\(\int \frac {\sqrt {c+d x}}{1-x^2} \, dx\) [651]

Optimal result

Integrand size = 19, antiderivative size = 58 \[ \int \frac {\sqrt {c+d x}}{1-x^2} \, dx=-\sqrt {c-d} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c-d}}\right )+\sqrt {c+d} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c+d}}\right ) \]

[Out]

-arctanh((d*x+c)^(1/2)/(c-d)^(1/2))*(c-d)^(1/2)+arctanh((d*x+c)^(1/2)/(c+d)^(1/2))*(c+d)^(1/2)

Rubi [A] (verified)

Time = 0.05 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {714, 1144, 212} \[ \int \frac {\sqrt {c+d x}}{1-x^2} \, dx=\sqrt {c+d} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c+d}}\right )-\sqrt {c-d} \text {arctanh}\left (\frac {\sqrt {c+d x}}{\sqrt {c-d}}\right ) \]

[In]

Int[Sqrt[c + d*x]/(1 - x^2),x]

[Out]

-(Sqrt[c - d]*ArcTanh[Sqrt[c + d*x]/Sqrt[c - d]]) + Sqrt[c + d]*ArcTanh[Sqrt[c + d*x]/Sqrt[c + d]]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 714

Int[Sqrt[(d_) + (e_.)*(x_)]/((a_) + (c_.)*(x_)^2), x_Symbol] :> Dist[2*e, Subst[Int[x^2/(c*d^2 + a*e^2 - 2*c*d
*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0]

Rule 1144

Int[((d_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Dist[(
d^2/2)*(b/q + 1), Int[(d*x)^(m - 2)/(b/2 + q/2 + c*x^2), x], x] - Dist[(d^2/2)*(b/q - 1), Int[(d*x)^(m - 2)/(b
/2 - q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b^2 - 4*a*c, 0] && GeQ[m, 2]

Rubi steps \begin{align*} \text {integral}& = (2 d) \text {Subst}\left (\int \frac {x^2}{-c^2+d^2+2 c x^2-x^4} \, dx,x,\sqrt {c+d x}\right ) \\ & = -\left ((c-d) \text {Subst}\left (\int \frac {1}{c-d-x^2} \, dx,x,\sqrt {c+d x}\right )\right )+(c+d) \text {Subst}\left (\int \frac {1}{c+d-x^2} \, dx,x,\sqrt {c+d x}\right ) \\ & = -\sqrt {c-d} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c-d}}\right )+\sqrt {c+d} \tanh ^{-1}\left (\frac {\sqrt {c+d x}}{\sqrt {c+d}}\right ) \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.14 \[ \int \frac {\sqrt {c+d x}}{1-x^2} \, dx=\sqrt {-c-d} \arctan \left (\frac {\sqrt {c+d x}}{\sqrt {-c-d}}\right )-\sqrt {-c+d} \arctan \left (\frac {\sqrt {c+d x}}{\sqrt {-c+d}}\right ) \]

[In]

Integrate[Sqrt[c + d*x]/(1 - x^2),x]

[Out]

Sqrt[-c - d]*ArcTan[Sqrt[c + d*x]/Sqrt[-c - d]] - Sqrt[-c + d]*ArcTan[Sqrt[c + d*x]/Sqrt[-c + d]]

Maple [A] (verified)

Time = 2.91 (sec) , antiderivative size = 47, normalized size of antiderivative = 0.81

method	result	size
pseudoelliptic	\(-\sqrt {-c +d}\, \arctan \left (\frac {\sqrt {d x +c}}{\sqrt {-c +d}}\right )+\operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c +d}}\right ) \sqrt {c +d}\)	\(47\)
derivativedivides	\(-2 d \left (-\frac {\sqrt {c +d}\, \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c +d}}\right )}{2 d}+\frac {\sqrt {-c +d}\, \arctan \left (\frac {\sqrt {d x +c}}{\sqrt {-c +d}}\right )}{2 d}\right )\)	\(57\)
default	\(-2 d \left (-\frac {\sqrt {c +d}\, \operatorname {arctanh}\left (\frac {\sqrt {d x +c}}{\sqrt {c +d}}\right )}{2 d}+\frac {\sqrt {-c +d}\, \arctan \left (\frac {\sqrt {d x +c}}{\sqrt {-c +d}}\right )}{2 d}\right )\)	\(57\)

[In]

int((d*x+c)^(1/2)/(-x^2+1),x,method=_RETURNVERBOSE)

[Out]

-(-c+d)^(1/2)*arctan((d*x+c)^(1/2)/(-c+d)^(1/2))+arctanh((d*x+c)^(1/2)/(c+d)^(1/2))*(c+d)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 295, normalized size of antiderivative = 5.09 \[ \int \frac {\sqrt {c+d x}}{1-x^2} \, dx=\left [\frac {1}{2} \, \sqrt {c - d} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c - d} + 2 \, c - d}{x + 1}\right ) + \frac {1}{2} \, \sqrt {c + d} \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c + d} + 2 \, c + d}{x - 1}\right ), -\sqrt {-c + d} \arctan \left (-\frac {\sqrt {d x + c} \sqrt {-c + d}}{c - d}\right ) + \frac {1}{2} \, \sqrt {c + d} \log \left (\frac {d x + 2 \, \sqrt {d x + c} \sqrt {c + d} + 2 \, c + d}{x - 1}\right ), -\sqrt {-c - d} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c - d}}{c + d}\right ) + \frac {1}{2} \, \sqrt {c - d} \log \left (\frac {d x - 2 \, \sqrt {d x + c} \sqrt {c - d} + 2 \, c - d}{x + 1}\right ), -\sqrt {-c + d} \arctan \left (-\frac {\sqrt {d x + c} \sqrt {-c + d}}{c - d}\right ) - \sqrt {-c - d} \arctan \left (\frac {\sqrt {d x + c} \sqrt {-c - d}}{c + d}\right )\right ] \]

[In]

integrate((d*x+c)^(1/2)/(-x^2+1),x, algorithm="fricas")

[Out]

[1/2*sqrt(c - d)*log((d*x - 2*sqrt(d*x + c)*sqrt(c - d) + 2*c - d)/(x + 1)) + 1/2*sqrt(c + d)*log((d*x + 2*sqr
t(d*x + c)*sqrt(c + d) + 2*c + d)/(x - 1)), -sqrt(-c + d)*arctan(-sqrt(d*x + c)*sqrt(-c + d)/(c - d)) + 1/2*sq
rt(c + d)*log((d*x + 2*sqrt(d*x + c)*sqrt(c + d) + 2*c + d)/(x - 1)), -sqrt(-c - d)*arctan(sqrt(d*x + c)*sqrt(
-c - d)/(c + d)) + 1/2*sqrt(c - d)*log((d*x - 2*sqrt(d*x + c)*sqrt(c - d) + 2*c - d)/(x + 1)), -sqrt(-c + d)*a
rctan(-sqrt(d*x + c)*sqrt(-c + d)/(c - d)) - sqrt(-c - d)*arctan(sqrt(d*x + c)*sqrt(-c - d)/(c + d))]

Sympy [A] (verification not implemented)

Time = 1.97 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.38 \[ \int \frac {\sqrt {c+d x}}{1-x^2} \, dx=\begin {cases} \frac {2 \left (\frac {d \left (c - d\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c + d}} \right )}}{2 \sqrt {- c + d}} - \frac {d \left (c + d\right ) \operatorname {atan}{\left (\frac {\sqrt {c + d x}}{\sqrt {- c - d}} \right )}}{2 \sqrt {- c - d}}\right )}{d} & \text {for}\: d \neq 0 \\\sqrt {c} \left (- \frac {\log {\left (x - 1 \right )}}{2} + \frac {\log {\left (x + 1 \right )}}{2}\right ) & \text {otherwise} \end {cases} \]

[In]

integrate((d*x+c)**(1/2)/(-x**2+1),x)

[Out]

Piecewise((2*(d*(c - d)*atan(sqrt(c + d*x)/sqrt(-c + d))/(2*sqrt(-c + d)) - d*(c + d)*atan(sqrt(c + d*x)/sqrt(
-c - d))/(2*sqrt(-c - d)))/d, Ne(d, 0)), (sqrt(c)*(-log(x - 1)/2 + log(x + 1)/2), True))

Maxima [F(-2)]

Exception generated. \[ \int \frac {\sqrt {c+d x}}{1-x^2} \, dx=\text {Exception raised: ValueError} \]

[In]

integrate((d*x+c)^(1/2)/(-x^2+1),x, algorithm="maxima")

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*c-4*d>0)', see `assume?` for
more detail

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.93 \[ \int \frac {\sqrt {c+d x}}{1-x^2} \, dx=-\sqrt {-c + d} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c + d}}\right ) + \sqrt {-c - d} \arctan \left (\frac {\sqrt {d x + c}}{\sqrt {-c - d}}\right ) \]

[In]

integrate((d*x+c)^(1/2)/(-x^2+1),x, algorithm="giac")

[Out]

-sqrt(-c + d)*arctan(sqrt(d*x + c)/sqrt(-c + d)) + sqrt(-c - d)*arctan(sqrt(d*x + c)/sqrt(-c - d))

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.79 \[ \int \frac {\sqrt {c+d x}}{1-x^2} \, dx=\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c+d}}\right )\,\sqrt {c+d}-\mathrm {atanh}\left (\frac {\sqrt {c+d\,x}}{\sqrt {c-d}}\right )\,\sqrt {c-d} \]

[In]

int(-(c + d*x)^(1/2)/(x^2 - 1),x)

[Out]

atanh((c + d*x)^(1/2)/(c + d)^(1/2))*(c + d)^(1/2) - atanh((c + d*x)^(1/2)/(c - d)^(1/2))*(c - d)^(1/2)