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

Optimal. Leaf size=167 \[ \frac{5 a b^2 \sqrt{(a+b x)^2+1}}{24 \left (a^2+1\right )^2 x^2}+\frac{\left (4-11 a^2\right ) b^3 \sqrt{(a+b x)^2+1}}{24 \left (a^2+1\right )^3 x}-\frac{a \left (3-2 a^2\right ) b^4 \tanh ^{-1}\left (\frac{a (a+b x)+1}{\sqrt{a^2+1} \sqrt{(a+b x)^2+1}}\right )}{8 \left (a^2+1\right )^{7/2}}-\frac{b \sqrt{(a+b x)^2+1}}{12 \left (a^2+1\right ) x^3}-\frac{\sinh ^{-1}(a+b x)}{4 x^4} \]

[Out]

-(b*Sqrt[1 + (a + b*x)^2])/(12*(1 + a^2)*x^3) + (5*a*b^2*Sqrt[1 + (a + b*x)^2])/(24*(1 + a^2)^2*x^2) + ((4 - 1
1*a^2)*b^3*Sqrt[1 + (a + b*x)^2])/(24*(1 + a^2)^3*x) - ArcSinh[a + b*x]/(4*x^4) - (a*(3 - 2*a^2)*b^4*ArcTanh[(
1 + a*(a + b*x))/(Sqrt[1 + a^2]*Sqrt[1 + (a + b*x)^2])])/(8*(1 + a^2)^(7/2))

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Rubi [A]  time = 0.228429, antiderivative size = 167, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {5865, 5801, 745, 835, 807, 725, 206} \[ \frac{5 a b^2 \sqrt{(a+b x)^2+1}}{24 \left (a^2+1\right )^2 x^2}+\frac{\left (4-11 a^2\right ) b^3 \sqrt{(a+b x)^2+1}}{24 \left (a^2+1\right )^3 x}-\frac{a \left (3-2 a^2\right ) b^4 \tanh ^{-1}\left (\frac{a (a+b x)+1}{\sqrt{a^2+1} \sqrt{(a+b x)^2+1}}\right )}{8 \left (a^2+1\right )^{7/2}}-\frac{b \sqrt{(a+b x)^2+1}}{12 \left (a^2+1\right ) x^3}-\frac{\sinh ^{-1}(a+b x)}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]/x^5,x]

[Out]

-(b*Sqrt[1 + (a + b*x)^2])/(12*(1 + a^2)*x^3) + (5*a*b^2*Sqrt[1 + (a + b*x)^2])/(24*(1 + a^2)^2*x^2) + ((4 - 1
1*a^2)*b^3*Sqrt[1 + (a + b*x)^2])/(24*(1 + a^2)^3*x) - ArcSinh[a + b*x]/(4*x^4) - (a*(3 - 2*a^2)*b^4*ArcTanh[(
1 + a*(a + b*x))/(Sqrt[1 + a^2]*Sqrt[1 + (a + b*x)^2])])/(8*(1 + a^2)^(7/2))

Rule 5865

Int[((a_.) + ArcSinh[(c_) + (d_.)*(x_)]*(b_.))^(n_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[
Int[((d*e - c*f)/d + (f*x)/d)^m*(a + b*ArcSinh[x])^n, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x
]

Rule 5801

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

Rule 745

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1)*(a + c*x^2)^(p
+ 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[c/((m + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^(m + 1)*Simp[d*(m + 1)
- e*(m + 2*p + 3)*x, x]*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, m, p}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[
m, -1] && ((LtQ[m, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]) || (SumSimplerQ[m, 1] && IntegerQ[p]) || ILtQ
[Simplify[m + 2*p + 3], 0])

Rule 835

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[((e*f - d*g)
*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/((m + 1)*(c*d^2 + a*e^2)), x] + Dist[1/((m + 1)*(c*d^2 + a*e^2)), Int[
(d + e*x)^(m + 1)*(a + c*x^2)^p*Simp[(c*d*f + a*e*g)*(m + 1) - c*(e*f - d*g)*(m + 2*p + 3)*x, x], x], x] /; Fr
eeQ[{a, c, d, e, f, g, p}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1] && (IntegerQ[m] || IntegerQ[p] || Integer
sQ[2*m, 2*p])

Rule 807

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> -Simp[((e*f - d*g
)*(d + e*x)^(m + 1)*(a + c*x^2)^(p + 1))/(2*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[(c*d*f + a*e*g)/(c*d^2 + a*e^2
), Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p}, x] && NeQ[c*d^2 + a*e^2, 0]
&& EqQ[Simplify[m + 2*p + 3], 0]

Rule 725

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

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{\sinh ^{-1}(a+b x)}{x^5} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{\left (-\frac{a}{b}+\frac{x}{b}\right )^5} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\sinh ^{-1}(a+b x)}{4 x^4}+\frac{1}{4} \operatorname{Subst}\left (\int \frac{1}{\left (-\frac{a}{b}+\frac{x}{b}\right )^4 \sqrt{1+x^2}} \, dx,x,a+b x\right )\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{12 \left (1+a^2\right ) x^3}-\frac{\sinh ^{-1}(a+b x)}{4 x^4}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\frac{3 a}{b}+\frac{2 x}{b}}{\left (-\frac{a}{b}+\frac{x}{b}\right )^3 \sqrt{1+x^2}} \, dx,x,a+b x\right )}{12 \left (1+a^2\right )}\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{12 \left (1+a^2\right ) x^3}+\frac{5 a b^2 \sqrt{1+(a+b x)^2}}{24 \left (1+a^2\right )^2 x^2}-\frac{\sinh ^{-1}(a+b x)}{4 x^4}+\frac{b^4 \operatorname{Subst}\left (\int \frac{-\frac{2 \left (2-3 a^2\right )}{b^2}+\frac{5 a x}{b^2}}{\left (-\frac{a}{b}+\frac{x}{b}\right )^2 \sqrt{1+x^2}} \, dx,x,a+b x\right )}{24 \left (1+a^2\right )^2}\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{12 \left (1+a^2\right ) x^3}+\frac{5 a b^2 \sqrt{1+(a+b x)^2}}{24 \left (1+a^2\right )^2 x^2}+\frac{\left (4-11 a^2\right ) b^3 \sqrt{1+(a+b x)^2}}{24 \left (1+a^2\right )^3 x}-\frac{\sinh ^{-1}(a+b x)}{4 x^4}+\frac{\left (a \left (3-2 a^2\right ) b^3\right ) \operatorname{Subst}\left (\int \frac{1}{\left (-\frac{a}{b}+\frac{x}{b}\right ) \sqrt{1+x^2}} \, dx,x,a+b x\right )}{8 \left (1+a^2\right )^3}\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{12 \left (1+a^2\right ) x^3}+\frac{5 a b^2 \sqrt{1+(a+b x)^2}}{24 \left (1+a^2\right )^2 x^2}+\frac{\left (4-11 a^2\right ) b^3 \sqrt{1+(a+b x)^2}}{24 \left (1+a^2\right )^3 x}-\frac{\sinh ^{-1}(a+b x)}{4 x^4}-\frac{\left (a \left (3-2 a^2\right ) b^3\right ) \operatorname{Subst}\left (\int \frac{1}{\frac{1}{b^2}+\frac{a^2}{b^2}-x^2} \, dx,x,\frac{\frac{1}{b}+\frac{a (a+b x)}{b}}{\sqrt{1+(a+b x)^2}}\right )}{8 \left (1+a^2\right )^3}\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{12 \left (1+a^2\right ) x^3}+\frac{5 a b^2 \sqrt{1+(a+b x)^2}}{24 \left (1+a^2\right )^2 x^2}+\frac{\left (4-11 a^2\right ) b^3 \sqrt{1+(a+b x)^2}}{24 \left (1+a^2\right )^3 x}-\frac{\sinh ^{-1}(a+b x)}{4 x^4}-\frac{a \left (3-2 a^2\right ) b^4 \tanh ^{-1}\left (\frac{1+a (a+b x)}{\sqrt{1+a^2} \sqrt{1+(a+b x)^2}}\right )}{8 \left (1+a^2\right )^{7/2}}\\ \end{align*}

Mathematica [A]  time = 0.200879, size = 179, normalized size = 1.07 \[ \frac{1}{8} \left (-\frac{b \sqrt{a^2+2 a b x+b^2 x^2+1} \left (a^2 \left (11 b^2 x^2+4\right )-5 a^3 b x+2 a^4-5 a b x-4 b^2 x^2+2\right )}{3 \left (a^2+1\right )^3 x^3}+\frac{a \left (2 a^2-3\right ) b^4 \log \left (\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )}{\left (a^2+1\right )^{7/2}}-\frac{a \left (2 a^2-3\right ) b^4 \log (x)}{\left (a^2+1\right )^{7/2}}-\frac{2 \sinh ^{-1}(a+b x)}{x^4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]/x^5,x]

[Out]

(-(b*Sqrt[1 + a^2 + 2*a*b*x + b^2*x^2]*(2 + 2*a^4 - 5*a*b*x - 5*a^3*b*x - 4*b^2*x^2 + a^2*(4 + 11*b^2*x^2)))/(
3*(1 + a^2)^3*x^3) - (2*ArcSinh[a + b*x])/x^4 - (a*(-3 + 2*a^2)*b^4*Log[x])/(1 + a^2)^(7/2) + (a*(-3 + 2*a^2)*
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))/8

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Maple [A]  time = 0.007, size = 275, normalized size = 1.7 \begin{align*} -{\frac{{\it Arcsinh} \left ( bx+a \right ) }{4\,{x}^{4}}}-{\frac{b}{ \left ( 12\,{a}^{2}+12 \right ){x}^{3}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{5\,{b}^{2}a}{24\, \left ({a}^{2}+1 \right ) ^{2}{x}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{5\,{b}^{3}{a}^{2}}{8\, \left ({a}^{2}+1 \right ) ^{3}x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{5\,{b}^{4}{a}^{3}}{8}\ln \left ({\frac{1}{bx} \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{7}{2}}}}-{\frac{3\,{b}^{4}a}{8}\ln \left ({\frac{1}{bx} \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{{b}^{3}}{6\, \left ({a}^{2}+1 \right ) ^{2}x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(b*x+a)/x^5,x)

[Out]

-1/4*arcsinh(b*x+a)/x^4-1/12*b/(a^2+1)/x^3*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+5/24*b^2*a/(a^2+1)^2/x^2*(b^2*x^2+2*a
*b*x+a^2+1)^(1/2)-5/8*b^3*a^2/(a^2+1)^3/x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+5/8*b^4*a^3/(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))/b/x)-3/8*b^4*a/(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))/b/x)+1/6*b^3/(a^2+1)^2/x*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)

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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(arcsinh(b*x+a)/x^5,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.0567, size = 801, normalized size = 4.8 \begin{align*} \frac{3 \,{\left (2 \, a^{3} - 3 \, a\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 ) -{\left (11 \, a^{4} + 7 \, a^{2} - 4\right )} b^{4} x^{4} + 6 \,{\left (a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1\right )} x^{4} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 6 \,{\left (a^{8} + 4 \, a^{6} -{\left (a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1\right )} x^{4} + 6 \, a^{4} + 4 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) -{\left ({\left (11 \, a^{4} + 7 \, a^{2} - 4\right )} b^{3} x^{3} - 5 \,{\left (a^{5} + 2 \, a^{3} + a\right )} b^{2} x^{2} + 2 \,{\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} b x\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{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(arcsinh(b*x+a)/x^5,x, algorithm="fricas")

[Out]

1/24*(3*(2*a^3 - 3*a)*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 + sqr
t(a^2 + 1)*a + 1) + (a*b*x + a^2 + 1)*sqrt(a^2 + 1) + a)/x) - (11*a^4 + 7*a^2 - 4)*b^4*x^4 + 6*(a^8 + 4*a^6 +
6*a^4 + 4*a^2 + 1)*x^4*log(-b*x - a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - 6*(a^8 + 4*a^6 - (a^8 + 4*a^6 + 6*a
^4 + 4*a^2 + 1)*x^4 + 6*a^4 + 4*a^2 + 1)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - ((11*a^4 + 7*a^2 -
4)*b^3*x^3 - 5*(a^5 + 2*a^3 + a)*b^2*x^2 + 2*(a^6 + 3*a^4 + 3*a^2 + 1)*b*x)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)
)/((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{\operatorname{asinh}{\left (a + b x \right )}}{x^{5}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(b*x+a)/x**5,x)

[Out]

Integral(asinh(a + b*x)/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(arcsinh(b*x+a)/x^5,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError