Integrand size = 23, antiderivative size = 225 \[ \int \frac {x^3 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {d+e x^2}} \, dx=-\frac {b x \sqrt {-1+c^2 x^2} \sqrt {d+e x^2}}{6 c e \sqrt {c^2 x^2}}-\frac {d \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^2}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}-\frac {2 b c d^{3/2} x \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {-1+c^2 x^2}}\right )}{3 e^2 \sqrt {c^2 x^2}}+\frac {b \left (3 c^2 d-e\right ) x \text {arctanh}\left (\frac {\sqrt {e} \sqrt {-1+c^2 x^2}}{c \sqrt {d+e x^2}}\right )}{6 c^2 e^{3/2} \sqrt {c^2 x^2}} \] Output:
-1/6*b*x*(c^2*x^2-1)^(1/2)*(e*x^2+d)^(1/2)/c/e/(c^2*x^2)^(1/2)-d*(e*x^2+d) ^(1/2)*(a+b*arcsec(c*x))/e^2+1/3*(e*x^2+d)^(3/2)*(a+b*arcsec(c*x))/e^2-2/3 *b*c*d^(3/2)*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)+1/6*b*(3*c^2*d-e)*x*arctanh(e^(1/2)*(c^2*x^2-1)^(1/2)/c/(e*x^2+ d)^(1/2))/c^2/e^(3/2)/(c^2*x^2)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 3 in optimal.
Time = 1.35 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.08 \[ \int \frac {x^3 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {d+e x^2}} \, dx=\frac {-4 b d^2 \sqrt {1+\frac {d}{e x^2}} \left (-1+c^2 x^2\right ) \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,\frac {1}{c^2 x^2},-\frac {d}{e x^2}\right )+b \left (3 c^2 d-e\right ) e \sqrt {1-\frac {1}{c^2 x^2}} x^4 \sqrt {1-c^2 x^2} \sqrt {1+\frac {e x^2}{d}} \operatorname {AppellF1}\left (1,\frac {1}{2},\frac {1}{2},2,c^2 x^2,-\frac {e x^2}{d}\right )-2 x \left (-1+c^2 x^2\right ) \left (d+e x^2\right ) \left (4 a c d+b e \sqrt {1-\frac {1}{c^2 x^2}} x-2 a c e x^2+2 b c \left (2 d-e x^2\right ) \sec ^{-1}(c x)\right )}{12 c e^2 x \left (-1+c^2 x^2\right ) \sqrt {d+e x^2}} \] Input:
Integrate[(x^3*(a + b*ArcSec[c*x]))/Sqrt[d + e*x^2],x]
Output:
(-4*b*d^2*Sqrt[1 + d/(e*x^2)]*(-1 + c^2*x^2)*AppellF1[1, 1/2, 1/2, 2, 1/(c ^2*x^2), -(d/(e*x^2))] + b*(3*c^2*d - e)*e*Sqrt[1 - 1/(c^2*x^2)]*x^4*Sqrt[ 1 - c^2*x^2]*Sqrt[1 + (e*x^2)/d]*AppellF1[1, 1/2, 1/2, 2, c^2*x^2, -((e*x^ 2)/d)] - 2*x*(-1 + c^2*x^2)*(d + e*x^2)*(4*a*c*d + b*e*Sqrt[1 - 1/(c^2*x^2 )]*x - 2*a*c*e*x^2 + 2*b*c*(2*d - e*x^2)*ArcSec[c*x]))/(12*c*e^2*x*(-1 + c ^2*x^2)*Sqrt[d + e*x^2])
Time = 0.51 (sec) , antiderivative size = 207, normalized size of antiderivative = 0.92, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.435, Rules used = {5761, 27, 435, 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.
\(\displaystyle \int \frac {x^3 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {d+e x^2}} \, dx\) |
\(\Big \downarrow \) 5761 |
\(\displaystyle -\frac {b c x \int -\frac {\left (2 d-e x^2\right ) \sqrt {e x^2+d}}{3 e^2 x \sqrt {c^2 x^2-1}}dx}{\sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b c x \int \frac {\left (2 d-e x^2\right ) \sqrt {e x^2+d}}{x \sqrt {c^2 x^2-1}}dx}{3 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^2}\) |
\(\Big \downarrow \) 435 |
\(\displaystyle \frac {b c x \int \frac {\left (2 d-e x^2\right ) \sqrt {e x^2+d}}{x^2 \sqrt {c^2 x^2-1}}dx^2}{6 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^2}\) |
\(\Big \downarrow \) 171 |
\(\displaystyle \frac {b c x \left (\frac {\int \frac {4 c^2 d^2+\left (3 c^2 d-e\right ) e x^2}{2 x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{c^2}-\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {b c x \left (\frac {\int \frac {4 c^2 d^2+\left (3 c^2 d-e\right ) e x^2}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^2}\) |
\(\Big \downarrow \) 175 |
\(\displaystyle \frac {b c x \left (\frac {4 c^2 d^2 \int \frac {1}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2+e \left (3 c^2 d-e\right ) \int \frac {1}{\sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^2}\) |
\(\Big \downarrow \) 66 |
\(\displaystyle \frac {b c x \left (\frac {4 c^2 d^2 \int \frac {1}{x^2 \sqrt {c^2 x^2-1} \sqrt {e x^2+d}}dx^2+2 e \left (3 c^2 d-e\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^2}\) |
\(\Big \downarrow \) 104 |
\(\displaystyle \frac {b c x \left (\frac {8 c^2 d^2 \int \frac {1}{-x^4-d}d\frac {\sqrt {e x^2+d}}{\sqrt {c^2 x^2-1}}+2 e \left (3 c^2 d-e\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^2}\) |
\(\Big \downarrow \) 217 |
\(\displaystyle \frac {b c x \left (\frac {2 e \left (3 c^2 d-e\right ) \int \frac {1}{c^2-e x^4}d\frac {\sqrt {c^2 x^2-1}}{\sqrt {e x^2+d}}-8 c^2 d^{3/2} \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e^2 \sqrt {c^2 x^2}}+\frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\left (d+e x^2\right )^{3/2} \left (a+b \sec ^{-1}(c x)\right )}{3 e^2}-\frac {d \sqrt {d+e x^2} \left (a+b \sec ^{-1}(c x)\right )}{e^2}+\frac {b c x \left (\frac {\frac {2 \sqrt {e} \left (3 c^2 d-e\right ) \text {arctanh}\left (\frac {\sqrt {e} \sqrt {c^2 x^2-1}}{c \sqrt {d+e x^2}}\right )}{c}-8 c^2 d^{3/2} \arctan \left (\frac {\sqrt {d+e x^2}}{\sqrt {d} \sqrt {c^2 x^2-1}}\right )}{2 c^2}-\frac {e \sqrt {c^2 x^2-1} \sqrt {d+e x^2}}{c^2}\right )}{6 e^2 \sqrt {c^2 x^2}}\) |
Input:
Int[(x^3*(a + b*ArcSec[c*x]))/Sqrt[d + e*x^2],x]
Output:
-((d*Sqrt[d + e*x^2]*(a + b*ArcSec[c*x]))/e^2) + ((d + e*x^2)^(3/2)*(a + b *ArcSec[c*x]))/(3*e^2) + (b*c*x*(-((e*Sqrt[-1 + c^2*x^2]*Sqrt[d + e*x^2])/ c^2) + (-8*c^2*d^(3/2)*ArcTan[Sqrt[d + e*x^2]/(Sqrt[d]*Sqrt[-1 + c^2*x^2]) ] + (2*(3*c^2*d - e)*Sqrt[e]*ArcTanh[(Sqrt[e]*Sqrt[-1 + c^2*x^2])/(c*Sqrt[ d + e*x^2])])/c)/(2*c^2)))/(6*e^2*Sqrt[c^2*x^2])
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]))
\[\int \frac {x^{3} \left (a +b \,\operatorname {arcsec}\left (c x \right )\right )}{\sqrt {e \,x^{2}+d}}d x\]
Input:
int(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^(1/2),x)
Output:
int(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^(1/2),x)
Time = 0.39 (sec) , antiderivative size = 1111, normalized size of antiderivative = 4.94 \[ \int \frac {x^3 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {d+e x^2}} \, dx=\text {Too large to display} \] Input:
integrate(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^(1/2),x, algorithm="fricas")
Output:
[1/24*(4*b*c^3*sqrt(-d)*d*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) - (3*b*c^2*d - b*e)*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*(2*a*c^3*e*x^2 - 4*a*c^3*d - sqrt(c^2*x^2 - 1)*b*c*e + 2*(b*c^3*e*x^2 - 2*b*c^3*d)*arcsec(c *x))*sqrt(e*x^2 + d))/(c^3*e^2), -1/24*(8*b*c^3*d^(3/2)*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)) + (3*b*c^2*d - b*e)*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*(2*a*c^3*e*x ^2 - 4*a*c^3*d - sqrt(c^2*x^2 - 1)*b*c*e + 2*(b*c^3*e*x^2 - 2*b*c^3*d)*arc sec(c*x))*sqrt(e*x^2 + d))/(c^3*e^2), 1/12*(2*b*c^3*sqrt(-d)*d*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) - (3*b*c^2*d - b*e)*sqrt(-e)*arctan(1/2*(2*c^2*e*x^2 + c^2*d - e)*sqrt(c^2*x^2 - 1)*sqr t(e*x^2 + d)*sqrt(-e)/(c^3*e^2*x^4 - c*d*e + (c^3*d*e - c*e^2)*x^2)) + 2*( 2*a*c^3*e*x^2 - 4*a*c^3*d - sqrt(c^2*x^2 - 1)*b*c*e + 2*(b*c^3*e*x^2 - 2*b *c^3*d)*arcsec(c*x))*sqrt(e*x^2 + d))/(c^3*e^2), -1/12*(4*b*c^3*d^(3/2)*ar ctan(-1/2*sqrt(c^2*x^2 - 1)*((c^2*d - e)*x^2 - 2*d)*sqrt(e*x^2 + d)*sqr...
\[ \int \frac {x^3 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {d+e x^2}} \, dx=\int \frac {x^{3} \left (a + b \operatorname {asec}{\left (c x \right )}\right )}{\sqrt {d + e x^{2}}}\, dx \] Input:
integrate(x**3*(a+b*asec(c*x))/(e*x**2+d)**(1/2),x)
Output:
Integral(x**3*(a + b*asec(c*x))/sqrt(d + e*x**2), x)
Exception generated. \[ \int \frac {x^3 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {d+e x^2}} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^(1/2),x, algorithm="maxima")
Output:
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
\[ \int \frac {x^3 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {d+e x^2}} \, dx=\int { \frac {{\left (b \operatorname {arcsec}\left (c x\right ) + a\right )} x^{3}}{\sqrt {e x^{2} + d}} \,d x } \] Input:
integrate(x^3*(a+b*arcsec(c*x))/(e*x^2+d)^(1/2),x, algorithm="giac")
Output:
integrate((b*arcsec(c*x) + a)*x^3/sqrt(e*x^2 + d), x)
Timed out. \[ \int \frac {x^3 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {d+e x^2}} \, dx=\int \frac {x^3\,\left (a+b\,\mathrm {acos}\left (\frac {1}{c\,x}\right )\right )}{\sqrt {e\,x^2+d}} \,d x \] Input:
int((x^3*(a + b*acos(1/(c*x))))/(d + e*x^2)^(1/2),x)
Output:
int((x^3*(a + b*acos(1/(c*x))))/(d + e*x^2)^(1/2), x)
\[ \int \frac {x^3 \left (a+b \sec ^{-1}(c x)\right )}{\sqrt {d+e x^2}} \, dx=\frac {-2 \sqrt {e \,x^{2}+d}\, a d +\sqrt {e \,x^{2}+d}\, a e \,x^{2}+3 \left (\int \frac {\mathit {asec} \left (c x \right ) x^{3}}{\sqrt {e \,x^{2}+d}}d x \right ) b \,e^{2}}{3 e^{2}} \] Input:
int(x^3*(a+b*asec(c*x))/(e*x^2+d)^(1/2),x)
Output:
( - 2*sqrt(d + e*x**2)*a*d + sqrt(d + e*x**2)*a*e*x**2 + 3*int((asec(c*x)* x**3)/sqrt(d + e*x**2),x)*b*e**2)/(3*e**2)