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

Optimal. Leaf size=99 \[ \frac {a^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{8 b^{3/2}}-\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a-b x}+\frac {1}{3} x^{3/2} (a-b x)^{3/2} \]

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Rubi [A]  time = 0.03, antiderivative size = 99, 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^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{8 b^{3/2}}-\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a-b x}+\frac {1}{3} x^{3/2} (a-b x)^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]*(a - b*x)^(3/2),x]

[Out]

-(a^2*Sqrt[x]*Sqrt[a - b*x])/(8*b) + (a*x^(3/2)*Sqrt[a - b*x])/4 + (x^(3/2)*(a - b*x)^(3/2))/3 + (a^3*ArcTan[(
Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/(8*b^(3/2))

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 \sqrt {x} (a-b x)^{3/2} \, dx &=\frac {1}{3} x^{3/2} (a-b x)^{3/2}+\frac {1}{2} a \int \sqrt {x} \sqrt {a-b x} \, dx\\ &=\frac {1}{4} a x^{3/2} \sqrt {a-b x}+\frac {1}{3} x^{3/2} (a-b x)^{3/2}+\frac {1}{8} a^2 \int \frac {\sqrt {x}}{\sqrt {a-b x}} \, dx\\ &=-\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a-b x}+\frac {1}{3} x^{3/2} (a-b x)^{3/2}+\frac {a^3 \int \frac {1}{\sqrt {x} \sqrt {a-b x}} \, dx}{16 b}\\ &=-\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a-b x}+\frac {1}{3} x^{3/2} (a-b x)^{3/2}+\frac {a^3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b x^2}} \, dx,x,\sqrt {x}\right )}{8 b}\\ &=-\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a-b x}+\frac {1}{3} x^{3/2} (a-b x)^{3/2}+\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}\\ &=-\frac {a^2 \sqrt {x} \sqrt {a-b x}}{8 b}+\frac {1}{4} a x^{3/2} \sqrt {a-b x}+\frac {1}{3} x^{3/2} (a-b x)^{3/2}+\frac {a^3 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{8 b^{3/2}}\\ \end {align*}

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Mathematica [A]  time = 0.12, size = 87, normalized size = 0.88 \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+14 a b x-8 b^2 x^2\right )\right )}{24 b^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]*(a - b*x)^(3/2),x]

[Out]

(Sqrt[a - b*x]*(Sqrt[b]*Sqrt[x]*(-3*a^2 + 14*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^(3/2))

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IntegrateAlgebraic [A]  time = 0.12, size = 91, normalized size = 0.92 \begin {gather*} \frac {a^3 \sqrt {-b} \log \left (\sqrt {a-b x}-\sqrt {-b} \sqrt {x}\right )}{8 b^2}+\frac {\sqrt {a-b x} \left (-3 a^2 \sqrt {x}+14 a b x^{3/2}-8 b^2 x^{5/2}\right )}{24 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - b*x]*(-3*a^2*Sqrt[x] + 14*a*b*x^(3/2) - 8*b^2*x^(5/2)))/(24*b) + (a^3*Sqrt[-b]*Log[-(Sqrt[-b]*Sqrt[x
]) + Sqrt[a - b*x]])/(8*b^2)

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fricas [A]  time = 1.16, size = 141, normalized size = 1.42 \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} - 14 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt {-b x + a} \sqrt {x}}{48 \, b^{2}}, -\frac {3 \, a^{3} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) + {\left (8 \, b^{3} x^{2} - 14 \, a b^{2} x + 3 \, a^{2} b\right )} \sqrt {-b x + a} \sqrt {x}}{24 \, b^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+a)^(3/2)*x^(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 - 14*a*b^2*x + 3*a^2
*b)*sqrt(-b*x + a)*sqrt(x))/b^2, -1/24*(3*a^3*sqrt(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) + (8*b^3*x^2 -
14*a*b^2*x + 3*a^2*b)*sqrt(-b*x + a)*sqrt(x))/b^2]

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

[Out]

Timed out

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maple [A]  time = 0.01, size = 102, normalized size = 1.03 \begin {gather*} \frac {\sqrt {-b x +a}\, a \,x^{\frac {3}{2}}}{4}+\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 {3}{2}} \sqrt {x}}-\frac {\sqrt {-b x +a}\, a^{2} \sqrt {x}}{8 b}+\frac {\left (-b x +a \right )^{\frac {3}{2}} x^{\frac {3}{2}}}{3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*x^(3/2)*(-b*x+a)^(3/2)+1/4*a*x^(3/2)*(-b*x+a)^(1/2)-1/8*a^2*x^(1/2)*(-b*x+a)^(1/2)/b+1/16*a^3/b^(3/2)*((-b
*x+a)*x)^(1/2)/x^(1/2)/(-b*x+a)^(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 = 133, normalized size = 1.34 \begin {gather*} -\frac {a^{3} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{8 \, b^{\frac {3}{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^{4} - \frac {3 \, {\left (b x - a\right )} b^{3}}{x} + \frac {3 \, {\left (b x - a\right )}^{2} b^{2}}{x^{2}} - \frac {{\left (b x - a\right )}^{3} b}{x^{3}}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(1/2)*(a - b*x)^(3/2),x)

[Out]

int(x^(1/2)*(a - b*x)^(3/2), x)

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sympy [A]  time = 5.54, size = 264, normalized size = 2.67 \begin {gather*} \begin {cases} \frac {i a^{\frac {5}{2}} \sqrt {x}}{8 b \sqrt {-1 + \frac {b x}{a}}} - \frac {17 i a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 \sqrt {-1 + \frac {b x}{a}}} + \frac {11 i \sqrt {a} b 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 {3}{2}}} - \frac {i b^{2} 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 \sqrt {1 - \frac {b x}{a}}} + \frac {17 a^{\frac {3}{2}} x^{\frac {3}{2}}}{24 \sqrt {1 - \frac {b x}{a}}} - \frac {11 \sqrt {a} b 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 {3}{2}}} + \frac {b^{2} 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((-b*x+a)**(3/2)*x**(1/2),x)

[Out]

Piecewise((I*a**(5/2)*sqrt(x)/(8*b*sqrt(-1 + b*x/a)) - 17*I*a**(3/2)*x**(3/2)/(24*sqrt(-1 + b*x/a)) + 11*I*sqr
t(a)*b*x**(5/2)/(12*sqrt(-1 + b*x/a)) - I*a**3*acosh(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(3/2)) - I*b**2*x**(7/2)/(
3*sqrt(a)*sqrt(-1 + b*x/a)), Abs(b*x/a) > 1), (-a**(5/2)*sqrt(x)/(8*b*sqrt(1 - b*x/a)) + 17*a**(3/2)*x**(3/2)/
(24*sqrt(1 - b*x/a)) - 11*sqrt(a)*b*x**(5/2)/(12*sqrt(1 - b*x/a)) + a**3*asin(sqrt(b)*sqrt(x)/sqrt(a))/(8*b**(
3/2)) + b**2*x**(7/2)/(3*sqrt(a)*sqrt(1 - b*x/a)), True))

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