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

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

[Out]

-1/3*arcsinh(b*x+a)/x^3+1/6*(-2*a^2+1)*b^3*arctanh((1+a*(b*x+a))/(a^2+1)^(1/2)/(1+(b*x+a)^2)^(1/2))/(a^2+1)^(5
/2)-1/6*b*(1+(b*x+a)^2)^(1/2)/(a^2+1)/x^2+1/2*a*b^2*(1+(b*x+a)^2)^(1/2)/(a^2+1)^2/x

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Rubi [A]  time = 0.15, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.600, 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 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])

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 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 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 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
]

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*}

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Mathematica [A]  time = 0.24, size = 149, normalized size = 1.16 \[ \frac {\left (2 a^2-1\right ) b^3 x^3 \log (x)-\sqrt {a^2+1} b x \left (a^2-3 a b x+1\right ) \sqrt {a^2+2 a b x+b^2 x^2+1}+\left (1-2 a^2\right ) b^3 x^3 \log \left (\sqrt {a^2+1} \sqrt {a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )-2 \left (a^2+1\right )^{5/2} \sinh ^{-1}(a+b x)}{6 \left (a^2+1\right )^{5/2} x^3} \]

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)

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fricas [B]  time = 0.72, size = 285, normalized size = 2.21 \[ \frac {{\left (2 \, a^{2} - 1\right )} \sqrt {a^{2} + 1} b^{3} x^{3} \log \left (-\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} - \sqrt {a^{2} + 1} a + 1\right )} - {\left (a b x + a^{2} + 1\right )} \sqrt {a^{2} + 1} + a}{x}\right ) + 3 \, {\left (a^{3} + a\right )} b^{3} x^{3} + 2 \, {\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} x^{3} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 2 \, {\left (a^{6} + 3 \, a^{4} - {\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} x^{3} + 3 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) + {\left (3 \, {\left (a^{3} + a\right )} b^{2} x^{2} - {\left (a^{4} + 2 \, a^{2} + 1\right )} b x\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{6 \, {\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} x^{3}} \]

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)

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giac [B]  time = 0.47, size = 381, normalized size = 2.95 \[ \frac {1}{6} \, b {\left (\frac {{\left (2 \, a^{2} b^{2} - b^{2}\right )} \log \left (\frac {{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} - 2 \, \sqrt {a^{2} + 1} \right |}}{{\left | -2 \, x {\left | b \right |} + 2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 2 \, \sqrt {a^{2} + 1} \right |}}\right )}{{\left (a^{4} + 2 \, a^{2} + 1\right )} \sqrt {a^{2} + 1}} - \frac {2 \, {\left (2 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{3} a^{2} b^{2} - 6 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} a^{4} b^{2} - 4 \, a^{5} b {\left | b \right |} - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{3} b^{2} - 7 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} a^{2} b^{2} - 8 \, a^{3} b {\left | b \right |} - {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} b^{2} - 4 \, a b {\left | b \right |}\right )}}{{\left (a^{4} + 2 \, a^{2} + 1\right )} {\left ({\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{2} - a^{2} - 1\right )}^{2}}\right )} - \frac {\log \left (b x + a + \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)/x^4,x, algorithm="giac")

[Out]

1/6*b*((2*a^2*b^2 - b^2)*log(abs(-2*x*abs(b) + 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*sqrt(a^2 + 1))/abs(-2*x
*abs(b) + 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1) + 2*sqrt(a^2 + 1)))/((a^4 + 2*a^2 + 1)*sqrt(a^2 + 1)) - 2*(2*(x*
abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^3*a^2*b^2 - 6*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*a^4*b
^2 - 4*a^5*b*abs(b) - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^3*b^2 - 7*(x*abs(b) - sqrt(b^2*x^2 + 2*a*
b*x + a^2 + 1))*a^2*b^2 - 8*a^3*b*abs(b) - (x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*b^2 - 4*a*b*abs(b))/
((a^4 + 2*a^2 + 1)*((x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2 - a^2 - 1)^2)) - 1/3*log(b*x + a + sqrt((
b*x + a)^2 + 1))/x^3

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maple [A]  time = 0.01, size = 203, normalized size = 1.57 \[ -\frac {\arcsinh \left (b x +a \right )}{3 x^{3}}-\frac {b \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{6 \left (a^{2}+1\right ) x^{2}}+\frac {b^{2} a \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{2 \left (a^{2}+1\right )^{2} x}-\frac {b^{3} a^{2} \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b x}\right )}{2 \left (a^{2}+1\right )^{\frac {5}{2}}}+\frac {b^{3} \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b x}\right )}{6 \left (a^{2}+1\right )^{\frac {3}{2}}} \]

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*a*b*x+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*a*b*x+2*(a^2+1)^(1/2)*(b^2*x^2+2*a*b*x+a^2+1)^(1/2))/b/x)

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maxima [B]  time = 0.36, size = 284, normalized size = 2.20 \[ -\frac {1}{6} \, {\left (\frac {3 \, a^{2} b^{2} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{{\left (a^{2} + 1\right )}^{\frac {5}{2}}} - \frac {b^{2} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{{\left (a^{2} + 1\right )}^{\frac {3}{2}}} - \frac {3 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a b}{{\left (a^{2} + 1\right )}^{2} x} + \frac {\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (a^{2} + 1\right )} x^{2}}\right )} b - \frac {\operatorname {arsinh}\left (b x + a\right )}{3 \, x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arcsinh(b*x+a)/x^4,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {asinh}\left (a+b\,x\right )}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a + b*x)/x^4,x)

[Out]

int(asinh(a + b*x)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}{\left (a + b x \right )}}{x^{4}}\, dx \]

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)

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