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

Optimal. Leaf size=74 \[ \frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 a^{5/2}}+\frac {3 b \sqrt {b x-a}}{4 a^2 x}+\frac {\sqrt {b x-a}}{2 a x^2} \]

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Rubi [A]  time = 0.02, antiderivative size = 74, 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, 205} \begin {gather*} \frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 a^{5/2}}+\frac {3 b \sqrt {b x-a}}{4 a^2 x}+\frac {\sqrt {b x-a}}{2 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^3*Sqrt[-a + b*x]),x]

[Out]

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

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

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Mathematica [C]  time = 0.01, size = 36, normalized size = 0.49 \begin {gather*} \frac {2 b^2 \sqrt {b x-a} \, _2F_1\left (\frac {1}{2},3;\frac {3}{2};1-\frac {b x}{a}\right )}{a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^3*Sqrt[-a + b*x]),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.07, size = 69, normalized size = 0.93 \begin {gather*} \frac {3 b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 a^{5/2}}+\frac {3 (b x-a)^{3/2}+5 a \sqrt {b x-a}}{4 a^2 x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.36, size = 128, normalized size = 1.73 \begin {gather*} \left [-\frac {3 \, \sqrt {-a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) - 2 \, {\left (3 \, a b x + 2 \, a^{2}\right )} \sqrt {b x - a}}{8 \, a^{3} x^{2}}, \frac {3 \, \sqrt {a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + {\left (3 \, a b x + 2 \, a^{2}\right )} \sqrt {b x - a}}{4 \, a^{3} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.97, size = 68, normalized size = 0.92 \begin {gather*} \frac {\frac {3 \, b^{3} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {5}{2}}} + \frac {3 \, {\left (b x - a\right )}^{\frac {3}{2}} b^{3} + 5 \, \sqrt {b x - a} a b^{3}}{a^{2} b^{2} x^{2}}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 59, normalized size = 0.80 \begin {gather*} \frac {3 b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{4 a^{\frac {5}{2}}}+\frac {3 \sqrt {b x -a}\, b}{4 a^{2} x}+\frac {\sqrt {b x -a}}{2 a \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.96, size = 86, normalized size = 1.16 \begin {gather*} \frac {3 \, b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{4 \, a^{\frac {5}{2}}} + \frac {3 \, {\left (b x - a\right )}^{\frac {3}{2}} b^{2} + 5 \, \sqrt {b x - a} a b^{2}}{4 \, {\left ({\left (b x - a\right )}^{2} a^{2} + 2 \, {\left (b x - a\right )} a^{3} + a^{4}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.05, size = 57, normalized size = 0.77 \begin {gather*} \frac {3\,b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4\,a^{5/2}}+\frac {5\,\sqrt {b\,x-a}}{4\,a\,x^2}+\frac {3\,{\left (b\,x-a\right )}^{3/2}}{4\,a^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^3*(b*x - a)^(1/2)),x)

[Out]

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

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sympy [A]  time = 4.20, size = 216, normalized size = 2.92 \begin {gather*} \begin {cases} \frac {i}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {i \sqrt {b}}{4 a x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {3 i b^{\frac {3}{2}}}{4 a^{2} \sqrt {x} \sqrt {\frac {a}{b x} - 1}} + \frac {3 i b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {5}{2}}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\- \frac {1}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {\sqrt {b}}{4 a x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {3 b^{\frac {3}{2}}}{4 a^{2} \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} - \frac {3 b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {5}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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