3.357 \(\int \frac{e^{\sinh ^{-1}(a+b x)}}{x^5} \, dx\)

Optimal. Leaf size=207 \[ \frac{\left (1-4 a^2\right ) b^2 \left (a^2+a b x+1\right ) \sqrt{a^2+2 a b x+b^2 x^2+1}}{8 \left (a^2+1\right )^3 x^2}+\frac{5 a b \left (a^2+2 a b x+b^2 x^2+1\right )^{3/2}}{12 \left (a^2+1\right )^2 x^3}-\frac{\left (a^2+2 a b x+b^2 x^2+1\right )^{3/2}}{4 \left (a^2+1\right ) x^4}+\frac{\left (1-4 a^2\right ) b^4 \tanh ^{-1}\left (\frac{a^2+a b x+1}{\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}}\right )}{8 \left (a^2+1\right )^{7/2}}-\frac{a}{4 x^4}-\frac{b}{3 x^3} \]

[Out]

-a/(4*x^4) - b/(3*x^3) + ((1 - 4*a^2)*b^2*(1 + a^2 + a*b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])/(8*(1 + a^2)^3*
x^2) - (1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(4*(1 + a^2)*x^4) + (5*a*b*(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(12
*(1 + a^2)^2*x^3) + ((1 - 4*a^2)*b^4*ArcTanh[(1 + a^2 + a*b*x)/(Sqrt[1 + a^2]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2
])])/(8*(1 + a^2)^(7/2))

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Rubi [A]  time = 0.170881, antiderivative size = 207, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.583, Rules used = {5907, 14, 744, 806, 720, 724, 206} \[ \frac{\left (1-4 a^2\right ) b^2 \left (a^2+a b x+1\right ) \sqrt{a^2+2 a b x+b^2 x^2+1}}{8 \left (a^2+1\right )^3 x^2}+\frac{5 a b \left (a^2+2 a b x+b^2 x^2+1\right )^{3/2}}{12 \left (a^2+1\right )^2 x^3}-\frac{\left (a^2+2 a b x+b^2 x^2+1\right )^{3/2}}{4 \left (a^2+1\right ) x^4}+\frac{\left (1-4 a^2\right ) b^4 \tanh ^{-1}\left (\frac{a^2+a b x+1}{\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}}\right )}{8 \left (a^2+1\right )^{7/2}}-\frac{a}{4 x^4}-\frac{b}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[E^ArcSinh[a + b*x]/x^5,x]

[Out]

-a/(4*x^4) - b/(3*x^3) + ((1 - 4*a^2)*b^2*(1 + a^2 + a*b*x)*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2])/(8*(1 + a^2)^3*
x^2) - (1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2)/(4*(1 + a^2)*x^4) + (5*a*b*(1 + a^2 + 2*a*b*x + b^2*x^2)^(3/2))/(12
*(1 + a^2)^2*x^3) + ((1 - 4*a^2)*b^4*ArcTanh[(1 + a^2 + a*b*x)/(Sqrt[1 + a^2]*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2
])])/(8*(1 + a^2)^(7/2))

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 744

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[1/((m + 1)*(c*d^2 - b*d*e + a*e^2)),
Int[(d + e*x)^(m + 1)*Simp[c*d*(m + 1) - b*e*(m + p + 2) - c*e*(m + 2*p + 3)*x, x]*(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] && NeQ[m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, b, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ
[p]) || ILtQ[Simplify[m + 2*p + 3], 0])

Rule 806

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Si
mp[((e*f - d*g)*(d + e*x)^(m + 1)*(a + b*x + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 - b*d*e + a*e^2)), x] - Dist[(b
*(e*f + d*g) - 2*(c*d*f + a*e*g))/(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, f, g, m, p}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && EqQ[Sim
plify[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^5} \, dx &=\int \frac{a+b x+\sqrt{1+(a+b x)^2}}{x^5} \, dx\\ &=\int \left (\frac{a}{x^5}+\frac{b}{x^4}+\frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x^5}\right ) \, dx\\ &=-\frac{a}{4 x^4}-\frac{b}{3 x^3}+\int \frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x^5} \, dx\\ &=-\frac{a}{4 x^4}-\frac{b}{3 x^3}-\frac{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 \left (1+a^2\right ) x^4}-\frac{\int \frac{\left (5 a b+b^2 x\right ) \sqrt{1+a^2+2 a b x+b^2 x^2}}{x^4} \, dx}{4 \left (1+a^2\right )}\\ &=-\frac{a}{4 x^4}-\frac{b}{3 x^3}-\frac{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 \left (1+a^2\right ) x^4}+\frac{5 a b \left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{12 \left (1+a^2\right )^2 x^3}-\frac{\left (\left (1-4 a^2\right ) b^2\right ) \int \frac{\sqrt{1+a^2+2 a b x+b^2 x^2}}{x^3} \, dx}{4 \left (1+a^2\right )^2}\\ &=-\frac{a}{4 x^4}-\frac{b}{3 x^3}+\frac{\left (1-4 a^2\right ) b^2 \left (1+a^2+a b x\right ) \sqrt{1+a^2+2 a b x+b^2 x^2}}{8 \left (1+a^2\right )^3 x^2}-\frac{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 \left (1+a^2\right ) x^4}+\frac{5 a b \left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{12 \left (1+a^2\right )^2 x^3}-\frac{\left (\left (1-4 a^2\right ) b^4\right ) \int \frac{1}{x \sqrt{1+a^2+2 a b x+b^2 x^2}} \, dx}{8 \left (1+a^2\right )^3}\\ &=-\frac{a}{4 x^4}-\frac{b}{3 x^3}+\frac{\left (1-4 a^2\right ) b^2 \left (1+a^2+a b x\right ) \sqrt{1+a^2+2 a b x+b^2 x^2}}{8 \left (1+a^2\right )^3 x^2}-\frac{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 \left (1+a^2\right ) x^4}+\frac{5 a b \left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{12 \left (1+a^2\right )^2 x^3}+\frac{\left (\left (1-4 a^2\right ) b^4\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 )}{4 \left (1+a^2\right )^3}\\ &=-\frac{a}{4 x^4}-\frac{b}{3 x^3}+\frac{\left (1-4 a^2\right ) b^2 \left (1+a^2+a b x\right ) \sqrt{1+a^2+2 a b x+b^2 x^2}}{8 \left (1+a^2\right )^3 x^2}-\frac{\left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{4 \left (1+a^2\right ) x^4}+\frac{5 a b \left (1+a^2+2 a b x+b^2 x^2\right )^{3/2}}{12 \left (1+a^2\right )^2 x^3}+\frac{\left (1-4 a^2\right ) b^4 \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 )}{8 \left (1+a^2\right )^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.73903, size = 192, normalized size = 0.93 \[ \frac{1}{24} \left (-\frac{\sqrt{a^2+2 a b x+b^2 x^2+1} \left (\frac{a \left (2 a^2-13\right ) b^3 x^3}{\left (a^2+1\right )^3}-\frac{\left (2 a^2-3\right ) b^2 x^2}{\left (a^2+1\right )^2}+\frac{2 a b x}{a^2+1}+6\right )}{x^4}-\frac{3 (2 a-1) (2 a+1) b^4 \log \left (\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )}{\left (a^2+1\right )^{7/2}}+\frac{3 (2 a-1) (2 a+1) b^4 \log (x)}{\left (a^2+1\right )^{7/2}}-\frac{6 a}{x^4}-\frac{8 b}{x^3}\right ) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcSinh[a + b*x]/x^5,x]

[Out]

((-6*a)/x^4 - (8*b)/x^3 - (Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(6 + (2*a*b*x)/(1 + a^2) - ((-3 + 2*a^2)*b^2*x^2)
/(1 + a^2)^2 + (a*(-13 + 2*a^2)*b^3*x^3)/(1 + a^2)^3))/x^4 + (3*(-1 + 2*a)*(1 + 2*a)*b^4*Log[x])/(1 + a^2)^(7/
2) - (3*(-1 + 2*a)*(1 + 2*a)*b^4*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)^(7/2))/24

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Maple [B]  time = 0.01, size = 841, normalized size = 4.1 \begin{align*} \text{result too large to display} \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^5,x)

[Out]

-1/4*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)/(a^2+1)/x^4+5/12*a*b*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)/(a^2+1)^2/x^3-5/8*a^2*b^
2/(a^2+1)^3/x^2*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)+5/8*a^3*b^3/(a^2+1)^4/x*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)-5/4*a^4*b^
4/(a^2+1)^4*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-5/8*a^5*b^5/(a^2+1)^4*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)+5/8*a^4*b^4/(a^2+1)^(7/2)*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)-5/8*a^3*b^5/(a^2+1)^4*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*x-5/8*a^3*b^5/(a^2+1)^4*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)+7/8*a^2*b^4/(a^2+1)^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+3/4*a^3*b^5/
(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)-3/4*a^2*b^4/(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)+1/8*b^2/(a^2+1)^2/x^2*(b^2*x^2+2*a*b*x+a^2+1
)^(3/2)-1/8*b^3/(a^2+1)^3*a/x*(b^2*x^2+2*a*b*x+a^2+1)^(3/2)+1/8*b^5/(a^2+1)^3*a*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)*
x+1/8*b^5/(a^2+1)^3*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)-1/8*b^4/(a^2+1)^2*
(b^2*x^2+2*a*b*x+a^2+1)^(1/2)-1/8*b^5/(a^2+1)^2*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)+1/8*b^4/(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)-1/4*a/x^
4-1/3*b/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^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.87863, size = 683, normalized size = 3.3 \begin{align*} \frac{3 \,{\left (4 \, a^{2} - 1\right )} \sqrt{a^{2} + 1} b^{4} x^{4} \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 ) - 6 \, a^{9} -{\left (2 \, a^{5} - 11 \, a^{3} - 13 \, a\right )} b^{4} x^{4} - 24 \, a^{7} - 36 \, a^{5} - 24 \, a^{3} - 8 \,{\left (a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1\right )} b x -{\left (6 \, a^{8} +{\left (2 \, a^{5} - 11 \, a^{3} - 13 \, a\right )} b^{3} x^{3} + 24 \, a^{6} -{\left (2 \, a^{6} + a^{4} - 4 \, a^{2} - 3\right )} b^{2} x^{2} + 36 \, a^{4} + 2 \,{\left (a^{7} + 3 \, a^{5} + 3 \, a^{3} + a\right )} b x + 24 \, a^{2} + 6\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 6 \, a}{24 \,{\left (a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1\right )} x^{4}} \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^5,x, algorithm="fricas")

[Out]

1/24*(3*(4*a^2 - 1)*sqrt(a^2 + 1)*b^4*x^4*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) - 6*a^9 - (2*a^5 - 11*a^3 - 13*a)*b^4*x^4 - 24*a^7 -
36*a^5 - 24*a^3 - 8*(a^8 + 4*a^6 + 6*a^4 + 4*a^2 + 1)*b*x - (6*a^8 + (2*a^5 - 11*a^3 - 13*a)*b^3*x^3 + 24*a^6
- (2*a^6 + a^4 - 4*a^2 - 3)*b^2*x^2 + 36*a^4 + 2*(a^7 + 3*a^5 + 3*a^3 + a)*b*x + 24*a^2 + 6)*sqrt(b^2*x^2 + 2
*a*b*x + a^2 + 1) - 6*a)/((a^8 + 4*a^6 + 6*a^4 + 4*a^2 + 1)*x^4)

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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^{5}}\, 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**5,x)

[Out]

Integral((a + b*x + sqrt(a**2 + 2*a*b*x + b**2*x**2 + 1))/x**5, 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^5,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError