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

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

[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)

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Rubi [A]  time = 0.0220824, antiderivative size = 71, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [C]  time = 0.0107771, 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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Maple [B]  time = 0.02, size = 114, normalized size = 1.6 \begin{align*}{\frac{1}{{b}^{2}}\sqrt{x}\sqrt{-bx+a}}+{ \left ( -{\frac{3\,a}{2}\arctan \left ({\sqrt{b} \left ( x-{\frac{a}{2\,b}} \right ){\frac{1}{\sqrt{-b{x}^{2}+ax}}}} \right ){b}^{-{\frac{5}{2}}}}-2\,{\frac{a}{{b}^{3}}\sqrt{-b \left ( x-{\frac{a}{b}} \right ) ^{2}-a \left ( x-{\frac{a}{b}} \right ) } \left ( x-{\frac{a}{b}} \right ) ^{-1}} \right ) \sqrt{x \left ( -bx+a \right ) }{\frac{1}{\sqrt{x}}}{\frac{1}{\sqrt{-bx+a}}}} \end{align*}

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.79657, size = 369, normalized size = 5.2 \begin{align*} \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 ] \end{align*}

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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Sympy [A]  time = 4.24528, size = 156, normalized size = 2.2 \begin{align*} \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}\: \frac{\left |{b x}\right |}{\left |{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} \end{align*}

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)/Abs(a) > 1), (3*sqrt(a)*sqrt(x)/(b**2*sqrt(1 - b*x/a)) - 3*a*asi
n(sqrt(b)*sqrt(x)/sqrt(a))/b**(5/2) - x**(3/2)/(sqrt(a)*b*sqrt(1 - b*x/a)), True))

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Giac [B]  time = 59.3328, size = 176, normalized size = 2.48 \begin{align*} -\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}} \end{align*}

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