3.5.98 \(\int x^{3/2} \sqrt {a-b x} \, dx\)

Optimal. Leaf size=102 \[ \frac {a^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{8 b^{5/2}}-\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b^2}-\frac {a x^{3/2} \sqrt {a-b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a-b x} \]

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Rubi [A]  time = 0.03, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {50, 63, 217, 203} \begin {gather*} -\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b^2}+\frac {a^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{8 b^{5/2}}-\frac {a x^{3/2} \sqrt {a-b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a-b x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^(3/2)*Sqrt[a - b*x],x]

[Out]

-(a^2*Sqrt[x]*Sqrt[a - b*x])/(8*b^2) - (a*x^(3/2)*Sqrt[a - b*x])/(12*b) + (x^(5/2)*Sqrt[a - b*x])/3 + (a^3*Arc
Tan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/(8*b^(5/2))

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 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int x^{3/2} \sqrt {a-b x} \, dx &=\frac {1}{3} x^{5/2} \sqrt {a-b x}+\frac {1}{6} a \int \frac {x^{3/2}}{\sqrt {a-b x}} \, dx\\ &=-\frac {a x^{3/2} \sqrt {a-b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a-b x}+\frac {a^2 \int \frac {\sqrt {x}}{\sqrt {a-b x}} \, dx}{8 b}\\ &=-\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b^2}-\frac {a x^{3/2} \sqrt {a-b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a-b x}+\frac {a^3 \int \frac {1}{\sqrt {x} \sqrt {a-b x}} \, dx}{16 b^2}\\ &=-\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b^2}-\frac {a x^{3/2} \sqrt {a-b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a-b x}+\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b x^2}} \, dx,x,\sqrt {x}\right )}{8 b^2}\\ &=-\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b^2}-\frac {a x^{3/2} \sqrt {a-b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a-b x}+\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{1+b x^2} \, dx,x,\frac {\sqrt {x}}{\sqrt {a-b x}}\right )}{8 b^2}\\ &=-\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b^2}-\frac {a x^{3/2} \sqrt {a-b x}}{12 b}+\frac {1}{3} x^{5/2} \sqrt {a-b x}+\frac {a^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{8 b^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 87, normalized size = 0.85 \begin {gather*} \frac {\sqrt {a-b x} \left (\frac {3 a^{5/2} \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {1-\frac {b x}{a}}}+\sqrt {b} \sqrt {x} \left (-3 a^2-2 a b x+8 b^2 x^2\right )\right )}{24 b^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)*Sqrt[a - b*x],x]

[Out]

(Sqrt[a - b*x]*(Sqrt[b]*Sqrt[x]*(-3*a^2 - 2*a*b*x + 8*b^2*x^2) + (3*a^(5/2)*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[a]])
/Sqrt[1 - (b*x)/a]))/(24*b^(5/2))

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IntegrateAlgebraic [A]  time = 0.12, size = 91, normalized size = 0.89 \begin {gather*} \frac {a^3 \sqrt {-b} \log \left (\sqrt {a-b x}-\sqrt {-b} \sqrt {x}\right )}{8 b^3}+\frac {\sqrt {a-b x} \left (-3 a^2 \sqrt {x}-2 a b x^{3/2}+8 b^2 x^{5/2}\right )}{24 b^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[x^(3/2)*Sqrt[a - b*x],x]

[Out]

(Sqrt[a - b*x]*(-3*a^2*Sqrt[x] - 2*a*b*x^(3/2) + 8*b^2*x^(5/2)))/(24*b^2) + (a^3*Sqrt[-b]*Log[-(Sqrt[-b]*Sqrt[
x]) + Sqrt[a - b*x]])/(8*b^3)

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fricas [A]  time = 1.02, size = 142, normalized size = 1.39 \begin {gather*} \left [-\frac {3 \, a^{3} \sqrt {-b} \log \left (-2 \, b x + 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) - 2 \, {\left (8 \, b^{3} x^{2} - 2 \, a b^{2} x - 3 \, a^{2} b\right )} \sqrt {-b x + a} \sqrt {x}}{48 \, b^{3}}, -\frac {3 \, a^{3} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) - {\left (8 \, b^{3} x^{2} - 2 \, a b^{2} x - 3 \, a^{2} b\right )} \sqrt {-b x + a} \sqrt {x}}{24 \, b^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(-b*x+a)^(1/2),x, algorithm="fricas")

[Out]

[-1/48*(3*a^3*sqrt(-b)*log(-2*b*x + 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) - 2*(8*b^3*x^2 - 2*a*b^2*x - 3*a^2*
b)*sqrt(-b*x + a)*sqrt(x))/b^3, -1/24*(3*a^3*sqrt(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) - (8*b^3*x^2 - 2
*a*b^2*x - 3*a^2*b)*sqrt(-b*x + a)*sqrt(x))/b^3]

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giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(-b*x+a)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.01, size = 108, normalized size = 1.06 \begin {gather*} \frac {\sqrt {\left (-b x +a \right ) x}\, a^{3} \arctan \left (\frac {\left (x -\frac {a}{2 b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+a x}}\right )}{16 \sqrt {-b x +a}\, b^{\frac {5}{2}} \sqrt {x}}+\frac {\sqrt {-b x +a}\, a^{2} \sqrt {x}}{8 b^{2}}-\frac {\left (-b x +a \right )^{\frac {3}{2}} x^{\frac {3}{2}}}{3 b}-\frac {\left (-b x +a \right )^{\frac {3}{2}} a \sqrt {x}}{4 b^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(-b*x+a)^(1/2),x)

[Out]

-1/3/b*x^(3/2)*(-b*x+a)^(3/2)-1/4*a/b^2*x^(1/2)*(-b*x+a)^(3/2)+1/8*a^2*x^(1/2)*(-b*x+a)^(1/2)/b^2+1/16*a^3/b^(
5/2)*((-b*x+a)*x)^(1/2)/(-b*x+a)^(1/2)/x^(1/2)*arctan((x-1/2*a/b)/(-b*x^2+a*x)^(1/2)*b^(1/2))

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maxima [A]  time = 2.99, size = 135, normalized size = 1.32 \begin {gather*} -\frac {a^{3} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{8 \, b^{\frac {5}{2}}} + \frac {\frac {3 \, \sqrt {-b x + a} a^{3} b^{2}}{\sqrt {x}} - \frac {8 \, {\left (-b x + a\right )}^{\frac {3}{2}} a^{3} b}{x^{\frac {3}{2}}} - \frac {3 \, {\left (-b x + a\right )}^{\frac {5}{2}} a^{3}}{x^{\frac {5}{2}}}}{24 \, {\left (b^{5} - \frac {3 \, {\left (b x - a\right )} b^{4}}{x} + \frac {3 \, {\left (b x - a\right )}^{2} b^{3}}{x^{2}} - \frac {{\left (b x - a\right )}^{3} b^{2}}{x^{3}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(3/2)*(-b*x+a)^(1/2),x, algorithm="maxima")

[Out]

-1/8*a^3*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x)))/b^(5/2) + 1/24*(3*sqrt(-b*x + a)*a^3*b^2/sqrt(x) - 8*(-b*x +
a)^(3/2)*a^3*b/x^(3/2) - 3*(-b*x + a)^(5/2)*a^3/x^(5/2))/(b^5 - 3*(b*x - a)*b^4/x + 3*(b*x - a)^2*b^3/x^2 - (
b*x - a)^3*b^2/x^3)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int x^{3/2}\,\sqrt {a-b\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)*(a - b*x)^(1/2),x)

[Out]

int(x^(3/2)*(a - b*x)^(1/2), x)

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sympy [A]  time = 6.33, size = 260, normalized size = 2.55 \begin {gather*} \begin {cases} \frac {i a^{\frac {5}{2}} \sqrt {x}}{8 b^{2} \sqrt {-1 + \frac {b x}{a}}} - \frac {i a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 b \sqrt {-1 + \frac {b x}{a}}} - \frac {5 i \sqrt {a} x^{\frac {5}{2}}}{12 \sqrt {-1 + \frac {b x}{a}}} - \frac {i a^{3} \operatorname {acosh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {5}{2}}} + \frac {i b x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {-1 + \frac {b x}{a}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\- \frac {a^{\frac {5}{2}} \sqrt {x}}{8 b^{2} \sqrt {1 - \frac {b x}{a}}} + \frac {a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 b \sqrt {1 - \frac {b x}{a}}} + \frac {5 \sqrt {a} x^{\frac {5}{2}}}{12 \sqrt {1 - \frac {b x}{a}}} + \frac {a^{3} \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{8 b^{\frac {5}{2}}} - \frac {b x^{\frac {7}{2}}}{3 \sqrt {a} \sqrt {1 - \frac {b x}{a}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(3/2)*(-b*x+a)**(1/2),x)

[Out]

Piecewise((I*a**(5/2)*sqrt(x)/(8*b**2*sqrt(-1 + b*x/a)) - I*a**(3/2)*x**(3/2)/(24*b*sqrt(-1 + b*x/a)) - 5*I*sq
rt(a)*x**(5/2)/(12*sqrt(-1 + b*x/a)) - I*a**3*acosh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(5/2)) + I*b*x**(7/2)/(3*sq
rt(a)*sqrt(-1 + b*x/a)), Abs(b*x/a) > 1), (-a**(5/2)*sqrt(x)/(8*b**2*sqrt(1 - b*x/a)) + a**(3/2)*x**(3/2)/(24*
b*sqrt(1 - b*x/a)) + 5*sqrt(a)*x**(5/2)/(12*sqrt(1 - b*x/a)) + a**3*asin(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(5/2))
- b*x**(7/2)/(3*sqrt(a)*sqrt(1 - b*x/a)), True))

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