Integrand size = 21, antiderivative size = 262 \[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=-\frac {b \left (7 c^2 d+3 e\right ) x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{40 c^3 \sqrt {c^2 x^2}}-\frac {b x \sqrt {-1+c^2 x^2} \left (d+e x^2\right )^{3/2}}{20 c \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}+\frac {b c d^{5/2} x \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{5 e \sqrt {c^2 x^2}}-\frac {b \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) x \text {arctanh}\left (\frac {\sqrt {e} \sqrt {-1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{40 c^4 \sqrt {e} \sqrt {c^2 x^2}} \]
1/5*(e*x^2+d)^(5/2)*(a+b*arcsec(c*x))/e+1/5*b*c*d^(5/2)*x*arctan((e*x^2+d) ^(1/2)/d^(1/2)/(c^2*x^2-1)^(1/2))/e/(c^2*x^2)^(1/2)-1/40*b*(15*c^4*d^2+10* c^2*d*e+3*e^2)*x*arctanh(e^(1/2)*(c^2*x^2-1)^(1/2)/c/(e*x^2+d)^(1/2))/c^4/ e^(1/2)/(c^2*x^2)^(1/2)-1/20*b*x*(e*x^2+d)^(3/2)*(c^2*x^2-1)^(1/2)/c/(c^2* x^2)^(1/2)-1/40*b*(7*c^2*d+3*e)*x*(c^2*x^2-1)^(1/2)*(e*x^2+d)^(1/2)/c^3/(c ^2*x^2)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.63 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.94 \[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\frac {\frac {16 a \left (d+e x^2\right )^3}{e}-\frac {2 b \sqrt {1-\frac {1}{c^2 x^2}} x \left (d+e x^2\right ) \left (3 e+c^2 \left (9 d+2 e x^2\right )\right )}{c^3}+\frac {b \left (\frac {8 c^2 d^3 \sqrt {1+\frac {d}{e x^2}} \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,\frac {1}{c^2 x^2},-\frac {d}{e x^2}\right )}{e}+\frac {\left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \sqrt {1-\frac {1}{c^2 x^2}} x^4 \sqrt {1+\frac {e x^2}{d}} \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,c^2 x^2,-\frac {e x^2}{d}\right )}{\sqrt {1-c^2 x^2}}\right )}{c^3 x}+\frac {16 b \left (d+e x^2\right )^3 \sec ^{-1}(c x)}{e}}{80 \sqrt {d+e x^2}} \]
((16*a*(d + e*x^2)^3)/e - (2*b*Sqrt[1 - 1/(c^2*x^2)]*x*(d + e*x^2)*(3*e + c^2*(9*d + 2*e*x^2)))/c^3 + (b*((8*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 + 3*e^2)*Sqrt[1 - 1/(c^2*x^2)]*x^4*Sqrt[1 + (e*x^2)/d]*AppellF1[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 )^3*ArcSec[c*x])/e)/(80*Sqrt[d + e*x^2])
Time = 0.47 (sec) , antiderivative size = 242, normalized size of antiderivative = 0.92, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {5759, 354, 113, 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.
| \(\displaystyle \int x \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right ) \, dx\) |
| \(\Big \downarrow \) 5759 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {b c x \int \frac {\left (e x^2+d\right )^{5/2}}{x \sqrt {c^2 x^2-1}}dx}{5 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {b c x \int \frac {\left (e x^2+d\right )^{5/2}}{x^2 \sqrt {c^2 x^2-1}}dx^2}{10 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 113 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {b c x \left (\frac {\int \frac {\sqrt {e x^2+d} \left (4 c^2 d^2+e \left (7 d c^2+3 e\right ) x^2\right )}{2 x^2 \sqrt {c^2 x^2-1}}dx^2}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{10 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {b c x \left (\frac {\int \frac {\sqrt {e x^2+d} \left (4 c^2 d^2+e \left (7 d c^2+3 e\right ) x^2\right )}{x^2 \sqrt {c^2 x^2-1}}dx^2}{4 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{10 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 171 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {b c x \left (\frac {\frac {\int \frac {8 d^3 c^4+e \left (15 d^2 c^4+10 d e c^2+3 e^2\right ) x^2}{2 x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{c^2}+\frac {e \sqrt {c^2 x^2-1} \left (7 c^2 d+3 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{10 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {b c x \left (\frac {\frac {\int \frac {8 d^3 c^4+e \left (15 d^2 c^4+10 d e c^2+3 e^2\right ) x^2}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (7 c^2 d+3 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{10 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 175 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {b c x \left (\frac {\frac {8 c^4 d^3 \int \frac {1}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2+e \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \int \frac {1}{\sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (7 c^2 d+3 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{10 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 66 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {b c x \left (\frac {\frac {8 c^4 d^3 \int \frac {1}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2+2 e \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (7 c^2 d+3 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{10 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 104 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {b c x \left (\frac {\frac {16 c^4 d^3 \int \frac {1}{-x^4-d}d\frac {\sqrt {e x^2+d}}{\sqrt {c^2 x^2-1}}+2 e \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (7 c^2 d+3 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{10 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 217 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {b c x \left (\frac {\frac {2 e \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}-16 c^4 d^{5/2} \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (7 c^2 d+3 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{10 e \sqrt {c^2 x^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\left (d+e x^2\right )^{5/2} \left (a+b \sec ^{-1}(c x)\right )}{5 e}-\frac {b c x \left (\frac {\frac {\frac {2 \sqrt {e} \left (15 c^4 d^2+10 c^2 d e+3 e^2\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{c}-16 c^4 d^{5/2} \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{2 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (7 c^2 d+3 e\right ) \sqrt {d+e x^2}}{c^2}}{4 c^2}+\frac {e \sqrt {c^2 x^2-1} \left (d+e x^2\right )^{3/2}}{2 c^2}\right )}{10 e \sqrt {c^2 x^2}}\) |
((d + e*x^2)^(5/2)*(a + b*ArcSec[c*x]))/(5*e) - (b*c*x*((e*Sqrt[-1 + c^2*x ^2]*(d + e*x^2)^(3/2))/(2*c^2) + ((e*(7*c^2*d + 3*e)*Sqrt[-1 + c^2*x^2]*Sq rt[d + e*x^2])/c^2 + (-16*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 + 3*e^2)*ArcTanh[(S qrt[e]*Sqrt[-1 + c^2*x^2])/(c*Sqrt[d + e*x^2])])/c)/(2*c^2))/(4*c^2)))/(10 *e*Sqrt[c^2*x^2])
3.2.22.3.1 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_), x_] :> Simp[b*(a + b*x)^(m - 1)*(c + d*x)^(n + 1)*((e + f*x)^(p + 1 )/(d*f*(m + n + p + 1))), x] + Simp[1/(d*f*(m + n + p + 1)) Int[(a + b*x) ^(m - 2)*(c + d*x)^n*(e + f*x)^p*Simp[a^2*d*f*(m + n + p + 1) - b*(b*c*e*(m - 1) + a*(d*e*(n + 1) + c*f*(p + 1))) + b*(a*d*f*(2*m + n + p) - b*(d*e*(m + n) + c*f*(m + p)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] & & GtQ[m, 1] && NeQ[m + n + p + 1, 0] && IntegersQ[2*m, 2*n, 2*p]
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_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_.) + ArcSec[(c_.)*(x_)]*(b_.))*(x_)*((d_.) + (e_.)*(x_)^2)^(p_.), x _Symbol] :> Simp[(d + e*x^2)^(p + 1)*((a + b*ArcSec[c*x])/(2*e*(p + 1))), x ] - Simp[b*c*(x/(2*e*(p + 1)*Sqrt[c^2*x^2])) Int[(d + e*x^2)^(p + 1)/(x*S qrt[c^2*x^2 - 1]), x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[p, -1]
\[\int x \left (e \,x^{2}+d \right )^{\frac {3}{2}} \left (a +b \,\operatorname {arcsec}\left (c x \right )\right )d x\]
Time = 1.09 (sec) , antiderivative size = 1377, normalized size of antiderivative = 5.26 \[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\text {Too large to display} \]
[1/160*(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 + 3*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*(8*a*c^5*e^2*x^4 + 16*a*c^5*d*e*x^2 + 8*a*c^5*d^2 + 8*(b*c^5*e^2*x ^4 + 2*b*c^5*d*e*x^2 + b*c^5*d^2)*arcsec(c*x) - (2*b*c^3*e^2*x^2 + 9*b*c^3 *d*e + 3*b*c*e^2)*sqrt(c^2*x^2 - 1))*sqrt(e*x^2 + d))/(c^5*e), 1/160*(16*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 + 3*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*(8*a*c^5*e^2*x^4 + 16*a*c^5*d*e*x ^2 + 8*a*c^5*d^2 + 8*(b*c^5*e^2*x^4 + 2*b*c^5*d*e*x^2 + b*c^5*d^2)*arcsec( c*x) - (2*b*c^3*e^2*x^2 + 9*b*c^3*d*e + 3*b*c*e^2)*sqrt(c^2*x^2 - 1))*sqrt (e*x^2 + d))/(c^5*e), 1/80*(4*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 + 3*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*d*e - c*e^2)*...
\[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int x \left (a + b \operatorname {asec}{\left (c x \right )}\right ) \left (d + e x^{2}\right )^{\frac {3}{2}}\, dx \]
\[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x \,d x } \]
1/5*(e*x^2 + d)^(5/2)*a/e + 1/5*((e^2*x^4 + 2*d*e*x^2 + d^2)*sqrt(e*x^2 + d)*arctan(sqrt(c*x + 1)*sqrt(c*x - 1)) - 5*e*integrate((5*(c^2*e^2*x^5 + ( c^2*d*e - e^2)*x^3 - d*e*x + (c^2*e^2*x^5 + (c^2*d*e - e^2)*x^3 - d*e*x)*e ^(log(c*x + 1) + log(c*x - 1)))*sqrt(e*x^2 + d)*log(x) + (5*c^2*e^2*x^5*lo g(c) + 5*(c^2*d*e*log(c) - e^2*log(c))*x^3 - 5*d*e*x*log(c) + ((5*c^2*log( c) + c^2)*e^2*x^5 + ((5*c^2*log(c) + 2*c^2)*d*e - 5*e^2*log(c))*x^3 + (c^2 *d^2 - 5*d*e*log(c))*x)*e^(log(c*x + 1) + log(c*x - 1)))*sqrt(e*x^2 + d))/ (c^2*e*x^2 + (c^2*e*x^2 - e)*e^(log(c*x + 1) + log(c*x - 1)) - e), x))*b/e
\[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int { {\left (e x^{2} + d\right )}^{\frac {3}{2}} {\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x \,d x } \]
Timed out. \[ \int x \left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right ) \, dx=\int x\,{\left (e\,x^2+d\right )}^{3/2}\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right ) \,d x \]