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

Optimal. Leaf size=98 \[ -\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(1-a) \sqrt {1-a^2}}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{(1-a) x} \]

[Out]

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

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Rubi [A]  time = 0.06, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6163, 94, 93, 208} \[ -\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {1-a} \sqrt {a+b x+1}}{\sqrt {a+1} \sqrt {-a-b x+1}}\right )}{(1-a) \sqrt {1-a^2}}-\frac {\sqrt {-a-b x+1} \sqrt {a+b x+1}}{(1-a) x} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcTanh[a + b*x]/x^2,x]

[Out]

-((Sqrt[1 - a - b*x]*Sqrt[1 + a + b*x])/((1 - a)*x)) - (2*b*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 94

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

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 88, normalized size = 0.90 \[ \frac {\sqrt {-((a+b x-1) (a+b x+1))}}{(a-1) x}-\frac {2 b \tanh ^{-1}\left (\frac {\sqrt {-a-1} \sqrt {-a-b x+1}}{\sqrt {a-1} \sqrt {a+b x+1}}\right )}{\sqrt {-a-1} (a-1)^{3/2}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcTanh[a + b*x]/x^2,x]

[Out]

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

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fricas [A]  time = 1.12, size = 282, normalized size = 2.88 \[ \left [-\frac {\sqrt {-a^{2} + 1} b x \log \left (\frac {{\left (2 \, a^{2} - 1\right )} b^{2} x^{2} + 2 \, a^{4} + 4 \, {\left (a^{3} - a\right )} b x - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {-a^{2} + 1} - 4 \, a^{2} + 2}{x^{2}}\right ) - 2 \, \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )}}{2 \, {\left (a^{3} - a^{2} - a + 1\right )} x}, -\frac {\sqrt {a^{2} - 1} b x \arctan \left (\frac {\sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a b x + a^{2} - 1\right )} \sqrt {a^{2} - 1}}{{\left (a^{2} - 1\right )} b^{2} x^{2} + a^{4} + 2 \, {\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1}\right ) - \sqrt {-b^{2} x^{2} - 2 \, a b x - a^{2} + 1} {\left (a^{2} - 1\right )}}{{\left (a^{3} - a^{2} - a + 1\right )} x}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.25, size = 190, normalized size = 1.94 \[ -\frac {2 \, b^{2} \arctan \left (\frac {\frac {{\left (\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left | b \right |} + b\right )} a}{b^{2} x + a b} - 1}{\sqrt {a^{2} - 1}}\right )}{\sqrt {a^{2} - 1} {\left (a {\left | b \right |} - {\left | b \right |}\right )}} + \frac {2 \, {\left (a b^{2} - \frac {{\left (\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left | b \right |} + b\right )} b^{2}}{b^{2} x + a b}\right )}}{{\left (a^{2} {\left | b \right |} - a {\left | b \right |}\right )} {\left (\frac {{\left (\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left | b \right |} + b\right )}^{2} a}{{\left (b^{2} x + a b\right )}^{2}} + a - \frac {2 \, {\left (\sqrt {-{\left (b x + a\right )}^{2} + 1} {\left | b \right |} + b\right )}}{b^{2} x + a b}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.04, size = 265, normalized size = 2.70 \[ -\frac {b \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\sqrt {-a^{2}+1}}-\frac {\sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{\left (-a^{2}+1\right ) x}-\frac {a b \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}}-\frac {a \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{\left (-a^{2}+1\right ) x}-\frac {a^{2} b \ln \left (\frac {-2 a^{2}+2-2 a b x +2 \sqrt {-a^{2}+1}\, \sqrt {-b^{2} x^{2}-2 a b x -a^{2}+1}}{x}\right )}{\left (-a^{2}+1\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: ValueError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a+1)/(1-(b*x+a)^2)^(1/2)/x^2,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(a-1>0)', see `assume?` for mor
e details)Is a-1 positive, negative or zero?

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mupad [B]  time = 1.69, size = 241, normalized size = 2.46 \[ \frac {\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{x\,\left (a^2-1\right )}-\frac {b\,\ln \left (\frac {\sqrt {1-a^2}\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{x}-\frac {a^2-1}{x}-a\,b\right )}{\sqrt {1-a^2}}+\frac {a^2\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a-1}{\sqrt {1-a^2}\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}\right )}{{\left (1-a^2\right )}^{3/2}}+\frac {a\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}{x\,\left (a^2-1\right )}+\frac {a\,b\,\mathrm {atanh}\left (\frac {a^2+b\,x\,a-1}{\sqrt {1-a^2}\,\sqrt {-a^2-2\,a\,b\,x-b^2\,x^2+1}}\right )}{{\left (1-a^2\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(1 - b^2*x^2 - 2*a*b*x - a^2)^(1/2)/(x*(a^2 - 1)) - (b*log(((1 - a^2)^(1/2)*(1 - b^2*x^2 - 2*a*b*x - a^2)^(1/2
))/x - (a^2 - 1)/x - a*b))/(1 - a^2)^(1/2) + (a^2*b*atanh((a^2 + a*b*x - 1)/((1 - a^2)^(1/2)*(1 - b^2*x^2 - 2*
a*b*x - a^2)^(1/2))))/(1 - a^2)^(3/2) + (a*(1 - b^2*x^2 - 2*a*b*x - a^2)^(1/2))/(x*(a^2 - 1)) + (a*b*atanh((a^
2 + a*b*x - 1)/((1 - a^2)^(1/2)*(1 - b^2*x^2 - 2*a*b*x - a^2)^(1/2))))/(1 - a^2)^(3/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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