3.487 \(\int \frac{1}{x^{3/2} (-a+b x)^3} \, dx\)

Optimal. Leaf size=84 \[ -\frac{5}{4 a^2 \sqrt{x} (a-b x)}-\frac{15 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{7/2}}+\frac{15}{4 a^3 \sqrt{x}}-\frac{1}{2 a \sqrt{x} (a-b x)^2} \]

[Out]

15/(4*a^3*Sqrt[x]) - 1/(2*a*Sqrt[x]*(a - b*x)^2) - 5/(4*a^2*Sqrt[x]*(a - b*x)) - (15*Sqrt[b]*ArcTanh[(Sqrt[b]*
Sqrt[x])/Sqrt[a]])/(4*a^(7/2))

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Rubi [A]  time = 0.0239046, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {51, 63, 208} \[ -\frac{5}{4 a^2 \sqrt{x} (a-b x)}-\frac{15 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{7/2}}+\frac{15}{4 a^3 \sqrt{x}}-\frac{1}{2 a \sqrt{x} (a-b x)^2} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^(3/2)*(-a + b*x)^3),x]

[Out]

15/(4*a^3*Sqrt[x]) - 1/(2*a*Sqrt[x]*(a - b*x)^2) - 5/(4*a^2*Sqrt[x]*(a - b*x)) - (15*Sqrt[b]*ArcTanh[(Sqrt[b]*
Sqrt[x])/Sqrt[a]])/(4*a^(7/2))

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 208

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

Rubi steps

\begin{align*} \int \frac{1}{x^{3/2} (-a+b x)^3} \, dx &=-\frac{1}{2 a \sqrt{x} (a-b x)^2}-\frac{5 \int \frac{1}{x^{3/2} (-a+b x)^2} \, dx}{4 a}\\ &=-\frac{1}{2 a \sqrt{x} (a-b x)^2}-\frac{5}{4 a^2 \sqrt{x} (a-b x)}+\frac{15 \int \frac{1}{x^{3/2} (-a+b x)} \, dx}{8 a^2}\\ &=\frac{15}{4 a^3 \sqrt{x}}-\frac{1}{2 a \sqrt{x} (a-b x)^2}-\frac{5}{4 a^2 \sqrt{x} (a-b x)}+\frac{(15 b) \int \frac{1}{\sqrt{x} (-a+b x)} \, dx}{8 a^3}\\ &=\frac{15}{4 a^3 \sqrt{x}}-\frac{1}{2 a \sqrt{x} (a-b x)^2}-\frac{5}{4 a^2 \sqrt{x} (a-b x)}+\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{-a+b x^2} \, dx,x,\sqrt{x}\right )}{4 a^3}\\ &=\frac{15}{4 a^3 \sqrt{x}}-\frac{1}{2 a \sqrt{x} (a-b x)^2}-\frac{5}{4 a^2 \sqrt{x} (a-b x)}-\frac{15 \sqrt{b} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{x}}{\sqrt{a}}\right )}{4 a^{7/2}}\\ \end{align*}

Mathematica [C]  time = 0.0053577, size = 24, normalized size = 0.29 \[ \frac{2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};\frac{b x}{a}\right )}{a^3 \sqrt{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^(3/2)*(-a + b*x)^3),x]

[Out]

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

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Maple [A]  time = 0.012, size = 58, normalized size = 0.7 \begin{align*} 2\,{\frac{1}{{a}^{3}\sqrt{x}}}+2\,{\frac{b}{{a}^{3}} \left ({\frac{1}{ \left ( bx-a \right ) ^{2}} \left ({\frac{7\,b{x}^{3/2}}{8}}-{\frac{9\,a\sqrt{x}}{8}} \right ) }-{\frac{15}{8\,\sqrt{ab}}{\it Artanh} \left ({\frac{b\sqrt{x}}{\sqrt{ab}}} \right ) } \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/a^3/x^(1/2)+2/a^3*b*((7/8*b*x^(3/2)-9/8*a*x^(1/2))/(b*x-a)^2-15/8/(a*b)^(1/2)*arctanh(b*x^(1/2)/(a*b)^(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(1/x^(3/2)/(b*x-a)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.66448, size = 466, normalized size = 5.55 \begin{align*} \left [\frac{15 \,{\left (b^{2} x^{3} - 2 \, a b x^{2} + a^{2} x\right )} \sqrt{\frac{b}{a}} \log \left (\frac{b x - 2 \, a \sqrt{x} \sqrt{\frac{b}{a}} + a}{b x - a}\right ) + 2 \,{\left (15 \, b^{2} x^{2} - 25 \, a b x + 8 \, a^{2}\right )} \sqrt{x}}{8 \,{\left (a^{3} b^{2} x^{3} - 2 \, a^{4} b x^{2} + a^{5} x\right )}}, \frac{15 \,{\left (b^{2} x^{3} - 2 \, a b x^{2} + a^{2} x\right )} \sqrt{-\frac{b}{a}} \arctan \left (\frac{a \sqrt{-\frac{b}{a}}}{b \sqrt{x}}\right ) +{\left (15 \, b^{2} x^{2} - 25 \, a b x + 8 \, a^{2}\right )} \sqrt{x}}{4 \,{\left (a^{3} b^{2} x^{3} - 2 \, a^{4} b x^{2} + a^{5} x\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x-a)^3,x, algorithm="fricas")

[Out]

[1/8*(15*(b^2*x^3 - 2*a*b*x^2 + a^2*x)*sqrt(b/a)*log((b*x - 2*a*sqrt(x)*sqrt(b/a) + a)/(b*x - a)) + 2*(15*b^2*
x^2 - 25*a*b*x + 8*a^2)*sqrt(x))/(a^3*b^2*x^3 - 2*a^4*b*x^2 + a^5*x), 1/4*(15*(b^2*x^3 - 2*a*b*x^2 + a^2*x)*sq
rt(-b/a)*arctan(a*sqrt(-b/a)/(b*sqrt(x))) + (15*b^2*x^2 - 25*a*b*x + 8*a^2)*sqrt(x))/(a^3*b^2*x^3 - 2*a^4*b*x^
2 + a^5*x)]

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Sympy [A]  time = 132.681, size = 802, normalized size = 9.55 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x**(3/2)/(b*x-a)**3,x)

[Out]

Piecewise((zoo/x**(7/2), Eq(a, 0) & Eq(b, 0)), (-2/(7*b**3*x**(7/2)), Eq(a, 0)), (2/(a**3*sqrt(x)), Eq(b, 0)),
(16*a**(5/2)*sqrt(1/b)/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x*
*(5/2)*sqrt(1/b)) - 50*a**(3/2)*b*x*sqrt(1/b)/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b
) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 30*sqrt(a)*b**2*x**2*sqrt(1/b)/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a
**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 15*a**2*sqrt(x)*log(-sqrt(a)*sqrt(1/b) +
sqrt(x))/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b
)) - 15*a**2*sqrt(x)*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*
sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 30*a*b*x**(3/2)*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(11/
2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) + 30*a*b*x**(3/2
)*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7
/2)*b**2*x**(5/2)*sqrt(1/b)) + 15*b**2*x**(5/2)*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(11/2)*sqrt(x)*sqrt(1/
b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)*sqrt(1/b)) - 15*b**2*x**(5/2)*log(sqrt(a)*sqr
t(1/b) + sqrt(x))/(8*a**(11/2)*sqrt(x)*sqrt(1/b) - 16*a**(9/2)*b*x**(3/2)*sqrt(1/b) + 8*a**(7/2)*b**2*x**(5/2)
*sqrt(1/b)), True))

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Giac [A]  time = 1.1446, size = 85, normalized size = 1.01 \begin{align*} \frac{15 \, b \arctan \left (\frac{b \sqrt{x}}{\sqrt{-a b}}\right )}{4 \, \sqrt{-a b} a^{3}} + \frac{2}{a^{3} \sqrt{x}} + \frac{7 \, b^{2} x^{\frac{3}{2}} - 9 \, a b \sqrt{x}}{4 \,{\left (b x - a\right )}^{2} a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/x^(3/2)/(b*x-a)^3,x, algorithm="giac")

[Out]

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