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

Optimal. Leaf size=64 \[ \sin ^{-1}(a+b x)-\frac{2 (a+1) \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{\sqrt{1-a^2}} \]

[Out]

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

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Rubi [A]  time = 0.0754015, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {6163, 105, 53, 619, 216, 93, 208} \[ \sin ^{-1}(a+b x)-\frac{2 (a+1) \tanh ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{-a-b x+1}}\right )}{\sqrt{1-a^2}} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a + b*x]/x,x]

[Out]

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

Rule 6163

Int[E^(ArcTanh[(c_.)*((a_) + (b_.)*(x_))]*(n_.))*((d_.) + (e_.)*(x_))^(m_.), x_Symbol] :> Int[((d + e*x)^m*(1
+ a*c + b*c*x)^(n/2))/(1 - a*c - b*c*x)^(n/2), x] /; FreeQ[{a, b, c, d, e, m, n}, x]

Rule 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

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]

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 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]

Rubi steps

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

Mathematica [A]  time = 0.10338, size = 100, normalized size = 1.56 \[ \frac{2 \tan ^{-1}\left (\frac{\sqrt{-a-b x+1}}{\sqrt{\frac{a-1}{a+1}} \sqrt{a+b x+1}}\right )}{\sqrt{\frac{a-1}{a+1}}}+\frac{2 \sqrt{-b} \sinh ^{-1}\left (\frac{\sqrt{-b} \sqrt{-a-b x+1}}{\sqrt{2} \sqrt{b}}\right )}{\sqrt{b}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[a + b*x]/x,x]

[Out]

(2*Sqrt[-b]*ArcSinh[(Sqrt[-b]*Sqrt[1 - a - b*x])/(Sqrt[2]*Sqrt[b])])/Sqrt[b] + (2*ArcTan[Sqrt[1 - a - b*x]/(Sq
rt[(-1 + a)/(1 + a)]*Sqrt[1 + a + b*x])])/Sqrt[(-1 + a)/(1 + a)]

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Maple [B]  time = 0.057, size = 168, normalized size = 2.6 \begin{align*}{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}}}}}-{\ln \left ({\frac{1}{x} \left ( -2\,{a}^{2}+2-2\,xab+2\,\sqrt{-{a}^{2}+1}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1} \right ) } \right ){\frac{1}{\sqrt{-{a}^{2}+1}}}}-{a\ln \left ({\frac{1}{x} \left ( -2\,{a}^{2}+2-2\,xab+2\,\sqrt{-{a}^{2}+1}\sqrt{-{b}^{2}{x}^{2}-2\,xab-{a}^{2}+1} \right ) } \right ){\frac{1}{\sqrt{-{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,x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \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,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.63945, size = 717, normalized size = 11.2 \begin{align*} \left [\frac{1}{2} \, \sqrt{-\frac{a + 1}{a - 1}} \log \left (\frac{{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \,{\left (a^{3} - a\right )} b x - 4 \, a^{2} - 2 \, \sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a^{3} +{\left (a^{2} - a\right )} b x - a^{2} - a + 1\right )} \sqrt{-\frac{a + 1}{a - 1}} + 2}{x^{2}}\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{\frac{a + 1}{a - 1}} \arctan \left (\frac{\sqrt{-b^{2} x^{2} - 2 \, a b x - a^{2} + 1}{\left (a b x + a^{2} - 1\right )} \sqrt{\frac{a + 1}{a - 1}}}{{\left (a + 1\right )} b^{2} x^{2} + a^{3} + 2 \,{\left (a^{2} + a\right )} b x + a^{2} - 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 )\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,x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{a + b x + 1}{x \sqrt{- \left (a + b x - 1\right ) \left (a + b x + 1\right )}}\, 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,x)

[Out]

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

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Giac [A]  time = 1.20983, size = 107, normalized size = 1.67 \begin{align*} -\frac{b \arcsin \left (-b x - a\right ) \mathrm{sgn}\left (b\right )}{{\left | b \right |}} + \frac{2 \,{\left (a b + b\right )} \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 | 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,x, algorithm="giac")

[Out]

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