\(\int \frac {x^{\frac {1-n}{2}+\frac {1}{2} (-3+n)}}{\sqrt {a+b x}} \, dx\) [370]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [A] (verification not implemented)
   Maxima [A] (verification not implemented)
   Giac [A] (verification not implemented)
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 29, antiderivative size = 23 \[ \int \frac {x^{\frac {1-n}{2}+\frac {1}{2} (-3+n)}}{\sqrt {a+b x}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {7, 65, 214} \[ \int \frac {x^{\frac {1-n}{2}+\frac {1}{2} (-3+n)}}{\sqrt {a+b x}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

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

[Out]

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

Rule 7

Int[(u_.)*(Px_)^(p_), x_Symbol] :> Int[u*Px^Simplify[p], x] /; PolyQ[Px, x] &&  !RationalQ[p] && FreeQ[p, x] &
& RationalQ[Simplify[p]]

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

Mathematica [A] (verified)

Time = 0.00 (sec) , antiderivative size = 23, normalized size of antiderivative = 1.00 \[ \int \frac {x^{\frac {1-n}{2}+\frac {1}{2} (-3+n)}}{\sqrt {a+b x}} \, dx=-\frac {2 \text {arctanh}\left (\frac {\sqrt {a+b x}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.06 (sec) , antiderivative size = 18, normalized size of antiderivative = 0.78

method result size
derivativedivides \(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\) \(18\)
default \(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\) \(18\)
pseudoelliptic \(-\frac {2 \,\operatorname {arctanh}\left (\frac {\sqrt {b x +a}}{\sqrt {a}}\right )}{\sqrt {a}}\) \(18\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.23 (sec) , antiderivative size = 56, normalized size of antiderivative = 2.43 \[ \int \frac {x^{\frac {1-n}{2}+\frac {1}{2} (-3+n)}}{\sqrt {a+b x}} \, dx=\left [\frac {\log \left (\frac {b x - 2 \, \sqrt {b x + a} \sqrt {a} + 2 \, a}{x}\right )}{\sqrt {a}}, \frac {2 \, \sqrt {-a} \arctan \left (\frac {\sqrt {b x + a} \sqrt {-a}}{a}\right )}{a}\right ] \]

[In]

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

[Out]

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

Sympy [A] (verification not implemented)

Time = 0.80 (sec) , antiderivative size = 24, normalized size of antiderivative = 1.04 \[ \int \frac {x^{\frac {1-n}{2}+\frac {1}{2} (-3+n)}}{\sqrt {a+b x}} \, dx=- \frac {2 \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} \sqrt {x}} \right )}}{\sqrt {a}} \]

[In]

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

[Out]

-2*asinh(sqrt(a)/(sqrt(b)*sqrt(x)))/sqrt(a)

Maxima [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 32, normalized size of antiderivative = 1.39 \[ \int \frac {x^{\frac {1-n}{2}+\frac {1}{2} (-3+n)}}{\sqrt {a+b x}} \, dx=\frac {\log \left (\frac {\sqrt {b x + a} - \sqrt {a}}{\sqrt {b x + a} + \sqrt {a}}\right )}{\sqrt {a}} \]

[In]

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

[Out]

log((sqrt(b*x + a) - sqrt(a))/(sqrt(b*x + a) + sqrt(a)))/sqrt(a)

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.91 \[ \int \frac {x^{\frac {1-n}{2}+\frac {1}{2} (-3+n)}}{\sqrt {a+b x}} \, dx=\frac {2 \, \arctan \left (\frac {\sqrt {b x + a}}{\sqrt {-a}}\right )}{\sqrt {-a}} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.00 (sec) , antiderivative size = 17, normalized size of antiderivative = 0.74 \[ \int \frac {x^{\frac {1-n}{2}+\frac {1}{2} (-3+n)}}{\sqrt {a+b x}} \, dx=-\frac {2\,\mathrm {atanh}\left (\frac {\sqrt {a+b\,x}}{\sqrt {a}}\right )}{\sqrt {a}} \]

[In]

int(1/(x*(a + b*x)^(1/2)),x)

[Out]

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