Integrand size = 26, antiderivative size = 268 \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\frac {b \sqrt {1-c^2 x^2} \sqrt {1+c^2 x^2}}{3 c^9 \sqrt {1-\frac {1}{c^2 x^2}} x}-\frac {b \sqrt {1-c^2 x^2} \left (1+c^2 x^2\right )^{3/2}}{18 c^9 \sqrt {1-\frac {1}{c^2 x^2}} x}+\frac {b \sqrt {1-c^2 x^2} \left (1+c^2 x^2\right )^{5/2}}{30 c^9 \sqrt {1-\frac {1}{c^2 x^2}} x}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}+\frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {b \sqrt {1-c^2 x^2} \text {arctanh}\left (\sqrt {1+c^2 x^2}\right )}{3 c^9 \sqrt {1-\frac {1}{c^2 x^2}} x} \]
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
Time = 0.28 (sec) , antiderivative size = 159, normalized size of antiderivative = 0.59 \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\frac {-15 a \sqrt {1-c^4 x^4} \left (2+c^4 x^4\right )+\frac {b c \sqrt {1-\frac {1}{c^2 x^2}} x \sqrt {1-c^4 x^4} \left (28+c^2 x^2+3 c^4 x^4\right )}{-1+c^2 x^2}-15 b \sqrt {1-c^4 x^4} \left (2+c^4 x^4\right ) \sec ^{-1}(c x)+30 b \arctan \left (\frac {c \sqrt {1-\frac {1}{c^2 x^2}} x}{\sqrt {1-c^4 x^4}}\right )}{90 c^8} \]
(-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)
Time = 1.23 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.58, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.346, Rules used = {5769, 27, 7272, 1388, 1579, 517, 25, 1584, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx\) |
| \(\Big \downarrow \) 5769 |
| \(\displaystyle -\frac {b \int -\frac {\sqrt {1-c^4 x^4} \left (c^4 x^4+2\right )}{6 c^8 \sqrt {1-\frac {1}{c^2 x^2}} x^2}dx}{c}+\frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b \int \frac {\sqrt {1-c^4 x^4} \left (c^4 x^4+2\right )}{\sqrt {1-\frac {1}{c^2 x^2}} x^2}dx}{6 c^9}+\frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}\) |
| \(\Big \downarrow \) 7272 |
| \(\displaystyle \frac {b \sqrt {1-c^2 x^2} \int \frac {\sqrt {1-c^4 x^4} \left (c^4 x^4+2\right )}{x \sqrt {1-c^2 x^2}}dx}{6 c^9 x \sqrt {1-\frac {1}{c^2 x^2}}}+\frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}\) |
| \(\Big \downarrow \) 1388 |
| \(\displaystyle \frac {b \sqrt {1-c^2 x^2} \int \frac {\sqrt {c^2 x^2+1} \left (c^4 x^4+2\right )}{x}dx}{6 c^9 x \sqrt {1-\frac {1}{c^2 x^2}}}+\frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}\) |
| \(\Big \downarrow \) 1579 |
| \(\displaystyle \frac {b \sqrt {1-c^2 x^2} \int \frac {\sqrt {c^2 x^2+1} \left (c^4 x^4+2\right )}{x^2}dx^2}{12 c^9 x \sqrt {1-\frac {1}{c^2 x^2}}}+\frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}\) |
| \(\Big \downarrow \) 517 |
| \(\displaystyle \frac {b \sqrt {1-c^2 x^2} \int -\frac {x^4 \left (c^4 x^8-2 c^4 x^4+3 c^4\right )}{1-x^4}d\sqrt {c^2 x^2+1}}{6 c^{13} x \sqrt {1-\frac {1}{c^2 x^2}}}+\frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {b \sqrt {1-c^2 x^2} \int \frac {x^4 \left (c^4 x^8-2 c^4 x^4+3 c^4\right )}{1-x^4}d\sqrt {c^2 x^2+1}}{6 c^{13} x \sqrt {1-\frac {1}{c^2 x^2}}}+\frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}\) |
| \(\Big \downarrow \) 1584 |
| \(\displaystyle -\frac {b \sqrt {1-c^2 x^2} \int \left (-c^4 x^8+c^4 x^4-2 c^4+\frac {2 c^4}{1-x^4}\right )d\sqrt {c^2 x^2+1}}{6 c^{13} x \sqrt {1-\frac {1}{c^2 x^2}}}+\frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\left (1-c^4 x^4\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{6 c^8}-\frac {\sqrt {1-c^4 x^4} \left (a+b \sec ^{-1}(c x)\right )}{2 c^8}+\frac {b \sqrt {1-c^2 x^2} \left (-2 c^4 \text {arctanh}\left (\sqrt {c^2 x^2+1}\right )+\frac {c^4 x^{10}}{5}-\frac {c^4 x^6}{3}+2 c^4 \sqrt {c^2 x^2+1}\right )}{6 c^{13} x \sqrt {1-\frac {1}{c^2 x^2}}}\) |
-1/2*(Sqrt[1 - c^4*x^4]*(a + b*ArcSec[c*x]))/c^8 + ((1 - c^4*x^4)^(3/2)*(a + b*ArcSec[c*x]))/(6*c^8) + (b*Sqrt[1 - c^2*x^2]*(-1/3*(c^4*x^6) + (c^4*x ^10)/5 + 2*c^4*Sqrt[1 + c^2*x^2] - 2*c^4*ArcTanh[Sqrt[1 + c^2*x^2]]))/(6*c ^13*Sqrt[1 - 1/(c^2*x^2)]*x)
3.2.71.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((e_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[2*(e^m/d^(m + 2*p + 1)) Subst[Int[x^(2*n + 1)*(-c + x^ 2)^m*(b*c^2 + a*d^2 - 2*b*c*x^2 + b*x^4)^p, x], x, Sqrt[c + d*x]], x] /; Fr eeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && ILtQ[m, 0] && IntegerQ[n + 1/2]
Int[(u_.)*((a_) + (c_.)*(x_)^(n2_.))^(p_.)*((d_) + (e_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[u*(d + e*x^n)^(p + q)*(a/d + (c/e)*x^n)^p, x] /; FreeQ[{a, c, d, e, n, p, q}, x] && EqQ[n2, 2*n] && EqQ[c*d^2 + a*e^2, 0] && (Integer Q[p] || (GtQ[a, 0] && GtQ[d, 0]))
Int[(x_)^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (c_.)*(x_)^4)^(p_.), x_S ymbol] :> Simp[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]
Int[((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.)*((a_) + (b_.)*(x_)^2 + ( c_.)*(x_)^4)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(f*x)^m*(d + e*x^2)^q* (a + b*x^2 + c*x^4)^p, x], x] /; FreeQ[{a, b, c, d, e, f, m, q}, x] && NeQ[ b^2 - 4*a*c, 0] && IGtQ[p, 0] && IGtQ[q, -2]
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*(u_), x_Symbol] :> With[{v = IntHide [u, x]}, Simp[(a + b*ArcSec[c*x]) v, x] - Simp[b/c Int[SimplifyIntegran d[v/(x^2*Sqrt[1 - 1/(c^2*x^2)]), x], x], x] /; InverseFunctionFreeQ[v, x]] /; FreeQ[{a, b, c}, x]
Int[(u_.)*((a_.) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[b^IntPart[p]*(( a + b*x^n)^FracPart[p]/(x^(n*FracPart[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]
\[\int \frac {x^{7} \left (a +b \,\operatorname {arcsec}\left (c x \right )\right )}{\sqrt {-c^{4} x^{4}+1}}d x\]
Time = 0.28 (sec) , antiderivative size = 181, normalized size of antiderivative = 0.68 \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\frac {{\left (3 \, b c^{4} x^{4} + b c^{2} x^{2} + 28 \, b\right )} \sqrt {-c^{4} x^{4} + 1} \sqrt {c^{2} x^{2} - 1} - 30 \, {\left (b c^{2} x^{2} - b\right )} \arctan \left (\frac {\sqrt {-c^{4} x^{4} + 1}}{\sqrt {c^{2} x^{2} - 1}}\right ) - 15 \, {\left (a c^{6} x^{6} - a c^{4} x^{4} + 2 \, a c^{2} x^{2} + {\left (b c^{6} x^{6} - b c^{4} x^{4} + 2 \, b c^{2} x^{2} - 2 \, b\right )} \operatorname {arcsec}\left (c x\right ) - 2 \, a\right )} \sqrt {-c^{4} x^{4} + 1}}{90 \, {\left (c^{10} x^{2} - c^{8}\right )}} \]
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)
Timed out. \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\text {Timed out} \]
\[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{7}}{\sqrt {-c^{4} x^{4} + 1}} \,d x } \]
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)*log(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)))*sqr t(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)*sqrt(c*x + 1)*sqrt(-c*x + 1)*c^8)
Exception generated. \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\text {Exception raised: TypeError} \]
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Value
Timed out. \[ \int \frac {x^7 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {1-c^4 x^4}} \, dx=\int \frac {x^7\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{\sqrt {1-c^4\,x^4}} \,d x \]