3.4.60 \(\int \frac {1}{x^2 \sqrt {-a+b x}} \, dx\) [360]

Optimal. Leaf size=44 \[ \frac {\sqrt {-a+b x}}{a x}+\frac {b \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[Out]

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

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Rubi [A]
time = 0.01, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {44, 65, 211} \begin {gather*} \frac {b \text {ArcTan}\left (\frac {\sqrt {b x-a}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {\sqrt {b x-a}}{a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/(x^2*Sqrt[-a + b*x]),x]

[Out]

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

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 44, normalized size = 1.00 \begin {gather*} \frac {\sqrt {-a+b x}}{a x}+\frac {b \tan ^{-1}\left (\frac {\sqrt {-a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/(x^2*Sqrt[-a + b*x]),x]

[Out]

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

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Maple [A]
time = 0.13, size = 44, normalized size = 1.00

method	result	size
derivativedivides	\(2 b \left (\frac {\sqrt {b x -a}}{2 a b x}+\frac {\arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}\right )\)	\(44\)
default	\(2 b \left (\frac {\sqrt {b x -a}}{2 a b x}+\frac {\arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}\right )\)	\(44\)
risch	\(-\frac {-b x +a}{a x \sqrt {b x -a}}+\frac {b \arctan \left (\frac {\sqrt {b x -a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}\)	\(44\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/x^2/(b*x-a)^(1/2),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 46, normalized size = 1.05 \begin {gather*} \frac {\sqrt {b x - a} b}{{\left (b x - a\right )} a + a^{2}} + \frac {b \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.50, size = 97, normalized size = 2.20 \begin {gather*} \left [-\frac {\sqrt {-a} b x \log \left (\frac {b x - 2 \, \sqrt {b x - a} \sqrt {-a} - 2 \, a}{x}\right ) - 2 \, \sqrt {b x - a} a}{2 \, a^{2} x}, \frac {\sqrt {a} b x \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right ) + \sqrt {b x - a} a}{a^{2} x}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 1.13, size = 121, normalized size = 2.75 \begin {gather*} \begin {cases} \frac {i \sqrt {b} \sqrt {\frac {a}{b x} - 1}}{a \sqrt {x}} + \frac {i b \operatorname {acosh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{a^{\frac {3}{2}}} & \text {for}\: \left |{\frac {a}{b x}}\right | > 1 \\- \frac {1}{\sqrt {b} x^{\frac {3}{2}} \sqrt {- \frac {a}{b x} + 1}} + \frac {\sqrt {b}}{a \sqrt {x} \sqrt {- \frac {a}{b x} + 1}} - \frac {b \operatorname {asin}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{a^{\frac {3}{2}}} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.84, size = 43, normalized size = 0.98 \begin {gather*} \frac {\frac {b^{2} \arctan \left (\frac {\sqrt {b x - a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}} + \frac {\sqrt {b x - a} b}{a x}}{b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.04, size = 36, normalized size = 0.82 \begin {gather*} \frac {\sqrt {b\,x-a}}{a\,x}+\frac {b\,\mathrm {atan}\left (\frac {\sqrt {b\,x-a}}{\sqrt {a}}\right )}{a^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(b*x - a)^(1/2)),x)

[Out]

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

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