\(\int \frac {x^7 (a+b \sec ^{-1}(c x))}{\sqrt {1-c^4 x^4}} \, dx\) [171]

Optimal result

Integrand size = 26, antiderivative size = 268 \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\frac {b \sqrt {1-c^2 x^2} \sqrt {1+c^2 x^2}}{3 c^9 \sqrt {1-\frac {1}{c^2 x^2}} x}-\frac {b \sqrt {1-c^2 x^2} \left (1+c^2 x^2\right )^{3/2}}{18 c^9 \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {b \sqrt {1-c^2 x^2} \left (1+c^2 x^2\right )^{5/2}}{30 c^9 \sqrt {1-\frac {1}{c^2 x^2}} x}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}+\frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {b \sqrt {1-c^2 x^2} \text {arctanh}\left (\sqrt {1+c^2 x^2}\right )}{3 c^9 \sqrt {1-\frac {1}{c^2 x^2}} x} \]

[Out]

1/6*(-c^4*x^4+1)^(3/2)*(a+b*arcsec(c*x))/c^8-1/18*b*(c^2*x^2+1)^(3/2)*(-c^2*x^2+1)^(1/2)/c^9/x/(1-1/c^2/x^2)^(
1/2)+1/30*b*(c^2*x^2+1)^(5/2)*(-c^2*x^2+1)^(1/2)/c^9/x/(1-1/c^2/x^2)^(1/2)-1/3*b*arctanh((c^2*x^2+1)^(1/2))*(-
c^2*x^2+1)^(1/2)/c^9/x/(1-1/c^2/x^2)^(1/2)+1/3*b*(-c^2*x^2+1)^(1/2)*(c^2*x^2+1)^(1/2)/c^9/x/(1-1/c^2/x^2)^(1/2
)-1/2*(a+b*arcsec(c*x))*(-c^4*x^4+1)^(1/2)/c^8

Rubi [A] (verified)

Time = 1.50 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.423, Rules used = {272, 45, 5354, 12, 6853, 6874, 862, 52, 65, 214, 797} \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}-\frac {b \sqrt {1-c^2 x^2} \text {arctanh}\left (\sqrt {c^2 x^2+1}\right )}{3 c^9 x \sqrt {1-\frac {1}{c^2 x^2}}}+\frac {b \sqrt {1-c^2 x^2} \left (c^2 x^2+1\right )^{5/2}}{30 c^9 x \sqrt {1-\frac {1}{c^2 x^2}}}-\frac {b \sqrt {1-c^2 x^2} \left (c^2 x^2+1\right )^{3/2}}{18 c^9 x \sqrt {1-\frac {1}{c^2 x^2}}}+\frac {b \sqrt {1-c^2 x^2} \sqrt {c^2 x^2+1}}{3 c^9 x \sqrt {1-\frac {1}{c^2 x^2}}} \]

[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 12

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

Rule 45

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 52

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 65

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 214

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

Rule 272

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 797

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/e)*x)^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]))

Rule 862

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/e)*x)^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 5354

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 6853

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

Rule 6874

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

Rubi steps \begin{align*} \text {integral}& = -\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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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} \text {arctanh}\left (\sqrt {1+c^2 x^2}\right )}{3 c^9 \sqrt {1-\frac {1}{c^2 x^2}} x} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.59 \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\frac {-15 a \sqrt {1-c^4 x^4} \left (2+c^4 x^4\right )+\frac {b c \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {1-c^4 x^4} \left (28+c^2 x^2+3 c^4 x^4\right )}{-1+c^2 x^2}-15 b \sqrt {1-c^4 x^4} \left (2+c^4 x^4\right ) \sec ^{-1}(c x)+30 b \arctan \left (\frac {c \sqrt {1-\frac {1}{c^2 x^2}} x}{\sqrt {1-c^4 x^4}}\right )}{90 c^8} \]

[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)

Maple [F]

\[\int \frac {x^{7} \left (a +b \,\operatorname {arcsec}\left (c x \right )\right )}{\sqrt {-c^{4} x^{4}+1}}d x\]

[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)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.68 \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\frac {{\left (3 \, b c^{4} x^{4} + b c^{2} x^{2} + 28 \, b\right )} \sqrt {-c^{4} x^{4} + 1} \sqrt {c^{2} x^{2} - 1} - 30 \, {\left (b c^{2} x^{2} - b\right )} \arctan \left (\frac {\sqrt {-c^{4} x^{4} + 1}}{\sqrt {c^{2} x^{2} - 1}}\right ) - 15 \, {\left (a c^{6} x^{6} - a c^{4} x^{4} + 2 \, a c^{2} x^{2} + {\left (b c^{6} x^{6} - b c^{4} x^{4} + 2 \, b c^{2} x^{2} - 2 \, b\right )} \operatorname {arcsec}\left (c x\right ) - 2 \, a\right )} \sqrt {-c^{4} x^{4} + 1}}{90 \, {\left (c^{10} x^{2} - c^{8}\right )}} \]

[In]

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

[Out]

1/90*((3*b*c^4*x^4 + b*c^2*x^2 + 28*b)*sqrt(-c^4*x^4 + 1)*sqrt(c^2*x^2 - 1) - 30*(b*c^2*x^2 - b)*arctan(sqrt(-
c^4*x^4 + 1)/sqrt(c^2*x^2 - 1)) - 15*(a*c^6*x^6 - a*c^4*x^4 + 2*a*c^2*x^2 + (b*c^6*x^6 - b*c^4*x^4 + 2*b*c^2*x
^2 - 2*b)*arcsec(c*x) - 2*a)*sqrt(-c^4*x^4 + 1))/(c^10*x^2 - c^8)

Sympy [F(-1)]

Timed out. \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\text {Timed out} \]

[In]

integrate(x**7*(a+b*asec(c*x))/(-c**4*x**4+1)**(1/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{7}}{\sqrt {-c^{4} x^{4} + 1}} \,d x } \]

[In]

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

[Out]

1/6*a*((-c^4*x^4 + 1)^(3/2)/c^8 - 3*sqrt(-c^4*x^4 + 1)/c^8) - 1/6*(6*sqrt(c^2*x^2 + 1)*sqrt(c*x + 1)*sqrt(-c*x
+ 1)*c^8*integrate((6*sqrt(c*x + 1)*c^6*x^7*log(c) + (c^4*x^5 + (6*c^6*log(c) + c^6)*x^7 + 2*c^2*x^3 + 2*x)*e
^(3/2*log(c*x + 1) + log(c*x - 1)) + 6*(c^6*x^7*e^(3/2*log(c*x + 1) + log(c*x - 1)) + sqrt(c*x + 1)*c^6*x^7)*l
og(x))/((c^6*e^(2*log(c*x + 1) + log(c*x - 1) + 1/2*log(-c*x + 1)) + c^6*e^(log(c*x + 1) + 1/2*log(-c*x + 1)))
*sqrt(c^2*x^2 + 1)), x) - (c^8*x^8 + c^4*x^4 - 2)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)))*b/(sqrt(c^2*x^2 + 1)*sq
rt(c*x + 1)*sqrt(-c*x + 1)*c^8)

Giac [F(-2)]

Exception generated. \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\text {Exception raised: TypeError} \]

[In]

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

[Out]

Exception raised: TypeError >> an error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value

Mupad [F(-1)]

Timed out. \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\int \frac {x^7\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{\sqrt {1-c^4\,x^4}} \,d x \]

[In]

int((x^7*(a + b*acos(1/(c*x))))/(1 - c^4*x^4)^(1/2),x)

[Out]

int((x^7*(a + b*acos(1/(c*x))))/(1 - c^4*x^4)^(1/2), x)