3.281 \(\int \frac{e^{\cosh ^{-1}(a+b x)}}{x^3} \, dx\)

Optimal. Leaf size=138 \[ -\frac{b^2 \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{a+b x-1}}\right )}{\left (1-a^2\right )^{3/2}}+\frac{b \sqrt{a+b x-1} \sqrt{a+b x+1}}{2 \left (1-a^2\right ) x}-\frac{\sqrt{a+b x-1} (a+b x+1)^{3/2}}{2 (a+1) x^2}-\frac{a}{2 x^2}-\frac{b}{x} \]

[Out]

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

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Rubi [A]  time = 0.108985, antiderivative size = 138, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {5909, 14, 94, 93, 205} \[ -\frac{b^2 \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{a+b x+1}}{\sqrt{a+1} \sqrt{a+b x-1}}\right )}{\left (1-a^2\right )^{3/2}}+\frac{b \sqrt{a+b x-1} \sqrt{a+b x+1}}{2 \left (1-a^2\right ) x}-\frac{\sqrt{a+b x-1} (a+b x+1)^{3/2}}{2 (a+1) x^2}-\frac{a}{2 x^2}-\frac{b}{x} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcCosh[a + b*x]/x^3,x]

[Out]

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

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 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 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^3} \, dx &=\int \frac{a+b x+\sqrt{-1+a+b x} \sqrt{1+a+b x}}{x^3} \, dx\\ &=\int \left (\frac{a}{x^3}+\frac{b}{x^2}+\frac{\sqrt{-1+a+b x} \sqrt{1+a+b x}}{x^3}\right ) \, dx\\ &=-\frac{a}{2 x^2}-\frac{b}{x}+\int \frac{\sqrt{-1+a+b x} \sqrt{1+a+b x}}{x^3} \, dx\\ &=-\frac{a}{2 x^2}-\frac{b}{x}-\frac{\sqrt{-1+a+b x} (1+a+b x)^{3/2}}{2 (1+a) x^2}+\frac{b \int \frac{\sqrt{1+a+b x}}{x^2 \sqrt{-1+a+b x}} \, dx}{2 (1+a)}\\ &=-\frac{a}{2 x^2}-\frac{b}{x}+\frac{b \sqrt{-1+a+b x} \sqrt{1+a+b x}}{2 \left (1-a^2\right ) x}-\frac{\sqrt{-1+a+b x} (1+a+b x)^{3/2}}{2 (1+a) x^2}+\frac{b^2 \int \frac{1}{x \sqrt{-1+a+b x} \sqrt{1+a+b x}} \, dx}{2 \left (1-a^2\right )}\\ &=-\frac{a}{2 x^2}-\frac{b}{x}+\frac{b \sqrt{-1+a+b x} \sqrt{1+a+b x}}{2 \left (1-a^2\right ) x}-\frac{\sqrt{-1+a+b x} (1+a+b x)^{3/2}}{2 (1+a) x^2}+\frac{b^2 \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^2}\\ &=-\frac{a}{2 x^2}-\frac{b}{x}+\frac{b \sqrt{-1+a+b x} \sqrt{1+a+b x}}{2 \left (1-a^2\right ) x}-\frac{\sqrt{-1+a+b x} (1+a+b x)^{3/2}}{2 (1+a) x^2}-\frac{b^2 \tan ^{-1}\left (\frac{\sqrt{1-a} \sqrt{1+a+b x}}{\sqrt{1+a} \sqrt{-1+a+b x}}\right )}{\left (1-a^2\right )^{3/2}}\\ \end{align*}

Mathematica [C]  time = 0.240234, size = 142, normalized size = 1.03 \[ \frac{1}{2} \left (-\frac{i b^2 \log \left (\frac{4 i \sqrt{1-a^2} \left (-i \sqrt{1-a^2} \sqrt{a+b x-1} \sqrt{a+b x+1}+a^2+a b x-1\right )}{b^2 x}\right )}{\left (1-a^2\right )^{3/2}}-\frac{\sqrt{a+b x-1} \sqrt{a+b x+1} \left (a^2+a b x-1\right )}{\left (a^2-1\right ) x^2}-\frac{a}{x^2}-\frac{2 b}{x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCosh[a + b*x]/x^3,x]

[Out]

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

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Maple [B]  time = 0.016, size = 236, normalized size = 1.7 \begin{align*}{\frac{1}{2\, \left ({a}^{2}-1 \right ) ^{2}{x}^{2}}\sqrt{bx+a-1}\sqrt{bx+a+1} \left ( \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 ){x}^{2}{b}^{2}-x{a}^{3}b\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}-{a}^{4}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}xab+2\,\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}{a}^{2}-\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1} \right ){\frac{1}{\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}-1}}}}-{\frac{b}{x}}-{\frac{a}{2\,{x}^{2}}} \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^3,x)

[Out]

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

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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^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.0167, size = 818, normalized size = 5.93 \begin{align*} \left [\frac{\sqrt{a^{2} - 1} b^{2} x^{2} \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 ) - a^{5} -{\left (a^{3} - a\right )} b^{2} x^{2} + 2 \, a^{3} - 2 \,{\left (a^{4} - 2 \, a^{2} + 1\right )} b x -{\left (a^{4} +{\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} - a}{2 \,{\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2}}, -\frac{2 \, \sqrt{-a^{2} + 1} b^{2} x^{2} \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 ) + a^{5} +{\left (a^{3} - a\right )} b^{2} x^{2} - 2 \, a^{3} + 2 \,{\left (a^{4} - 2 \, a^{2} + 1\right )} b x +{\left (a^{4} +{\left (a^{3} - a\right )} b x - 2 \, a^{2} + 1\right )} \sqrt{b x + a + 1} \sqrt{b x + a - 1} + a}{2 \,{\left (a^{4} - 2 \, a^{2} + 1\right )} x^{2}}\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^3,x, algorithm="fricas")

[Out]

[1/2*(sqrt(a^2 - 1)*b^2*x^2*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^5 - (a^3 - a)*b^2*x^2 + 2*a^3 - 2*(a^4 - 2*a^2 + 1)*b*x - (a^
4 + (a^3 - a)*b*x - 2*a^2 + 1)*sqrt(b*x + a + 1)*sqrt(b*x + a - 1) - a)/((a^4 - 2*a^2 + 1)*x^2), -1/2*(2*sqrt(
-a^2 + 1)*b^2*x^2*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^5 + (a^3 - a)*b^2*x^2 - 2*a^3 + 2*(a^4 - 2*a^2 + 1)*b*x + (a^4 + (a^3 - a)*b*x - 2*a^2 + 1)*sqrt(b*x + a
+ 1)*sqrt(b*x + a - 1) + a)/((a^4 - 2*a^2 + 1)*x^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \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**3,x)

[Out]

Timed out

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Giac [B]  time = 1.39922, size = 471, normalized size = 3.41 \begin{align*} \frac{\frac{2 \, b^{3} \arctan \left (\frac{{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2} - 2 \, a}{2 \, \sqrt{-a^{2} + 1}}\right )}{{\left (a^{2} - 1\right )} \sqrt{-a^{2} + 1}} - \frac{a b^{3} + 2 \, b^{3}}{a^{2} + 2 \, a + 1} + \frac{4 \,{\left (2 \, a^{2} b^{3}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{6} - 4 \, a^{3} b^{3}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{4} - b^{3}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{6} - 2 \, a b^{3}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{4} + 8 \, a^{2} b^{3}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2} + 4 \, b^{3}{\left (\sqrt{b x + a + 1} - \sqrt{b x + a - 1}\right )}^{2} - 8 \, a b^{3}\right )}}{{\left ({\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\right )}^{2}{\left (a^{2} - 1\right )}} - \frac{2 \,{\left (b x + a + 1\right )} b^{3} - a b^{3} - 2 \, b^{3}}{b^{2} x^{2}}}{2 \, 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^3,x, algorithm="giac")

[Out]

1/2*(2*b^3*arctan(1/2*((sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2 - 2*a)/sqrt(-a^2 + 1))/((a^2 - 1)*sqrt(-a^2 +
1)) - (a*b^3 + 2*b^3)/(a^2 + 2*a + 1) + 4*(2*a^2*b^3*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^6 - 4*a^3*b^3*(s
qrt(b*x + a + 1) - sqrt(b*x + a - 1))^4 - b^3*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^6 - 2*a*b^3*(sqrt(b*x +
a + 1) - sqrt(b*x + a - 1))^4 + 8*a^2*b^3*(sqrt(b*x + a + 1) - sqrt(b*x + a - 1))^2 + 4*b^3*(sqrt(b*x + a + 1)
- sqrt(b*x + a - 1))^2 - 8*a*b^3)/(((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)^2*(a^2 - 1)) - (2*(b*x + a + 1)*b^3 - a*b^3 - 2*b^3)/(b^2*x^2))/b