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]

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

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Rubi [A]  time = 0.06, antiderivative size = 46, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, 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 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])

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 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 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 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]

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*}

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Mathematica [A]  time = 0.15, 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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fricas [B]  time = 0.61, size = 266, normalized size = 5.78 \[ \frac {b x^{4} \log \left (\frac {{\left (b x^{4} + a\right )} \sqrt {\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} + 1}{b x^{4} + a}\right ) + a \log \left (-b x^{4} + {\left (b x^{4} + a\right )} \sqrt {\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} - a + 1\right ) - a \log \left (-b x^{4} + {\left (b x^{4} + a\right )} \sqrt {\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} - a - 1\right ) - \log \left (-b x^{4} + {\left (b x^{4} + a\right )} \sqrt {\frac {b^{2} x^{8} + 2 \, a b x^{4} + a^{2} + 1}{b^{2} x^{8} + 2 \, a b x^{4} + a^{2}}} - a\right )}{4 \, b} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{3} \operatorname {arcsch}\left (b x^{4} + a\right )\,{d x} \]

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)

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maple [A]  time = 0.08, size = 63, normalized size = 1.37 \[ \frac {\mathrm {arccsch}\left (b \,x^{4}+a \right ) x^{4}}{4}+\frac {\mathrm {arccsch}\left (b \,x^{4}+a \right ) a}{4 b}+\frac {\ln \left (b \,x^{4}+a +\left (b \,x^{4}+a \right ) \sqrt {1+\frac {1}{\left (b \,x^{4}+a \right )^{2}}}\right )}{4 b} \]

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 = 0.31, size = 57, normalized size = 1.24 \[ \frac {2 \, {\left (b x^{4} + a\right )} \operatorname {arcsch}\left (b x^{4} + a\right ) + \log \left (\sqrt {\frac {1}{{\left (b x^{4} + a\right )}^{2}} + 1} + 1\right ) - \log \left (\sqrt {\frac {1}{{\left (b x^{4} + a\right )}^{2}} + 1} - 1\right )}{8 \, b} \]

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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mupad [B]  time = 2.71, size = 42, normalized size = 0.91 \[ \frac {\mathrm {atanh}\left (\sqrt {\frac {1}{{\left (b\,x^4+a\right )}^2}+1}\right )}{4\,b}+\frac {\mathrm {asinh}\left (\frac {1}{b\,x^4+a}\right )\,\left (b\,x^4+a\right )}{4\,b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*asinh(1/(a + b*x^4)),x)

[Out]

atanh((1/(a + b*x^4)^2 + 1)^(1/2))/(4*b) + (asinh(1/(a + b*x^4))*(a + b*x^4))/(4*b)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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