3.2 \(\int x^2 \text{csch}^{-1}(a+b x) \, dx\)

Optimal. Leaf size=110 \[ -\frac{\left (1-6 a^2\right ) \tanh ^{-1}\left (\sqrt{\frac{1}{(a+b x)^2}+1}\right )}{6 b^3}+\frac{a^3 \text{csch}^{-1}(a+b x)}{3 b^3}-\frac{5 a (a+b x) \sqrt{\frac{1}{(a+b x)^2}+1}}{6 b^3}+\frac{x (a+b x) \sqrt{\frac{1}{(a+b x)^2}+1}}{6 b^2}+\frac{1}{3} x^3 \text{csch}^{-1}(a+b x) \]

[Out]

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

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Rubi [A]  time = 0.0952465, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.6, Rules used = {6322, 5469, 3782, 3770, 3767, 8} \[ -\frac{\left (1-6 a^2\right ) \tanh ^{-1}\left (\sqrt{\frac{1}{(a+b x)^2}+1}\right )}{6 b^3}+\frac{a^3 \text{csch}^{-1}(a+b x)}{3 b^3}-\frac{5 a (a+b x) \sqrt{\frac{1}{(a+b x)^2}+1}}{6 b^3}+\frac{x (a+b x) \sqrt{\frac{1}{(a+b x)^2}+1}}{6 b^2}+\frac{1}{3} x^3 \text{csch}^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcCsch[a + b*x],x]

[Out]

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

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]

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 3782

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Simp[(b^2*Cot[c + d*x]*(a + b*Csc[c + d*x])^(n
- 2))/(d*(n - 1)), x] + Dist[1/(n - 1), Int[(a + b*Csc[c + d*x])^(n - 3)*Simp[a^3*(n - 1) + (b*(b^2*(n - 2) +
3*a^2*(n - 1)))*Csc[c + d*x] + (a*b^2*(3*n - 4))*Csc[c + d*x]^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] && GtQ[n, 2] && IntegerQ[2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int x^2 \text{csch}^{-1}(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int x \coth (x) \text{csch}(x) (-a+\text{csch}(x))^2 \, dx,x,\text{csch}^{-1}(a+b x)\right )}{b^3}\\ &=\frac{1}{3} x^3 \text{csch}^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int (-a+\text{csch}(x))^3 \, dx,x,\text{csch}^{-1}(a+b x)\right )}{3 b^3}\\ &=\frac{x (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{6 b^2}+\frac{1}{3} x^3 \text{csch}^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int \left (-2 a^3-\left (1-6 a^2\right ) \text{csch}(x)-5 a \text{csch}^2(x)\right ) \, dx,x,\text{csch}^{-1}(a+b x)\right )}{6 b^3}\\ &=\frac{x (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{6 b^2}+\frac{a^3 \text{csch}^{-1}(a+b x)}{3 b^3}+\frac{1}{3} x^3 \text{csch}^{-1}(a+b x)+\frac{(5 a) \operatorname{Subst}\left (\int \text{csch}^2(x) \, dx,x,\text{csch}^{-1}(a+b x)\right )}{6 b^3}+\frac{\left (1-6 a^2\right ) \operatorname{Subst}\left (\int \text{csch}(x) \, dx,x,\text{csch}^{-1}(a+b x)\right )}{6 b^3}\\ &=\frac{x (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{6 b^2}+\frac{a^3 \text{csch}^{-1}(a+b x)}{3 b^3}+\frac{1}{3} x^3 \text{csch}^{-1}(a+b x)-\frac{\left (1-6 a^2\right ) \tanh ^{-1}\left (\sqrt{1+\frac{1}{(a+b x)^2}}\right )}{6 b^3}-\frac{(5 i a) \operatorname{Subst}\left (\int 1 \, dx,x,-i (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}\right )}{6 b^3}\\ &=-\frac{5 a (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{6 b^3}+\frac{x (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{6 b^2}+\frac{a^3 \text{csch}^{-1}(a+b x)}{3 b^3}+\frac{1}{3} x^3 \text{csch}^{-1}(a+b x)-\frac{\left (1-6 a^2\right ) \tanh ^{-1}\left (\sqrt{1+\frac{1}{(a+b x)^2}}\right )}{6 b^3}\\ \end{align*}

Mathematica [A]  time = 0.176214, size = 129, normalized size = 1.17 \[ \frac{\left (-5 a^2-4 a b x+b^2 x^2\right ) \sqrt{\frac{a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}}+\left (6 a^2-1\right ) \log \left ((a+b x) \left (\sqrt{\frac{a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}}+1\right )\right )+2 a^3 \sinh ^{-1}\left (\frac{1}{a+b x}\right )+2 b^3 x^3 \text{csch}^{-1}(a+b x)}{6 b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcCsch[a + b*x],x]

[Out]

((-5*a^2 - 4*a*b*x + b^2*x^2)*Sqrt[(1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2] + 2*b^3*x^3*ArcCsch[a + b*x] + 2
*a^3*ArcSinh[(a + b*x)^(-1)] + (-1 + 6*a^2)*Log[(a + b*x)*(1 + Sqrt[(1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2]
)])/(6*b^3)

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Maple [A]  time = 0.219, size = 170, normalized size = 1.6 \begin{align*}{\frac{1}{{b}^{3}} \left ({\frac{{\rm arccsch} \left (bx+a\right ) \left ( bx+a \right ) ^{3}}{3}}-{\rm arccsch} \left (bx+a\right ) \left ( bx+a \right ) ^{2}a+{\rm arccsch} \left (bx+a\right ) \left ( bx+a \right ){a}^{2}-{\frac{{\rm arccsch} \left (bx+a\right ){a}^{3}}{3}}+{\frac{1}{6\,bx+6\,a}\sqrt{1+ \left ( bx+a \right ) ^{2}} \left ( 2\,{a}^{3}{\it Artanh} \left ({\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}} \right ) +6\,{\it Arcsinh} \left ( bx+a \right ){a}^{2}+ \left ( bx+a \right ) \sqrt{1+ \left ( bx+a \right ) ^{2}}-6\,a\sqrt{1+ \left ( bx+a \right ) ^{2}}-{\it Arcsinh} \left ( bx+a \right ) \right ){\frac{1}{\sqrt{{\frac{1+ \left ( bx+a \right ) ^{2}}{ \left ( bx+a \right ) ^{2}}}}}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arccsch(b*x+a),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{{\left (3 i \, a^{2} - i\right )}{\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 )}}{6 \, b^{3}} + \frac{2 \, b^{3} x^{3} \log \left (\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 1\right ) + 2 \, b x +{\left (a^{3} - 3 \, a\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \,{\left (b^{3} x^{3} + a^{3}\right )} \log \left (b x + a\right )}{6 \, b^{3}} + \int \frac{b^{2} x^{4} + a b x^{3}}{3 \,{\left (b^{2} x^{2} + 2 \, a b x + a^{2} +{\left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right )}^{\frac{3}{2}} + 1\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccsch(b*x+a),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 3.31905, size = 686, normalized size = 6.24 \begin{align*} \frac{2 \, b^{3} x^{3} \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 ) + 2 \, a^{3} \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 ) - 2 \, a^{3} \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 (6 \, a^{2} - 1\right )} \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\right ) +{\left (b^{2} x^{2} - 4 \, a b x - 5 \, 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}}}}{6 \, b^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccsch(b*x+a),x, algorithm="fricas")

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{acsch}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*acsch(b*x+a),x)

[Out]

Integral(x**2*acsch(a + b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \operatorname{arcsch}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arccsch(b*x+a),x, algorithm="giac")

[Out]

integrate(x^2*arccsch(b*x + a), x)