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

Optimal. Leaf size=71 \[ -\frac {3 a \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{b^{5/2}}+\frac {3 \sqrt {x} \sqrt {a-b x}}{b^2}+\frac {2 x^{3/2}}{b \sqrt {a-b x}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {47, 50, 63, 217, 203} \[ \frac {3 \sqrt {x} \sqrt {a-b x}}{b^2}-\frac {3 a \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a-b x}}\right )}{b^{5/2}}+\frac {2 x^{3/2}}{b \sqrt {a-b x}} \]

Antiderivative was successfully verified.

[In]

Int[x^(3/2)/(a - b*x)^(3/2),x]

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 51, normalized size = 0.72 \[ \frac {2 x^{5/2} \sqrt {1-\frac {b x}{a}} \, _2F_1\left (\frac {3}{2},\frac {5}{2};\frac {7}{2};\frac {b x}{a}\right )}{5 a \sqrt {a-b x}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(3/2)/(a - b*x)^(3/2),x]

[Out]

(2*x^(5/2)*Sqrt[1 - (b*x)/a]*Hypergeometric2F1[3/2, 5/2, 7/2, (b*x)/a])/(5*a*Sqrt[a - b*x])

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fricas [A]  time = 0.45, size = 152, normalized size = 2.14 \[ \left [-\frac {3 \, {\left (a b x - a^{2}\right )} \sqrt {-b} \log \left (-2 \, b x - 2 \, \sqrt {-b x + a} \sqrt {-b} \sqrt {x} + a\right ) - 2 \, {\left (b^{2} x - 3 \, a b\right )} \sqrt {-b x + a} \sqrt {x}}{2 \, {\left (b^{4} x - a b^{3}\right )}}, \frac {3 \, {\left (a b x - a^{2}\right )} \sqrt {b} \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right ) + {\left (b^{2} x - 3 \, a b\right )} \sqrt {-b x + a} \sqrt {x}}{b^{4} x - a b^{3}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 111.13, size = 130, normalized size = 1.83 \[ -\frac {{\left (\frac {8 \, a^{2} \sqrt {-b}}{{\left (\sqrt {-b x + a} \sqrt {-b} - \sqrt {{\left (b x - a\right )} b + a b}\right )}^{2} - a b} + \frac {3 \, a \log \left ({\left (\sqrt {-b x + a} \sqrt {-b} - \sqrt {{\left (b x - a\right )} b + a b}\right )}^{2}\right )}{\sqrt {-b}} - \frac {2 \, \sqrt {{\left (b x - a\right )} b + a b} \sqrt {-b x + a}}{b}\right )} {\left | b \right |}}{2 \, b^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(8*a^2*sqrt(-b)/((sqrt(-b*x + a)*sqrt(-b) - sqrt((b*x - a)*b + a*b))^2 - a*b) + 3*a*log((sqrt(-b*x + a)*s
qrt(-b) - sqrt((b*x - a)*b + a*b))^2)/sqrt(-b) - 2*sqrt((b*x - a)*b + a*b)*sqrt(-b*x + a)/b)*abs(b)/b^3

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maple [B]  time = 0.03, size = 114, normalized size = 1.61 \[ \frac {\left (-\frac {3 a \arctan \left (\frac {\left (x -\frac {a}{2 b}\right ) \sqrt {b}}{\sqrt {-b \,x^{2}+a x}}\right )}{2 b^{\frac {5}{2}}}-\frac {2 \sqrt {-\left (x -\frac {a}{b}\right ) a -\left (x -\frac {a}{b}\right )^{2} b}\, a}{\left (x -\frac {a}{b}\right ) b^{3}}\right ) \sqrt {\left (-b x +a \right ) x}}{\sqrt {-b x +a}\, \sqrt {x}}+\frac {\sqrt {-b x +a}\, \sqrt {x}}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.95, size = 75, normalized size = 1.06 \[ \frac {2 \, a b - \frac {3 \, {\left (b x - a\right )} a}{x}}{\frac {\sqrt {-b x + a} b^{3}}{\sqrt {x}} + \frac {{\left (-b x + a\right )}^{\frac {3}{2}} b^{2}}{x^{\frac {3}{2}}}} + \frac {3 \, a \arctan \left (\frac {\sqrt {-b x + a}}{\sqrt {b} \sqrt {x}}\right )}{b^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(3/2)/(a - b*x)^(3/2),x)

[Out]

int(x^(3/2)/(a - b*x)^(3/2), x)

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sympy [A]  time = 3.70, size = 155, normalized size = 2.18 \[ \begin {cases} - \frac {3 i \sqrt {a} \sqrt {x}}{b^{2} \sqrt {-1 + \frac {b x}{a}}} + \frac {3 i a \operatorname {acosh}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {5}{2}}} + \frac {i x^{\frac {3}{2}}}{\sqrt {a} b \sqrt {-1 + \frac {b x}{a}}} & \text {for}\: \left |{\frac {b x}{a}}\right | > 1 \\\frac {3 \sqrt {a} \sqrt {x}}{b^{2} \sqrt {1 - \frac {b x}{a}}} - \frac {3 a \operatorname {asin}{\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}} \right )}}{b^{\frac {5}{2}}} - \frac {x^{\frac {3}{2}}}{\sqrt {a} b \sqrt {1 - \frac {b x}{a}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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