3.188 \(\int \frac {x^3 (a+b \text {sech}^{-1}(c x))}{\sqrt {1-c^4 x^4}} \, dx\)

Optimal. Leaf size=159 \[ -\frac {\sqrt {1-c^4 x^4} \left (a+b \text {sech}^{-1}(c x)\right )}{2 c^4}-\frac {b \sqrt {1-c^2 x^2} \sqrt {c^2 x^2+1}}{2 c^5 x \sqrt {\frac {1}{c x}-1} \sqrt {\frac {1}{c x}+1}}+\frac {b \sqrt {1-c^2 x^2} \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )}{2 c^5 x \sqrt {\frac {1}{c x}-1} \sqrt {\frac {1}{c x}+1}} \]

[Out]

1/2*b*arctanh((c^2*x^2+1)^(1/2))*(-c^2*x^2+1)^(1/2)/c^5/x/(-1+1/c/x)^(1/2)/(1+1/c/x)^(1/2)-1/2*b*(-c^2*x^2+1)^
(1/2)*(c^2*x^2+1)^(1/2)/c^5/x/(-1+1/c/x)^(1/2)/(1+1/c/x)^(1/2)-1/2*(a+b*arcsech(c*x))*(-c^4*x^4+1)^(1/2)/c^4

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Rubi [A]  time = 0.16, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 8, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.308, Rules used = {261, 6309, 12, 1252, 848, 50, 63, 208} \[ -\frac {\sqrt {1-c^4 x^4} \left (a+b \text {sech}^{-1}(c x)\right )}{2 c^4}-\frac {b \sqrt {1-c^2 x^2} \sqrt {c^2 x^2+1}}{2 c^5 x \sqrt {\frac {1}{c x}-1} \sqrt {\frac {1}{c x}+1}}+\frac {b \sqrt {1-c^2 x^2} \tanh ^{-1}\left (\sqrt {c^2 x^2+1}\right )}{2 c^5 x \sqrt {\frac {1}{c x}-1} \sqrt {\frac {1}{c x}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSech[c*x]))/Sqrt[1 - c^4*x^4],x]

[Out]

-(b*Sqrt[1 - c^2*x^2]*Sqrt[1 + c^2*x^2])/(2*c^5*Sqrt[-1 + 1/(c*x)]*Sqrt[1 + 1/(c*x)]*x) - (Sqrt[1 - c^4*x^4]*(
a + b*ArcSech[c*x]))/(2*c^4) + (b*Sqrt[1 - c^2*x^2]*ArcTanh[Sqrt[1 + c^2*x^2]])/(2*c^5*Sqrt[-1 + 1/(c*x)]*Sqrt
[1 + 1/(c*x)]*x)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && 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 261

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

Rule 848

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[(d + e*x)
^(m + p)*(f + g*x)^n*(a/d + (c*x)/e)^p, x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[e*f - d*g, 0] && EqQ[c
*d^2 + a*e^2, 0] && (IntegerQ[p] || (GtQ[a, 0] && GtQ[d, 0] && EqQ[m + p, 0]))

Rule 1252

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_Symbol] :> Dist[1/2, Subst[Int[x^((m
- 1)/2)*(d + e*x)^q*(a + c*x^2)^p, x], x, x^2], x] /; FreeQ[{a, c, d, e, p, q}, x] && IntegerQ[(m + 1)/2]

Rule 6309

Int[((a_.) + ArcSech[(c_.)*(x_)]*(b_.))*(u_), x_Symbol] :> With[{v = IntHide[u, x]}, Dist[a + b*ArcSech[c*x],
v, x] + Dist[(b*Sqrt[1 - c^2*x^2])/(c*x*Sqrt[-1 + 1/(c*x)]*Sqrt[1 + 1/(c*x)]), Int[SimplifyIntegrand[v/(x*Sqrt
[1 - c^2*x^2]), x], x], x] /; InverseFunctionFreeQ[v, x]] /; FreeQ[{a, b, c}, x]

Rubi steps

\begin {align*} \int \frac {x^3 \left (a+b \text {sech}^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx &=-\frac {\sqrt {1-c^4 x^4} \left (a+b \text {sech}^{-1}(c x)\right )}{2 c^4}+\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int -\frac {\sqrt {1-c^4 x^4}}{2 c^4 x \sqrt {1-c^2 x^2}} \, dx}{c \sqrt {-1+\frac {1}{c x}} \sqrt {1+\frac {1}{c x}} x}\\ &=-\frac {\sqrt {1-c^4 x^4} \left (a+b \text {sech}^{-1}(c x)\right )}{2 c^4}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \int \frac {\sqrt {1-c^4 x^4}}{x \sqrt {1-c^2 x^2}} \, dx}{2 c^5 \sqrt {-1+\frac {1}{c x}} \sqrt {1+\frac {1}{c x}} x}\\ &=-\frac {\sqrt {1-c^4 x^4} \left (a+b \text {sech}^{-1}(c x)\right )}{2 c^4}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1-c^4 x^2}}{x \sqrt {1-c^2 x}} \, dx,x,x^2\right )}{4 c^5 \sqrt {-1+\frac {1}{c x}} \sqrt {1+\frac {1}{c x}} x}\\ &=-\frac {\sqrt {1-c^4 x^4} \left (a+b \text {sech}^{-1}(c x)\right )}{2 c^4}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {\sqrt {1+c^2 x}}{x} \, dx,x,x^2\right )}{4 c^5 \sqrt {-1+\frac {1}{c x}} \sqrt {1+\frac {1}{c x}} x}\\ &=-\frac {b \sqrt {1-c^2 x^2} \sqrt {1+c^2 x^2}}{2 c^5 \sqrt {-1+\frac {1}{c x}} \sqrt {1+\frac {1}{c x}} x}-\frac {\sqrt {1-c^4 x^4} \left (a+b \text {sech}^{-1}(c x)\right )}{2 c^4}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {1+c^2 x}} \, dx,x,x^2\right )}{4 c^5 \sqrt {-1+\frac {1}{c x}} \sqrt {1+\frac {1}{c x}} x}\\ &=-\frac {b \sqrt {1-c^2 x^2} \sqrt {1+c^2 x^2}}{2 c^5 \sqrt {-1+\frac {1}{c x}} \sqrt {1+\frac {1}{c x}} x}-\frac {\sqrt {1-c^4 x^4} \left (a+b \text {sech}^{-1}(c x)\right )}{2 c^4}-\frac {\left (b \sqrt {1-c^2 x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{-\frac {1}{c^2}+\frac {x^2}{c^2}} \, dx,x,\sqrt {1+c^2 x^2}\right )}{2 c^7 \sqrt {-1+\frac {1}{c x}} \sqrt {1+\frac {1}{c x}} x}\\ &=-\frac {b \sqrt {1-c^2 x^2} \sqrt {1+c^2 x^2}}{2 c^5 \sqrt {-1+\frac {1}{c x}} \sqrt {1+\frac {1}{c x}} x}-\frac {\sqrt {1-c^4 x^4} \left (a+b \text {sech}^{-1}(c x)\right )}{2 c^4}+\frac {b \sqrt {1-c^2 x^2} \tanh ^{-1}\left (\sqrt {1+c^2 x^2}\right )}{2 c^5 \sqrt {-1+\frac {1}{c x}} \sqrt {1+\frac {1}{c x}} x}\\ \end {align*}

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Mathematica [A]  time = 0.35, size = 140, normalized size = 0.88 \[ -\frac {a \sqrt {1-c^4 x^4}+\frac {b \sqrt {1-c^4 x^4}}{\sqrt {\frac {1-c x}{c x+1}} (c x+1)}+b \log \left (-\sqrt {\frac {1-c x}{c x+1}} \sqrt {1-c^4 x^4}-c x+1\right )+b \sqrt {1-c^4 x^4} \text {sech}^{-1}(c x)-b \log (x (1-c x))}{2 c^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*ArcSech[c*x]))/Sqrt[1 - c^4*x^4],x]

[Out]

-1/2*(a*Sqrt[1 - c^4*x^4] + (b*Sqrt[1 - c^4*x^4])/(Sqrt[(1 - c*x)/(1 + c*x)]*(1 + c*x)) + b*Sqrt[1 - c^4*x^4]*
ArcSech[c*x] - b*Log[x*(1 - c*x)] + b*Log[1 - c*x - Sqrt[(1 - c*x)/(1 + c*x)]*Sqrt[1 - c^4*x^4]])/c^4

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fricas [B]  time = 0.67, size = 279, normalized size = 1.75 \[ \frac {2 \, \sqrt {-c^{4} x^{4} + 1} b c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 2 \, \sqrt {-c^{4} x^{4} + 1} {\left (b c^{2} x^{2} - b\right )} \log \left (\frac {c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} + 1}{c x}\right ) - {\left (b c^{2} x^{2} - b\right )} \log \left (\frac {c^{2} x^{2} + \sqrt {-c^{4} x^{4} + 1} c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c^{2} x^{2} - 1}\right ) + {\left (b c^{2} x^{2} - b\right )} \log \left (-\frac {c^{2} x^{2} - \sqrt {-c^{4} x^{4} + 1} c x \sqrt {-\frac {c^{2} x^{2} - 1}{c^{2} x^{2}}} - 1}{c^{2} x^{2} - 1}\right ) - 2 \, \sqrt {-c^{4} x^{4} + 1} {\left (a c^{2} x^{2} - a\right )}}{4 \, {\left (c^{6} x^{2} - c^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsech(c*x))/(-c^4*x^4+1)^(1/2),x, algorithm="fricas")

[Out]

1/4*(2*sqrt(-c^4*x^4 + 1)*b*c*x*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) - 2*sqrt(-c^4*x^4 + 1)*(b*c^2*x^2 - b)*log((c*x
*sqrt(-(c^2*x^2 - 1)/(c^2*x^2)) + 1)/(c*x)) - (b*c^2*x^2 - b)*log((c^2*x^2 + sqrt(-c^4*x^4 + 1)*c*x*sqrt(-(c^2
*x^2 - 1)/(c^2*x^2)) - 1)/(c^2*x^2 - 1)) + (b*c^2*x^2 - b)*log(-(c^2*x^2 - sqrt(-c^4*x^4 + 1)*c*x*sqrt(-(c^2*x
^2 - 1)/(c^2*x^2)) - 1)/(c^2*x^2 - 1)) - 2*sqrt(-c^4*x^4 + 1)*(a*c^2*x^2 - a))/(c^6*x^2 - c^4)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {arsech}\left (c x\right ) + a\right )} x^{3}}{\sqrt {-c^{4} x^{4} + 1}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsech(c*x))/(-c^4*x^4+1)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsech(c*x) + a)*x^3/sqrt(-c^4*x^4 + 1), x)

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maple [F]  time = 3.33, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (a +b \,\mathrm {arcsech}\left (c x \right )\right )}{\sqrt {-c^{4} x^{4}+1}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arcsech(c*x))/(-c^4*x^4+1)^(1/2),x)

[Out]

int(x^3*(a+b*arcsech(c*x))/(-c^4*x^4+1)^(1/2),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, b {\left (\frac {{\left (c^{4} x^{4} - 1\right )} \log \left (\sqrt {c x + 1} \sqrt {-c x + 1} + 1\right )}{\sqrt {c^{2} x^{2} + 1} \sqrt {c x + 1} \sqrt {-c x + 1} c^{4}} - 2 \, \int \frac {2 \, c^{2} x^{5} \log \relax (c) + 4 \, c^{2} x^{5} \log \left (\sqrt {x}\right ) + {\left (4 \, c^{2} x^{5} \log \left (\sqrt {x}\right ) + {\left (c^{2} x^{2} {\left (2 \, \log \relax (c) + 1\right )} + 1\right )} x^{3}\right )} e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )}}{2 \, {\left (c^{2} x^{2} e^{\left (\log \left (c x + 1\right ) + \log \left (-c x + 1\right )\right )} + c^{2} x^{2} e^{\left (\frac {1}{2} \, \log \left (c x + 1\right ) + \frac {1}{2} \, \log \left (-c x + 1\right )\right )}\right )} \sqrt {c^{2} x^{2} + 1}}\,{d x}\right )} - \frac {\sqrt {-c^{4} x^{4} + 1} a}{2 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arcsech(c*x))/(-c^4*x^4+1)^(1/2),x, algorithm="maxima")

[Out]

1/2*b*((c^4*x^4 - 1)*log(sqrt(c*x + 1)*sqrt(-c*x + 1) + 1)/(sqrt(c^2*x^2 + 1)*sqrt(c*x + 1)*sqrt(-c*x + 1)*c^4
) - 2*integrate(1/2*(2*c^2*x^5*log(c) + 4*c^2*x^5*log(sqrt(x)) + (4*c^2*x^5*log(sqrt(x)) + (c^2*x^2*(2*log(c)
+ 1) + 1)*x^3)*e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1)))/((c^2*x^2*e^(log(c*x + 1) + log(-c*x + 1)) + c^2*x^2*
e^(1/2*log(c*x + 1) + 1/2*log(-c*x + 1)))*sqrt(c^2*x^2 + 1)), x)) - 1/2*sqrt(-c^4*x^4 + 1)*a/c^4

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\mathrm {acosh}\left (\frac {1}{c\,x}\right )\right )}{\sqrt {1-c^4\,x^4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*acosh(1/(c*x))))/(1 - c^4*x^4)^(1/2),x)

[Out]

int((x^3*(a + b*acosh(1/(c*x))))/(1 - c^4*x^4)^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {x^{3} \left (a + b \operatorname {asech}{\left (c x \right )}\right )}{\sqrt {- \left (c x - 1\right ) \left (c x + 1\right ) \left (c^{2} x^{2} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*asech(c*x))/(-c**4*x**4+1)**(1/2),x)

[Out]

Integral(x**3*(a + b*asech(c*x))/sqrt(-(c*x - 1)*(c*x + 1)*(c**2*x**2 + 1)), x)

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