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

Optimal. Leaf size=86 \[ \frac{1}{5} x^5 \left (a+b \text{csch}^{-1}(c x)\right )+\frac{b x^4 \sqrt{\frac{1}{c^2 x^2}+1}}{20 c}-\frac{3 b x^2 \sqrt{\frac{1}{c^2 x^2}+1}}{40 c^3}+\frac{3 b \tanh ^{-1}\left (\sqrt{\frac{1}{c^2 x^2}+1}\right )}{40 c^5} \]

[Out]

(-3*b*Sqrt[1 + 1/(c^2*x^2)]*x^2)/(40*c^3) + (b*Sqrt[1 + 1/(c^2*x^2)]*x^4)/(20*c) + (x^5*(a + b*ArcCsch[c*x]))/
5 + (3*b*ArcTanh[Sqrt[1 + 1/(c^2*x^2)]])/(40*c^5)

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Rubi [A]  time = 0.047877, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.417, Rules used = {6284, 266, 51, 63, 208} \[ \frac{1}{5} x^5 \left (a+b \text{csch}^{-1}(c x)\right )+\frac{b x^4 \sqrt{\frac{1}{c^2 x^2}+1}}{20 c}-\frac{3 b x^2 \sqrt{\frac{1}{c^2 x^2}+1}}{40 c^3}+\frac{3 b \tanh ^{-1}\left (\sqrt{\frac{1}{c^2 x^2}+1}\right )}{40 c^5} \]

Antiderivative was successfully verified.

[In]

Int[x^4*(a + b*ArcCsch[c*x]),x]

[Out]

(-3*b*Sqrt[1 + 1/(c^2*x^2)]*x^2)/(40*c^3) + (b*Sqrt[1 + 1/(c^2*x^2)]*x^4)/(20*c) + (x^5*(a + b*ArcCsch[c*x]))/
5 + (3*b*ArcTanh[Sqrt[1 + 1/(c^2*x^2)]])/(40*c^5)

Rule 6284

Int[((a_.) + ArcCsch[(c_.)*(x_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcCsch[c*
x]))/(d*(m + 1)), x] + Dist[(b*d)/(c*(m + 1)), Int[(d*x)^(m - 1)/Sqrt[1 + 1/(c^2*x^2)], x], x] /; FreeQ[{a, b,
c, d, m}, x] && NeQ[m, -1]

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 51

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

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 208

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

Rubi steps

\begin{align*} \int x^4 \left (a+b \text{csch}^{-1}(c x)\right ) \, dx &=\frac{1}{5} x^5 \left (a+b \text{csch}^{-1}(c x)\right )+\frac{b \int \frac{x^3}{\sqrt{1+\frac{1}{c^2 x^2}}} \, dx}{5 c}\\ &=\frac{1}{5} x^5 \left (a+b \text{csch}^{-1}(c x)\right )-\frac{b \operatorname{Subst}\left (\int \frac{1}{x^3 \sqrt{1+\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{10 c}\\ &=\frac{b \sqrt{1+\frac{1}{c^2 x^2}} x^4}{20 c}+\frac{1}{5} x^5 \left (a+b \text{csch}^{-1}(c x)\right )+\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{x^2 \sqrt{1+\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{40 c^3}\\ &=-\frac{3 b \sqrt{1+\frac{1}{c^2 x^2}} x^2}{40 c^3}+\frac{b \sqrt{1+\frac{1}{c^2 x^2}} x^4}{20 c}+\frac{1}{5} x^5 \left (a+b \text{csch}^{-1}(c x)\right )-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{1+\frac{x}{c^2}}} \, dx,x,\frac{1}{x^2}\right )}{80 c^5}\\ &=-\frac{3 b \sqrt{1+\frac{1}{c^2 x^2}} x^2}{40 c^3}+\frac{b \sqrt{1+\frac{1}{c^2 x^2}} x^4}{20 c}+\frac{1}{5} x^5 \left (a+b \text{csch}^{-1}(c x)\right )-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{-c^2+c^2 x^2} \, dx,x,\sqrt{1+\frac{1}{c^2 x^2}}\right )}{40 c^3}\\ &=-\frac{3 b \sqrt{1+\frac{1}{c^2 x^2}} x^2}{40 c^3}+\frac{b \sqrt{1+\frac{1}{c^2 x^2}} x^4}{20 c}+\frac{1}{5} x^5 \left (a+b \text{csch}^{-1}(c x)\right )+\frac{3 b \tanh ^{-1}\left (\sqrt{1+\frac{1}{c^2 x^2}}\right )}{40 c^5}\\ \end{align*}

Mathematica [A]  time = 0.0468538, size = 97, normalized size = 1.13 \[ \frac{a x^5}{5}+b \sqrt{\frac{c^2 x^2+1}{c^2 x^2}} \left (\frac{x^4}{20 c}-\frac{3 x^2}{40 c^3}\right )+\frac{3 b \log \left (x \left (\sqrt{\frac{c^2 x^2+1}{c^2 x^2}}+1\right )\right )}{40 c^5}+\frac{1}{5} b x^5 \text{csch}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^4*(a + b*ArcCsch[c*x]),x]

[Out]

(a*x^5)/5 + b*Sqrt[(1 + c^2*x^2)/(c^2*x^2)]*((-3*x^2)/(40*c^3) + x^4/(20*c)) + (b*x^5*ArcCsch[c*x])/5 + (3*b*L
og[x*(1 + Sqrt[(1 + c^2*x^2)/(c^2*x^2)])])/(40*c^5)

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Maple [A]  time = 0.184, size = 108, normalized size = 1.3 \begin{align*}{\frac{1}{{c}^{5}} \left ({\frac{{c}^{5}{x}^{5}a}{5}}+b \left ({\frac{{c}^{5}{x}^{5}{\rm arccsch} \left (cx\right )}{5}}+{\frac{1}{40\,cx}\sqrt{{c}^{2}{x}^{2}+1} \left ( 2\,{c}^{3}{x}^{3}\sqrt{{c}^{2}{x}^{2}+1}-3\,cx\sqrt{{c}^{2}{x}^{2}+1}+3\,{\it Arcsinh} \left ( cx \right ) \right ){\frac{1}{\sqrt{{\frac{{c}^{2}{x}^{2}+1}{{c}^{2}{x}^{2}}}}}}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^5*(1/5*c^5*x^5*a+b*(1/5*c^5*x^5*arccsch(c*x)+1/40*(c^2*x^2+1)^(1/2)*(2*c^3*x^3*(c^2*x^2+1)^(1/2)-3*c*x*(c^
2*x^2+1)^(1/2)+3*arcsinh(c*x))/((c^2*x^2+1)/c^2/x^2)^(1/2)/c/x))

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Maxima [A]  time = 1.00616, size = 173, normalized size = 2.01 \begin{align*} \frac{1}{5} \, a x^{5} + \frac{1}{80} \,{\left (16 \, x^{5} \operatorname{arcsch}\left (c x\right ) - \frac{\frac{2 \,{\left (3 \,{\left (\frac{1}{c^{2} x^{2}} + 1\right )}^{\frac{3}{2}} - 5 \, \sqrt{\frac{1}{c^{2} x^{2}} + 1}\right )}}{c^{4}{\left (\frac{1}{c^{2} x^{2}} + 1\right )}^{2} - 2 \, c^{4}{\left (\frac{1}{c^{2} x^{2}} + 1\right )} + c^{4}} - \frac{3 \, \log \left (\sqrt{\frac{1}{c^{2} x^{2}} + 1} + 1\right )}{c^{4}} + \frac{3 \, \log \left (\sqrt{\frac{1}{c^{2} x^{2}} + 1} - 1\right )}{c^{4}}}{c}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/5*a*x^5 + 1/80*(16*x^5*arccsch(c*x) - (2*(3*(1/(c^2*x^2) + 1)^(3/2) - 5*sqrt(1/(c^2*x^2) + 1))/(c^4*(1/(c^2*
x^2) + 1)^2 - 2*c^4*(1/(c^2*x^2) + 1) + c^4) - 3*log(sqrt(1/(c^2*x^2) + 1) + 1)/c^4 + 3*log(sqrt(1/(c^2*x^2) +
1) - 1)/c^4)/c)*b

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Fricas [B]  time = 2.3142, size = 443, normalized size = 5.15 \begin{align*} \frac{8 \, a c^{5} x^{5} + 8 \, b c^{5} \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x + 1\right ) - 8 \, b c^{5} \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x - 1\right ) - 3 \, b \log \left (c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x\right ) + 8 \,{\left (b c^{5} x^{5} - b c^{5}\right )} \log \left (\frac{c x \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right ) +{\left (2 \, b c^{4} x^{4} - 3 \, b c^{2} x^{2}\right )} \sqrt{\frac{c^{2} x^{2} + 1}{c^{2} x^{2}}}}{40 \, c^{5}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/40*(8*a*c^5*x^5 + 8*b*c^5*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x + 1) - 8*b*c^5*log(c*x*sqrt((c^2*x^2 +
1)/(c^2*x^2)) - c*x - 1) - 3*b*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x) + 8*(b*c^5*x^5 - b*c^5)*log((c*x*
sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)) + (2*b*c^4*x^4 - 3*b*c^2*x^2)*sqrt((c^2*x^2 + 1)/(c^2*x^2)))/c^5

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{4} \left (a + b \operatorname{acsch}{\left (c x \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**4*(a + b*acsch(c*x)), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccsch(c*x) + a)*x^4, x)