Integrand size = 23, antiderivative size = 294 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=-\frac {b \left (c^2 d+9 e\right ) x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{120 c^3 e \sqrt {c^2 x^2}}-\frac {b x \sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c e \sqrt {c^2 x^2}}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {2 b c d^{5/2} x \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{15 e^2 \sqrt {c^2 x^2}}+\frac {b \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) x \text {arctanh}\left (\frac {\sqrt {e} \sqrt {-1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{120 c^4 e^{3/2} \sqrt {c^2 x^2}} \]
-1/3*d*(e*x^2+d)^(3/2)*(a+b*arcsec(c*x))/e^2+1/5*(e*x^2+d)^(5/2)*(a+b*arcs ec(c*x))/e^2-2/15*b*c*d^(5/2)*x*arctan((e*x^2+d)^(1/2)/d^(1/2)/(c^2*x^2-1) ^(1/2))/e^2/(c^2*x^2)^(1/2)+1/120*b*(15*c^4*d^2-10*c^2*d*e-9*e^2)*x*arctan h(e^(1/2)*(c^2*x^2-1)^(1/2)/c/(e*x^2+d)^(1/2))/c^4/e^(3/2)/(c^2*x^2)^(1/2) -1/20*b*x*(e*x^2+d)^(3/2)*(c^2*x^2-1)^(1/2)/c/e/(c^2*x^2)^(1/2)-1/120*b*(c ^2*d+9*e)*x*(c^2*x^2-1)^(1/2)*(e*x^2+d)^(1/2)/c^3/e/(c^2*x^2)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.70 (sec) , antiderivative size = 263, normalized size of antiderivative = 0.89 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {16 a \left (d+e x^2\right )^2 \left (-2 d+3 e x^2\right )-\frac {2 b e \sqrt {1-\frac {1}{c^2 x^2}} x \left (d+e x^2\right ) \left (9 e+c^2 \left (7 d+6 e x^2\right )\right )}{c^3}+\frac {b \left (-16 c^2 d^3 \sqrt {1+\frac {d}{e x^2}} \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,\frac {1}{c^2 x^2},-\frac {d}{e x^2}\right )+\frac {e \left (-15 c^4 d^2+10 c^2 d e+9 e^2\right ) \sqrt {1-\frac {1}{c^2 x^2}} x^4 \sqrt {1+\frac {e x^2}{d}} \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,c^2 x^2,-\frac {e x^2}{d}\right )}{\sqrt {1-c^2 x^2}}\right )}{c^3 x}+16 b \left (d+e x^2\right )^2 \left (-2 d+3 e x^2\right ) \sec ^{-1}(c x)}{240 e^2 \sqrt {d+e x^2}} \]
(16*a*(d + e*x^2)^2*(-2*d + 3*e*x^2) - (2*b*e*Sqrt[1 - 1/(c^2*x^2)]*x*(d + e*x^2)*(9*e + c^2*(7*d + 6*e*x^2)))/c^3 + (b*(-16*c^2*d^3*Sqrt[1 + d/(e*x ^2)]*AppellF1[1, 1/2, 1/2, 2, 1/(c^2*x^2), -(d/(e*x^2))] + (e*(-15*c^4*d^2 + 10*c^2*d*e + 9*e^2)*Sqrt[1 - 1/(c^2*x^2)]*x^4*Sqrt[1 + (e*x^2)/d]*Appel lF1[1, 1/2, 1/2, 2, c^2*x^2, -((e*x^2)/d)])/Sqrt[1 - c^2*x^2]))/(c^3*x) + 16*b*(d + e*x^2)^2*(-2*d + 3*e*x^2)*ArcSec[c*x])/(240*e^2*Sqrt[d + e*x^2])
Time = 0.58 (sec) , antiderivative size = 269, normalized size of antiderivative = 0.91, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.522, Rules used = {5761, 27, 435, 171, 27, 171, 27, 175, 66, 104, 217, 221}
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.
| \(\displaystyle \int x^3 \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx\) |
| \(\Big \downarrow \) 5761 |
| \(\displaystyle -\frac {b c x \int -\frac {\left (2 d-3 e x^2\right ) \left (e x^2+d\right )^{3/2}}{15 e^2 x \sqrt {c^2 x^2-1}}dx}{\sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c x \int \frac {\left (2 d-3 e x^2\right ) \left (e x^2+d\right )^{3/2}}{x \sqrt {c^2 x^2-1}}dx}{15 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}\) |
| \(\Big \downarrow \) 435 |
| \(\displaystyle \frac {b c x \int \frac {\left (2 d-3 e x^2\right ) \left (e x^2+d\right )^{3/2}}{x^2 \sqrt {c^2 x^2-1}}dx^2}{30 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {b c x \left (\frac {\int \frac {\sqrt {e x^2+d} \left (8 c^2 d^2-e \left (d c^2+9 e\right ) x^2\right )}{2 x^2 \sqrt {c^2 x^2-1}}dx^2}{2 c^2}-\frac {3 e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{30 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c x \left (\frac {\int \frac {\sqrt {e x^2+d} \left (8 c^2 d^2-e \left (d c^2+9 e\right ) x^2\right )}{x^2 \sqrt {c^2 x^2-1}}dx^2}{4 c^2}-\frac {3 e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{30 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {b c x \left (\frac {\frac {\int \frac {16 d^3 c^4+e \left (15 d^2 c^4-10 d e c^2-9 e^2\right ) x^2}{2 x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{c^2}-\frac {e \sqrt {c^2 x^2-1} \left (c^2 d+9 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}-\frac {3 e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{30 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {b c x \left (\frac {\frac {\int \frac {16 d^3 c^4+e \left (15 d^2 c^4-10 d e c^2-9 e^2\right ) x^2}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \left (c^2 d+9 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}-\frac {3 e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{30 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {b c x \left (\frac {\frac {16 c^4 d^3 \int \frac {1}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2+e \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) \int \frac {1}{\sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \left (c^2 d+9 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}-\frac {3 e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{30 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {b c x \left (\frac {\frac {16 c^4 d^3 \int \frac {1}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2+2 e \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \left (c^2 d+9 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}-\frac {3 e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{30 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {b c x \left (\frac {\frac {32 c^4 d^3 \int \frac {1}{-x^4-d}d\frac {\sqrt {e x^2+d}}{\sqrt {c^2 x^2-1}}+2 e \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \left (c^2 d+9 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}-\frac {3 e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{30 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {b c x \left (\frac {\frac {2 e \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}-32 c^4 d^{5/2} \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \left (c^2 d+9 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}-\frac {3 e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{30 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e^2}-\frac {d \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}+\frac {b c x \left (\frac {\frac {\frac {2 \sqrt {e} \left (15 c^4 d^2-10 c^2 d e-9 e^2\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{c}-32 c^4 d^{5/2} \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \left (c^2 d+9 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}-\frac {3 e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{30 e^2 \sqrt {c^2 x^2}}\) |
-1/3*(d*(d + e*x^2)^(3/2)*(a + b*ArcSec[c*x]))/e^2 + ((d + e*x^2)^(5/2)*(a + b*ArcSec[c*x]))/(5*e^2) + (b*c*x*((-3*e*Sqrt[-1 + c^2*x^2]*(d + e*x^2)^ (3/2))/(2*c^2) + (-((e*(c^2*d + 9*e)*Sqrt[-1 + c^2*x^2]*Sqrt[d + e*x^2])/c ^2) + (-32*c^4*d^(5/2)*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[-1 + c^2*x^2]) ] + (2*Sqrt[e]*(15*c^4*d^2 - 10*c^2*d*e - 9*e^2)*ArcTanh[(Sqrt[e]*Sqrt[-1 + c^2*x^2])/(c*Sqrt[d + e*x^2])])/c)/(2*c^2))/(4*c^2)))/(30*e^2*Sqrt[c^2*x ^2])
3.2.12.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && !GtQ[c - a*(d/b), 0]
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x _)), x_] :> With[{q = Denominator[m]}, Simp[q Subst[Int[x^(q*(m + 1) - 1) /(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^(1/q)], x] ] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && L tQ[-1, m, 0] && SimplerQ[a + b*x, c + d*x]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[h*(a + b*x)^m*(c + d*x)^(n + 1)*(( e + f*x)^(p + 1)/(d*f*(m + n + p + 2))), x] + Simp[1/(d*f*(m + n + p + 2)) Int[(a + b*x)^(m - 1)*(c + d*x)^n*(e + f*x)^p*Simp[a*d*f*g*(m + n + p + 2 ) - h*(b*c*e*m + a*(d*e*(n + 1) + c*f*(p + 1))) + (b*d*f*g*(m + n + p + 2) + h*(a*d*f*m - b*(d*e*(m + n + 1) + c*f*(m + p + 1))))*x, x], x], x] /; Fre eQ[{a, b, c, d, e, f, g, h, n, p}, x] && GtQ[m, 0] && NeQ[m + n + p + 2, 0] && IntegersQ[2*m, 2*n, 2*p]
Int[(((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^(p_)*((g_.) + (h_.)*(x_ )))/((a_.) + (b_.)*(x_)), x_] :> Simp[h/b Int[(c + d*x)^n*(e + f*x)^p, x] , x] + Simp[(b*g - a*h)/b Int[(c + d*x)^n*((e + f*x)^p/(a + b*x)), x], x] /; FreeQ[{a, b, c, d, e, f, g, h, n, p}, x]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(-(Rt[-a, 2]*Rt[-b, 2])^( -1))*ArcTan[Rt[-b, 2]*(x/Rt[-a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] & & (LtQ[a, 0] || LtQ[b, 0])
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]
Int[(x_)^(m_.)*((a_.) + (b_.)*(x_)^2)^(p_.)*((c_.) + (d_.)*(x_)^2)^(q_.)*(( e_.) + (f_.)*(x_)^2)^(r_.), x_Symbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2) *(a + b*x)^p*(c + d*x)^q*(e + f*x)^r, x], x, x^2], x] /; FreeQ[{a, b, c, d, e, f, p, q, r}, x] && IntegerQ[(m - 1)/2]
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*((f_.)*(x_))^(m_.)*((d_.) + (e_.)*(x _)^2)^(p_.), x_Symbol] :> With[{u = IntHide[(f*x)^m*(d + e*x^2)^p, x]}, Sim p[(a + b*ArcSec[c*x]) u, x] - Simp[b*c*(x/Sqrt[c^2*x^2]) Int[SimplifyIn tegrand[u/(x*Sqrt[c^2*x^2 - 1]), x], x], x]] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && ((IGtQ[p, 0] && !(ILtQ[(m - 1)/2, 0] && GtQ[m + 2*p + 3, 0])) | | (IGtQ[(m + 1)/2, 0] && !(ILtQ[p, 0] && GtQ[m + 2*p + 3, 0])) || (ILtQ[(m + 2*p + 1)/2, 0] && !ILtQ[(m - 1)/2, 0]))
\[\int x^{3} \left (a +b \,\operatorname {arcsec}\left (c x \right )\right ) \sqrt {e \,x^{2}+d}d x\]
Time = 1.09 (sec) , antiderivative size = 1383, normalized size of antiderivative = 4.70 \[ \int x^3 \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\text {Too large to display} \]
[1/480*(16*b*c^5*sqrt(-d)*d^2*log(((c^4*d^2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^ 2*d^2 - d*e)*x^2 + 4*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(-d) + 8*d^2)/x^4) - (15*b*c^4*d^2 - 10*b*c^2*d*e - 9*b*e^2)*sqrt (e)*log(8*c^4*e^2*x^4 + c^4*d^2 - 6*c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^3*e*x^2 + c^3*d - c*e)*sqrt(c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(e) + e^2) + 4*(24*a*c^5*e^2*x^4 + 8*a*c^5*d*e*x^2 - 16*a*c^5*d^2 + 8*(3*b*c^5*e ^2*x^4 + b*c^5*d*e*x^2 - 2*b*c^5*d^2)*arcsec(c*x) - (6*b*c^3*e^2*x^2 + 7*b *c^3*d*e + 9*b*c*e^2)*sqrt(c^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^5*e^2), -1/48 0*(32*b*c^5*d^(5/2)*arctan(-1/2*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)* sqrt(e*x^2 + d)*sqrt(d)/(c^2*d*e*x^4 + (c^2*d^2 - d*e)*x^2 - d^2)) + (15*b *c^4*d^2 - 10*b*c^2*d*e - 9*b*e^2)*sqrt(e)*log(8*c^4*e^2*x^4 + c^4*d^2 - 6 *c^2*d*e + 8*(c^4*d*e - c^2*e^2)*x^2 - 4*(2*c^3*e*x^2 + c^3*d - c*e)*sqrt( c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(e) + e^2) - 4*(24*a*c^5*e^2*x^4 + 8*a*c^ 5*d*e*x^2 - 16*a*c^5*d^2 + 8*(3*b*c^5*e^2*x^4 + b*c^5*d*e*x^2 - 2*b*c^5*d^ 2)*arcsec(c*x) - (6*b*c^3*e^2*x^2 + 7*b*c^3*d*e + 9*b*c*e^2)*sqrt(c^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^5*e^2), 1/240*(8*b*c^5*sqrt(-d)*d^2*log(((c^4*d^ 2 - 6*c^2*d*e + e^2)*x^4 - 8*(c^2*d^2 - d*e)*x^2 + 4*sqrt(c^2*x^2 - 1)*((c ^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqrt(-d) + 8*d^2)/x^4) - (15*b*c^4*d^ 2 - 10*b*c^2*d*e - 9*b*e^2)*sqrt(-e)*arctan(1/2*(2*c^2*e*x^2 + c^2*d - e)* sqrt(c^2*x^2 - 1)*sqrt(e*x^2 + d)*sqrt(-e)/(c^3*e^2*x^4 - c*d*e + (c^3*...
\[ \int x^3 \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int x^{3} \left (a + b \operatorname {asec}{\left (c x \right )}\right ) \sqrt {d + e x^{2}}\, dx \]
Exception generated. \[ \int x^3 \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int x^3 \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { \sqrt {e x^{2} + d} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{3} \,d x } \]
Timed out. \[ \int x^3 \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int x^3\,\sqrt {e\,x^2+d}\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \]