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

Optimal. Leaf size=126 \[ -\frac{\sqrt{1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^4}+\frac{b x \sqrt{1-c^4 x^4}}{2 c^3 \sqrt{c^2 x^2} \sqrt{c^2 x^2-1}}-\frac{b x \tan ^{-1}\left (\frac{\sqrt{1-c^4 x^4}}{\sqrt{c^2 x^2-1}}\right )}{2 c^3 \sqrt{c^2 x^2}} \]

[Out]

(b*x*Sqrt[1 - c^4*x^4])/(2*c^3*Sqrt[c^2*x^2]*Sqrt[-1 + c^2*x^2]) - (Sqrt[1 - c^4*x^4]*(a + b*ArcSec[c*x]))/(2*
c^4) - (b*x*ArcTan[Sqrt[1 - c^4*x^4]/Sqrt[-1 + c^2*x^2]])/(2*c^3*Sqrt[c^2*x^2])

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Rubi [A]  time = 0.207959, antiderivative size = 135, normalized size of antiderivative = 1.07, number of steps used = 8, number of rules used = 9, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.346, Rules used = {261, 5246, 12, 1572, 1252, 848, 50, 63, 208} \[ -\frac{\sqrt{1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^4}+\frac{b \sqrt{1-c^2 x^2} \sqrt{c^2 x^2+1}}{2 c^5 x \sqrt{1-\frac{1}{c^2 x^2}}}-\frac{b \sqrt{1-c^2 x^2} \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{2 c^5 x \sqrt{1-\frac{1}{c^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*ArcSec[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^2*x^2)]*x) - (Sqrt[1 - c^4*x^4]*(a + b*ArcSec[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^2*x^2)]*x)

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 5246

Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*(u_), x_Symbol] :> With[{v = IntHide[u, x]}, Dist[a + b*ArcSec[c*x], v,
x] - Dist[b/c, Int[SimplifyIntegrand[v/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x], x] /; InverseFunctionFreeQ[v, x]]
/; FreeQ[{a, b, c}, 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 1572

Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^(mn_.))^(q_)*((a_) + (c_.)*(x_)^(n2_.))^(p_.), x_Symbol] :> Dist[(e^IntPart[
q]*(d + e*x^mn)^FracPart[q])/(x^(mn*FracPart[q])*(1 + d/(x^mn*e))^FracPart[q]), Int[x^(m + mn*q)*(1 + d/(x^mn*
e))^q*(a + c*x^n2)^p, x], x] /; FreeQ[{a, c, d, e, m, mn, p, q}, x] && EqQ[n2, -2*mn] &&  !IntegerQ[p] &&  !In
tegerQ[q] && PosQ[n2]

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 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 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]

Rubi steps

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

Mathematica [A]  time = 0.213105, size = 118, normalized size = 0.94 \[ \frac{\frac{\sqrt{1-c^4 x^4} \left (-a c^2 x^2+a+b c x \sqrt{1-\frac{1}{c^2 x^2}}\right )}{c^2 x^2-1}+b \tan ^{-1}\left (\frac{c x \sqrt{1-\frac{1}{c^2 x^2}}}{\sqrt{1-c^4 x^4}}\right )-b \sqrt{1-c^4 x^4} \sec ^{-1}(c x)}{2 c^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.641, size = 0, normalized size = 0. \begin{align*} \int{{x}^{3} \left ( a+b{\rm arcsec} \left (cx\right ) \right ){\frac{1}{\sqrt{-{c}^{4}{x}^{4}+1}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [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*(a+b*arcsec(c*x))/(-c^4*x^4+1)^(1/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: UnboundLocalError

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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*(a+b*asec(c*x))/(-c**4*x**4+1)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b \operatorname{arcsec}\left (c x\right ) + a\right )} x^{3}}{\sqrt{-c^{4} x^{4} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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