3.70 \(\int x^3 \text{csch}^{-1}(a+b x^4) \, dx\)

Optimal. Leaf size=46 \[ \frac{\tanh ^{-1}\left (\sqrt{\frac{1}{\left (a+b x^4\right )^2}+1}\right )}{4 b}+\frac{\left (a+b x^4\right ) \text{csch}^{-1}\left (a+b x^4\right )}{4 b} \]

[Out]

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

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Rubi [A]  time = 0.0582378, antiderivative size = 46, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.5, Rules used = {6715, 6314, 372, 266, 63, 207} \[ \frac{\tanh ^{-1}\left (\sqrt{\frac{1}{\left (a+b x^4\right )^2}+1}\right )}{4 b}+\frac{\left (a+b x^4\right ) \text{csch}^{-1}\left (a+b x^4\right )}{4 b} \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCsch[a + b*x^4],x]

[Out]

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

Rule 6715

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rule 6314

Int[ArcCsch[(c_) + (d_.)*(x_)], x_Symbol] :> Simp[((c + d*x)*ArcCsch[c + d*x])/d, x] + Int[1/((c + d*x)*Sqrt[1
+ 1/(c + d*x)^2]), x] /; FreeQ[{c, d}, x]

Rule 372

Int[(u_)^(m_.)*((a_) + (b_.)*(v_)^(n_))^(p_.), x_Symbol] :> Dist[u^m/(Coefficient[v, x, 1]*v^m), Subst[Int[x^m
*(a + b*x^n)^p, x], x, v], x] /; FreeQ[{a, b, m, n, p}, x] && LinearPairQ[u, v, x]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.136193, size = 74, normalized size = 1.61 \[ \frac{\frac{\sqrt{\left (a+b x^4\right )^2+1} \sinh ^{-1}\left (a+b x^4\right )}{\sqrt{\frac{1}{\left (a+b x^4\right )^2}+1}}+\left (a+b x^4\right )^2 \text{csch}^{-1}\left (a+b x^4\right )}{4 b \left (a+b x^4\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCsch[a + b*x^4],x]

[Out]

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

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Maple [A]  time = 0.25, size = 63, normalized size = 1.4 \begin{align*}{\frac{{\rm arccsch} \left (b{x}^{4}+a\right ){x}^{4}}{4}}+{\frac{{\rm arccsch} \left (b{x}^{4}+a\right )a}{4\,b}}+{\frac{1}{4\,b}\ln \left ( b{x}^{4}+a+ \left ( b{x}^{4}+a \right ) \sqrt{1+ \left ( b{x}^{4}+a \right ) ^{-2}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccsch(b*x^4+a),x)

[Out]

1/4*arccsch(b*x^4+a)*x^4+1/4/b*arccsch(b*x^4+a)*a+1/4/b*ln(b*x^4+a+(b*x^4+a)*(1+1/(b*x^4+a)^2)^(1/2))

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Maxima [A]  time = 1.01065, size = 77, normalized size = 1.67 \begin{align*} \frac{2 \,{\left (b x^{4} + a\right )} \operatorname{arcsch}\left (b x^{4} + a\right ) + \log \left (\sqrt{\frac{1}{{\left (b x^{4} + a\right )}^{2}} + 1} + 1\right ) - \log \left (\sqrt{\frac{1}{{\left (b x^{4} + a\right )}^{2}} + 1} - 1\right )}{8 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccsch(b*x^4+a),x, algorithm="maxima")

[Out]

1/8*(2*(b*x^4 + a)*arccsch(b*x^4 + a) + log(sqrt(1/(b*x^4 + a)^2 + 1) + 1) - log(sqrt(1/(b*x^4 + a)^2 + 1) - 1
))/b

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Fricas [B]  time = 2.81386, size = 570, normalized size = 12.39 \begin{align*} \frac{b x^{4} \log \left (\frac{{\left (b x^{4} + a\right )} \sqrt{\frac{b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} + 1}{b x^{4} + a}\right ) + a \log \left (-b x^{4} +{\left (b x^{4} + a\right )} \sqrt{\frac{b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} - a + 1\right ) - a \log \left (-b x^{4} +{\left (b x^{4} + a\right )} \sqrt{\frac{b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} - a - 1\right ) - \log \left (-b x^{4} +{\left (b x^{4} + a\right )} \sqrt{\frac{b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} - a\right )}{4 \, b} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccsch(b*x^4+a),x, algorithm="fricas")

[Out]

1/4*(b*x^4*log(((b*x^4 + a)*sqrt((b^2*x^8 + 2*a*b*x^4 + a^2 + 1)/(b^2*x^8 + 2*a*b*x^4 + a^2)) + 1)/(b*x^4 + a)
) + a*log(-b*x^4 + (b*x^4 + a)*sqrt((b^2*x^8 + 2*a*b*x^4 + a^2 + 1)/(b^2*x^8 + 2*a*b*x^4 + a^2)) - a + 1) - a*
log(-b*x^4 + (b*x^4 + a)*sqrt((b^2*x^8 + 2*a*b*x^4 + a^2 + 1)/(b^2*x^8 + 2*a*b*x^4 + a^2)) - a - 1) - log(-b*x
^4 + (b*x^4 + a)*sqrt((b^2*x^8 + 2*a*b*x^4 + a^2 + 1)/(b^2*x^8 + 2*a*b*x^4 + a^2)) - a))/b

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acsch(b*x**4+a),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{3} \operatorname{arcsch}\left (b x^{4} + a\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccsch(b*x^4+a),x, algorithm="giac")

[Out]

integrate(x^3*arccsch(b*x^4 + a), x)