3.2.86 \(\int \frac {1}{x^2 \sqrt {a+b x}} \, dx\) [186]

Optimal. Leaf size=41 \[ -\frac {\sqrt {a+b x}}{a x}+\frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}} \]

[Out]

b*arctanh((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 = 41, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, Rules used = {44, 65, 214} \begin {gather*} \frac {b \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b x}}{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*ArcTanh[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 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[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 \tanh ^{-1}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{a^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 41, normalized size = 1.00 \begin {gather*} -\frac {\sqrt {a+b x}}{a x}+\frac {b \tanh ^{-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*ArcTanh[Sqrt[a + b*x]/Sqrt[a]])/a^(3/2)

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Maple [A]
time = 0.03, size = 40, normalized size = 0.98

method	result	size
risch	\(\frac {b \arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{a^{\frac {3}{2}}}-\frac {\sqrt {b x +a}}{a x}\)	\(34\)
derivativedivides	\(2 b \left (-\frac {\sqrt {b x +a}}{2 a b x}+\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}\right )\)	\(40\)
default	\(2 b \left (-\frac {\sqrt {b x +a}}{2 a b x}+\frac {\arctanh \left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{2 a^{\frac {3}{2}}}\right )\)	\(40\)

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

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Maxima [A]
time = 3.32, size = 60, normalized size = 1.46 \begin {gather*} -\frac {\sqrt {b x + a} b}{{\left (b x + a\right )} a - a^{2}} - \frac {b \log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{2 \, 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) - 1/2*b*log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/a^(3/2)

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Fricas [A]
time = 0.62, size = 93, normalized size = 2.27 \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}}{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*ar
ctan(sqrt(b*x + a)*sqrt(-a)/a) + sqrt(b*x + a)*a)/(a^2*x)]

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Sympy [A]
time = 1.13, size = 44, normalized size = 1.07 \begin {gather*} - \frac {\sqrt {b} \sqrt {\frac {a}{b x} + 1}}{a \sqrt {x}} + \frac {b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \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)

[Out]

-sqrt(b)*sqrt(a/(b*x) + 1)/(a*sqrt(x)) + b*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/a**(3/2)

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Giac [A]
time = 0.48, size = 47, normalized size = 1.15 \begin {gather*} -\frac {\frac {b^{2} \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a} + \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))/(sqrt(-a)*a) + sqrt(b*x + a)*b/(a*x))/b

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Mupad [B]
time = 0.16, size = 33, normalized size = 0.80 \begin {gather*} \frac {b\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{a^{3/2}}-\frac {\sqrt {a+b\,x}}{a\,x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(x^2*(a + b*x)^(1/2)),x)

[Out]

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

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