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

Optimal. Leaf size=100 \[ 2 \sqrt{1-a^2} \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{a+b x-1}}\right )+\sqrt{a+b x-1} \sqrt{a+b x+1}+2 a \sinh ^{-1}\left (\frac{\sqrt{a+b x-1}}{\sqrt{2}}\right )+a \log (x)+b x \]

[Out]

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

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Rubi [A]  time = 0.091578, antiderivative size = 100, 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, 101, 157, 63, 215, 93, 205} \[ 2 \sqrt{1-a^2} \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{a+b x-1}}\right )+\sqrt{a+b x-1} \sqrt{a+b x+1}+2 a \sinh ^{-1}\left (\frac{\sqrt{a+b x-1}}{\sqrt{2}}\right )+a \log (x)+b x \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCosh[a + b*x]/x,x]

[Out]

b*x + Sqrt[-1 + a + b*x]*Sqrt[1 + a + b*x] + 2*a*ArcSinh[Sqrt[-1 + a + b*x]/Sqrt[2]] + 2*Sqrt[1 - a^2]*ArcTan[
(Sqrt[1 - a]*Sqrt[1 + a + b*x])/(Sqrt[1 + a]*Sqrt[-1 + a + b*x])] + a*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 101

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((a +
b*x)^m*(c + d*x)^n*(e + f*x)^(p + 1))/(f*(m + n + p + 1)), x] - Dist[1/(f*(m + n + p + 1)), Int[(a + b*x)^(m -
1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[c*m*(b*e - a*f) + a*n*(d*e - c*f) + (d*m*(b*e - a*f) + b*n*(d*e - c*f))
*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && GtQ[m, 0] && GtQ[n, 0] && NeQ[m + n + p + 1, 0] && (Integ
ersQ[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} \, dx &=\int \frac{a+b x+\sqrt{-1+a+b x} \sqrt{1+a+b x}}{x} \, dx\\ &=\int \left (b+\frac{a}{x}+\frac{\sqrt{-1+a+b x} \sqrt{1+a+b x}}{x}\right ) \, dx\\ &=b x+a \log (x)+\int \frac{\sqrt{-1+a+b x} \sqrt{1+a+b x}}{x} \, dx\\ &=b x+\sqrt{-1+a+b x} \sqrt{1+a+b x}+a \log (x)-\int \frac{1-a^2-a b x}{x \sqrt{-1+a+b x} \sqrt{1+a+b x}} \, dx\\ &=b x+\sqrt{-1+a+b x} \sqrt{1+a+b x}+a \log (x)-\left (1-a^2\right ) \int \frac{1}{x \sqrt{-1+a+b x} \sqrt{1+a+b x}} \, dx+(a b) \int \frac{1}{\sqrt{-1+a+b x} \sqrt{1+a+b x}} \, dx\\ &=b x+\sqrt{-1+a+b x} \sqrt{1+a+b x}+a \log (x)+(2 a) \operatorname{Subst}\left (\int \frac{1}{\sqrt{2+x^2}} \, dx,x,\sqrt{-1+a+b x}\right )-\left (2 \left (1-a^2\right )\right ) \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 x+\sqrt{-1+a+b x} \sqrt{1+a+b x}+2 a \sinh ^{-1}\left (\frac{\sqrt{-1+a+b x}}{\sqrt{2}}\right )+2 \sqrt{1-a^2} \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{1+a+b x}}{\sqrt{1+a} \sqrt{-1+a+b x}}\right )+a \log (x)\\ \end{align*}

Mathematica [C]  time = 0.0918227, size = 141, normalized size = 1.41 \[ i \sqrt{1-a^2} \log \left (\frac{2 \sqrt{a+b x-1} \sqrt{a+b x+1}}{\left (a^2-1\right ) x}+\frac{2 i \left (a^2+a b x-1\right )}{\sqrt{1-a^2} \left (a^2-1\right ) x}\right )+\sqrt{a+b x-1} \sqrt{a+b x+1}+a \log \left (\sqrt{a+b x-1} \sqrt{a+b x+1}+a+b x\right )+a \log (x)+b x \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCosh[a + b*x]/x,x]

[Out]

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

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Maple [C]  time = 0.013, size = 156, normalized size = 1.6 \begin{align*}{{\it csgn} \left ( b \right ) \sqrt{bx+a-1}\sqrt{bx+a+1} \left ( -{\it csgn} \left ( b \right ) \ln \left ( 2\,{\frac{xab+\sqrt{{a}^{2}-1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}+{a}^{2}-1}{x}} \right ) \sqrt{{a}^{2}-1}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{\it csgn} \left ( b \right ) +\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 ) a \right ){\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}}}}+bx+a\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,x)

[Out]

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

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.07392, size = 647, normalized size = 6.47 \begin{align*} \left [b x - a \log \left (-b x + \sqrt{b x + a + 1} \sqrt{b x + a - 1} - a\right ) + a \log \left (x\right ) + \sqrt{a^{2} - 1} \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 ) + \sqrt{b x + a + 1} \sqrt{b x + a - 1}, b x - a \log \left (-b x + \sqrt{b x + a + 1} \sqrt{b x + a - 1} - a\right ) + a \log \left (x\right ) + 2 \, \sqrt{-a^{2} + 1} \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 ) + \sqrt{b x + a + 1} \sqrt{b x + a - 1}\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,x, algorithm="fricas")

[Out]

[b*x - a*log(-b*x + sqrt(b*x + a + 1)*sqrt(b*x + a - 1) - a) + a*log(x) + sqrt(a^2 - 1)*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) + sqr
t(b*x + a + 1)*sqrt(b*x + a - 1), b*x - a*log(-b*x + sqrt(b*x + a + 1)*sqrt(b*x + a - 1) - a) + a*log(x) + 2*s
qrt(-a^2 + 1)*arctan(-(sqrt(-a^2 + 1)*b*x - sqrt(-a^2 + 1)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1))/(a^2 - 1)) + s
qrt(b*x + a + 1)*sqrt(b*x + a - 1)]

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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}\, 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,x)

[Out]

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

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Giac [A]  time = 1.23964, size = 159, normalized size = 1.59 \begin{align*} b x - a \log \left ({\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2}\right ) + a \log \left ({\left | b x \right |}\right ) - a \log \left ({\left | -a - 1 \right |}\right ) - \frac{2 \,{\left (a^{2} - 1\right )} \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}} + \sqrt{b x + a + 1} \sqrt{b x + a - 1} + a + 1 \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,x, algorithm="giac")

[Out]

b*x - a*log((sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2) + a*log(abs(b*x)) - a*log(abs(-a - 1)) - 2*(a^2 - 1)*ar
ctan(1/2*((sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2 - 2*a)/sqrt(-a^2 + 1))/sqrt(-a^2 + 1) + sqrt(b*x + a + 1)*
sqrt(b*x + a - 1) + a + 1