3.873 \(\int \frac {e^{\tanh ^{-1}(a+b x)}}{x (1-a^2-2 a b x-b^2 x^2)} \, dx\)

Optimal. Leaf size=93 \[ \frac {\sqrt {a+b x+1}}{(1-a) \sqrt {-a-b x+1}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(1-a) \sqrt {1-a^2}} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.13, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {6164, 96, 93, 208} \[ \frac {\sqrt {a+b x+1}}{(1-a) \sqrt {-a-b x+1}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(1-a) \sqrt {1-a^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a + b*x]/(x*(1 - a^2 - 2*a*b*x - b^2*x^2)),x]

[Out]

Sqrt[1 + a + b*x]/((1 - a)*Sqrt[1 - a - b*x]) - (2*ArcTanh[(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[1
- a - b*x])])/((1 - a)*Sqrt[1 - a^2])

Rule 93

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]

Rule 96

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(a +
b*x)^(m + 1)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), x] + Dist[(a*d*f*(m + 1)
+ b*c*f*(n + 1) + b*d*e*(p + 1))/((m + 1)*(b*c - a*d)*(b*e - a*f)), Int[(a + b*x)^(m + 1)*(c + d*x)^n*(e + f*
x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[Simplify[m + n + p + 3], 0] && (LtQ[m, -1] || Sum
SimplerQ[m, 1])

Rule 208

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

Rule 6164

Int[E^(ArcTanh[(a_) + (b_.)*(x_)]*(n_.))*(u_.)*((c_) + (d_.)*(x_) + (e_.)*(x_)^2)^(p_.), x_Symbol] :> Dist[(c/
(1 - a^2))^p, Int[u*(1 - a - b*x)^(p - n/2)*(1 + a + b*x)^(p + n/2), x], x] /; FreeQ[{a, b, c, d, e, n, p}, x]
&& EqQ[b*d - 2*a*e, 0] && EqQ[b^2*c + e*(1 - a^2), 0] && (IntegerQ[p] || GtQ[c/(1 - a^2), 0])

Rubi steps

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

________________________________________________________________________________________

Mathematica [A]  time = 0.25, size = 118, normalized size = 1.27 \[ -\frac {-\frac {\sqrt {-a^2-2 a b x-b^2 x^2+1}}{a+b x-1}-\frac {\log \left (\sqrt {1-a^2} \sqrt {-a^2-2 a b x-b^2 x^2+1}-a^2-a b x+1\right )}{\sqrt {1-a^2}}+\frac {\log (x)}{\sqrt {1-a^2}}}{a-1} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[a + b*x]/(x*(1 - a^2 - 2*a*b*x - b^2*x^2)),x]

[Out]

-((-(Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]/(-1 + a + b*x)) + Log[x]/Sqrt[1 - a^2] - Log[1 - a^2 - a*b*x + Sqrt[1 -
a^2]*Sqrt[1 - a^2 - 2*a*b*x - b^2*x^2]]/Sqrt[1 - a^2])/(-1 + a))

________________________________________________________________________________________

fricas [A]  time = 0.62, size = 316, normalized size = 3.40 \[ \left [-\frac {\sqrt {-a^{2} + 1} {\left (b x + a - 1\right )} \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )}}{2 \, {\left (a^{4} - 2 \, a^{3} + {\left (a^{3} - a^{2} - a + 1\right )} b x + 2 \, a - 1\right )}}, -\frac {\sqrt {a^{2} - 1} {\left (b x + a - 1\right )} \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) - \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )}}{a^{4} - 2 \, a^{3} + {\left (a^{3} - a^{2} - a + 1\right )} b x + 2 \, a - 1}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.37, size = 112, normalized size = 1.20 \[ -\frac {2 \, b \arctan \left (\frac {\frac {{\left (\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{\sqrt {a^{2} - 1} {\left (a {\left | b \right |} - {\left | b \right |}\right )}} - \frac {2 \, b}{{\left (a {\left | b \right |} - {\left | b \right |}\right )} {\left (\frac {\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left | b \right |} + b}{b^{2} x + a b} - 1\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [B]  time = 0.04, size = 391, normalized size = 4.20 \[ \frac {2 b \left (-2 b^{2} x -2 a b \right )}{\left (-4 b^{2} \left (-a^{2}+1\right )-4 a^{2} b^{2}\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {1}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {a b x}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {a^{2}}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}+\frac {a}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {a^{2} b x}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}+\frac {a^{3}}{\left (-a^{2}+1\right ) \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}-\frac {\ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right ) a}{\left (-a^{2}+1\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {b x + a + 1}{{\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} \sqrt {-{\left (b x + a\right )}^{2} + 1} x}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int -\frac {a+b\,x+1}{x\,\sqrt {1-{\left (a+b\,x\right )}^2}\,\left (a^2+2\,a\,b\,x+b^2\,x^2-1\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - \int \frac {1}{a x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} + b x^{2} \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1} - x \sqrt {- a^{2} - 2 a b x - b^{2} x^{2} + 1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-Integral(1/(a*x*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1) + b*x**2*sqrt(-a**2 - 2*a*b*x - b**2*x**2 + 1) - x*sqrt
(-a**2 - 2*a*b*x - b**2*x**2 + 1)), x)

________________________________________________________________________________________