3.6 \(\int \frac {\text {csch}^{-1}(a+b x)}{x^3} \, dx\)

Optimal. Leaf size=114 \[ \frac {b^2 \text {csch}^{-1}(a+b x)}{2 a^2}-\frac {\left (2 a^2+1\right ) b^2 \tanh ^{-1}\left (\frac {\tanh \left (\frac {1}{2} \text {csch}^{-1}(a+b x)\right )+a}{\sqrt {a^2+1}}\right )}{a^2 \left (a^2+1\right )^{3/2}}+\frac {b (a+b x) \sqrt {\frac {1}{(a+b x)^2}+1}}{2 a \left (a^2+1\right ) x}-\frac {\text {csch}^{-1}(a+b x)}{2 x^2} \]

[Out]

1/2*b^2*arccsch(b*x+a)/a^2-1/2*arccsch(b*x+a)/x^2-(2*a^2+1)*b^2*arctanh((a+tanh(1/2*arccsch(b*x+a)))/(a^2+1)^(
1/2))/a^2/(a^2+1)^(3/2)+1/2*b*(b*x+a)*(1+1/(b*x+a)^2)^(1/2)/a/(a^2+1)/x

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Rubi [A]  time = 0.21, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.800, Rules used = {6322, 5469, 3785, 3919, 3831, 2660, 618, 206} \[ \frac {b^2 \text {csch}^{-1}(a+b x)}{2 a^2}-\frac {\left (2 a^2+1\right ) b^2 \tanh ^{-1}\left (\frac {\tanh \left (\frac {1}{2} \text {csch}^{-1}(a+b x)\right )+a}{\sqrt {a^2+1}}\right )}{a^2 \left (a^2+1\right )^{3/2}}+\frac {b (a+b x) \sqrt {\frac {1}{(a+b x)^2}+1}}{2 a \left (a^2+1\right ) x}-\frac {\text {csch}^{-1}(a+b x)}{2 x^2} \]

Antiderivative was successfully verified.

[In]

Int[ArcCsch[a + b*x]/x^3,x]

[Out]

(b*(a + b*x)*Sqrt[1 + (a + b*x)^(-2)])/(2*a*(1 + a^2)*x) + (b^2*ArcCsch[a + b*x])/(2*a^2) - ArcCsch[a + b*x]/(
2*x^2) - ((1 + 2*a^2)*b^2*ArcTanh[(a + Tanh[ArcCsch[a + b*x]/2])/Sqrt[1 + a^2]])/(a^2*(1 + a^2)^(3/2))

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 618

Int[((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> Dist[-2, Subst[Int[1/Simp[b^2 - 4*a*c - x^2, x], x]
, x, b + 2*c*x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 2660

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{e = FreeFactors[Tan[(c + d*x)/2], x]}, Dis
t[(2*e)/d, Subst[Int[1/(a + 2*b*e*x + a*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] &&
NeQ[a^2 - b^2, 0]

Rule 3785

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n +
1))/(a*d*(n + 1)*(a^2 - b^2)), x] + Dist[1/(a*(n + 1)*(a^2 - b^2)), Int[(a + b*Csc[c + d*x])^(n + 1)*Simp[(a^
2 - b^2)*(n + 1) - a*b*(n + 1)*Csc[c + d*x] + b^2*(n + 2)*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x]
&& NeQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 3831

Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[1/b, Int[1/(1 + (a*Sin[e
+ f*x])/b), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]

Rule 3919

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(c*x)/a,
x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[
b*c - a*d, 0]

Rule 5469

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

Rule 6322

Int[((a_.) + ArcCsch[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> -Dist[(d^(m + 1)
)^(-1), Subst[Int[(a + b*x)^p*Csch[x]*Coth[x]*(d*e - c*f + f*Csch[x])^m, x], x, ArcCsch[c + d*x]], x] /; FreeQ
[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && IntegerQ[m]

Rubi steps

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

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Mathematica [A]  time = 0.48, size = 220, normalized size = 1.93 \[ \frac {1}{2} \left (\frac {b (a+b x) \sqrt {\frac {a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}}}{a \left (a^2+1\right ) x}-\frac {\left (2 a^2+1\right ) b^2 \log \left (\sqrt {a^2+1} a \sqrt {\frac {a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}}+\sqrt {a^2+1} b x \sqrt {\frac {a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}}+a^2+a b x+1\right )}{a^2 \left (a^2+1\right )^{3/2}}+\frac {\left (2 a^2+1\right ) b^2 \log (x)}{a^2 \left (a^2+1\right )^{3/2}}+\frac {b^2 \sinh ^{-1}\left (\frac {1}{a+b x}\right )}{a^2}-\frac {\text {csch}^{-1}(a+b x)}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[ArcCsch[a + b*x]/x^3,x]

[Out]

((b*(a + b*x)*Sqrt[(1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2])/(a*(1 + a^2)*x) - ArcCsch[a + b*x]/x^2 + (b^2*A
rcSinh[(a + b*x)^(-1)])/a^2 + ((1 + 2*a^2)*b^2*Log[x])/(a^2*(1 + a^2)^(3/2)) - ((1 + 2*a^2)*b^2*Log[1 + a^2 +
a*b*x + a*Sqrt[1 + a^2]*Sqrt[(1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2] + Sqrt[1 + a^2]*b*x*Sqrt[(1 + a^2 + 2*
a*b*x + b^2*x^2)/(a + b*x)^2]])/(a^2*(1 + a^2)^(3/2)))/2

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fricas [B]  time = 0.51, size = 461, normalized size = 4.04 \[ \frac {{\left (2 \, a^{2} + 1\right )} \sqrt {a^{2} + 1} b^{2} x^{2} \log \left (-\frac {a^{2} b x + a^{3} - {\left (a b x + a^{2} + {\left (a b x + a^{2}\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1\right )} \sqrt {a^{2} + 1} + {\left (a^{3} + {\left (a^{2} + 1\right )} b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + a}{x}\right ) + {\left (a^{4} + 2 \, a^{2} + 1\right )} b^{2} x^{2} \log \left (-b x + {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a + 1\right ) - {\left (a^{4} + 2 \, a^{2} + 1\right )} b^{2} x^{2} \log \left (-b x + {\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} - a - 1\right ) + {\left (a^{3} + a\right )} b^{2} x^{2} - {\left (a^{6} + 2 \, a^{4} + a^{2}\right )} \log \left (\frac {{\left (b x + a\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}} + 1}{b x + a}\right ) + {\left ({\left (a^{3} + a\right )} b^{2} x^{2} + {\left (a^{4} + a^{2}\right )} b x\right )} \sqrt {\frac {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}{b^{2} x^{2} + 2 \, a b x + a^{2}}}}{2 \, {\left (a^{6} + 2 \, a^{4} + a^{2}\right )} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsch(b*x+a)/x^3,x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {arcsch}\left (b x + a\right )}{x^{3}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsch(b*x+a)/x^3,x, algorithm="giac")

[Out]

integrate(arccsch(b*x + a)/x^3, x)

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maple [B]  time = 0.06, size = 453, normalized size = 3.97 \[ -\frac {\mathrm {arccsch}\left (b x +a \right )}{2 x^{2}}+\frac {b^{2} \sqrt {1+\left (b x +a \right )^{2}}\, \arctanh \left (\frac {1}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{2 \sqrt {\frac {1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) \left (a^{2}+1\right )}-\frac {b^{2} \sqrt {1+\left (b x +a \right )^{2}}\, a^{2} \ln \left (\frac {2 \sqrt {a^{2}+1}\, \sqrt {1+\left (b x +a \right )^{2}}+2 a \left (b x +a \right )+2}{b x}\right )}{\sqrt {\frac {1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) \left (a^{2}+1\right )^{\frac {5}{2}}}+\frac {b^{2} \sqrt {1+\left (b x +a \right )^{2}}\, \arctanh \left (\frac {1}{\sqrt {1+\left (b x +a \right )^{2}}}\right )}{2 \sqrt {\frac {1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a^{2} \left (a^{2}+1\right )}+\frac {b \left (1+\left (b x +a \right )^{2}\right )}{2 \sqrt {\frac {1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a \left (a^{2}+1\right ) x}-\frac {3 b^{2} \sqrt {1+\left (b x +a \right )^{2}}\, \ln \left (\frac {2 \sqrt {a^{2}+1}\, \sqrt {1+\left (b x +a \right )^{2}}+2 a \left (b x +a \right )+2}{b x}\right )}{2 \sqrt {\frac {1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) \left (a^{2}+1\right )^{\frac {5}{2}}}-\frac {b^{2} \sqrt {1+\left (b x +a \right )^{2}}\, \ln \left (\frac {2 \sqrt {a^{2}+1}\, \sqrt {1+\left (b x +a \right )^{2}}+2 a \left (b x +a \right )+2}{b x}\right )}{2 \sqrt {\frac {1+\left (b x +a \right )^{2}}{\left (b x +a \right )^{2}}}\, \left (b x +a \right ) a^{2} \left (a^{2}+1\right )^{\frac {5}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arccsch(b*x+a)/x^3,x)

[Out]

-1/2*arccsch(b*x+a)/x^2+1/2*b^2*(1+(b*x+a)^2)^(1/2)/((1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)/(a^2+1)*arctanh(1/
(1+(b*x+a)^2)^(1/2))-b^2*(1+(b*x+a)^2)^(1/2)/((1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)*a^2/(a^2+1)^(5/2)*ln(2*((
a^2+1)^(1/2)*(1+(b*x+a)^2)^(1/2)+a*(b*x+a)+1)/b/x)+1/2*b^2*(1+(b*x+a)^2)^(1/2)/((1+(b*x+a)^2)/(b*x+a)^2)^(1/2)
/(b*x+a)/a^2/(a^2+1)*arctanh(1/(1+(b*x+a)^2)^(1/2))+1/2*b*(1+(b*x+a)^2)/((1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a
)/a/(a^2+1)/x-3/2*b^2*(1+(b*x+a)^2)^(1/2)/((1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a)/(a^2+1)^(5/2)*ln(2*((a^2+1)^
(1/2)*(1+(b*x+a)^2)^(1/2)+a*(b*x+a)+1)/b/x)-1/2*b^2*(1+(b*x+a)^2)^(1/2)/((1+(b*x+a)^2)/(b*x+a)^2)^(1/2)/(b*x+a
)/a^2/(a^2+1)^(5/2)*ln(2*((a^2+1)^(1/2)*(1+(b*x+a)^2)^(1/2)+a*(b*x+a)+1)/b/x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {i \, a b^{2} {\left (\log \left (\frac {i \, {\left (b^{2} x + a b\right )}}{b} + 1\right ) - \log \left (-\frac {i \, {\left (b^{2} x + a b\right )}}{b} + 1\right )\right )}}{2 \, {\left (a^{4} + 2 \, a^{2} + 1\right )}} + \frac {{\left (3 \, a^{2} b^{2} + b^{2}\right )} \log \relax (x)}{2 \, {\left (a^{6} + 2 \, a^{4} + a^{2}\right )}} + \frac {{\left (a^{4} b^{2} - a^{2} b^{2}\right )} x^{2} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) + 2 \, {\left (a^{3} b + a b\right )} x + 2 \, {\left (a^{6} + 2 \, a^{4} - {\left (a^{4} b^{2} + 2 \, a^{2} b^{2} + b^{2}\right )} x^{2} + a^{2}\right )} \log \left (b x + a\right ) - 2 \, {\left (a^{6} + 2 \, a^{4} + a^{2}\right )} \log \left (\sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 1\right )}{4 \, {\left (a^{6} + 2 \, a^{4} + a^{2}\right )} x^{2}} - \int \frac {b^{2} x + a b}{2 \, {\left (b^{2} x^{4} + 2 \, a b x^{3} + {\left (a^{2} + 1\right )} x^{2} + {\left (b^{2} x^{4} + 2 \, a b x^{3} + {\left (a^{2} + 1\right )} x^{2}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arccsch(b*x+a)/x^3,x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\mathrm {asinh}\left (\frac {1}{a+b\,x}\right )}{x^3} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(asinh(1/(a + b*x))/x^3,x)

[Out]

int(asinh(1/(a + b*x))/x^3, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {acsch}{\left (a + b x \right )}}{x^{3}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(acsch(b*x+a)/x**3,x)

[Out]

Integral(acsch(a + b*x)/x**3, x)

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