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

Optimal. Leaf size=44 \[ \frac{(1-a) \sqrt{a+b x+1}}{b^2 \sqrt{-a-b x+1}}-\frac{\sin ^{-1}(a+b x)}{b^2} \]

[Out]

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

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Rubi [A]  time = 0.0822346, antiderivative size = 44, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used = {6164, 78, 53, 619, 216} \[ \frac{(1-a) \sqrt{a+b x+1}}{b^2 \sqrt{-a-b x+1}}-\frac{\sin ^{-1}(a+b x)}{b^2} \]

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]

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

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])

Rule 78

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

Rule 53

Int[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_.) + (d_.)*(x_)]), x_Symbol] :> Int[1/Sqrt[a*c - b*(a - c)*x - b^2*x^2]
, x] /; FreeQ[{a, b, c, d}, x] && EqQ[b + d, 0] && GtQ[a + c, 0]

Rule 619

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Dist[1/(2*c*((-4*c)/(b^2 - 4*a*c))^p), Subst[Int[Si
mp[1 - x^2/(b^2 - 4*a*c), x]^p, x], x, b + 2*c*x], x] /; FreeQ[{a, b, c, p}, x] && GtQ[4*a - b^2/c, 0]

Rule 216

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

Rubi steps

\begin{align*} \int \frac{e^{\tanh ^{-1}(a+b x)} x}{1-a^2-2 a b x-b^2 x^2} \, dx &=\int \frac{x}{(1-a-b x)^{3/2} \sqrt{1+a+b x}} \, dx\\ &=\frac{(1-a) \sqrt{1+a+b x}}{b^2 \sqrt{1-a-b x}}-\frac{\int \frac{1}{\sqrt{1-a-b x} \sqrt{1+a+b x}} \, dx}{b}\\ &=\frac{(1-a) \sqrt{1+a+b x}}{b^2 \sqrt{1-a-b x}}-\frac{\int \frac{1}{\sqrt{(1-a) (1+a)-2 a b x-b^2 x^2}} \, dx}{b}\\ &=\frac{(1-a) \sqrt{1+a+b x}}{b^2 \sqrt{1-a-b x}}+\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{1-\frac{x^2}{4 b^2}}} \, dx,x,-2 a b-2 b^2 x\right )}{2 b^3}\\ &=\frac{(1-a) \sqrt{1+a+b x}}{b^2 \sqrt{1-a-b x}}-\frac{\sin ^{-1}(a+b x)}{b^2}\\ \end{align*}

Mathematica [A]  time = 0.145452, size = 49, normalized size = 1.11 \[ -\frac{\sin ^{-1}(a+b x)-\frac{(a-1) \sqrt{-a^2-2 a b x-b^2 x^2+1}}{a+b x-1}}{b^2} \]

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]

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

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Maple [B]  time = 0.037, size = 160, normalized size = 3.6 \begin{align*}{\frac{x}{b}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{\frac{1}{b}\arctan \left ({\sqrt{{b}^{2}} \left ( x+{\frac{a}{b}} \right ){\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}} \right ){\frac{1}{\sqrt{{b}^{2}}}}}+{\frac{1}{{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{\frac{ax}{b}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}}-{\frac{{a}^{2}}{{b}^{2}}{\frac{1}{\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1}}}} \end{align*}

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]

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

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Maxima [B]  time = 1.6145, size = 594, normalized size = 13.5 \begin{align*} -\frac{b^{2}{\left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{b^{4} x + a b^{3} + \sqrt{b^{2}} b^{2}} - \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{b^{4} x + a b^{3} - \sqrt{b^{2}} b^{2}} - \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{\sqrt{b^{2}} b^{3} x + a \sqrt{b^{2}} b^{2} + b^{3}} - \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} a}{\sqrt{b^{2}} b^{3} x + a \sqrt{b^{2}} b^{2} - b^{3}} - \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{4} x + a b^{3} + \sqrt{b^{2}} b^{2}} + \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{b^{4} x + a b^{3} - \sqrt{b^{2}} b^{2}} + \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{\sqrt{b^{2}} b^{3} x + a \sqrt{b^{2}} b^{2} + b^{3}} + \frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}}{\sqrt{b^{2}} b^{3} x + a \sqrt{b^{2}} b^{2} - b^{3}} + \frac{2 \, \arcsin \left (\sqrt{b^{2}} x + \frac{a \sqrt{b^{2}}}{b}\right )}{b^{3}}\right )}}{2 \, \sqrt{a^{2} b^{2} -{\left (a^{2} - 1\right )} b^{2}}} \end{align*}

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]

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

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Fricas [B]  time = 1.63749, size = 225, normalized size = 5.11 \begin{align*} \frac{{\left (b x + a - 1\right )} \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (b x + a\right )}}{b^{2} x^{2} + 2 \, a b x + a^{2} - 1}\right ) + \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a - 1\right )}}{b^{3} x +{\left (a - 1\right )} b^{2}} \end{align*}

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]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} - \int \frac{x}{a \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} + b x \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1} - \sqrt{- a^{2} - 2 a b x - b^{2} x^{2} + 1}}\, dx \end{align*}

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

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Giac [A]  time = 1.19682, size = 89, normalized size = 2.02 \begin{align*} \frac{\arcsin \left (-b x - a\right ) \mathrm{sgn}\left (b\right )}{b{\left | b \right |}} - \frac{2 \,{\left (a - 1\right )}}{b{\left (\frac{\sqrt{-{\left (b x + a\right )}^{2} + 1}{\left | b \right |} + b}{b^{2} x + a b} - 1\right )}{\left | b \right |}} \end{align*}

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]

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