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

Optimal. Leaf size=71 \[ \frac {b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 a^{3/2}}-\frac {\sqrt {b x-a}}{2 x^2}+\frac {b \sqrt {b x-a}}{4 a x} \]

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Rubi [A]  time = 0.02, antiderivative size = 71, 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, 205} \begin {gather*} \frac {b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 a^{3/2}}-\frac {\sqrt {b x-a}}{2 x^2}+\frac {b \sqrt {b x-a}}{4 a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sqrt[-a + b*x]/x^3,x]

[Out]

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

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Mathematica [C]  time = 0.01, size = 38, normalized size = 0.54 \begin {gather*} \frac {2 b^2 (b x-a)^{3/2} \, _2F_1\left (\frac {3}{2},3;\frac {5}{2};1-\frac {b x}{a}\right )}{3 a^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[-a + b*x]/x^3,x]

[Out]

(2*b^2*(-a + b*x)^(3/2)*Hypergeometric2F1[3/2, 3, 5/2, 1 - (b*x)/a])/(3*a^3)

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IntegrateAlgebraic [A]  time = 0.08, size = 60, normalized size = 0.85 \begin {gather*} \frac {b^2 \tan ^{-1}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{4 a^{3/2}}-\frac {(2 a-b x) \sqrt {b x-a}}{4 a x^2} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 1.09, size = 124, normalized size = 1.75 \begin {gather*} \left [-\frac {\sqrt {-a} b^{2} x^{2} \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) - 2 \, {\left (a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{8 \, a^{2} x^{2}}, \frac {\sqrt {a} b^{2} x^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + {\left (a b x - 2 \, a^{2}\right )} \sqrt {b x - a}}{4 \, a^{2} x^{2}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 1.08, size = 66, normalized size = 0.93 \begin {gather*} \frac {\frac {b^{3} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {{\left (b x - a\right )}^{\frac {3}{2}} b^{3} - \sqrt {b x - a} a b^{3}}{a b^{2} x^{2}}}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.01, size = 55, normalized size = 0.77 \begin {gather*} \frac {b^{2} \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{4 a^{\frac {3}{2}}}-\frac {\sqrt {b x -a}}{4 x^{2}}+\frac {\left (b x -a \right )^{\frac {3}{2}}}{4 a \,x^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 2.96, size = 83, normalized size = 1.17 \begin {gather*} \frac {b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{4 \, a^{\frac {3}{2}}} + \frac {{\left (b x - a\right )}^{\frac {3}{2}} b^{2} - \sqrt {b x - a} a b^{2}}{4 \, {\left ({\left (b x - a\right )}^{2} a + 2 \, {\left (b x - a\right )} a^{2} + a^{3}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.10, size = 54, normalized size = 0.76 \begin {gather*} \frac {b^2\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{4\,a^{3/2}}-\frac {\sqrt {b\,x-a}}{4\,x^2}+\frac {{\left (b\,x-a\right )}^{3/2}}{4\,a\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x - a)^(1/2)/x^3,x)

[Out]

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

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sympy [A]  time = 4.16, size = 207, normalized size = 2.92 \begin {gather*} \begin {cases} - \frac {i a}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {\frac {a}{b x} - 1}} + \frac {3 i \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {\frac {a}{b x} - 1}} - \frac {i b^{\frac {3}{2}}}{4 a \sqrt {x} \sqrt {\frac {a}{b x} - 1}} + \frac {i b^{2} \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {3}{2}}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\\frac {a}{2 \sqrt {b} x^{\frac {5}{2}} \sqrt {- \frac {a}{b x} + 1}} - \frac {3 \sqrt {b}}{4 x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {b^{\frac {3}{2}}}{4 a \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} - \frac {b^{2} \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{4 a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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