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

Optimal. Leaf size=147 \[ -\frac{\left (2-17 a^2\right ) (a+b x) \sqrt{\frac{1}{(a+b x)^2}+1}}{12 b^4}+\frac{\left (1-2 a^2\right ) a \tanh ^{-1}\left (\sqrt{\frac{1}{(a+b x)^2}+1}\right )}{2 b^4}-\frac{a^4 \text{csch}^{-1}(a+b x)}{4 b^4}+\frac{x^2 (a+b x) \sqrt{\frac{1}{(a+b x)^2}+1}}{12 b^2}-\frac{a (a+b x)^2 \sqrt{\frac{1}{(a+b x)^2}+1}}{3 b^4}+\frac{1}{4} x^4 \text{csch}^{-1}(a+b x) \]

[Out]

-((2 - 17*a^2)*(a + b*x)*Sqrt[1 + (a + b*x)^(-2)])/(12*b^4) + (x^2*(a + b*x)*Sqrt[1 + (a + b*x)^(-2)])/(12*b^2
) - (a*(a + b*x)^2*Sqrt[1 + (a + b*x)^(-2)])/(3*b^4) - (a^4*ArcCsch[a + b*x])/(4*b^4) + (x^4*ArcCsch[a + b*x])
/4 + (a*(1 - 2*a^2)*ArcTanh[Sqrt[1 + (a + b*x)^(-2)]])/(2*b^4)

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Rubi [A]  time = 0.152366, antiderivative size = 147, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 10, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.7, Rules used = {6322, 5469, 3782, 4048, 3770, 3767, 8} \[ -\frac{\left (2-17 a^2\right ) (a+b x) \sqrt{\frac{1}{(a+b x)^2}+1}}{12 b^4}+\frac{\left (1-2 a^2\right ) a \tanh ^{-1}\left (\sqrt{\frac{1}{(a+b x)^2}+1}\right )}{2 b^4}-\frac{a^4 \text{csch}^{-1}(a+b x)}{4 b^4}+\frac{x^2 (a+b x) \sqrt{\frac{1}{(a+b x)^2}+1}}{12 b^2}-\frac{a (a+b x)^2 \sqrt{\frac{1}{(a+b x)^2}+1}}{3 b^4}+\frac{1}{4} x^4 \text{csch}^{-1}(a+b x) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCsch[a + b*x],x]

[Out]

-((2 - 17*a^2)*(a + b*x)*Sqrt[1 + (a + b*x)^(-2)])/(12*b^4) + (x^2*(a + b*x)*Sqrt[1 + (a + b*x)^(-2)])/(12*b^2
) - (a*(a + b*x)^2*Sqrt[1 + (a + b*x)^(-2)])/(3*b^4) - (a^4*ArcCsch[a + b*x])/(4*b^4) + (x^4*ArcCsch[a + b*x])
/4 + (a*(1 - 2*a^2)*ArcTanh[Sqrt[1 + (a + b*x)^(-2)]])/(2*b^4)

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 4048

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) +
(a_)), x_Symbol] :> -Simp[(b*C*Csc[e + f*x]*Cot[e + f*x])/(2*f), x] + Dist[1/2, Int[Simp[2*A*a + (2*B*a + b*(
2*A + C))*Csc[e + f*x] + 2*(a*C + B*b)*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x]

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^3 \text{csch}^{-1}(a+b x) \, dx &=-\frac{\operatorname{Subst}\left (\int x \coth (x) \text{csch}(x) (-a+\text{csch}(x))^3 \, dx,x,\text{csch}^{-1}(a+b x)\right )}{b^4}\\ &=\frac{1}{4} x^4 \text{csch}^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int (-a+\text{csch}(x))^4 \, dx,x,\text{csch}^{-1}(a+b x)\right )}{4 b^4}\\ &=\frac{x^2 (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{12 b^2}+\frac{1}{4} x^4 \text{csch}^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int (-a+\text{csch}(x)) \left (-3 a^3-\left (2-9 a^2\right ) \text{csch}(x)-8 a \text{csch}^2(x)\right ) \, dx,x,\text{csch}^{-1}(a+b x)\right )}{12 b^4}\\ &=\frac{x^2 (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{12 b^2}-\frac{a (a+b x)^2 \sqrt{1+\frac{1}{(a+b x)^2}}}{3 b^4}+\frac{1}{4} x^4 \text{csch}^{-1}(a+b x)-\frac{\operatorname{Subst}\left (\int \left (6 a^4+12 a \left (1-2 a^2\right ) \text{csch}(x)-2 \left (2-17 a^2\right ) \text{csch}^2(x)\right ) \, dx,x,\text{csch}^{-1}(a+b x)\right )}{24 b^4}\\ &=\frac{x^2 (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{12 b^2}-\frac{a (a+b x)^2 \sqrt{1+\frac{1}{(a+b x)^2}}}{3 b^4}-\frac{a^4 \text{csch}^{-1}(a+b x)}{4 b^4}+\frac{1}{4} x^4 \text{csch}^{-1}(a+b x)+\frac{\left (2-17 a^2\right ) \operatorname{Subst}\left (\int \text{csch}^2(x) \, dx,x,\text{csch}^{-1}(a+b x)\right )}{12 b^4}-\frac{\left (a \left (1-2 a^2\right )\right ) \operatorname{Subst}\left (\int \text{csch}(x) \, dx,x,\text{csch}^{-1}(a+b x)\right )}{2 b^4}\\ &=\frac{x^2 (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{12 b^2}-\frac{a (a+b x)^2 \sqrt{1+\frac{1}{(a+b x)^2}}}{3 b^4}-\frac{a^4 \text{csch}^{-1}(a+b x)}{4 b^4}+\frac{1}{4} x^4 \text{csch}^{-1}(a+b x)+\frac{a \left (1-2 a^2\right ) \tanh ^{-1}\left (\sqrt{1+\frac{1}{(a+b x)^2}}\right )}{2 b^4}-\frac{\left (i \left (2-17 a^2\right )\right ) \operatorname{Subst}\left (\int 1 \, dx,x,-i (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}\right )}{12 b^4}\\ &=-\frac{\left (2-17 a^2\right ) (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{12 b^4}+\frac{x^2 (a+b x) \sqrt{1+\frac{1}{(a+b x)^2}}}{12 b^2}-\frac{a (a+b x)^2 \sqrt{1+\frac{1}{(a+b x)^2}}}{3 b^4}-\frac{a^4 \text{csch}^{-1}(a+b x)}{4 b^4}+\frac{1}{4} x^4 \text{csch}^{-1}(a+b x)+\frac{a \left (1-2 a^2\right ) \tanh ^{-1}\left (\sqrt{1+\frac{1}{(a+b x)^2}}\right )}{2 b^4}\\ \end{align*}

Mathematica [A]  time = 0.281149, size = 149, normalized size = 1.01 \[ \frac{\sqrt{\frac{a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}} \left (9 a^2 b x+13 a^3-3 a b^2 x^2-2 a+b^3 x^3-2 b x\right )+6 \left (1-2 a^2\right ) a \log \left ((a+b x) \left (\sqrt{\frac{a^2+2 a b x+b^2 x^2+1}{(a+b x)^2}}+1\right )\right )-3 a^4 \sinh ^{-1}\left (\frac{1}{a+b x}\right )+3 b^4 x^4 \text{csch}^{-1}(a+b x)}{12 b^4} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCsch[a + b*x],x]

[Out]

(Sqrt[(1 + a^2 + 2*a*b*x + b^2*x^2)/(a + b*x)^2]*(-2*a + 13*a^3 - 2*b*x + 9*a^2*b*x - 3*a*b^2*x^2 + b^3*x^3) +
3*b^4*x^4*ArcCsch[a + b*x] - 3*a^4*ArcSinh[(a + b*x)^(-1)] + 6*a*(1 - 2*a^2)*Log[(a + b*x)*(1 + Sqrt[(1 + a^2
+ 2*a*b*x + b^2*x^2)/(a + b*x)^2])])/(12*b^4)

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Maple [A]  time = 0.249, size = 227, normalized size = 1.5 \begin{align*}{\frac{1}{{b}^{4}} \left ({\frac{{\rm arccsch} \left (bx+a\right ) \left ( bx+a \right ) ^{4}}{4}}-{\rm arccsch} \left (bx+a\right ) \left ( bx+a \right ) ^{3}a+{\frac{3\,{\rm arccsch} \left (bx+a\right ) \left ( bx+a \right ) ^{2}{a}^{2}}{2}}-{\rm arccsch} \left (bx+a\right ) \left ( bx+a \right ){a}^{3}+{\frac{{\rm arccsch} \left (bx+a\right ){a}^{4}}{4}}-{\frac{1}{12\,bx+12\,a}\sqrt{1+ \left ( bx+a \right ) ^{2}} \left ( 3\,{a}^{4}{\it Artanh} \left ({\frac{1}{\sqrt{1+ \left ( bx+a \right ) ^{2}}}} \right ) +12\,{a}^{3}{\it Arcsinh} \left ( bx+a \right ) - \left ( bx+a \right ) ^{2}\sqrt{1+ \left ( bx+a \right ) ^{2}}+6\,a \left ( bx+a \right ) \sqrt{1+ \left ( bx+a \right ) ^{2}}-18\,{a}^{2}\sqrt{1+ \left ( bx+a \right ) ^{2}}-6\,a{\it Arcsinh} \left ( bx+a \right ) +2\,\sqrt{1+ \left ( bx+a \right ) ^{2}} \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^3*arccsch(b*x+a),x)

[Out]

1/b^4*(1/4*arccsch(b*x+a)*(b*x+a)^4-arccsch(b*x+a)*(b*x+a)^3*a+3/2*arccsch(b*x+a)*(b*x+a)^2*a^2-arccsch(b*x+a)
*(b*x+a)*a^3+1/4*arccsch(b*x+a)*a^4-1/12*(1+(b*x+a)^2)^(1/2)*(3*a^4*arctanh(1/(1+(b*x+a)^2)^(1/2))+12*a^3*arcs
inh(b*x+a)-(b*x+a)^2*(1+(b*x+a)^2)^(1/2)+6*a*(b*x+a)*(1+(b*x+a)^2)^(1/2)-18*a^2*(1+(b*x+a)^2)^(1/2)-6*a*arcsin
h(b*x+a)+2*(1+(b*x+a)^2)^(1/2))/((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 (-i \, a^{3} + i \, a\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 )}}{2 \, b^{4}} + \frac{2 \, b^{4} x^{4} \log \left (\sqrt{b^{2} x^{2} + 2 \, a b x + a^{2} + 1} + 1\right ) + b^{2} x^{2} - 6 \, a b x -{\left (a^{4} - 6 \, a^{2} + 1\right )} \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - 2 \,{\left (b^{4} x^{4} - a^{4}\right )} \log \left (b x + a\right )}{8 \, b^{4}} + \int \frac{b^{2} x^{5} + a b x^{4}}{4 \,{\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^3*arccsch(b*x+a),x, algorithm="maxima")

[Out]

-1/2*(-I*a^3 + I*a)*(log(I*(b^2*x + a*b)/b + 1) - log(-I*(b^2*x + a*b)/b + 1))/b^4 + 1/8*(2*b^4*x^4*log(sqrt(b
^2*x^2 + 2*a*b*x + a^2 + 1) + 1) + b^2*x^2 - 6*a*b*x - (a^4 - 6*a^2 + 1)*log(b^2*x^2 + 2*a*b*x + a^2 + 1) - 2*
(b^4*x^4 - a^4)*log(b*x + a))/b^4 + integrate(1/4*(b^2*x^5 + a*b*x^4)/(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 = 3.16608, size = 729, normalized size = 4.96 \begin{align*} \frac{3 \, b^{4} x^{4} \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 ) - 3 \, a^{4} \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 ) + 3 \, a^{4} \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 ) + 6 \,{\left (2 \, a^{3} - a\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^{3} x^{3} - 3 \, a b^{2} x^{2} + 13 \, a^{3} +{\left (9 \, a^{2} - 2\right )} b x - 2 \, 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}}}}{12 \, b^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*b^4*x^4*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)) -
3*a^4*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) + 3*a^4*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*(2*a^3 - a)*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^3*x^3 - 3*a*b^2*x^2 +
13*a^3 + (9*a^2 - 2)*b*x - 2*a)*sqrt((b^2*x^2 + 2*a*b*x + a^2 + 1)/(b^2*x^2 + 2*a*b*x + a^2)))/b^4

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{acsch}{\left (a + b x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**3*acsch(a + b*x), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arcsch}\left (b x + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(x^3*arccsch(b*x + a), x)