3.555 \(\int \frac{(a-b x)^{5/2}}{x^{3/2}} \, dx\)

Optimal. Leaf size=93 \[ -\frac{15}{4} a^2 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a-b x}}\right )-\frac{2 (a-b x)^{5/2}}{\sqrt{x}}-\frac{5}{2} b \sqrt{x} (a-b x)^{3/2}-\frac{15}{4} a b \sqrt{x} \sqrt{a-b x} \]

[Out]

(-15*a*b*Sqrt[x]*Sqrt[a - b*x])/4 - (5*b*Sqrt[x]*(a - b*x)^(3/2))/2 - (2*(a - b*x)^(5/2))/Sqrt[x] - (15*a^2*Sq
rt[b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/4

________________________________________________________________________________________

Rubi [A]  time = 0.0281738, antiderivative size = 93, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{15}{4} a^2 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a-b x}}\right )-\frac{2 (a-b x)^{5/2}}{\sqrt{x}}-\frac{5}{2} b \sqrt{x} (a-b x)^{3/2}-\frac{15}{4} a b \sqrt{x} \sqrt{a-b x} \]

Antiderivative was successfully verified.

[In]

Int[(a - b*x)^(5/2)/x^(3/2),x]

[Out]

(-15*a*b*Sqrt[x]*Sqrt[a - b*x])/4 - (5*b*Sqrt[x]*(a - b*x)^(3/2))/2 - (2*(a - b*x)^(5/2))/Sqrt[x] - (15*a^2*Sq
rt[b]*ArcTan[(Sqrt[b]*Sqrt[x])/Sqrt[a - b*x]])/4

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{(a-b x)^{5/2}}{x^{3/2}} \, dx &=-\frac{2 (a-b x)^{5/2}}{\sqrt{x}}-(5 b) \int \frac{(a-b x)^{3/2}}{\sqrt{x}} \, dx\\ &=-\frac{5}{2} b \sqrt{x} (a-b x)^{3/2}-\frac{2 (a-b x)^{5/2}}{\sqrt{x}}-\frac{1}{4} (15 a b) \int \frac{\sqrt{a-b x}}{\sqrt{x}} \, dx\\ &=-\frac{15}{4} a b \sqrt{x} \sqrt{a-b x}-\frac{5}{2} b \sqrt{x} (a-b x)^{3/2}-\frac{2 (a-b x)^{5/2}}{\sqrt{x}}-\frac{1}{8} \left (15 a^2 b\right ) \int \frac{1}{\sqrt{x} \sqrt{a-b x}} \, dx\\ &=-\frac{15}{4} a b \sqrt{x} \sqrt{a-b x}-\frac{5}{2} b \sqrt{x} (a-b x)^{3/2}-\frac{2 (a-b x)^{5/2}}{\sqrt{x}}-\frac{1}{4} \left (15 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a-b x^2}} \, dx,x,\sqrt{x}\right )\\ &=-\frac{15}{4} a b \sqrt{x} \sqrt{a-b x}-\frac{5}{2} b \sqrt{x} (a-b x)^{3/2}-\frac{2 (a-b x)^{5/2}}{\sqrt{x}}-\frac{1}{4} \left (15 a^2 b\right ) \operatorname{Subst}\left (\int \frac{1}{1+b x^2} \, dx,x,\frac{\sqrt{x}}{\sqrt{a-b x}}\right )\\ &=-\frac{15}{4} a b \sqrt{x} \sqrt{a-b x}-\frac{5}{2} b \sqrt{x} (a-b x)^{3/2}-\frac{2 (a-b x)^{5/2}}{\sqrt{x}}-\frac{15}{4} a^2 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a-b x}}\right )\\ \end{align*}

Mathematica [C]  time = 0.0120249, size = 49, normalized size = 0.53 \[ -\frac{2 a^2 \sqrt{a-b x} \, _2F_1\left (-\frac{5}{2},-\frac{1}{2};\frac{1}{2};\frac{b x}{a}\right )}{\sqrt{x} \sqrt{1-\frac{b x}{a}}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a - b*x)^(5/2)/x^(3/2),x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.013, size = 88, normalized size = 1. \begin{align*} -{\frac{-2\,{b}^{2}{x}^{2}+9\,abx+8\,{a}^{2}}{4}\sqrt{-bx+a}{\frac{1}{\sqrt{x}}}}-{\frac{15\,{a}^{2}}{8}\sqrt{b}\arctan \left ({\sqrt{b} \left ( x-{\frac{a}{2\,b}} \right ){\frac{1}{\sqrt{-b{x}^{2}+ax}}}} \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((-b*x+a)^(5/2)/x^(3/2),x)

[Out]

-1/4*(-b*x+a)^(1/2)*(-2*b^2*x^2+9*a*b*x+8*a^2)/x^(1/2)-15/8*a^2*b^(1/2)*arctan(b^(1/2)*(x-1/2/b*a)/(-b*x^2+a*x
)^(1/2))*(x*(-b*x+a))^(1/2)/x^(1/2)/(-b*x+a)^(1/2)

________________________________________________________________________________________

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((-b*x+a)^(5/2)/x^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.86165, size = 354, normalized size = 3.81 \begin{align*} \left [\frac{15 \, a^{2} \sqrt{-b} x \log \left (-2 \, b x + 2 \, \sqrt{-b x + a} \sqrt{-b} \sqrt{x} + a\right ) + 2 \,{\left (2 \, b^{2} x^{2} - 9 \, a b x - 8 \, a^{2}\right )} \sqrt{-b x + a} \sqrt{x}}{8 \, x}, \frac{15 \, a^{2} \sqrt{b} x \arctan \left (\frac{\sqrt{-b x + a}}{\sqrt{b} \sqrt{x}}\right ) +{\left (2 \, b^{2} x^{2} - 9 \, a b x - 8 \, a^{2}\right )} \sqrt{-b x + a} \sqrt{x}}{4 \, x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+a)^(5/2)/x^(3/2),x, algorithm="fricas")

[Out]

[1/8*(15*a^2*sqrt(-b)*x*log(-2*b*x + 2*sqrt(-b*x + a)*sqrt(-b)*sqrt(x) + a) + 2*(2*b^2*x^2 - 9*a*b*x - 8*a^2)*
sqrt(-b*x + a)*sqrt(x))/x, 1/4*(15*a^2*sqrt(b)*x*arctan(sqrt(-b*x + a)/(sqrt(b)*sqrt(x))) + (2*b^2*x^2 - 9*a*b
*x - 8*a^2)*sqrt(-b*x + a)*sqrt(x))/x]

________________________________________________________________________________________

Sympy [A]  time = 12.6059, size = 269, normalized size = 2.89 \begin{align*} \begin{cases} \frac{2 i a^{\frac{5}{2}}}{\sqrt{x} \sqrt{-1 + \frac{b x}{a}}} + \frac{i a^{\frac{3}{2}} b \sqrt{x}}{4 \sqrt{-1 + \frac{b x}{a}}} - \frac{11 i \sqrt{a} b^{2} x^{\frac{3}{2}}}{4 \sqrt{-1 + \frac{b x}{a}}} + \frac{15 i a^{2} \sqrt{b} \operatorname{acosh}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{4} + \frac{i b^{3} x^{\frac{5}{2}}}{2 \sqrt{a} \sqrt{-1 + \frac{b x}{a}}} & \text{for}\: \frac{\left |{b x}\right |}{\left |{a}\right |} > 1 \\- \frac{2 a^{\frac{5}{2}}}{\sqrt{x} \sqrt{1 - \frac{b x}{a}}} - \frac{a^{\frac{3}{2}} b \sqrt{x}}{4 \sqrt{1 - \frac{b x}{a}}} + \frac{11 \sqrt{a} b^{2} x^{\frac{3}{2}}}{4 \sqrt{1 - \frac{b x}{a}}} - \frac{15 a^{2} \sqrt{b} \operatorname{asin}{\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}} \right )}}{4} - \frac{b^{3} x^{\frac{5}{2}}}{2 \sqrt{a} \sqrt{1 - \frac{b x}{a}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+a)**(5/2)/x**(3/2),x)

[Out]

Piecewise((2*I*a**(5/2)/(sqrt(x)*sqrt(-1 + b*x/a)) + I*a**(3/2)*b*sqrt(x)/(4*sqrt(-1 + b*x/a)) - 11*I*sqrt(a)*
b**2*x**(3/2)/(4*sqrt(-1 + b*x/a)) + 15*I*a**2*sqrt(b)*acosh(sqrt(b)*sqrt(x)/sqrt(a))/4 + I*b**3*x**(5/2)/(2*s
qrt(a)*sqrt(-1 + b*x/a)), Abs(b*x)/Abs(a) > 1), (-2*a**(5/2)/(sqrt(x)*sqrt(1 - b*x/a)) - a**(3/2)*b*sqrt(x)/(4
*sqrt(1 - b*x/a)) + 11*sqrt(a)*b**2*x**(3/2)/(4*sqrt(1 - b*x/a)) - 15*a**2*sqrt(b)*asin(sqrt(b)*sqrt(x)/sqrt(a
))/4 - b**3*x**(5/2)/(2*sqrt(a)*sqrt(1 - b*x/a)), True))

________________________________________________________________________________________

Giac [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-b*x+a)^(5/2)/x^(3/2),x, algorithm="giac")

[Out]

Timed out