt {x}}{b^{3} \sqrt {- b x + 2}} + \frac {5 \operatorname {asin}{\left (\frac {\sqrt {2} \sqrt {b} \sqrt {x}}{2} \right )}}{b^{\frac {7}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(5/2)/(-b*x+2)**(1/2),x)

[Out]

Piecewise((-I*x**(7/2)/(3*sqrt(b*x - 2)) - I*x**(5/2)/(6*b*sqrt(b*x - 2)) - 5*I*x**(3/2)/(6*b**2*sqrt(b*x - 2)
) + 5*I*sqrt(x)/(b**3*sqrt(b*x - 2)) - 5*I*acosh(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/2), Abs(b*x)/2 > 1), (x**(7/
2)/(3*sqrt(-b*x + 2)) + x**(5/2)/(6*b*sqrt(-b*x + 2)) + 5*x**(3/2)/(6*b**2*sqrt(-b*x + 2)) - 5*sqrt(x)/(b**3*s
qrt(-b*x + 2)) + 5*asin(sqrt(2)*sqrt(b)*sqrt(x)/2)/b**(7/2), True))

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