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

Optimal. Leaf size=86 \[ \frac{15}{4} b^2 \sqrt{b x-a}-\frac{15}{4} \sqrt{a} b^2 \tan ^{-1}\left (\frac{\sqrt{b x-a}}{\sqrt{a}}\right )-\frac{(b x-a)^{5/2}}{2 x^2}-\frac{5 b (b x-a)^{3/2}}{4 x} \]

[Out]

(15*b^2*Sqrt[-a + b*x])/4 - (5*b*(-a + b*x)^(3/2))/(4*x) - (-a + b*x)^(5/2)/(2*x^2) - (15*Sqrt[a]*b^2*ArcTan[S
qrt[-a + b*x]/Sqrt[a]])/4

________________________________________________________________________________________

Rubi [A]  time = 0.022194, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.267, Rules used = {47, 50, 63, 205} \[ \frac{15}{4} b^2 \sqrt{b x-a}-\frac{15}{4} \sqrt{a} b^2 \tan ^{-1}\left (\frac{\sqrt{b x-a}}{\sqrt{a}}\right )-\frac{(b x-a)^{5/2}}{2 x^2}-\frac{5 b (b x-a)^{3/2}}{4 x} \]

Antiderivative was successfully verified.

[In]

Int[(-a + b*x)^(5/2)/x^3,x]

[Out]

(15*b^2*Sqrt[-a + b*x])/4 - (5*b*(-a + b*x)^(3/2))/(4*x) - (-a + b*x)^(5/2)/(2*x^2) - (15*Sqrt[a]*b^2*ArcTan[S
qrt[-a + b*x]/Sqrt[a]])/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 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [C]  time = 0.026345, size = 38, normalized size = 0.44 \[ \frac{2 b^2 (b x-a)^{7/2} \, _2F_1\left (3,\frac{7}{2};\frac{9}{2};1-\frac{b x}{a}\right )}{7 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(-a + b*x)^(5/2)/x^3,x]

[Out]

(2*b^2*(-a + b*x)^(7/2)*Hypergeometric2F1[3, 7/2, 9/2, 1 - (b*x)/a])/(7*a^3)

________________________________________________________________________________________

Maple [A]  time = 0.01, size = 70, normalized size = 0.8 \begin{align*} 2\,{b}^{2}\sqrt{bx-a}+{\frac{9\,a}{4\,{x}^{2}} \left ( bx-a \right ) ^{{\frac{3}{2}}}}+{\frac{7\,{a}^{2}}{4\,{x}^{2}}\sqrt{bx-a}}-{\frac{15\,{b}^{2}}{4}\arctan \left ({\sqrt{bx-a}{\frac{1}{\sqrt{a}}}} \right ) \sqrt{a}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x-a)^(5/2)/x^3,x)

[Out]

2*b^2*(b*x-a)^(1/2)+9/4*a/x^2*(b*x-a)^(3/2)+7/4/x^2*a^2*(b*x-a)^(1/2)-15/4*b^2*arctan((b*x-a)^(1/2)/a^(1/2))*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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 1.62388, size = 319, normalized size = 3.71 \begin{align*} \left [\frac{15 \, \sqrt{-a} b^{2} x^{2} \log \left (\frac{b x - 2 \, \sqrt{b x - a} \sqrt{-a} - 2 \, a}{x}\right ) + 2 \,{\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt{b x - a}}{8 \, x^{2}}, -\frac{15 \, \sqrt{a} b^{2} x^{2} \arctan \left (\frac{\sqrt{b x - a}}{\sqrt{a}}\right ) -{\left (8 \, b^{2} x^{2} + 9 \, a b x - 2 \, a^{2}\right )} \sqrt{b x - a}}{4 \, x^{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [A]  time = 5.67359, size = 270, normalized size = 3.14 \begin{align*} \begin{cases} - \frac{15 i \sqrt{a} b^{2} \operatorname{acosh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{4} - \frac{i a^{3}}{2 \sqrt{b} x^{\frac{5}{2}} \sqrt{\frac{a}{b x} - 1}} + \frac{11 i a^{2} \sqrt{b}}{4 x^{\frac{3}{2}} \sqrt{\frac{a}{b x} - 1}} - \frac{i a b^{\frac{3}{2}}}{4 \sqrt{x} \sqrt{\frac{a}{b x} - 1}} - \frac{2 i b^{\frac{5}{2}} \sqrt{x}}{\sqrt{\frac{a}{b x} - 1}} & \text{for}\: \frac{\left |{a}\right |}{\left |{b}\right | \left |{x}\right |} > 1 \\\frac{15 \sqrt{a} b^{2} \operatorname{asin}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{4} + \frac{a^{3}}{2 \sqrt{b} x^{\frac{5}{2}} \sqrt{- \frac{a}{b x} + 1}} - \frac{11 a^{2} \sqrt{b}}{4 x^{\frac{3}{2}} \sqrt{- \frac{a}{b x} + 1}} + \frac{a b^{\frac{3}{2}}}{4 \sqrt{x} \sqrt{- \frac{a}{b x} + 1}} + \frac{2 b^{\frac{5}{2}} \sqrt{x}}{\sqrt{- \frac{a}{b x} + 1}} & \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,x)

[Out]

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

________________________________________________________________________________________

Giac [A]  time = 1.19511, size = 112, normalized size = 1.3 \begin{align*} -\frac{15 \, \sqrt{a} b^{3} \arctan \left (\frac{\sqrt{b x - a}}{\sqrt{a}}\right ) - 8 \, \sqrt{b x - a} b^{3} - \frac{9 \,{\left (b x - a\right )}^{\frac{3}{2}} a b^{3} + 7 \, \sqrt{b x - a} a^{2} b^{3}}{b^{2} x^{2}}}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(15*sqrt(a)*b^3*arctan(sqrt(b*x - a)/sqrt(a)) - 8*sqrt(b*x - a)*b^3 - (9*(b*x - a)^(3/2)*a*b^3 + 7*sqrt(b
*x - a)*a^2*b^3)/(b^2*x^2))/b