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

Optimal. Leaf size=95 \[ -\frac{15 b^2}{4 a^3 \sqrt{b x-a}}-\frac{15 b^2 \tan ^{-1}\left (\frac{\sqrt{b x-a}}{\sqrt{a}}\right )}{4 a^{7/2}}+\frac{5 b}{4 a^2 x \sqrt{b x-a}}+\frac{1}{2 a x^2 \sqrt{b x-a}} \]

[Out]

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

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Rubi [A]  time = 0.0231135, antiderivative size = 93, normalized size of antiderivative = 0.98, 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, 205} \[ -\frac{15 b^2 \tan ^{-1}\left (\frac{\sqrt{b x-a}}{\sqrt{a}}\right )}{4 a^{7/2}}-\frac{5 \sqrt{b x-a}}{2 a^2 x^2}-\frac{15 b \sqrt{b x-a}}{4 a^3 x}-\frac{2}{a x^2 \sqrt{b x-a}} \]

Antiderivative was successfully verified.

[In]

Int[1/(x^3*(-a + b*x)^(3/2)),x]

[Out]

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

Mathematica [C]  time = 0.0108433, size = 36, normalized size = 0.38 \[ -\frac{2 b^2 \, _2F_1\left (-\frac{1}{2},3;\frac{1}{2};1-\frac{b x}{a}\right )}{a^3 \sqrt{b x-a}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*(-a + b*x)^(3/2)),x]

[Out]

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

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Maple [A]  time = 0.011, size = 75, normalized size = 0.8 \begin{align*} -2\,{\frac{{b}^{2}}{{a}^{3}\sqrt{bx-a}}}-{\frac{7}{4\,{a}^{3}{x}^{2}} \left ( bx-a \right ) ^{{\frac{3}{2}}}}-{\frac{9}{4\,{a}^{2}{x}^{2}}\sqrt{bx-a}}-{\frac{15\,{b}^{2}}{4}\arctan \left ({\sqrt{bx-a}{\frac{1}{\sqrt{a}}}} \right ){a}^{-{\frac{7}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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/(b*x-a)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.62984, size = 420, normalized size = 4.42 \begin{align*} \left [-\frac{15 \,{\left (b^{3} x^{3} - a b^{2} x^{2}\right )} \sqrt{-a} \log \left (\frac{b x + 2 \, \sqrt{b x - a} \sqrt{-a} - 2 \, a}{x}\right ) + 2 \,{\left (15 \, a b^{2} x^{2} - 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt{b x - a}}{8 \,{\left (a^{4} b x^{3} - a^{5} x^{2}\right )}}, -\frac{15 \,{\left (b^{3} x^{3} - a b^{2} x^{2}\right )} \sqrt{a} \arctan \left (\frac{\sqrt{b x - a}}{\sqrt{a}}\right ) +{\left (15 \, a b^{2} x^{2} - 5 \, a^{2} b x - 2 \, a^{3}\right )} \sqrt{b x - a}}{4 \,{\left (a^{4} b x^{3} - a^{5} x^{2}\right )}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 7.71199, size = 230, normalized size = 2.42 \begin{align*} \begin{cases} - \frac{i}{2 a \sqrt{b} x^{\frac{5}{2}} \sqrt{\frac{a}{b x} - 1}} - \frac{5 i \sqrt{b}}{4 a^{2} x^{\frac{3}{2}} \sqrt{\frac{a}{b x} - 1}} + \frac{15 i b^{\frac{3}{2}}}{4 a^{3} \sqrt{x} \sqrt{\frac{a}{b x} - 1}} - \frac{15 i b^{2} \operatorname{acosh}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{4 a^{\frac{7}{2}}} & \text{for}\: \frac{\left |{a}\right |}{\left |{b}\right | \left |{x}\right |} > 1 \\\frac{1}{2 a \sqrt{b} x^{\frac{5}{2}} \sqrt{- \frac{a}{b x} + 1}} + \frac{5 \sqrt{b}}{4 a^{2} x^{\frac{3}{2}} \sqrt{- \frac{a}{b x} + 1}} - \frac{15 b^{\frac{3}{2}}}{4 a^{3} \sqrt{x} \sqrt{- \frac{a}{b x} + 1}} + \frac{15 b^{2} \operatorname{asin}{\left (\frac{\sqrt{a}}{\sqrt{b} \sqrt{x}} \right )}}{4 a^{\frac{7}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((-I/(2*a*sqrt(b)*x**(5/2)*sqrt(a/(b*x) - 1)) - 5*I*sqrt(b)/(4*a**2*x**(3/2)*sqrt(a/(b*x) - 1)) + 15*
I*b**(3/2)/(4*a**3*sqrt(x)*sqrt(a/(b*x) - 1)) - 15*I*b**2*acosh(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(7/2)), Abs(a
)/(Abs(b)*Abs(x)) > 1), (1/(2*a*sqrt(b)*x**(5/2)*sqrt(-a/(b*x) + 1)) + 5*sqrt(b)/(4*a**2*x**(3/2)*sqrt(-a/(b*x
) + 1)) - 15*b**(3/2)/(4*a**3*sqrt(x)*sqrt(-a/(b*x) + 1)) + 15*b**2*asin(sqrt(a)/(sqrt(b)*sqrt(x)))/(4*a**(7/2
)), True))

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Giac [A]  time = 1.18303, size = 109, normalized size = 1.15 \begin{align*} -\frac{15 \, b^{2} \arctan \left (\frac{\sqrt{b x - a}}{\sqrt{a}}\right )}{4 \, a^{\frac{7}{2}}} - \frac{2 \, b^{2}}{\sqrt{b x - a} a^{3}} - \frac{7 \,{\left (b x - a\right )}^{\frac{3}{2}} b^{2} + 9 \, \sqrt{b x - a} a b^{2}}{4 \, a^{3} b^{2} x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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