\(\int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx\) [212]

Optimal result

Integrand size = 28, antiderivative size = 186 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\frac {2 a^3 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{11 d e^8}+\frac {6 a^3 \sin (c+d x)}{55 d e^5 (e \sec (c+d x))^{5/2}}+\frac {2 a^3 \sin (c+d x)}{11 d e^7 \sqrt {e \sec (c+d x)}}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}-\frac {12 i \left (a^3+i a^3 \tan (c+d x)\right )}{55 d e^2 (e \sec (c+d x))^{11/2}} \] Output:

2/11*a^3*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))*(e*sec(d* 
x+c))^(1/2)/d/e^8+6/55*a^3*sin(d*x+c)/d/e^5/(e*sec(d*x+c))^(5/2)+2/11*a^3* 
sin(d*x+c)/d/e^7/(e*sec(d*x+c))^(1/2)-2/15*I*(a+I*a*tan(d*x+c))^3/d/(e*sec 
(d*x+c))^(15/2)-12/55*I*(a^3+I*a^3*tan(d*x+c))/d/e^2/(e*sec(d*x+c))^(11/2)
 

Mathematica [A] (verified)

Time = 2.50 (sec) , antiderivative size = 170, normalized size of antiderivative = 0.91 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\frac {a^3 \sqrt {e \sec (c+d x)} \left (-332 i \cos (c+d x)-154 i \cos (3 (c+d x))+22 i \cos (5 (c+d x))-114 \sin (c+d x)+240 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) (\cos (3 (c+d x))-i \sin (3 (c+d x)))-81 \sin (3 (c+d x))+33 \sin (5 (c+d x))\right ) (\cos (3 (c+2 d x))+i \sin (3 (c+2 d x)))}{1320 d e^8 (\cos (d x)+i \sin (d x))^3} \] Input:

Integrate[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(15/2),x]
 

Output:

(a^3*Sqrt[e*Sec[c + d*x]]*((-332*I)*Cos[c + d*x] - (154*I)*Cos[3*(c + d*x) 
] + (22*I)*Cos[5*(c + d*x)] - 114*Sin[c + d*x] + 240*Sqrt[Cos[c + d*x]]*El 
lipticF[(c + d*x)/2, 2]*(Cos[3*(c + d*x)] - I*Sin[3*(c + d*x)]) - 81*Sin[3 
*(c + d*x)] + 33*Sin[5*(c + d*x)])*(Cos[3*(c + 2*d*x)] + I*Sin[3*(c + 2*d* 
x)]))/(1320*d*e^8*(Cos[d*x] + I*Sin[d*x])^3)
 

Rubi [A] (verified)

Time = 0.92 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.09, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3978, 3042, 3977, 3042, 4256, 3042, 4256, 3042, 4258, 3042, 3120}

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.

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}}dx\)

\(\Big \downarrow \) 3978

\(\displaystyle \frac {3 a \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{11/2}}dx}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 a \int \frac {(i \tan (c+d x) a+a)^2}{(e \sec (c+d x))^{11/2}}dx}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\)

\(\Big \downarrow \) 3977

\(\displaystyle \frac {3 a \left (\frac {7 a^2 \int \frac {1}{(e \sec (c+d x))^{7/2}}dx}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 a \left (\frac {7 a^2 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {3 a \left (\frac {7 a^2 \left (\frac {5 \int \frac {1}{(e \sec (c+d x))^{3/2}}dx}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 a \left (\frac {7 a^2 \left (\frac {5 \int \frac {1}{\left (e \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\)

\(\Big \downarrow \) 4256

\(\displaystyle \frac {3 a \left (\frac {7 a^2 \left (\frac {5 \left (\frac {\int \sqrt {e \sec (c+d x)}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 a \left (\frac {7 a^2 \left (\frac {5 \left (\frac {\int \sqrt {e \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {3 a \left (\frac {7 a^2 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {3 a \left (\frac {7 a^2 \left (\frac {5 \left (\frac {\sqrt {\cos (c+d x)} \sqrt {e \sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {3 a \left (\frac {7 a^2 \left (\frac {5 \left (\frac {2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {e \sec (c+d x)}}{3 d e^2}+\frac {2 \sin (c+d x)}{3 d e \sqrt {e \sec (c+d x)}}\right )}{7 e^2}+\frac {2 \sin (c+d x)}{7 d e (e \sec (c+d x))^{5/2}}\right )}{11 e^2}-\frac {4 i \left (a^2+i a^2 \tan (c+d x)\right )}{11 d (e \sec (c+d x))^{11/2}}\right )}{5 e^2}-\frac {2 i (a+i a \tan (c+d x))^3}{15 d (e \sec (c+d x))^{15/2}}\)

Input:

Int[(a + I*a*Tan[c + d*x])^3/(e*Sec[c + d*x])^(15/2),x]
 

Output:

(((-2*I)/15)*(a + I*a*Tan[c + d*x])^3)/(d*(e*Sec[c + d*x])^(15/2)) + (3*a* 
((7*a^2*((2*Sin[c + d*x])/(7*d*e*(e*Sec[c + d*x])^(5/2)) + (5*((2*Sqrt[Cos 
[c + d*x]]*EllipticF[(c + d*x)/2, 2]*Sqrt[e*Sec[c + d*x]])/(3*d*e^2) + (2* 
Sin[c + d*x])/(3*d*e*Sqrt[e*Sec[c + d*x]])))/(7*e^2)))/(11*e^2) - (((4*I)/ 
11)*(a^2 + I*a^2*Tan[c + d*x]))/(d*(e*Sec[c + d*x])^(11/2))))/(5*e^2)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3120

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 
)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3977

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x 
_)])^(n_), x_Symbol] :> Simp[2*b*(d*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^( 
n - 1)/(f*m)), x] - Simp[b^2*((m + 2*n - 2)/(d^2*m))   Int[(d*Sec[e + f*x]) 
^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] 
&& EqQ[a^2 + b^2, 0] && GtQ[n, 1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || 
 (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILtQ[m, 0] & 
& LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) 
&& IntegerQ[2*m]
 

rule 3978

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/( 
a*f*m)), x] + Simp[a*((m + n)/(m*d^2))   Int[(d*Sec[e + f*x])^(m + 2)*(a + 
b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b 
^2, 0] && GtQ[n, 0] && LtQ[m, -1] && IntegersQ[2*m, 2*n]
 

rule 4256

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Csc[c + d*x])^(n + 1)/(b*d*n)), x] + Simp[(n + 1)/(b^2*n)   Int[(b*Csc[c 
+ d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && IntegerQ[2* 
n]
 

rule 4258

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]
 

Maple [A] (verified)

Time = 32.80 (sec) , antiderivative size = 151, normalized size of antiderivative = 0.81

method	result	size
default	\(\frac {a^{3} \left (\frac {2 \sin \left (d x +c \right ) \left (44 \cos \left (d x +c \right )^{6}+7 \cos \left (d x +c \right )^{4}+9 \cos \left (d x +c \right )^{2}+15\right )}{165}+\frac {2 i \left (-44 \cos \left (d x +c \right )^{7}+15 \cos \left (d x +c \right )^{5}\right )}{165}+\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-15-15 \sec \left (d x +c \right )\right )}{165}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e^{7}}\)	\(151\)
parts	\(\frac {a^{3} \left (\frac {2 \sin \left (d x +c \right ) \left (77 \cos \left (d x +c \right )^{6}+91 \cos \left (d x +c \right )^{4}+117 \cos \left (d x +c \right )^{2}+195\right )}{1155}+\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-195-195 \sec \left (d x +c \right )\right )}{1155}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e^{7}}-\frac {i a^{3} \left (\frac {2 \cos \left (d x +c \right )^{7}}{15}-\frac {2 \cos \left (d x +c \right )^{5}}{11}\right )}{d \,e^{7} \sqrt {e \sec \left (d x +c \right )}}-\frac {2 i a^{3}}{5 d \left (e \sec \left (d x +c \right )\right )^{\frac {15}{2}}}-\frac {3 a^{3} \left (\frac {2 \sin \left (d x +c \right ) \left (-77 \cos \left (d x +c \right )^{6}+14 \cos \left (d x +c \right )^{4}+18 \cos \left (d x +c \right )^{2}+30\right )}{1155}+\frac {2 i \sqrt {\frac {1}{\cos \left (d x +c \right )+1}}\, \sqrt {\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \operatorname {EllipticF}\left (i \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right ), i\right ) \left (-30-30 \sec \left (d x +c \right )\right )}{1155}\right )}{d \sqrt {e \sec \left (d x +c \right )}\, e^{7}}\)	\(317\)

Input:

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x,method=_RETURNVERBOSE)
 

Output:

a^3/d*(2/165*sin(d*x+c)*(44*cos(d*x+c)^6+7*cos(d*x+c)^4+9*cos(d*x+c)^2+15) 
+2/165*I*(-44*cos(d*x+c)^7+15*cos(d*x+c)^5)+2/165*I*(1/(cos(d*x+c)+1))^(1/ 
2)*EllipticF(I*(csc(d*x+c)-cot(d*x+c)),I)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2 
)*(-15-15*sec(d*x+c)))/(e*sec(d*x+c))^(1/2)/e^7
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 156, normalized size of antiderivative = 0.84 \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\frac {{\left (-480 i \, \sqrt {2} a^{3} \sqrt {e} e^{\left (2 i \, d x + 2 i \, c\right )} {\rm weierstrassPInverse}\left (-4, 0, e^{\left (i \, d x + i \, c\right )}\right ) + \sqrt {2} {\left (-11 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 73 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 218 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 446 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 235 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 55 i \, a^{3}\right )} \sqrt {\frac {e}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (\frac {1}{2} i \, d x + \frac {1}{2} i \, c\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2640 \, d e^{8}} \] Input:

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x, algorithm="fricas" 
)
 

Output:

1/2640*(-480*I*sqrt(2)*a^3*sqrt(e)*e^(2*I*d*x + 2*I*c)*weierstrassPInverse 
(-4, 0, e^(I*d*x + I*c)) + sqrt(2)*(-11*I*a^3*e^(10*I*d*x + 10*I*c) - 73*I 
*a^3*e^(8*I*d*x + 8*I*c) - 218*I*a^3*e^(6*I*d*x + 6*I*c) - 446*I*a^3*e^(4* 
I*d*x + 4*I*c) - 235*I*a^3*e^(2*I*d*x + 2*I*c) + 55*I*a^3)*sqrt(e/(e^(2*I* 
d*x + 2*I*c) + 1))*e^(1/2*I*d*x + 1/2*I*c))*e^(-2*I*d*x - 2*I*c)/(d*e^8)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(d*x+c))**3/(e*sec(d*x+c))**(15/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}} \,d x } \] Input:

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x, algorithm="maxima" 
)
 

Output:

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(15/2), x)
 

Giac [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\int { \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}}{\left (e \sec \left (d x + c\right )\right )^{\frac {15}{2}}} \,d x } \] Input:

integrate((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x, algorithm="giac")
 

Output:

integrate((I*a*tan(d*x + c) + a)^3/(e*sec(d*x + c))^(15/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3}{{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^{15/2}} \,d x \] Input:

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(15/2),x)
 

Output:

int((a + a*tan(c + d*x)*1i)^3/(e/cos(c + d*x))^(15/2), x)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (c+d x))^3}{(e \sec (c+d x))^{15/2}} \, dx=\frac {\sqrt {e}\, a^{3} \left (\int \frac {\sqrt {\sec \left (d x +c \right )}}{\sec \left (d x +c \right )^{8}}d x -\left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{3}}{\sec \left (d x +c \right )^{8}}d x \right ) i -3 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{8}}d x \right )+3 \left (\int \frac {\sqrt {\sec \left (d x +c \right )}\, \tan \left (d x +c \right )}{\sec \left (d x +c \right )^{8}}d x \right ) i \right )}{e^{8}} \] Input:

int((a+I*a*tan(d*x+c))^3/(e*sec(d*x+c))^(15/2),x)
 

Output:

(sqrt(e)*a**3*(int(sqrt(sec(c + d*x))/sec(c + d*x)**8,x) - int((sqrt(sec(c 
 + d*x))*tan(c + d*x)**3)/sec(c + d*x)**8,x)*i - 3*int((sqrt(sec(c + d*x)) 
*tan(c + d*x)**2)/sec(c + d*x)**8,x) + 3*int((sqrt(sec(c + d*x))*tan(c + d 
*x))/sec(c + d*x)**8,x)*i))/e**8