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

Optimal. Leaf size=89 \[ \sqrt{a^2+2 a b x+b^2 x^2+1}-\sqrt{a^2+1} \tanh ^{-1}\left (\frac{a^2+a b x+1}{\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}}\right )+a \sinh ^{-1}(a+b x)+a \log (x)+b x \]

[Out]

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

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Rubi [A]  time = 0.120651, antiderivative size = 89, 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 = {5907, 14, 734, 843, 619, 215, 724, 206} \[ \sqrt{a^2+2 a b x+b^2 x^2+1}-\sqrt{a^2+1} \tanh ^{-1}\left (\frac{a^2+a b x+1}{\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}}\right )+a \sinh ^{-1}(a+b x)+a \log (x)+b x \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]/x,x]

[Out]

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

Rule 5907

Int[E^(ArcSinh[u_]*(n_.))*(x_)^(m_.), x_Symbol] :> Int[x^m*(u + Sqrt[1 + u^2])^n, x] /; RationalQ[m] && Intege
rQ[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 734

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m + 1)*(
a + b*x + c*x^2)^p)/(e*(m + 2*p + 1)), x] - Dist[p/(e*(m + 2*p + 1)), Int[(d + e*x)^m*Simp[b*d - 2*a*e + (2*c*
d - b*e)*x, x]*(a + b*x + c*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ
[c*d^2 - b*d*e + a*e^2, 0] && NeQ[2*c*d - b*e, 0] && GtQ[p, 0] && NeQ[m + 2*p + 1, 0] && ( !RationalQ[m] || Lt
Q[m, 1]) &&  !ILtQ[m + 2*p, 0] && IntQuadraticQ[a, b, c, d, e, m, p, x]

Rule 843

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Dis
t[g/e, Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] + Dist[(e*f - d*g)/e, Int[(d + e*x)^m*(a + b*x + c*x^
2)^p, x], x] /; FreeQ[{a, b, c, d, e, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0]
&&  !IGtQ[m, 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 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 724

Int[1/(((d_.) + (e_.)*(x_))*Sqrt[(a_.) + (b_.)*(x_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[-2, Subst[Int[1/(4*c*d
^2 - 4*b*d*e + 4*a*e^2 - x^2), x], x, (2*a*e - b*d - (2*c*d - b*e)*x)/Sqrt[a + b*x + c*x^2]], x] /; FreeQ[{a,
b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[2*c*d - b*e, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{e^{\sinh ^{-1}(a+b x)}}{x} \, dx &=\int \frac{a+b x+\sqrt{1+(a+b x)^2}}{x} \, dx\\ &=\int \left (b+\frac{a}{x}+\frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x}\right ) \, dx\\ &=b x+a \log (x)+\int \frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x} \, dx\\ &=b x+\sqrt{1+a^2+2 a b x+b^2 x^2}+a \log (x)-\frac{1}{2} \int \frac{-2 \left (1+a^2\right )-2 a b x}{x \sqrt{1+a^2+2 a b x+b^2 x^2}} \, dx\\ &=b x+\sqrt{1+a^2+2 a b x+b^2 x^2}+a \log (x)-\left (-1-a^2\right ) \int \frac{1}{x \sqrt{1+a^2+2 a b x+b^2 x^2}} \, dx+(a b) \int \frac{1}{\sqrt{1+a^2+2 a b x+b^2 x^2}} \, dx\\ &=b x+\sqrt{1+a^2+2 a b x+b^2 x^2}+a \log (x)-\left (2 \left (1+a^2\right )\right ) \operatorname{Subst}\left (\int \frac{1}{4 \left (1+a^2\right )-x^2} \, dx,x,\frac{2 \left (1+a^2\right )+2 a b x}{\sqrt{1+a^2+2 a b x+b^2 x^2}}\right )+\frac{a \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}\\ &=b x+\sqrt{1+a^2+2 a b x+b^2 x^2}+a \sinh ^{-1}(a+b x)-\sqrt{1+a^2} \tanh ^{-1}\left (\frac{1+a^2+a b x}{\sqrt{1+a^2} \sqrt{1+a^2+2 a b x+b^2 x^2}}\right )+a \log (x)\\ \end{align*}

Mathematica [A]  time = 0.0740409, size = 99, normalized size = 1.11 \[ \sqrt{a^2+2 a b x+b^2 x^2+1}-\sqrt{a^2+1} \log \left (\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )+\left (\sqrt{a^2+1}+a\right ) \log (x)+a \sinh ^{-1}(a+b x)+b x \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSinh[a + b*x]/x,x]

[Out]

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

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Maple [A]  time = 0.004, size = 126, normalized size = 1.4 \begin{align*} \sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}+{ab\ln \left ({({b}^{2}x+ab){\frac{1}{\sqrt{{b}^{2}}}}}+\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1} \right ){\frac{1}{\sqrt{{b}^{2}}}}}-\sqrt{{a}^{2}+1}\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 ) +bx+a\ln \left ( x \right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.62354, size = 335, normalized size = 3.76 \begin{align*} b x - a \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + a \log \left (x\right ) + \sqrt{a^{2} + 1} \log \left (-\frac{a^{2} b x + a^{3} + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (a^{2} - \sqrt{a^{2} + 1} a + 1\right )} -{\left (a b x + a^{2} + 1\right )} \sqrt{a^{2} + 1} + a}{x}\right ) + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*x - a*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) + a*log(x) + sqrt(a^2 + 1)*log(-(a^2*b*x + a^3 + sqr
t(b^2*x^2 + 2*a*b*x + a^2 + 1)*(a^2 - sqrt(a^2 + 1)*a + 1) - (a*b*x + a^2 + 1)*sqrt(a^2 + 1) + a)/x) + sqrt(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 + \sqrt{a^{2} + 2 a b x + b^{2} x^{2} + 1}}{x}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: NotImplementedError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: NotImplementedError