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

Optimal. Leaf size=75 \[ \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{3/2}}+\frac {\sqrt {x}}{4 a b (a-b x)}-\frac {\sqrt {x}}{2 b (a-b x)^2} \]

________________________________________________________________________________________

Rubi [A]  time = 0.02, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.267, Rules used = {47, 51, 63, 208} \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{3/2}}+\frac {\sqrt {x}}{4 a b (a-b x)}-\frac {\sqrt {x}}{2 b (a-b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[x]/(-a + b*x)^3,x]

[Out]

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

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

________________________________________________________________________________________

Mathematica [C]  time = 0.01, size = 26, normalized size = 0.35 \begin {gather*} -\frac {2 x^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};\frac {b x}{a}\right )}{3 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[x]/(-a + b*x)^3,x]

[Out]

(-2*x^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, (b*x)/a])/(3*a^3)

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 0.10, size = 60, normalized size = 0.80 \begin {gather*} \frac {\tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {x}}{\sqrt {a}}\right )}{4 a^{3/2} b^{3/2}}-\frac {\sqrt {x} (a+b x)}{4 a b (a-b x)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 1.05, size = 183, normalized size = 2.44 \begin {gather*} \left [\frac {{\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 (a b^{2} x + a^{2} b\right )} \sqrt {x}}{8 \, {\left (a^{2} b^{4} x^{2} - 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}, -\frac {{\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 (a b^{2} x + a^{2} b\right )} \sqrt {x}}{4 \, {\left (a^{2} b^{4} x^{2} - 2 \, a^{3} b^{3} x + a^{4} b^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.93, size = 55, normalized size = 0.73 \begin {gather*} -\frac {\arctan \left (\frac {b \sqrt {x}}{\sqrt {-a b}}\right )}{4 \, \sqrt {-a b} a b} - \frac {b x^{\frac {3}{2}} + a \sqrt {x}}{4 \, {\left (b x - a\right )}^{2} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.01, size = 54, normalized size = 0.72 \begin {gather*} \frac {\arctanh \left (\frac {b \sqrt {x}}{\sqrt {a b}}\right )}{4 \sqrt {a b}\, a b}+\frac {-\frac {x^{\frac {3}{2}}}{4 a}-\frac {\sqrt {x}}{4 b}}{\left (b x -a \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*(-1/8/a*x^(3/2)-1/8/b*x^(1/2))/(b*x-a)^2+1/4/b/a/(a*b)^(1/2)*arctanh(1/(a*b)^(1/2)*b*x^(1/2))

________________________________________________________________________________________

maxima [A]  time = 2.85, size = 80, normalized size = 1.07 \begin {gather*} -\frac {b x^{\frac {3}{2}} + a \sqrt {x}}{4 \, {\left (a b^{3} x^{2} - 2 \, a^{2} b^{2} x + a^{3} b\right )}} - \frac {\log \left (\frac {b \sqrt {x} - \sqrt {a b}}{b \sqrt {x} + \sqrt {a b}}\right )}{8 \, \sqrt {a b} a b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.14, size = 57, normalized size = 0.76 \begin {gather*} \frac {\mathrm {atanh}\left (\frac {\sqrt {b}\,\sqrt {x}}{\sqrt {a}}\right )}{4\,a^{3/2}\,b^{3/2}}-\frac {\frac {x^{3/2}}{4\,a}+\frac {\sqrt {x}}{4\,b}}{a^2-2\,a\,b\,x+b^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(-x^(1/2)/(a - b*x)^3,x)

[Out]

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

________________________________________________________________________________________

sympy [A]  time = 15.19, size = 668, normalized size = 8.91 \begin {gather*} \begin {cases} \frac {\tilde {\infty }}{x^{\frac {3}{2}}} & \text {for}\: a = 0 \wedge b = 0 \\- \frac {2 x^{\frac {3}{2}}}{3 a^{3}} & \text {for}\: b = 0 \\- \frac {2}{3 b^{3} x^{\frac {3}{2}}} & \text {for}\: a = 0 \\- \frac {2 a^{\frac {3}{2}} b \sqrt {x} \sqrt {\frac {1}{b}}}{8 a^{\frac {7}{2}} b^{2} \sqrt {\frac {1}{b}} - 16 a^{\frac {5}{2}} b^{3} x \sqrt {\frac {1}{b}} + 8 a^{\frac {3}{2}} b^{4} x^{2} \sqrt {\frac {1}{b}}} - \frac {2 \sqrt {a} b^{2} x^{\frac {3}{2}} \sqrt {\frac {1}{b}}}{8 a^{\frac {7}{2}} b^{2} \sqrt {\frac {1}{b}} - 16 a^{\frac {5}{2}} b^{3} x \sqrt {\frac {1}{b}} + 8 a^{\frac {3}{2}} b^{4} x^{2} \sqrt {\frac {1}{b}}} - \frac {a^{2} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {7}{2}} b^{2} \sqrt {\frac {1}{b}} - 16 a^{\frac {5}{2}} b^{3} x \sqrt {\frac {1}{b}} + 8 a^{\frac {3}{2}} b^{4} x^{2} \sqrt {\frac {1}{b}}} + \frac {a^{2} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {7}{2}} b^{2} \sqrt {\frac {1}{b}} - 16 a^{\frac {5}{2}} b^{3} x \sqrt {\frac {1}{b}} + 8 a^{\frac {3}{2}} b^{4} x^{2} \sqrt {\frac {1}{b}}} + \frac {2 a b x \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {7}{2}} b^{2} \sqrt {\frac {1}{b}} - 16 a^{\frac {5}{2}} b^{3} x \sqrt {\frac {1}{b}} + 8 a^{\frac {3}{2}} b^{4} x^{2} \sqrt {\frac {1}{b}}} - \frac {2 a b x \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {7}{2}} b^{2} \sqrt {\frac {1}{b}} - 16 a^{\frac {5}{2}} b^{3} x \sqrt {\frac {1}{b}} + 8 a^{\frac {3}{2}} b^{4} x^{2} \sqrt {\frac {1}{b}}} - \frac {b^{2} x^{2} \log {\left (- \sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {7}{2}} b^{2} \sqrt {\frac {1}{b}} - 16 a^{\frac {5}{2}} b^{3} x \sqrt {\frac {1}{b}} + 8 a^{\frac {3}{2}} b^{4} x^{2} \sqrt {\frac {1}{b}}} + \frac {b^{2} x^{2} \log {\left (\sqrt {a} \sqrt {\frac {1}{b}} + \sqrt {x} \right )}}{8 a^{\frac {7}{2}} b^{2} \sqrt {\frac {1}{b}} - 16 a^{\frac {5}{2}} b^{3} x \sqrt {\frac {1}{b}} + 8 a^{\frac {3}{2}} b^{4} x^{2} \sqrt {\frac {1}{b}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo/x**(3/2), Eq(a, 0) & Eq(b, 0)), (-2*x**(3/2)/(3*a**3), Eq(b, 0)), (-2/(3*b**3*x**(3/2)), Eq(a,
0)), (-2*a**(3/2)*b*sqrt(x)*sqrt(1/b)/(8*a**(7/2)*b**2*sqrt(1/b) - 16*a**(5/2)*b**3*x*sqrt(1/b) + 8*a**(3/2)*b
**4*x**2*sqrt(1/b)) - 2*sqrt(a)*b**2*x**(3/2)*sqrt(1/b)/(8*a**(7/2)*b**2*sqrt(1/b) - 16*a**(5/2)*b**3*x*sqrt(1
/b) + 8*a**(3/2)*b**4*x**2*sqrt(1/b)) - a**2*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(7/2)*b**2*sqrt(1/b) - 16
*a**(5/2)*b**3*x*sqrt(1/b) + 8*a**(3/2)*b**4*x**2*sqrt(1/b)) + a**2*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(7/
2)*b**2*sqrt(1/b) - 16*a**(5/2)*b**3*x*sqrt(1/b) + 8*a**(3/2)*b**4*x**2*sqrt(1/b)) + 2*a*b*x*log(-sqrt(a)*sqrt
(1/b) + sqrt(x))/(8*a**(7/2)*b**2*sqrt(1/b) - 16*a**(5/2)*b**3*x*sqrt(1/b) + 8*a**(3/2)*b**4*x**2*sqrt(1/b)) -
2*a*b*x*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(7/2)*b**2*sqrt(1/b) - 16*a**(5/2)*b**3*x*sqrt(1/b) + 8*a**(3/
2)*b**4*x**2*sqrt(1/b)) - b**2*x**2*log(-sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(7/2)*b**2*sqrt(1/b) - 16*a**(5/2)
*b**3*x*sqrt(1/b) + 8*a**(3/2)*b**4*x**2*sqrt(1/b)) + b**2*x**2*log(sqrt(a)*sqrt(1/b) + sqrt(x))/(8*a**(7/2)*b
**2*sqrt(1/b) - 16*a**(5/2)*b**3*x*sqrt(1/b) + 8*a**(3/2)*b**4*x**2*sqrt(1/b)), True))

________________________________________________________________________________________