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

Optimal. Leaf size=99 \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2+1}}{x}-\frac{a b \tanh ^{-1}\left (\frac{a^2+a b x+1}{\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}}\right )}{\sqrt{a^2+1}}+b \sinh ^{-1}(a+b x)-\frac{a}{x}+b \log (x) \]

[Out]

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

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Rubi [A]  time = 0.105395, antiderivative size = 99, 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, 732, 843, 619, 215, 724, 206} \[ -\frac{\sqrt{a^2+2 a b x+b^2 x^2+1}}{x}-\frac{a b \tanh ^{-1}\left (\frac{a^2+a b x+1}{\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}}\right )}{\sqrt{a^2+1}}+b \sinh ^{-1}(a+b x)-\frac{a}{x}+b \log (x) \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]/x^2,x]

[Out]

-(a/x) - Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]/x + b*ArcSinh[a + b*x] - (a*b*ArcTanh[(1 + a^2 + a*b*x)/(Sqrt[1 + a
^2]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])])/Sqrt[1 + a^2] + b*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 732

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 + 1)), x] - Dist[p/(e*(m + 1)), Int[(d + e*x)^(m + 1)*(b + 2*c*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] && (IntegerQ[p] || LtQ[m, -1]) && NeQ[m, -1] &&  !ILtQ[m + 2*p + 1, 0] && IntQua
draticQ[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^2} \, dx &=\int \frac{a+b x+\sqrt{1+(a+b x)^2}}{x^2} \, dx\\ &=\int \left (\frac{a}{x^2}+\frac{b}{x}+\frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x^2}\right ) \, dx\\ &=-\frac{a}{x}+b \log (x)+\int \frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x^2} \, dx\\ &=-\frac{a}{x}-\frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x}+b \log (x)+\frac{1}{2} \int \frac{2 a b+2 b^2 x}{x \sqrt{1+a^2+2 a b x+b^2 x^2}} \, dx\\ &=-\frac{a}{x}-\frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x}+b \log (x)+(a b) \int \frac{1}{x \sqrt{1+a^2+2 a b x+b^2 x^2}} \, dx+b^2 \int \frac{1}{\sqrt{1+a^2+2 a b x+b^2 x^2}} \, dx\\ &=-\frac{a}{x}-\frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x}+b \log (x)+\frac{1}{2} \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 a b) \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}{x}-\frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x}+b \sinh ^{-1}(a+b x)-\frac{a b \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 )}{\sqrt{1+a^2}}+b \log (x)\\ \end{align*}

Mathematica [A]  time = 0.158718, size = 110, normalized size = 1.11 \[ b \sinh ^{-1}(a+b x)-\frac{\sqrt{a^2+2 a b x+b^2 x^2+1}+\frac{a b x \log \left (\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )}{\sqrt{a^2+1}}+\left (-\frac{a}{\sqrt{a^2+1}}-1\right ) b x \log (x)+a}{x} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSinh[a + b*x]/x^2,x]

[Out]

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

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Maple [B]  time = 0.009, size = 267, normalized size = 2.7 \begin{align*} -{\frac{1}{ \left ({a}^{2}+1 \right ) x} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}+2\,{\frac{ab\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}{{a}^{2}+1}}+{\frac{{a}^{2}{b}^{2}}{{a}^{2}+1}\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}}}}}-{ab\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 ){\frac{1}{\sqrt{{a}^{2}+1}}}}+{\frac{{b}^{2}x}{{a}^{2}+1}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{b}^{2}}{{a}^{2}+1}\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}}}}}-{\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+(1+(b*x+a)^2)^(1/2))/x^2,x)

[Out]

-1/(a^2+1)/x*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)+2*a*b/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+a^2*b^2/(a^2+1)*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*b/(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^2/(a^2+1)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x+b^2/(a^2+1)*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/x+b*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^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.72342, size = 437, normalized size = 4.41 \begin{align*} \frac{\sqrt{a^{2} + 1} a b x \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 ) -{\left (a^{2} + 1\right )} b x \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) +{\left (a^{2} + 1\right )} b x \log \left (x\right ) - a^{3} -{\left (a^{2} + 1\right )} b x - \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{\left (a^{2} + 1\right )} - a}{{\left (a^{2} + 1\right )} x} \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^2,x, algorithm="fricas")

[Out]

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

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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^{2}}\, 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**2,x)

[Out]

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

[Out]

Exception raised: NotImplementedError