3.356 \(\int \frac{e^{\sinh ^{-1}(a+b x)}}{x^4} \, dx\)

Optimal. Leaf size=156 \[ \frac{a b \left (a^2+a b x+1\right ) \sqrt{a^2+2 a b x+b^2 x^2+1}}{2 \left (a^2+1\right )^2 x^2}-\frac{\left (a^2+2 a b x+b^2 x^2+1\right )^{3/2}}{3 \left (a^2+1\right ) x^3}+\frac{a b^3 \tanh ^{-1}\left (\frac{a^2+a b x+1}{\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}}\right )}{2 \left (a^2+1\right )^{5/2}}-\frac{a}{3 x^3}-\frac{b}{2 x^2} \]

[Out]

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

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Rubi [A]  time = 0.108295, antiderivative size = 156, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {5907, 14, 730, 720, 724, 206} \[ \frac{a b \left (a^2+a b x+1\right ) \sqrt{a^2+2 a b x+b^2 x^2+1}}{2 \left (a^2+1\right )^2 x^2}-\frac{\left (a^2+2 a b x+b^2 x^2+1\right )^{3/2}}{3 \left (a^2+1\right ) x^3}+\frac{a b^3 \tanh ^{-1}\left (\frac{a^2+a b x+1}{\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}}\right )}{2 \left (a^2+1\right )^{5/2}}-\frac{a}{3 x^3}-\frac{b}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]/x^4,x]

[Out]

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

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 730

Int[((d_.) + (e_.)*(x_))^(m_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)
*(a + b*x + c*x^2)^(p + 1))/((m + 1)*(c*d^2 - b*d*e + a*e^2)), x] + Dist[(2*c*d - b*e)/(2*(c*d^2 - b*d*e + a*e
^2)), Int[(d + e*x)^(m + 1)*(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e, m, p}, 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 + 3, 0]

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^4} \, dx &=\int \frac{a+b x+\sqrt{1+(a+b x)^2}}{x^4} \, dx\\ &=\int \left (\frac{a}{x^4}+\frac{b}{x^3}+\frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x^4}\right ) \, dx\\ &=-\frac{a}{3 x^3}-\frac{b}{2 x^2}+\int \frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x^4} \, dx\\ &=-\frac{a}{3 x^3}-\frac{b}{2 x^2}-\frac{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 \left (1+a^2\right ) x^3}-\frac{(a b) \int \frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x^3} \, dx}{1+a^2}\\ &=-\frac{a}{3 x^3}-\frac{b}{2 x^2}+\frac{a b \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 )^2 x^2}-\frac{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 \left (1+a^2\right ) x^3}-\frac{\left (a b^3\right ) \int \frac{1}{x \sqrt{1+a^2+2 a b x+b^2 x^2}} \, dx}{2 \left (1+a^2\right )^2}\\ &=-\frac{a}{3 x^3}-\frac{b}{2 x^2}+\frac{a b \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 )^2 x^2}-\frac{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 \left (1+a^2\right ) x^3}+\frac{\left (a b^3\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 )}{\left (1+a^2\right )^2}\\ &=-\frac{a}{3 x^3}-\frac{b}{2 x^2}+\frac{a b \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 )^2 x^2}-\frac{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{3 \left (1+a^2\right ) x^3}+\frac{a b^3 \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 )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.106884, size = 162, normalized size = 1.04 \[ \frac{1}{6} \left (-\frac{\sqrt{a^2+2 a b x+b^2 x^2+1} \left (a^2 \left (4-b^2 x^2\right )+a^3 b x+2 a^4+a b x+2 b^2 x^2+2\right )}{\left (a^2+1\right )^2 x^3}+\frac{3 a b^3 \log \left (\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )}{\left (a^2+1\right )^{5/2}}-\frac{3 a b^3 \log (x)}{\left (a^2+1\right )^{5/2}}-\frac{2 a}{x^3}-\frac{3 b}{x^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSinh[a + b*x]/x^4,x]

[Out]

((-2*a)/x^3 - (3*b)/x^2 - (Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(2 + 2*a^4 + a*b*x + a^3*b*x + 2*b^2*x^2 + a^2*(4
- b^2*x^2)))/((1 + a^2)^2*x^3) - (3*a*b^3*Log[x])/(1 + a^2)^(5/2) + (3*a*b^3*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)^(5/2))/6

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Maple [B]  time = 0.009, size = 501, normalized size = 3.2 \begin{align*} -{\frac{1}{ \left ( 3\,{a}^{2}+3 \right ){x}^{3}} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{ab}{2\, \left ({a}^{2}+1 \right ) ^{2}{x}^{2}} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}-{\frac{{a}^{2}{b}^{2}}{2\, \left ({a}^{2}+1 \right ) ^{3}x} \left ({b}^{2}{x}^{2}+2\,xab+{a}^{2}+1 \right ) ^{{\frac{3}{2}}}}+{\frac{{a}^{3}{b}^{3}}{ \left ({a}^{2}+1 \right ) ^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{a}^{4}{b}^{4}}{2\, \left ({a}^{2}+1 \right ) ^{3}}\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}^{3}{b}^{3}}{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{5}{2}}}}+{\frac{{a}^{2}{b}^{4}x}{2\, \left ({a}^{2}+1 \right ) ^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{a}^{2}{b}^{4}}{2\, \left ({a}^{2}+1 \right ) ^{3}}\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{b}^{3}}{2\, \left ({a}^{2}+1 \right ) ^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{a}^{2}{b}^{4}}{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{b}^{3}}{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{b}{2\,{x}^{2}}}-{\frac{a}{3\,{x}^{3}}} \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^4,x)

[Out]

-1/3*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)/(a^2+1)/x^3+1/2*a*b/(a^2+1)^2/x^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-1/2*a^2*b^2
/(a^2+1)^3/x*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)+a^3*b^3/(a^2+1)^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/2*a^4*b^4/(a^2+1)
^3*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^3*b^3/(a^2+1)^(5/2)*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^2*b^4/(a^2+1)^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)
*x+1/2*a^2*b^4/(a^2+1)^3*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*b^3/(a^2+
1)^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/2*a^2*b^4/(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*b^3/(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*b/x^2-1/3*a/x^3

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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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.64109, size = 533, normalized size = 3.42 \begin{align*} \frac{3 \, \sqrt{a^{2} + 1} a b^{3} x^{3} \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 ) - 2 \, a^{7} +{\left (a^{4} - a^{2} - 2\right )} b^{3} x^{3} - 6 \, a^{5} - 6 \, a^{3} - 3 \,{\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} b x -{\left (2 \, a^{6} -{\left (a^{4} - a^{2} - 2\right )} b^{2} x^{2} + 6 \, a^{4} +{\left (a^{5} + 2 \, a^{3} + a\right )} b x + 6 \, a^{2} + 2\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 2 \, a}{6 \,{\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} x^{3}} \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^4,x, algorithm="fricas")

[Out]

1/6*(3*sqrt(a^2 + 1)*a*b^3*x^3*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) - 2*a^7 + (a^4 - a^2 - 2)*b^3*x^3 - 6*a^5 - 6*a^3 - 3*(a^6 + 3*
a^4 + 3*a^2 + 1)*b*x - (2*a^6 - (a^4 - a^2 - 2)*b^2*x^2 + 6*a^4 + (a^5 + 2*a^3 + a)*b*x + 6*a^2 + 2)*sqrt(b^2*
x^2 + 2*a*b*x + a^2 + 1) - 2*a)/((a^6 + 3*a^4 + 3*a^2 + 1)*x^3)

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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^{4}}\, 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**4,x)

[Out]

Integral((a + b*x + sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1))/x**4, 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^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError