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

Optimal. Leaf size=109 \[ -\frac{2 a b \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{a+b x-1}}\right )}{\sqrt{1-a^2}}-\frac{\sqrt{a+b x-1} \sqrt{a+b x+1}}{x}+2 b \sinh ^{-1}\left (\frac{\sqrt{a+b x-1}}{\sqrt{2}}\right )-\frac{a}{x}+b \log (x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0737779, antiderivative size = 109, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.667, Rules used = {5909, 14, 97, 157, 63, 215, 93, 205} \[ -\frac{2 a b \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{a+b x-1}}\right )}{\sqrt{1-a^2}}-\frac{\sqrt{a+b x-1} \sqrt{a+b x+1}}{x}+2 b \sinh ^{-1}\left (\frac{\sqrt{a+b x-1}}{\sqrt{2}}\right )-\frac{a}{x}+b \log (x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCosh[a + b*x]/x^2,x]

[Out]

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

Rule 5909

Int[E^(ArcCosh[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(u + Sqrt[-1 + u]*Sqrt[1 + u])^n, x] /; RationalQ[m
] && IntegerQ[n] && PolynomialQ[u, x]

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 97

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)/(b*(m + 1)), x] - Dist[1/(b*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^(n
- 1)*(e + f*x)^(p - 1)*Simp[d*e*n + c*f*p + d*f*(n + p)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && LtQ[m
, -1] && GtQ[n, 0] && GtQ[p, 0] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])

Rule 157

Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_)))/((a_.) + (b_.)*(x_)), x_Symbol]
:> Dist[h/b, Int[(c + d*x)^n*(e + f*x)^p, x], x] + Dist[(b*g - a*h)/b, Int[((c + d*x)^n*(e + f*x)^p)/(a + b*x
), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[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 205

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

Rubi steps

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

Mathematica [C]  time = 0.151557, size = 140, normalized size = 1.28 \[ -\frac{i a b \log \left (\frac{2 \left (\sqrt{a+b x-1} \sqrt{a+b x+1}+\frac{i \left (a^2+a b x-1\right )}{\sqrt{1-a^2}}\right )}{a b x}\right )}{\sqrt{1-a^2}}-\frac{\sqrt{a+b x-1} \sqrt{a+b x+1}}{x}+b \log \left (\sqrt{a+b x-1} \sqrt{a+b x+1}+a+b x\right )-\frac{a}{x}+b \log (x) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCosh[a + b*x]/x^2,x]

[Out]

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

________________________________________________________________________________________

Maple [C]  time = 0.014, size = 237, normalized size = 2.2 \begin{align*}{\frac{{\it csgn} \left ( b \right ) }{ \left ({a}^{2}-1 \right ) x} \left ( -{\it csgn} \left ( b \right ) \sqrt{{a}^{2}-1}\ln \left ( 2\,{\frac{xab+\sqrt{{a}^{2}-1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}+{a}^{2}-1}{x}} \right ) xab+\ln \left ( \left ( \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) +bx+a \right ){\it csgn} \left ( b \right ) \right ) x{a}^{2}b-\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ){a}^{2}-\ln \left ( \left ( \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) +bx+a \right ){\it csgn} \left ( b \right ) \right ) xb+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) \right ) \sqrt{bx+a-1}\sqrt{bx+a+1}{\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}}}}-{\frac{a}{x}}+b\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [A]  time = 2.22424, size = 852, normalized size = 7.82 \begin{align*} \left [\frac{\sqrt{a^{2} - 1} a b x \log \left (\frac{a^{2} b x + a^{3} +{\left (a^{2} - \sqrt{a^{2} - 1} a - 1\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} -{\left (a b x + a^{2} - 1\right )} \sqrt{a^{2} - 1} - a}{x}\right ) -{\left (a^{2} - 1\right )} b x \log \left (-b x + \sqrt{b x + a + 1} \sqrt{b x + a - 1} - a\right ) +{\left (a^{2} - 1\right )} b x \log \left (x\right ) - a^{3} -{\left (a^{2} - 1\right )} b x -{\left (a^{2} - 1\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} + a}{{\left (a^{2} - 1\right )} x}, \frac{2 \, \sqrt{-a^{2} + 1} a b x \arctan \left (-\frac{\sqrt{-a^{2} + 1} b x - \sqrt{-a^{2} + 1} \sqrt{b x + a + 1} \sqrt{b x + a - 1}}{a^{2} - 1}\right ) -{\left (a^{2} - 1\right )} b x \log \left (-b x + \sqrt{b x + a + 1} \sqrt{b x + a - 1} - a\right ) +{\left (a^{2} - 1\right )} b x \log \left (x\right ) - a^{3} -{\left (a^{2} - 1\right )} b x -{\left (a^{2} - 1\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} + a}{{\left (a^{2} - 1\right )} x}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Giac [B]  time = 1.28769, size = 317, normalized size = 2.91 \begin{align*} -\frac{\frac{2 \, a b^{2} \arctan \left (\frac{{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2} - 2 \, a}{2 \, \sqrt{-a^{2} + 1}}\right )}{\sqrt{-a^{2} + 1}} + b^{2} \log \left ({\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2}\right ) - b^{2} \log \left ({\left | b x \right |}\right ) - \frac{4 \,{\left (a b^{2}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2} - 2 \, b^{2}\right )}}{{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{4} - 4 \, a{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2} + 4} + \frac{a b^{2} \log \left ({\left | -a - 1 \right |}\right ) + b^{2} \log \left ({\left | -a - 1 \right |}\right ) - b^{2}}{a + 1} + \frac{{\left (b x + a + 1\right )} b^{2} - b^{2}}{b x}}{b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(2*a*b^2*arctan(1/2*((sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2 - 2*a)/sqrt(-a^2 + 1))/sqrt(-a^2 + 1) + b^2*lo
g((sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2) - b^2*log(abs(b*x)) - 4*(a*b^2*(sqrt(b*x + a + 1) - sqrt(b*x + a
- 1))^2 - 2*b^2)/((sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^4 - 4*a*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2 +
4) + (a*b^2*log(abs(-a - 1)) + b^2*log(abs(-a - 1)) - b^2)/(a + 1) + ((b*x + a + 1)*b^2 - b^2)/(b*x))/b