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

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

[Out]

-1/4*arcsinh(b*x+a)/x^4-1/8*a*(-2*a^2+3)*b^4*arctanh((1+a*(b*x+a))/(a^2+1)^(1/2)/(1+(b*x+a)^2)^(1/2))/(a^2+1)^
(7/2)-1/12*b*(1+(b*x+a)^2)^(1/2)/(a^2+1)/x^3+5/24*a*b^2*(1+(b*x+a)^2)^(1/2)/(a^2+1)^2/x^2+1/24*(-11*a^2+4)*b^3
*(1+(b*x+a)^2)^(1/2)/(a^2+1)^3/x

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Rubi [A]  time = 0.23, antiderivative size = 167, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.700, 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 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 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 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^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*}

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Mathematica [A]  time = 0.21, size = 179, normalized size = 1.07 \[ \frac {1}{8} \left (-\frac {a \left (2 a^2-3\right ) b^4 \log (x)}{\left (a^2+1\right )^{7/2}}+\frac {a \left (2 a^2-3\right ) b^4 \log \left (\sqrt {a^2+1} \sqrt {a^2+2 a b x+b^2 x^2+1}+a^2+a b x+1\right )}{\left (a^2+1\right )^{7/2}}-\frac {b \sqrt {a^2+2 a b x+b^2 x^2+1} \left (2 a^4-5 a^3 b x+a^2 \left (11 b^2 x^2+4\right )-5 a b x-4 b^2 x^2+2\right )}{3 \left (a^2+1\right )^3 x^3}-\frac {2 \sinh ^{-1}(a+b x)}{x^4}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcSinh[a + b*x]/x^5,x]

[Out]

(-1/3*(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)
))/((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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fricas [B]  time = 0.71, size = 343, normalized size = 2.05 \[ \frac {3 \, {\left (2 \, a^{3} - 3 \, a\right )} \sqrt {a^{2} + 1} b^{4} x^{4} \log \left (-\frac {a^{2} b x + a^{3} + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} {\left (a^{2} + \sqrt {a^{2} + 1} a + 1\right )} + {\left (a b x + a^{2} + 1\right )} \sqrt {a^{2} + 1} + a}{x}\right ) - {\left (11 \, a^{4} + 7 \, a^{2} - 4\right )} b^{4} x^{4} + 6 \, {\left (a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1\right )} x^{4} \log \left (-b x - a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - 6 \, {\left (a^{8} + 4 \, a^{6} - {\left (a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1\right )} x^{4} + 6 \, a^{4} + 4 \, a^{2} + 1\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - {\left ({\left (11 \, a^{4} + 7 \, a^{2} - 4\right )} b^{3} x^{3} - 5 \, {\left (a^{5} + 2 \, a^{3} + a\right )} b^{2} x^{2} + 2 \, {\left (a^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} b x\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{24 \, {\left (a^{8} + 4 \, a^{6} + 6 \, a^{4} + 4 \, a^{2} + 1\right )} x^{4}} \]

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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giac [B]  time = 0.57, size = 709, normalized size = 4.25 \[ -\frac {1}{24} \, b {\left (\frac {3 \, {\left (2 \, a^{3} b^{3} - 3 \, a b^{3}\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^{6} + 3 \, a^{4} + 3 \, a^{2} + 1\right )} \sqrt {a^{2} + 1}} - \frac {2 \, {\left (6 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{5} a^{3} b^{3} - 16 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{3} a^{5} b^{3} + 42 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} a^{7} b^{3} + 12 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{2} a^{6} b^{2} {\left | b \right |} + 20 \, a^{8} b^{2} {\left | b \right |} - 9 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{5} a b^{3} + 8 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{3} a^{3} b^{3} + 93 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} a^{5} b^{3} + 36 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{2} a^{4} b^{2} {\left | b \right |} + 56 \, a^{6} b^{2} {\left | b \right |} + 24 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{3} a b^{3} + 60 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} a^{3} b^{3} + 36 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{2} a^{2} b^{2} {\left | b \right |} + 48 \, a^{4} b^{2} {\left | b \right |} + 9 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )} a b^{3} + 12 \, {\left (x {\left | b \right |} - \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}^{2} b^{2} {\left | b \right |} + 8 \, a^{2} b^{2} {\left | b \right |} - 4 \, b^{2} {\left | b \right |}\right )}}{{\left (a^{6} + 3 \, a^{4} + 3 \, 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 )}^{3}}\right )} - \frac {\log \left (b x + a + \sqrt {{\left (b x + a\right )}^{2} + 1}\right )}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*b*(3*(2*a^3*b^3 - 3*a*b^3)*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^6 + 3*a^4 + 3*a^2 + 1)*sqrt(a^2
+ 1)) - 2*(6*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^5*a^3*b^3 - 16*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x
+ a^2 + 1))^3*a^5*b^3 + 42*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*a^7*b^3 + 12*(x*abs(b) - sqrt(b^2*x^
2 + 2*a*b*x + a^2 + 1))^2*a^6*b^2*abs(b) + 20*a^8*b^2*abs(b) - 9*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)
)^5*a*b^3 + 8*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^3*a^3*b^3 + 93*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x
+ a^2 + 1))*a^5*b^3 + 36*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2*a^4*b^2*abs(b) + 56*a^6*b^2*abs(b)
+ 24*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^3*a*b^3 + 60*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)
)*a^3*b^3 + 36*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2*a^2*b^2*abs(b) + 48*a^4*b^2*abs(b) + 9*(x*abs(
b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))*a*b^3 + 12*(x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))^2*b^2*abs(b)
+ 8*a^2*b^2*abs(b) - 4*b^2*abs(b))/((a^6 + 3*a^4 + 3*a^2 + 1)*((x*abs(b) - sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))
^2 - a^2 - 1)^3)) - 1/4*log(b*x + a + sqrt((b*x + a)^2 + 1))/x^4

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maple [A]  time = 0.01, size = 275, normalized size = 1.65 \[ -\frac {\arcsinh \left (b x +a \right )}{4 x^{4}}-\frac {b \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{12 \left (a^{2}+1\right ) x^{3}}+\frac {5 b^{2} a \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{24 \left (a^{2}+1\right )^{2} x^{2}}-\frac {5 b^{3} a^{2} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{8 \left (a^{2}+1\right )^{3} x}+\frac {5 b^{4} a^{3} \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b x}\right )}{8 \left (a^{2}+1\right )^{\frac {7}{2}}}-\frac {3 b^{4} a \ln \left (\frac {2 a^{2}+2+2 a b x +2 \sqrt {a^{2}+1}\, \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{b x}\right )}{8 \left (a^{2}+1\right )^{\frac {5}{2}}}+\frac {b^{3} \sqrt {b^{2} x^{2}+2 a b x +a^{2}+1}}{6 \left (a^{2}+1\right )^{2} x} \]

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

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maxima [B]  time = 0.36, size = 357, normalized size = 2.14 \[ \frac {1}{24} \, {\left (\frac {15 \, a^{3} b^{3} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{{\left (a^{2} + 1\right )}^{\frac {7}{2}}} - \frac {9 \, a b^{3} \operatorname {arsinh}\left (\frac {2 \, a b x}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2 \, a^{2}}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}} + \frac {2}{\sqrt {-4 \, a^{2} b^{2} + 4 \, {\left (a^{2} + 1\right )} b^{2}} {\left | x \right |}}\right )}{{\left (a^{2} + 1\right )}^{\frac {5}{2}}} - \frac {15 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a^{2} b^{2}}{{\left (a^{2} + 1\right )}^{3} x} + \frac {4 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} b^{2}}{{\left (a^{2} + 1\right )}^{2} x} + \frac {5 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} a b}{{\left (a^{2} + 1\right )}^{2} x^{2}} - \frac {2 \, \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{{\left (a^{2} + 1\right )} x^{3}}\right )} b - \frac {\operatorname {arsinh}\left (b x + a\right )}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(15*a^3*b^3*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)^(7/2) - 9*a*b^3*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) - 15*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a^2*b^2/((a^2 + 1)^3
*x) + 4*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*b^2/((a^2 + 1)^2*x) + 5*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)*a*b/((a^2
+ 1)^2*x^2) - 2*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)/((a^2 + 1)*x^3))*b - 1/4*arcsinh(b*x + a)/x^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {asinh}\left (a+b\,x\right )}{x^5} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(a + b*x)/x^5,x)

[Out]

int(asinh(a + b*x)/x^5, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {asinh}{\left (a + b x \right )}}{x^{5}}\, dx \]

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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