3.171 \(\int \frac{x^7 (a+b \sec ^{-1}(c x))}{\sqrt{1-c^4 x^4}} \, dx\)

Optimal. Leaf size=268 \[ \frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac{\sqrt{1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}+\frac{b \sqrt{1-c^2 x^2} \left (c^2 x^2+1\right )^{5/2}}{30 c^9 x \sqrt{1-\frac{1}{c^2 x^2}}}-\frac{b \sqrt{1-c^2 x^2} \left (c^2 x^2+1\right )^{3/2}}{18 c^9 x \sqrt{1-\frac{1}{c^2 x^2}}}+\frac{b \sqrt{1-c^2 x^2} \sqrt{c^2 x^2+1}}{3 c^9 x \sqrt{1-\frac{1}{c^2 x^2}}}-\frac{b \sqrt{1-c^2 x^2} \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{3 c^9 x \sqrt{1-\frac{1}{c^2 x^2}}} \]

[Out]

(b*Sqrt[1 - c^2*x^2]*Sqrt[1 + c^2*x^2])/(3*c^9*Sqrt[1 - 1/(c^2*x^2)]*x) - (b*Sqrt[1 - c^2*x^2]*(1 + c^2*x^2)^(
3/2))/(18*c^9*Sqrt[1 - 1/(c^2*x^2)]*x) + (b*Sqrt[1 - c^2*x^2]*(1 + c^2*x^2)^(5/2))/(30*c^9*Sqrt[1 - 1/(c^2*x^2
)]*x) - (Sqrt[1 - c^4*x^4]*(a + b*ArcSec[c*x]))/(2*c^8) + ((1 - c^4*x^4)^(3/2)*(a + b*ArcSec[c*x]))/(6*c^8) -
(b*Sqrt[1 - c^2*x^2]*ArcTanh[Sqrt[1 + c^2*x^2]])/(3*c^9*Sqrt[1 - 1/(c^2*x^2)]*x)

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Rubi [A]  time = 2.01752, antiderivative size = 268, normalized size of antiderivative = 1., number of steps used = 13, number of rules used = 11, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.423, Rules used = {266, 43, 5246, 12, 6721, 6742, 848, 50, 63, 208, 783} \[ \frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac{\sqrt{1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}+\frac{b \sqrt{1-c^2 x^2} \left (c^2 x^2+1\right )^{5/2}}{30 c^9 x \sqrt{1-\frac{1}{c^2 x^2}}}-\frac{b \sqrt{1-c^2 x^2} \left (c^2 x^2+1\right )^{3/2}}{18 c^9 x \sqrt{1-\frac{1}{c^2 x^2}}}+\frac{b \sqrt{1-c^2 x^2} \sqrt{c^2 x^2+1}}{3 c^9 x \sqrt{1-\frac{1}{c^2 x^2}}}-\frac{b \sqrt{1-c^2 x^2} \tanh ^{-1}\left (\sqrt{c^2 x^2+1}\right )}{3 c^9 x \sqrt{1-\frac{1}{c^2 x^2}}} \]

Antiderivative was successfully verified.

[In]

Int[(x^7*(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])/(3*c^9*Sqrt[1 - 1/(c^2*x^2)]*x) - (b*Sqrt[1 - c^2*x^2]*(1 + c^2*x^2)^(
3/2))/(18*c^9*Sqrt[1 - 1/(c^2*x^2)]*x) + (b*Sqrt[1 - c^2*x^2]*(1 + c^2*x^2)^(5/2))/(30*c^9*Sqrt[1 - 1/(c^2*x^2
)]*x) - (Sqrt[1 - c^4*x^4]*(a + b*ArcSec[c*x]))/(2*c^8) + ((1 - c^4*x^4)^(3/2)*(a + b*ArcSec[c*x]))/(6*c^8) -
(b*Sqrt[1 - c^2*x^2]*ArcTanh[Sqrt[1 + c^2*x^2]])/(3*c^9*Sqrt[1 - 1/(c^2*x^2)]*x)

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 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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 6721

Int[(u_.)*((a_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[(b^IntPart[p]*(a + b*x^n)^FracPart[p])/(x^(n*FracP
art[p])*(1 + a/(x^n*b))^FracPart[p]), Int[u*x^(n*p)*(1 + a/(x^n*b))^p, x], x] /; FreeQ[{a, b, p}, x] &&  !Inte
gerQ[p] && ILtQ[n, 0] &&  !RationalFunctionQ[u, x] && IntegerQ[p + 1/2]

Rule 6742

Int[u_, x_Symbol] :> With[{v = ExpandIntegrand[u, x]}, Int[v, x] /; SumQ[v]]

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]

Rule 783

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

Rubi steps

\begin{align*} \int \frac{x^7 \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^8}+\frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac{b \int \frac{\left (-2-c^4 x^4\right ) \sqrt{1-c^4 x^4}}{6 c^8 \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^8}+\frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac{b \int \frac{\left (-2-c^4 x^4\right ) \sqrt{1-c^4 x^4}}{\sqrt{1-\frac{1}{c^2 x^2}} x^2} \, dx}{6 c^9}\\ &=-\frac{\sqrt{1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}+\frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac{\left (b \sqrt{1-c^2 x^2}\right ) \int \frac{\left (-2-c^4 x^4\right ) \sqrt{1-c^4 x^4}}{x \sqrt{1-c^2 x^2}} \, dx}{6 c^9 \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^8}+\frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-c^4 x^2} \left (2+c^4 x^2\right )}{x \sqrt{1-c^2 x}} \, dx,x,x^2\right )}{12 c^9 \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^8}+\frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (\frac{2 \sqrt{1-c^4 x^2}}{x \sqrt{1-c^2 x}}+\frac{c^4 x \sqrt{1-c^4 x^2}}{\sqrt{1-c^2 x}}\right ) \, dx,x,x^2\right )}{12 c^9 \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^8}+\frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}+\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 )}{6 c^9 \sqrt{1-\frac{1}{c^2 x^2}} x}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \frac{x \sqrt{1-c^4 x^2}}{\sqrt{1-c^2 x}} \, dx,x,x^2\right )}{12 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^8}+\frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}+\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 )}{6 c^9 \sqrt{1-\frac{1}{c^2 x^2}} x}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int x \sqrt{1+c^2 x} \, dx,x,x^2\right )}{12 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}}{3 c^9 \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^8}+\frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}+\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 )}{6 c^9 \sqrt{1-\frac{1}{c^2 x^2}} x}+\frac{\left (b \sqrt{1-c^2 x^2}\right ) \operatorname{Subst}\left (\int \left (-\frac{\sqrt{1+c^2 x}}{c^2}+\frac{\left (1+c^2 x\right )^{3/2}}{c^2}\right ) \, dx,x,x^2\right )}{12 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}}{3 c^9 \sqrt{1-\frac{1}{c^2 x^2}} x}-\frac{b \sqrt{1-c^2 x^2} \left (1+c^2 x^2\right )^{3/2}}{18 c^9 \sqrt{1-\frac{1}{c^2 x^2}} x}+\frac{b \sqrt{1-c^2 x^2} \left (1+c^2 x^2\right )^{5/2}}{30 c^9 \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^8}+\frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}+\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 )}{3 c^{11} \sqrt{1-\frac{1}{c^2 x^2}} x}\\ &=\frac{b \sqrt{1-c^2 x^2} \sqrt{1+c^2 x^2}}{3 c^9 \sqrt{1-\frac{1}{c^2 x^2}} x}-\frac{b \sqrt{1-c^2 x^2} \left (1+c^2 x^2\right )^{3/2}}{18 c^9 \sqrt{1-\frac{1}{c^2 x^2}} x}+\frac{b \sqrt{1-c^2 x^2} \left (1+c^2 x^2\right )^{5/2}}{30 c^9 \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^8}+\frac{\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac{b \sqrt{1-c^2 x^2} \tanh ^{-1}\left (\sqrt{1+c^2 x^2}\right )}{3 c^9 \sqrt{1-\frac{1}{c^2 x^2}} x}\\ \end{align*}

Mathematica [A]  time = 0.316417, size = 159, normalized size = 0.59 \[ \frac{-15 a \sqrt{1-c^4 x^4} \left (c^4 x^4+2\right )+\frac{b c x \sqrt{1-\frac{1}{c^2 x^2}} \sqrt{1-c^4 x^4} \left (3 c^4 x^4+c^2 x^2+28\right )}{c^2 x^2-1}+30 b \tan ^{-1}\left (\frac{c x \sqrt{1-\frac{1}{c^2 x^2}}}{\sqrt{1-c^4 x^4}}\right )-15 b \sqrt{1-c^4 x^4} \left (c^4 x^4+2\right ) \sec ^{-1}(c x)}{90 c^8} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^7*(a + b*ArcSec[c*x]))/Sqrt[1 - c^4*x^4],x]

[Out]

(-15*a*Sqrt[1 - c^4*x^4]*(2 + c^4*x^4) + (b*c*Sqrt[1 - 1/(c^2*x^2)]*x*Sqrt[1 - c^4*x^4]*(28 + c^2*x^2 + 3*c^4*
x^4))/(-1 + c^2*x^2) - 15*b*Sqrt[1 - c^4*x^4]*(2 + c^4*x^4)*ArcSec[c*x] + 30*b*ArcTan[(c*Sqrt[1 - 1/(c^2*x^2)]
*x)/Sqrt[1 - c^4*x^4]])/(90*c^8)

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Maple [F]  time = 4.461, size = 0, normalized size = 0. \begin{align*} \int{{x}^{7} \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^7*(a+b*arcsec(c*x))/(-c^4*x^4+1)^(1/2),x)

[Out]

int(x^7*(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^7*(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^7*(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**7*(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^{7}}{\sqrt{-c^{4} x^{4} + 1}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7*(a+b*arcsec(c*x))/(-c^4*x^4+1)^(1/2),x, algorithm="giac")

[Out]

integrate((b*arcsec(c*x) + a)*x^7/sqrt(-c^4*x^4 + 1), x)