3.5.86 \(\int \frac {1}{\sqrt {x} (-a+b x)^3} \, dx\)

Optimal. Leaf size=72 \[ -\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{5/2} \sqrt {b}}-\frac {3 \sqrt {x}}{4 a^2 (a-b x)}-\frac {\sqrt {x}}{2 a (a-b x)^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {51, 63, 208} \begin {gather*} -\frac {3 \sqrt {x}}{4 a^2 (a-b x)}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{5/2} \sqrt {b}}-\frac {\sqrt {x}}{2 a (a-b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(Sqrt[x]*(-a + b*x)^3),x]

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 24, normalized size = 0.33 \begin {gather*} -\frac {2 \sqrt {x} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};\frac {b x}{a}\right )}{a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(Sqrt[x]*(-a + b*x)^3),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.09, size = 64, normalized size = 0.89 \begin {gather*} \frac {3 b x^{3/2}-5 a \sqrt {x}}{4 a^2 (a-b x)^2}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{5/2} \sqrt {b}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.00, size = 185, normalized size = 2.57 \begin {gather*} \left [\frac {3 \, {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \sqrt {a b} \log \left (\frac {b x + a - 2 \, \sqrt {a b} \sqrt {x}}{b x - a}\right ) + 2 \, {\left (3 \, a b^{2} x - 5 \, a^{2} b\right )} \sqrt {x}}{8 \, {\left (a^{3} b^{3} x^{2} - 2 \, a^{4} b^{2} x + a^{5} b\right )}}, \frac {3 \, {\left (b^{2} x^{2} - 2 \, a b x + a^{2}\right )} \sqrt {-a b} \arctan \left (\frac {\sqrt {-a b}}{b \sqrt {x}}\right ) + {\left (3 \, a b^{2} x - 5 \, a^{2} b\right )} \sqrt {x}}{4 \, {\left (a^{3} b^{3} x^{2} - 2 \, a^{4} b^{2} x + a^{5} b\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.15, size = 51, normalized size = 0.71 \begin {gather*} \frac {3 \, \arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{4 \, \sqrt {-a b} a^{2}} + \frac {3 \, b x^{\frac {3}{2}} - 5 \, a \sqrt {x}}{4 \, {\left (b x - a\right )}^{2} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

3/4*arctan(b*sqrt(x)/sqrt(-a*b))/(sqrt(-a*b)*a^2) + 1/4*(3*b*x^(3/2) - 5*a*sqrt(x))/((b*x - a)^2*a^2)

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maple [A]  time = 0.01, size = 63, normalized size = 0.88 \begin {gather*} -\frac {\sqrt {x}}{2 \left (b x -a \right )^{2} a}-\frac {3 \left (\frac {\arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{2 \sqrt {a b}\, a}-\frac {\sqrt {x}}{2 \left (b x -a \right ) a}\right )}{2 a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.88, size = 77, normalized size = 1.07 \begin {gather*} \frac {3 \, b x^{\frac {3}{2}} - 5 \, a \sqrt {x}}{4 \, {\left (a^{2} b^{2} x^{2} - 2 \, a^{3} b x + a^{4}\right )}} + \frac {3 \, \log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(3*b*x^(3/2) - 5*a*sqrt(x))/(a^2*b^2*x^2 - 2*a^3*b*x + a^4) + 3/8*log((b*sqrt(x) - sqrt(a*b))/(b*sqrt(x) +
sqrt(a*b)))/(sqrt(a*b)*a^2)

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mupad [B]  time = 0.13, size = 58, normalized size = 0.81 \begin {gather*} -\frac {\frac {5\,\sqrt {x}}{4\,a}-\frac {3\,b\,x^{3/2}}{4\,a^2}}{a^2-2\,a\,b\,x+b^2\,x^2}-\frac {3\,\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,a^{5/2}\,\sqrt {b}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-1/(x^(1/2)*(a - b*x)^3),x)

[Out]

- ((5*x^(1/2))/(4*a) - (3*b*x^(3/2))/(4*a^2))/(a^2 + b^2*x^2 - 2*a*b*x) - (3*atanh((b^(1/2)*x^(1/2))/a^(1/2)))
/(4*a^(5/2)*b^(1/2))

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sympy [A]  time = 25.67, size = 660, normalized size = 9.17 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {5}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2 \sqrt {x}}{a^{3}} & \text {for}\: b = 0 \\- \frac {2}{5 b^{3} x^{\frac {5}{2}}} & \text {for}\: a = 0 \\- \frac {10 a^{\frac {3}{2}} b \sqrt {x} \sqrt {\frac {1}{b}}}{8 a^{\frac {9}{2}} b \sqrt {\frac {1}{b}} - 16 a^{\frac {7}{2}} b^{2} x \sqrt {\frac {1}{b}} + 8 a^{\frac {5}{2}} b^{3} x^{2} \sqrt {\frac {1}{b}}} + \frac {6 \sqrt {a} b^{2} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{8 a^{\frac {9}{2}} b \sqrt {\frac {1}{b}} - 16 a^{\frac {7}{2}} b^{2} x \sqrt {\frac {1}{b}} + 8 a^{\frac {5}{2}} b^{3} x^{2} \sqrt {\frac {1}{b}}} + \frac {3 a^{2} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {9}{2}} b \sqrt {\frac {1}{b}} - 16 a^{\frac {7}{2}} b^{2} x \sqrt {\frac {1}{b}} + 8 a^{\frac {5}{2}} b^{3} x^{2} \sqrt {\frac {1}{b}}} - \frac {3 a^{2} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {9}{2}} b \sqrt {\frac {1}{b}} - 16 a^{\frac {7}{2}} b^{2} x \sqrt {\frac {1}{b}} + 8 a^{\frac {5}{2}} b^{3} x^{2} \sqrt {\frac {1}{b}}} - \frac {6 a b x \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {9}{2}} b \sqrt {\frac {1}{b}} - 16 a^{\frac {7}{2}} b^{2} x \sqrt {\frac {1}{b}} + 8 a^{\frac {5}{2}} b^{3} x^{2} \sqrt {\frac {1}{b}}} + \frac {6 a b x \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {9}{2}} b \sqrt {\frac {1}{b}} - 16 a^{\frac {7}{2}} b^{2} x \sqrt {\frac {1}{b}} + 8 a^{\frac {5}{2}} b^{3} x^{2} \sqrt {\frac {1}{b}}} + \frac {3 b^{2} x^{2} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {9}{2}} b \sqrt {\frac {1}{b}} - 16 a^{\frac {7}{2}} b^{2} x \sqrt {\frac {1}{b}} + 8 a^{\frac {5}{2}} b^{3} x^{2} \sqrt {\frac {1}{b}}} - \frac {3 b^{2} x^{2} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {9}{2}} b \sqrt {\frac {1}{b}} - 16 a^{\frac {7}{2}} b^{2} x \sqrt {\frac {1}{b}} + 8 a^{\frac {5}{2}} b^{3} x^{2} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**(5/2), Eq(a, 0) & Eq(b, 0)), (-2*sqrt(x)/a**3, Eq(b, 0)), (-2/(5*b**3*x**(5/2)), Eq(a, 0)),
(-10*a**(3/2)*b*sqrt(x)*sqrt(1/b)/(8*a**(9/2)*b*sqrt(1/b) - 16*a**(7/2)*b**2*x*sqrt(1/b) + 8*a**(5/2)*b**3*x**
2*sqrt(1/b)) + 6*sqrt(a)*b**2*x**(3/2)*sqrt(1/b)/(8*a**(9/2)*b*sqrt(1/b) - 16*a**(7/2)*b**2*x*sqrt(1/b) + 8*a*
*(5/2)*b**3*x**2*sqrt(1/b)) + 3*a**2*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(9/2)*b*sqrt(1/b) - 16*a**(7/2)*b
**2*x*sqrt(1/b) + 8*a**(5/2)*b**3*x**2*sqrt(1/b)) - 3*a**2*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(9/2)*b*sqrt
(1/b) - 16*a**(7/2)*b**2*x*sqrt(1/b) + 8*a**(5/2)*b**3*x**2*sqrt(1/b)) - 6*a*b*x*log(-sqrt(a)*sqrt(1/b) + sqrt
(x))/(8*a**(9/2)*b*sqrt(1/b) - 16*a**(7/2)*b**2*x*sqrt(1/b) + 8*a**(5/2)*b**3*x**2*sqrt(1/b)) + 6*a*b*x*log(sq
rt(a)*sqrt(1/b) + sqrt(x))/(8*a**(9/2)*b*sqrt(1/b) - 16*a**(7/2)*b**2*x*sqrt(1/b) + 8*a**(5/2)*b**3*x**2*sqrt(
1/b)) + 3*b**2*x**2*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(9/2)*b*sqrt(1/b) - 16*a**(7/2)*b**2*x*sqrt(1/b) +
8*a**(5/2)*b**3*x**2*sqrt(1/b)) - 3*b**2*x**2*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(9/2)*b*sqrt(1/b) - 16*a
**(7/2)*b**2*x*sqrt(1/b) + 8*a**(5/2)*b**3*x**2*sqrt(1/b)), True))

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