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

Optimal. Leaf size=74 \[ \frac {3 a^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{4 \sqrt {b}}+\frac {3}{4} a \sqrt {x} \sqrt {a-b x}+\frac {1}{2} \sqrt {x} (a-b x)^{3/2} \]

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Rubi [A]  time = 0.02, antiderivative size = 74, 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 = {50, 63, 217, 203} \begin {gather*} \frac {3 a^2 \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{4 \sqrt {b}}+\frac {3}{4} a \sqrt {x} \sqrt {a-b x}+\frac {1}{2} \sqrt {x} (a-b x)^{3/2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a - b*x)^(3/2)/Sqrt[x],x]

[Out]

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

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

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Mathematica [A]  time = 0.12, size = 71, normalized size = 0.96 \begin {gather*} \frac {1}{4} \sqrt {a-b x} \left (\frac {3 a^{3/2} \sin ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {1-\frac {b x}{a}}}+\sqrt {x} (5 a-2 b x)\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x)^(3/2)/Sqrt[x],x]

[Out]

(Sqrt[a - b*x]*(Sqrt[x]*(5*a - 2*b*x) + (3*a^(3/2)*ArcSin[(Sqrt[b]*Sqrt[x])/Sqrt[a]])/(Sqrt[b]*Sqrt[1 - (b*x)/
a])))/4

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IntegrateAlgebraic [A]  time = 0.10, size = 75, normalized size = 1.01 \begin {gather*} \frac {3 a^2 \sqrt {-b} \log \left (\sqrt {a-b x}-\sqrt {-b} \sqrt {x}\right )}{4 b}+\frac {1}{4} \sqrt {a-b x} \left (5 a \sqrt {x}-2 b x^{3/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a - b*x]*(5*a*Sqrt[x] - 2*b*x^(3/2)))/4 + (3*a^2*Sqrt[-b]*Log[-(Sqrt[-b]*Sqrt[x]) + Sqrt[a - b*x]])/(4*b
)

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fricas [A]  time = 1.43, size = 119, normalized size = 1.61 \begin {gather*} \left [-\frac {3 \, a^{2} \sqrt {-b} \log \left (-2 \, b x + 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) + 2 \, {\left (2 \, b^{2} x - 5 \, a b\right )} \sqrt {-b x + a} \sqrt {x}}{8 \, b}, -\frac {3 \, a^{2} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) + {\left (2 \, b^{2} x - 5 \, a b\right )} \sqrt {-b x + a} \sqrt {x}}{4 \, b}\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]

[-1/8*(3*a^2*sqrt(-b)*log(-2*b*x + 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) + 2*(2*b^2*x - 5*a*b)*sqrt(-b*x + a)
*sqrt(x))/b, -1/4*(3*a^2*sqrt(b)*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) + (2*b^2*x - 5*a*b)*sqrt(-b*x + a)*s
qrt(x))/b]

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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 = 0.00, size = 83, normalized size = 1.12 \begin {gather*} \frac {3 \sqrt {\left (-b x +a \right ) x}\, a^{2} \arctan \left (\frac {\left (x -\frac {a}{2 b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+a x}}\right )}{8 \sqrt {-b x +a}\, \sqrt {b}\, \sqrt {x}}+\frac {3 \sqrt {-b x +a}\, a \sqrt {x}}{4}+\frac {\left (-b x +a \right )^{\frac {3}{2}} \sqrt {x}}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-b*x+a)^(3/2)*x^(1/2)+3/4*a*x^(1/2)*(-b*x+a)^(1/2)+3/8*a^2*((-b*x+a)*x)^(1/2)/(-b*x+a)^(1/2)/x^(1/2)/b^(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 = 2.90, size = 93, normalized size = 1.26 \begin {gather*} -\frac {3 \, a^{2} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{4 \, \sqrt {b}} + \frac {\frac {3 \, \sqrt {-b x + a} a^{2} b}{\sqrt {x}} + \frac {5 \, {\left (-b x + a\right )}^{\frac {3}{2}} a^{2}}{x^{\frac {3}{2}}}}{4 \, {\left (b^{2} - \frac {2 \, {\left (b x - a\right )} b}{x} + \frac {{\left (b x - a\right )}^{2}}{x^{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="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a - b*x)^(3/2)/x^(1/2),x)

[Out]

int((a - b*x)^(3/2)/x^(1/2), x)

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sympy [A]  time = 3.21, size = 190, normalized size = 2.57 \begin {gather*} \begin {cases} - \frac {5 i a^{\frac {3}{2}} \sqrt {x}}{4 \sqrt {-1 + \frac {b x}{a}}} + \frac {7 i \sqrt {a} b x^{\frac {3}{2}}}{4 \sqrt {-1 + \frac {b x}{a}}} - \frac {3 i a^{2} \operatorname {acosh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 \sqrt {b}} - \frac {i b^{2} x^{\frac {5}{2}}}{2 \sqrt {a} \sqrt {-1 + \frac {b x}{a}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\\frac {5 a^{\frac {3}{2}} \sqrt {x} \sqrt {1 - \frac {b x}{a}}}{4} - \frac {\sqrt {a} b x^{\frac {3}{2}} \sqrt {1 - \frac {b x}{a}}}{2} + \frac {3 a^{2} \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{4 \sqrt {b}} & \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((-5*I*a**(3/2)*sqrt(x)/(4*sqrt(-1 + b*x/a)) + 7*I*sqrt(a)*b*x**(3/2)/(4*sqrt(-1 + b*x/a)) - 3*I*a**2
*acosh(sqrt(b)*sqrt(x)/sqrt(a))/(4*sqrt(b)) - I*b**2*x**(5/2)/(2*sqrt(a)*sqrt(-1 + b*x/a)), Abs(b*x/a) > 1), (
5*a**(3/2)*sqrt(x)*sqrt(1 - b*x/a)/4 - sqrt(a)*b*x**(3/2)*sqrt(1 - b*x/a)/2 + 3*a**2*asin(sqrt(b)*sqrt(x)/sqrt
(a))/(4*sqrt(b)), True))

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