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

Optimal. Leaf size=129 \[ \frac{a b^2 \sqrt{(a+b x)^2+1}}{2 \left (a^2+1\right )^2 x}+\frac{\left (1-2 a^2\right ) b^3 \tanh ^{-1}\left (\frac{a (a+b x)+1}{\sqrt{a^2+1} \sqrt{(a+b x)^2+1}}\right )}{6 \left (a^2+1\right )^{5/2}}-\frac{b \sqrt{(a+b x)^2+1}}{6 \left (a^2+1\right ) x^2}-\frac{\sinh ^{-1}(a+b x)}{3 x^3} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.154325, antiderivative size = 129, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {5865, 5801, 745, 807, 725, 206} \[ \frac{a b^2 \sqrt{(a+b x)^2+1}}{2 \left (a^2+1\right )^2 x}+\frac{\left (1-2 a^2\right ) b^3 \tanh ^{-1}\left (\frac{a (a+b x)+1}{\sqrt{a^2+1} \sqrt{(a+b x)^2+1}}\right )}{6 \left (a^2+1\right )^{5/2}}-\frac{b \sqrt{(a+b x)^2+1}}{6 \left (a^2+1\right ) x^2}-\frac{\sinh ^{-1}(a+b x)}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[ArcSinh[a + b*x]/x^4,x]

[Out]

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

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 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^4} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\sinh ^{-1}(x)}{\left (-\frac{a}{b}+\frac{x}{b}\right )^4} \, dx,x,a+b x\right )}{b}\\ &=-\frac{\sinh ^{-1}(a+b x)}{3 x^3}+\frac{1}{3} \operatorname{Subst}\left (\int \frac{1}{\left (-\frac{a}{b}+\frac{x}{b}\right )^3 \sqrt{1+x^2}} \, dx,x,a+b x\right )\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{6 \left (1+a^2\right ) x^2}-\frac{\sinh ^{-1}(a+b x)}{3 x^3}-\frac{b^2 \operatorname{Subst}\left (\int \frac{\frac{2 a}{b}+\frac{x}{b}}{\left (-\frac{a}{b}+\frac{x}{b}\right )^2 \sqrt{1+x^2}} \, dx,x,a+b x\right )}{6 \left (1+a^2\right )}\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{6 \left (1+a^2\right ) x^2}+\frac{a b^2 \sqrt{1+(a+b x)^2}}{2 \left (1+a^2\right )^2 x}-\frac{\sinh ^{-1}(a+b x)}{3 x^3}-\frac{\left (\left (1-2 a^2\right ) b^2\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 )}{6 \left (1+a^2\right )^2}\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{6 \left (1+a^2\right ) x^2}+\frac{a b^2 \sqrt{1+(a+b x)^2}}{2 \left (1+a^2\right )^2 x}-\frac{\sinh ^{-1}(a+b x)}{3 x^3}+\frac{\left (\left (1-2 a^2\right ) b^2\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 )}{6 \left (1+a^2\right )^2}\\ &=-\frac{b \sqrt{1+(a+b x)^2}}{6 \left (1+a^2\right ) x^2}+\frac{a b^2 \sqrt{1+(a+b x)^2}}{2 \left (1+a^2\right )^2 x}-\frac{\sinh ^{-1}(a+b x)}{3 x^3}+\frac{\left (1-2 a^2\right ) b^3 \tanh ^{-1}\left (\frac{1+a (a+b x)}{\sqrt{1+a^2} \sqrt{1+(a+b x)^2}}\right )}{6 \left (1+a^2\right )^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.217009, size = 149, normalized size = 1.16 \[ \frac{-\sqrt{a^2+1} b x \left (a^2-3 a b x+1\right ) \sqrt{a^2+2 a b x+b^2 x^2+1}+\left (2 a^2-1\right ) b^3 x^3 \log (x)+\left (1-2 a^2\right ) b^3 x^3 \log \left (\sqrt{a^2+1} \sqrt{a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )-2 \left (a^2+1\right )^{5/2} \sinh ^{-1}(a+b x)}{6 \left (a^2+1\right )^{5/2} x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]/x^4,x]

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.009, size = 203, normalized size = 1.6 \begin{align*} -{\frac{{\it Arcsinh} \left ( bx+a \right ) }{3\,{x}^{3}}}-{\frac{b}{ \left ( 6\,{a}^{2}+6 \right ){x}^{2}}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}+{\frac{{b}^{2}a}{2\, \left ({a}^{2}+1 \right ) ^{2}x}\sqrt{{b}^{2}{x}^{2}+2\,xab+{a}^{2}+1}}-{\frac{{b}^{3}{a}^{2}}{2}\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}\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{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arcsinh(b*x+a)/x^4,x)

[Out]

-1/3*arcsinh(b*x+a)/x^3-1/6*b/(a^2+1)/x^2*(b^2*x^2+2*a*b*x+a^2+1)^(1/2)+1/2*b^2*a/(a^2+1)^2/x*(b^2*x^2+2*a*b*x
+a^2+1)^(1/2)-1/2*b^3*a^2/(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)^(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))/b/x)

________________________________________________________________________________________

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^4,x, algorithm="maxima")

[Out]

Exception raised: ValueError

________________________________________________________________________________________

Fricas [B]  time = 2.90232, size = 672, normalized size = 5.21 \begin{align*} \frac{{\left (2 \, a^{2} - 1\right )} \sqrt{a^{2} + 1} 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 ) + 3 \,{\left (a^{3} + a\right )} b^{3} x^{3} + 2 \,{\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} x^{3} \log \left (-b x - a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 2 \,{\left (a^{6} + 3 \, a^{4} -{\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} x^{3} + 3 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) +{\left (3 \,{\left (a^{3} + a\right )} b^{2} x^{2} -{\left (a^{4} + 2 \, a^{2} + 1\right )} b x\right )} \sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{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(arcsinh(b*x+a)/x^4,x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\operatorname{asinh}{\left (a + b x \right )}}{x^{4}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(asinh(b*x+a)/x**4,x)

[Out]

Integral(asinh(a + b*x)/x**4, x)

________________________________________________________________________________________

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^4,x, algorithm="giac")

[Out]

Exception raised: NotImplementedError