Integrand size = 28, antiderivative size = 175 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3 \, dx=-\frac {22 a^3 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}+\frac {22 i a^3 (e \sec (c+d x))^{3/2}}{15 d}+\frac {22 a^3 e \sqrt {e \sec (c+d x)} \sin (c+d x)}{5 d}+\frac {2 i a (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^2}{7 d}+\frac {22 i (e \sec (c+d x))^{3/2} \left (a^3+i a^3 \tan (c+d x)\right )}{35 d} \] Output:
-22/5*a^3*e^2*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d/cos(d*x+c)^(1/2)/(e* sec(d*x+c))^(1/2)+22/15*I*a^3*(e*sec(d*x+c))^(3/2)/d+22/5*a^3*e*(e*sec(d*x +c))^(1/2)*sin(d*x+c)/d+2/7*I*a*(e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^2/ d+22/35*I*(e*sec(d*x+c))^(3/2)*(a^3+I*a^3*tan(d*x+c))/d
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 3.19 (sec) , antiderivative size = 129, normalized size of antiderivative = 0.74 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3 \, dx=\frac {a^3 (e \sec (c+d x))^{3/2} (1+i \tan (c+d x)) \left (-116 i-308 i \cos (2 (c+d x))+77 i e^{-2 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{5/2} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-e^{2 i (c+d x)}\right )+77 \sec (c+d x) \sin (3 (c+d x))+17 \tan (c+d x)\right )}{210 d} \] Input:
Integrate[(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^3,x]
Output:
(a^3*(e*Sec[c + d*x])^(3/2)*(1 + I*Tan[c + d*x])*(-116*I - (308*I)*Cos[2*( c + d*x)] + ((77*I)*(1 + E^((2*I)*(c + d*x)))^(5/2)*Hypergeometric2F1[1/2, 3/4, 7/4, -E^((2*I)*(c + d*x))])/E^((2*I)*(c + d*x)) + 77*Sec[c + d*x]*Si n[3*(c + d*x)] + 17*Tan[c + d*x]))/(210*d)
Time = 0.84 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3979, 3042, 3979, 3042, 3967, 3042, 4255, 3042, 4258, 3042, 3119}
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.
| \(\displaystyle \int (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int (a+i a \tan (c+d x))^3 (e \sec (c+d x))^{3/2}dx\) |
| \(\Big \downarrow \) 3979 |
| \(\displaystyle \frac {11}{7} a \int (e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)^2dx+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11}{7} a \int (e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)^2dx+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d}\) |
| \(\Big \downarrow \) 3979 |
| \(\displaystyle \frac {11}{7} a \left (\frac {7}{5} a \int (e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11}{7} a \left (\frac {7}{5} a \int (e \sec (c+d x))^{3/2} (i \tan (c+d x) a+a)dx+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d}\) |
| \(\Big \downarrow \) 3967 |
| \(\displaystyle \frac {11}{7} a \left (\frac {7}{5} a \left (a \int (e \sec (c+d x))^{3/2}dx+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11}{7} a \left (\frac {7}{5} a \left (a \int \left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle \frac {11}{7} a \left (\frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \sec (c+d x)}}dx\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11}{7} a \left (\frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-e^2 \int \frac {1}{\sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d}\) |
| \(\Big \downarrow \) 4258 |
| \(\displaystyle \frac {11}{7} a \left (\frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\cos (c+d x)}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {11}{7} a \left (\frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {e^2 \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )+\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle \frac {11}{7} a \left (\frac {2 i \left (a^2+i a^2 \tan (c+d x)\right ) (e \sec (c+d x))^{3/2}}{5 d}+\frac {7}{5} a \left (a \left (\frac {2 e \sin (c+d x) \sqrt {e \sec (c+d x)}}{d}-\frac {2 e^2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d \sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)}}\right )+\frac {2 i a (e \sec (c+d x))^{3/2}}{3 d}\right )\right )+\frac {2 i a (a+i a \tan (c+d x))^2 (e \sec (c+d x))^{3/2}}{7 d}\) |
Input:
Int[(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^3,x]
Output:
(((2*I)/7)*a*(e*Sec[c + d*x])^(3/2)*(a + I*a*Tan[c + d*x])^2)/d + (11*a*(( 7*a*((((2*I)/3)*a*(e*Sec[c + d*x])^(3/2))/d + a*((-2*e^2*EllipticE[(c + d* x)/2, 2])/(d*Sqrt[Cos[c + d*x]]*Sqrt[e*Sec[c + d*x]]) + (2*e*Sqrt[e*Sec[c + d*x]]*Sin[c + d*x])/d)))/5 + (((2*I)/5)*(e*Sec[c + d*x])^(3/2)*(a^2 + I* a^2*Tan[c + d*x]))/d))/7
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* (c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)]), x_Symbol] :> Simp[b*((d*Sec[e + f*x])^m/(f*m)), x] + Simp[a Int[(d *Sec[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f, m}, x] && (IntegerQ[2*m] || NeQ[a^2 + b^2, 0])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_), x_Symbol] :> Simp[b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] + Simp[a*((m + 2*n - 2)/(m + n - 1)) Int[(d*Se c[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && NeQ[m + n - 1, 0] && IntegersQ [2*m, 2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] )^n*Sin[c + d*x]^n Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]
Time = 8.32 (sec) , antiderivative size = 242, normalized size of antiderivative = 1.38
| method | result | size |
| default | \(-\frac {2 a^{3} \sqrt {e \sec \left (d x +c \right )}\, \left (-231 \sin \left (d x +c \right )+63 \tan \left (d x +c \right )+63 \sec \left (d x +c \right ) \tan \left (d x +c \right )+5 i \left (3 \sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}-28 \sec \left (d x +c \right )-28\right )+231 i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \operatorname {EllipticF}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+231 i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \operatorname {EllipticE}\left (i \left (\cot \left (d x +c \right )-\csc \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\right ) e}{105 d \left (\cos \left (d x +c \right )+1\right )}\) | \(242\) |
| parts | \(\frac {2 a^{3} \left (i \left (\cos \left (d x +c \right )^{2}+2 \cos \left (d x +c \right )+1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+i \left (-\cos \left (d x +c \right )^{2}-2 \cos \left (d x +c \right )-1\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}+\sin \left (d x +c \right )\right ) \sqrt {e \sec \left (d x +c \right )}\, e}{d \left (\cos \left (d x +c \right )+1\right )}-\frac {2 i a^{3} \left (\frac {\left (e \sec \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {e^{2} \left (e \sec \left (d x +c \right )\right )^{\frac {3}{2}}}{3}\right )}{e^{2} d}+\frac {2 i a^{3} \left (e \sec \left (d x +c \right )\right )^{\frac {3}{2}}}{d}+\frac {6 a^{3} \sqrt {e \sec \left (d x +c \right )}\, e \left (2 \sin \left (d x +c \right )-\tan \left (d x +c \right )-\sec \left (d x +c \right ) \tan \left (d x +c \right )+i \left (2 \cos \left (d x +c \right )^{2}+4 \cos \left (d x +c \right )+2\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )+i \left (-2 \cos \left (d x +c \right )^{2}-4 \cos \left (d x +c \right )-2\right ) \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticE}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right )\right )}{5 d \left (\cos \left (d x +c \right )+1\right )}\) | \(455\) |
Input:
int((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
-2/105*a^3/d*(e*sec(d*x+c))^(1/2)/(cos(d*x+c)+1)*(-231*sin(d*x+c)+63*tan(d *x+c)+63*sec(d*x+c)*tan(d*x+c)+5*I*(3*sec(d*x+c)^3+3*sec(d*x+c)^2-28*sec(d *x+c)-28)+231*I*(cos(d*x+c)^2+2*cos(d*x+c)+1)*EllipticF(I*(cot(d*x+c)-csc( d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2)+231* I*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*EllipticE(I*(cot(d*x+c)-csc(d*x+c)),I)*(c os(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(cos(d*x+c)+1))^(1/2))*e
Time = 0.08 (sec) , antiderivative size = 208, normalized size of antiderivative = 1.19 \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3 \, dx=-\frac {2 \, {\left (\sqrt {2} {\left (231 i \, a^{3} e e^{\left (7 i \, d x + 7 i \, c\right )} + 287 i \, a^{3} e e^{\left (5 i \, d x + 5 i \, c\right )} + 253 i \, a^{3} e e^{\left (3 i \, d x + 3 i \, c\right )} + 77 i \, a^{3} e e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )} + 231 \, \sqrt {2} {\left (i \, a^{3} e e^{\left (6 i \, d x + 6 i \, c\right )} + 3 i \, a^{3} e e^{\left (4 i \, d x + 4 i \, c\right )} + 3 i \, a^{3} e e^{\left (2 i \, d x + 2 i \, c\right )} + i \, a^{3} e\right )} \sqrt {e} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right )\right )\right )}}{105 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \] Input:
integrate((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
-2/105*(sqrt(2)*(231*I*a^3*e*e^(7*I*d*x + 7*I*c) + 287*I*a^3*e*e^(5*I*d*x + 5*I*c) + 253*I*a^3*e*e^(3*I*d*x + 3*I*c) + 77*I*a^3*e*e^(I*d*x + I*c))*s qrt(e/(e^(2*I*d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c) + 231*sqrt(2)*(I* a^3*e*e^(6*I*d*x + 6*I*c) + 3*I*a^3*e*e^(4*I*d*x + 4*I*c) + 3*I*a^3*e*e^(2 *I*d*x + 2*I*c) + I*a^3*e)*sqrt(e)*weierstrassZeta(-4, 0, weierstrassPInve rse(-4, 0, e^(I*d*x + I*c))))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4* I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)
\[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3 \, dx=- i a^{3} \left (\int i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}\, dx + \int \left (- 3 \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan {\left (c + d x \right )}\right )\, dx + \int \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{3}{\left (c + d x \right )}\, dx + \int \left (- 3 i \left (e \sec {\left (c + d x \right )}\right )^{\frac {3}{2}} \tan ^{2}{\left (c + d x \right )}\right )\, dx\right ) \] Input:
integrate((e*sec(d*x+c))**(3/2)*(a+I*a*tan(d*x+c))**3,x)
Output:
-I*a**3*(Integral(I*(e*sec(c + d*x))**(3/2), x) + Integral(-3*(e*sec(c + d *x))**(3/2)*tan(c + d*x), x) + Integral((e*sec(c + d*x))**(3/2)*tan(c + d* x)**3, x) + Integral(-3*I*(e*sec(c + d*x))**(3/2)*tan(c + d*x)**2, x))
\[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \,d x } \] Input:
integrate((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
integrate((e*sec(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a)^3, x)
\[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3 \, dx=\int { \left (e \sec \left (d x + c\right )\right )^{\frac {3}{2}} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \,d x } \] Input:
integrate((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
integrate((e*sec(d*x + c))^(3/2)*(I*a*tan(d*x + c) + a)^3, x)
Timed out. \[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3 \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \] Input:
int((e/cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)^3,x)
Output:
int((e/cos(c + d*x))^(3/2)*(a + a*tan(c + d*x)*1i)^3, x)
\[ \int (e \sec (c+d x))^{3/2} (a+i a \tan (c+d x))^3 \, dx=\frac {\sqrt {e}\, a^{3} e \left (-6 \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2} i +50 \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right ) i -63 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right ) \tan \left (d x +c \right )^{2}d x \right ) d +21 \left (\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \right ) d \right )}{21 d} \] Input:
int((e*sec(d*x+c))^(3/2)*(a+I*a*tan(d*x+c))^3,x)
Output:
(sqrt(e)*a**3*e*( - 6*sqrt(sec(c + d*x))*sec(c + d*x)*tan(c + d*x)**2*i + 50*sqrt(sec(c + d*x))*sec(c + d*x)*i - 63*int(sqrt(sec(c + d*x))*sec(c + d *x)*tan(c + d*x)**2,x)*d + 21*int(sqrt(sec(c + d*x))*sec(c + d*x),x)*d))/( 21*d)