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

Optimal. Leaf size=116 \[ -\frac{\left (a^2+a b x+1\right ) \sqrt{a^2+2 a b x+b^2 x^2+1}}{2 \left (a^2+1\right ) x^2}-\frac{b^2 \tanh ^{-1}\left (\frac{a^2+a b x+1}{\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}}\right )}{2 \left (a^2+1\right )^{3/2}}-\frac{a}{2 x^2}-\frac{b}{x} \]

[Out]

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

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Rubi [A]  time = 0.0858755, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {5907, 14, 720, 724, 206} \[ -\frac{\left (a^2+a b x+1\right ) \sqrt{a^2+2 a b x+b^2 x^2+1}}{2 \left (a^2+1\right ) x^2}-\frac{b^2 \tanh ^{-1}\left (\frac{a^2+a b x+1}{\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}}\right )}{2 \left (a^2+1\right )^{3/2}}-\frac{a}{2 x^2}-\frac{b}{x} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]/x^3,x]

[Out]

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

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 720

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

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

Mathematica [A]  time = 0.19399, size = 129, normalized size = 1.11 \[ \frac{1}{2} \left (-\frac{\left (a^2+a b x+1\right ) \sqrt{a^2+2 a b x+b^2 x^2+1}}{\left (a^2+1\right ) x^2}-\frac{b^2 \log \left (\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )}{\left (a^2+1\right )^{3/2}}+\frac{b^2 \log (x)}{\left (a^2+1\right )^{3/2}}-\frac{a}{x^2}-\frac{2 b}{x}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSinh[a + b*x]/x^3,x]

[Out]

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

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Maple [B]  time = 0.009, size = 457, normalized size = 3.9 \begin{align*} -{\frac{1}{ \left ( 2\,{a}^{2}+2 \right ){x}^{2}} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{ab}{2\, \left ({a}^{2}+1 \right ) ^{2}x} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{{a}^{2}{b}^{2}}{ \left ({a}^{2}+1 \right ) ^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{a}^{3}{b}^{3}}{2\, \left ({a}^{2}+1 \right ) ^{2}}\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}^{2}{b}^{2}}{2}\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 ) \left ({a}^{2}+1 \right ) ^{-{\frac{3}{2}}}}-{\frac{a{b}^{3}x}{2\, \left ({a}^{2}+1 \right ) ^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{a{b}^{3}}{2\, \left ({a}^{2}+1 \right ) ^{2}}\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{{b}^{2}}{2\,{a}^{2}+2}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{a{b}^{3}}{2\,{a}^{2}+2}\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{{b}^{2}}{2}\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}{x}}-{\frac{a}{2\,{x}^{2}}} \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^3,x)

[Out]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.63687, size = 423, normalized size = 3.65 \begin{align*} \frac{\sqrt{a^{2} + 1} b^{2} x^{2} \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 ) - 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^{2} x^{2} + 2 \, a b x + a^{2} + 1} - a}{2 \,{\left (a^{4} + 2 \, a^{2} + 1\right )} x^{2}} \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^3,x, algorithm="fricas")

[Out]

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

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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^{3}}\, 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**3,x)

[Out]

Integral((a + b*x + sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1))/x**3, 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^3,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError