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

Optimal. Leaf size=62 \[ \frac {1}{3} x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {b x^2 \sqrt {\frac {1}{c^2 x^2}+1}}{6 c}-\frac {b \tanh ^{-1}\left (\sqrt {\frac {1}{c^2 x^2}+1}\right )}{6 c^3} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 5, 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} \[ \frac {1}{3} x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {b x^2 \sqrt {\frac {1}{c^2 x^2}+1}}{6 c}-\frac {b \tanh ^{-1}\left (\sqrt {\frac {1}{c^2 x^2}+1}\right )}{6 c^3} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcCsch[c*x]),x]

[Out]

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

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]

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 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]

Rubi steps

\begin {align*} \int x^2 \left (a+b \text {csch}^{-1}(c x)\right ) \, dx &=\frac {1}{3} x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {b \int \frac {x}{\sqrt {1+\frac {1}{c^2 x^2}}} \, dx}{3 c}\\ &=\frac {1}{3} x^3 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {b \operatorname {Subst}\left (\int \frac {1}{x^2 \sqrt {1+\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{6 c}\\ &=\frac {b \sqrt {1+\frac {1}{c^2 x^2}} x^2}{6 c}+\frac {1}{3} x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {b \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+\frac {x}{c^2}}} \, dx,x,\frac {1}{x^2}\right )}{12 c^3}\\ &=\frac {b \sqrt {1+\frac {1}{c^2 x^2}} x^2}{6 c}+\frac {1}{3} x^3 \left (a+b \text {csch}^{-1}(c x)\right )+\frac {b \operatorname {Subst}\left (\int \frac {1}{-c^2+c^2 x^2} \, dx,x,\sqrt {1+\frac {1}{c^2 x^2}}\right )}{6 c}\\ &=\frac {b \sqrt {1+\frac {1}{c^2 x^2}} x^2}{6 c}+\frac {1}{3} x^3 \left (a+b \text {csch}^{-1}(c x)\right )-\frac {b \tanh ^{-1}\left (\sqrt {1+\frac {1}{c^2 x^2}}\right )}{6 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 85, normalized size = 1.37 \[ \frac {a x^3}{3}+\frac {b x^2 \sqrt {\frac {c^2 x^2+1}{c^2 x^2}}}{6 c}-\frac {b \log \left (x \left (\sqrt {\frac {c^2 x^2+1}{c^2 x^2}}+1\right )\right )}{6 c^3}+\frac {1}{3} b x^3 \text {csch}^{-1}(c x) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*(a + b*ArcCsch[c*x]),x]

[Out]

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

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fricas [B]  time = 1.58, size = 186, normalized size = 3.00 \[ \frac {2 \, a c^{3} x^{3} + b c^{2} x^{2} \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 2 \, b c^{3} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x + 1\right ) - 2 \, b c^{3} \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x - 1\right ) + b \log \left (c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} - c x\right ) + 2 \, {\left (b c^{3} x^{3} - b c^{3}\right )} \log \left (\frac {c x \sqrt {\frac {c^{2} x^{2} + 1}{c^{2} x^{2}}} + 1}{c x}\right )}{6 \, c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(2*a*c^3*x^3 + b*c^2*x^2*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 2*b*c^3*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c
*x + 1) - 2*b*c^3*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) - c*x - 1) + b*log(c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) -
c*x) + 2*(b*c^3*x^3 - b*c^3)*log((c*x*sqrt((c^2*x^2 + 1)/(c^2*x^2)) + 1)/(c*x)))/c^3

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \operatorname {arcsch}\left (c x\right ) + a\right )} x^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*arccsch(c*x) + a)*x^2, x)

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maple [A]  time = 0.05, size = 88, normalized size = 1.42 \[ \frac {\frac {a \,c^{3} x^{3}}{3}+b \left (\frac {c^{3} x^{3} \mathrm {arccsch}\left (c x \right )}{3}+\frac {\sqrt {c^{2} x^{2}+1}\, \left (c x \sqrt {c^{2} x^{2}+1}-\arcsinh \left (c x \right )\right )}{6 \sqrt {\frac {c^{2} x^{2}+1}{c^{2} x^{2}}}\, c x}\right )}{c^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 96, normalized size = 1.55 \[ \frac {1}{3} \, a x^{3} + \frac {1}{12} \, {\left (4 \, x^{3} \operatorname {arcsch}\left (c x\right ) + \frac {\frac {2 \, \sqrt {\frac {1}{c^{2} x^{2}} + 1}}{c^{2} {\left (\frac {1}{c^{2} x^{2}} + 1\right )} - c^{2}} - \frac {\log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} + 1\right )}{c^{2}} + \frac {\log \left (\sqrt {\frac {1}{c^{2} x^{2}} + 1} - 1\right )}{c^{2}}}{c}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*a*x^3 + 1/12*(4*x^3*arccsch(c*x) + (2*sqrt(1/(c^2*x^2) + 1)/(c^2*(1/(c^2*x^2) + 1) - c^2) - log(sqrt(1/(c^
2*x^2) + 1) + 1)/c^2 + log(sqrt(1/(c^2*x^2) + 1) - 1)/c^2)/c)*b

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int x^2\,\left (a+b\,\mathrm {asinh}\left (\frac {1}{c\,x}\right )\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*asinh(1/(c*x))),x)

[Out]

int(x^2*(a + b*asinh(1/(c*x))), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x^{2} \left (a + b \operatorname {acsch}{\left (c x \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(x**2*(a + b*acsch(c*x)), x)

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