\(\int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx\) [1167]

Optimal result

Integrand size = 32, antiderivative size = 181 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=-\frac {4 i \sqrt {2} a^{5/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{(c-i d)^{5/2} f}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 (i c+d) f (c+d \tan (e+f x))^{3/2}}+\frac {4 i a^2 \sqrt {a+i a \tan (e+f x)}}{(c-i d)^2 f \sqrt {c+d \tan (e+f x)}} \] Output:

-4*I*2^(1/2)*a^(5/2)*arctanh(2^(1/2)*a^(1/2)*(c+d*tan(f*x+e))^(1/2)/(c-I*d 
)^(1/2)/(a+I*a*tan(f*x+e))^(1/2))/(c-I*d)^(5/2)/f-2/3*a*(a+I*a*tan(f*x+e)) 
^(3/2)/(I*c+d)/f/(c+d*tan(f*x+e))^(3/2)+4*I*a^2*(a+I*a*tan(f*x+e))^(1/2)/( 
c-I*d)^2/f/(c+d*tan(f*x+e))^(1/2)
 

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(1618\) vs. \(2(181)=362\).

Time = 7.25 (sec) , antiderivative size = 1618, normalized size of antiderivative = 8.94 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:

Integrate[(a + I*a*Tan[e + f*x])^(5/2)/(c + d*Tan[e + f*x])^(5/2),x]
 

Output:

(-2*d*(a + I*a*Tan[e + f*x])^(5/2))/(3*(c^2 + d^2)*f*(c + d*Tan[e + f*x])^ 
(3/2)) + (2*((-2*(-(a*c*d) + (a*(3*c + (5*I)*d)*d)/2)*(a + I*a*Tan[e + f*x 
])^(5/2))/((c^2 + d^2)*f*Sqrt[c + d*Tan[e + f*x]]) + (2*((((-3*I)/8)*(c + 
(5*I)*d)*(I*a*c - a*d)^3*Sqrt[(I*a)/(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/ 
(I*a*c - a*d))]*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d))^3*Sqr 
t[(I*a*(c + d*Tan[e + f*x]))/(I*a*c - a*d)]*Sqrt[1 + (I*a*d*(a + I*a*Tan[e 
 + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a* 
d)))]*(((2*I)*a*d*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c 
 - a*d)) - (I*a^2*d)/(I*a*c - a*d))) + (4*a^2*d^2*(a + I*a*Tan[e + f*x])^2 
)/(3*(I*a*c - a*d)^2*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d))^ 
2) - (2*(-1)^(1/4)*Sqrt[a]*Sqrt[d]*ArcSinh[((-1)^(1/4)*Sqrt[a]*Sqrt[d]*Sqr 
t[a + I*a*Tan[e + f*x]])/(Sqrt[I*a*c - a*d]*Sqrt[-((a^2*c)/(I*a*c - a*d)) 
- (I*a^2*d)/(I*a*c - a*d)])]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[I*a*c - a*d 
]*Sqrt[-((a^2*c)/(I*a*c - a*d)) - (I*a^2*d)/(I*a*c - a*d)]*Sqrt[1 + (I*a*d 
*(a + I*a*Tan[e + f*x]))/((I*a*c - a*d)*(-((a^2*c)/(I*a*c - a*d)) - (I*a^2 
*d)/(I*a*c - a*d)))])))/(a*d^2*f*Sqrt[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan 
[e + f*x]]) + (a*(a^3*(c + (5*I)*d)*d + (I/4)*a^3*(3*c^2 + (10*I)*c*d - 23 
*d^2))*(2*a*((-2*Sqrt[2]*ArcTan[(Sqrt[-(a*c) + I*a*d]*Sqrt[a + I*a*Tan[e + 
 f*x]])/(Sqrt[2]*a*Sqrt[c + d*Tan[e + f*x]])])/Sqrt[-(a*c) + I*a*d] - (2*( 
-1)^(3/4)*Sqrt[c + I*d]*Sqrt[(c/(c + I*d) + (I*d)/(c + I*d))^(-1)]*Sqrt...
 

Rubi [A] (verified)

Time = 0.76 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.219, Rules used = {3042, 4028, 3042, 4028, 3042, 4027, 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.

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}}dx\)

\(\Big \downarrow \) 4028

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4028

\(\displaystyle \frac {2 a \left (\frac {2 a \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c-i d}-\frac {2 a \sqrt {a+i a \tan (e+f x)}}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\right )}{c-i d}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 a \left (\frac {2 a \int \frac {\sqrt {i \tan (e+f x) a+a}}{\sqrt {c+d \tan (e+f x)}}dx}{c-i d}-\frac {2 a \sqrt {a+i a \tan (e+f x)}}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\right )}{c-i d}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 4027

\(\displaystyle \frac {2 a \left (-\frac {4 i a^3 \int \frac {1}{a (c-i d)-\frac {2 a^2 (c+d \tan (e+f x))}{i \tan (e+f x) a+a}}d\frac {\sqrt {c+d \tan (e+f x)}}{\sqrt {i \tan (e+f x) a+a}}}{f (c-i d)}-\frac {2 a \sqrt {a+i a \tan (e+f x)}}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\right )}{c-i d}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 a \left (-\frac {2 i \sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{f (c-i d)^{3/2}}-\frac {2 a \sqrt {a+i a \tan (e+f x)}}{f (d+i c) \sqrt {c+d \tan (e+f x)}}\right )}{c-i d}-\frac {2 a (a+i a \tan (e+f x))^{3/2}}{3 f (d+i c) (c+d \tan (e+f x))^{3/2}}\)

Input:

Int[(a + I*a*Tan[e + f*x])^(5/2)/(c + d*Tan[e + f*x])^(5/2),x]
 

Output:

(-2*a*(a + I*a*Tan[e + f*x])^(3/2))/(3*(I*c + d)*f*(c + d*Tan[e + f*x])^(3 
/2)) + (2*a*(((-2*I)*Sqrt[2]*a^(3/2)*ArcTanh[(Sqrt[2]*Sqrt[a]*Sqrt[c + d*T 
an[e + f*x]])/(Sqrt[c - I*d]*Sqrt[a + I*a*Tan[e + f*x]])])/((c - I*d)^(3/2 
)*f) - (2*a*Sqrt[a + I*a*Tan[e + f*x]])/((I*c + d)*f*Sqrt[c + d*Tan[e + f* 
x]])))/(c - I*d)
 

Defintions of rubi rules used

rule 221

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]
 

rule 3042

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

rule 4027

Int[Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*tan[(e_.) 
 + (f_.)*(x_)]], x_Symbol] :> Simp[-2*a*(b/f)   Subst[Int[1/(a*c - b*d - 2* 
a^2*x^2), x], x, Sqrt[c + d*Tan[e + f*x]]/Sqrt[a + b*Tan[e + f*x]]], x] /; 
FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && N 
eQ[c^2 + d^2, 0]
 

rule 4028

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + 
(f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*b*(a + b*Tan[e + f*x])^(m - 1)*((c + 
 d*Tan[e + f*x])^(n + 1)/(f*(m - 1)*(a*c - b*d))), x] + Simp[2*(a^2/(a*c - 
b*d))   Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1), x], 
x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0 
] && NeQ[c^2 + d^2, 0] && EqQ[m + n, 0] && GtQ[m, 1/2]
 

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 2967 vs. \(2 (146 ) = 292\).

Time = 0.62 (sec) , antiderivative size = 2968, normalized size of antiderivative = 16.40

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(2968\)
default	\(\text {Expression too large to display}\)	\(2968\)

Input:

int((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x,method=_RETURNVERBOS 
E)
 

Output:

1/3/f*2^(1/2)*(a*(1+I*tan(f*x+e)))^(1/2)*(6*I*2^(1/2)*ln(1/2*(2*I*a*d*tan( 
f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a 
*d)/(I*a*d)^(1/2))*a*c^3*d^2*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2-18*I*2^(1/2)* 
ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^( 
1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d^4*(-a*(I*d-c))^(1/2)*tan(f*x+ 
e)^2+12*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*( 
1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c^4*d*(-a*(I*d- 
c))^(1/2)*tan(f*x+e)-36*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c 
+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a 
*c^2*d^3*(-a*(I*d-c))^(1/2)*tan(f*x+e)-18*I*2^(1/2)*(-a*(I*d-c))^(1/2)*(I* 
a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^2*d^2*tan(f*x+e)+ 
6*I*2^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c+2*(a*(c+d*tan(f*x+e))*(1+I*ta 
n(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^(1/2))*a*c*d*(-a*(I*d-c))^(1/2 
)*tan(f*x+e)-3*2^(1/2)*(-a*(I*d-c))^(1/2)*ln(1/2*(2*I*a*d*tan(f*x+e)+I*a*c 
+2*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*(I*a*d)^(1/2)+a*d)/(I*a*d)^ 
(1/2))*a*c^2-3*I*ln((3*a*c+I*a*tan(f*x+e)*c-I*a*d+3*a*d*tan(f*x+e)+2*2^(1/ 
2)*(-a*(I*d-c))^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2))/(I+tan( 
f*x+e)))*a*c^2*(I*a*d)^(1/2)-7*2^(1/2)*(-a*(I*d-c))^(1/2)*(I*a*d)^(1/2)*(a 
*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^4+2^(1/2)*(-a*(I*d-c))^(1/2)*( 
I*a*d)^(1/2)*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*d^4-3*ln((3*a*...
 

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 849 vs. \(2 (137) = 274\).

Time = 0.10 (sec) , antiderivative size = 849, normalized size of antiderivative = 4.69 \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="fr 
icas")
 

Output:

-1/6*(8*sqrt(2)*(4*(-I*a^2*c - a^2*d)*e^(5*I*f*x + 5*I*e) + (-7*I*a^2*c - 
a^2*d)*e^(3*I*f*x + 3*I*e) + 3*(-I*a^2*c + a^2*d)*e^(I*f*x + I*e))*sqrt((( 
c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/ 
(e^(2*I*f*x + 2*I*e) + 1)) - 3*((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + 
 d^4)*f*e^(4*I*f*x + 4*I*e) + 2*(c^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2 
*I*f*x + 2*I*e) + (c^4 + 2*c^2*d^2 + d^4)*f)*sqrt(-32*I*a^5/((I*c^5 + 5*c^ 
4*d - 10*I*c^3*d^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(1/4*((I*c^3 + 
 3*c^2*d - 3*I*c*d^2 - d^3)*sqrt(-32*I*a^5/((I*c^5 + 5*c^4*d - 10*I*c^3*d^ 
2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*f*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2 
*e^(2*I*f*x + 2*I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d) 
/(e^(2*I*f*x + 2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - 
 I*e)/a^2) + 3*((c^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f 
*x + 4*I*e) + 2*(c^4 - 2*I*c^3*d - 2*I*c*d^3 - d^4)*f*e^(2*I*f*x + 2*I*e) 
+ (c^4 + 2*c^2*d^2 + d^4)*f)*sqrt(-32*I*a^5/((I*c^5 + 5*c^4*d - 10*I*c^3*d 
^2 - 10*c^2*d^3 + 5*I*c*d^4 + d^5)*f^2))*log(1/4*((-I*c^3 - 3*c^2*d + 3*I* 
c*d^2 + d^3)*sqrt(-32*I*a^5/((I*c^5 + 5*c^4*d - 10*I*c^3*d^2 - 10*c^2*d^3 
+ 5*I*c*d^4 + d^5)*f^2))*f*e^(I*f*x + I*e) + 4*sqrt(2)*(a^2*e^(2*I*f*x + 2 
*I*e) + a^2)*sqrt(((c - I*d)*e^(2*I*f*x + 2*I*e) + c + I*d)/(e^(2*I*f*x + 
2*I*e) + 1))*sqrt(a/(e^(2*I*f*x + 2*I*e) + 1)))*e^(-I*f*x - I*e)/a^2))/((c 
^4 - 4*I*c^3*d - 6*c^2*d^2 + 4*I*c*d^3 + d^4)*f*e^(4*I*f*x + 4*I*e) + 2...
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {\left (i a \left (\tan {\left (e + f x \right )} - i\right )\right )^{\frac {5}{2}}}{\left (c + d \tan {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \] Input:

integrate((a+I*a*tan(f*x+e))**(5/2)/(c+d*tan(f*x+e))**(5/2),x)
 

Output:

Integral((I*a*(tan(e + f*x) - I))**(5/2)/(c + d*tan(e + f*x))**(5/2), x)
 

Maxima [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Timed out} \] Input:

integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="ma 
xima")
 

Output:

Timed out
 

Giac [F(-2)]

Exception generated. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x, algorithm="gi 
ac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeRecursive assumption sageVARc>=(-sageVARd*t_nostep) ignoredDo 
ne
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\int \frac {{\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2}}{{\left (c+d\,\mathrm {tan}\left (e+f\,x\right )\right )}^{5/2}} \,d x \] Input:

int((a + a*tan(e + f*x)*1i)^(5/2)/(c + d*tan(e + f*x))^(5/2),x)
 

Output:

int((a + a*tan(e + f*x)*1i)^(5/2)/(c + d*tan(e + f*x))^(5/2), x)
 

Reduce [F]

\[ \int \frac {(a+i a \tan (e+f x))^{5/2}}{(c+d \tan (e+f x))^{5/2}} \, dx=\text {too large to display} \] Input:

int((a+I*a*tan(f*x+e))^(5/2)/(c+d*tan(f*x+e))^(5/2),x)
 

Output:

(2*sqrt(a)*a**2*( - sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan( 
e + f*x)*c**3 - 3*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e 
+ f*x)*c**2*d*i - 9*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan( 
e + f*x)*c*d**2 + 5*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan( 
e + f*x)*d**3*i - 5*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c**3 
*i - 9*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c**2*d + 3*sqrt(t 
an(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c*d**2*i - sqrt(tan(e + f*x)*i 
 + 1)*sqrt(tan(e + f*x)*d + c)*d**3 - 486*int((sqrt(tan(e + f*x)*i + 1)*sq 
rt(tan(e + f*x)*d + c)*tan(e + f*x))/(81*tan(e + f*x)**3*c**8*d**3*i - 432 
*tan(e + f*x)**3*c**7*d**4 - 1008*tan(e + f*x)**3*c**6*d**5*i + 1368*tan(e 
 + f*x)**3*c**5*d**6 + 1202*tan(e + f*x)**3*c**4*d**7*i - 704*tan(e + f*x) 
**3*c**3*d**8 - 264*tan(e + f*x)**3*c**2*d**9*i + 56*tan(e + f*x)**3*c*d** 
10 + 5*tan(e + f*x)**3*d**11*i + 243*tan(e + f*x)**2*c**9*d**2*i - 1296*ta 
n(e + f*x)**2*c**8*d**3 - 3024*tan(e + f*x)**2*c**7*d**4*i + 4104*tan(e + 
f*x)**2*c**6*d**5 + 3606*tan(e + f*x)**2*c**5*d**6*i - 2112*tan(e + f*x)** 
2*c**4*d**7 - 792*tan(e + f*x)**2*c**3*d**8*i + 168*tan(e + f*x)**2*c**2*d 
**9 + 15*tan(e + f*x)**2*c*d**10*i + 243*tan(e + f*x)*c**10*d*i - 1296*tan 
(e + f*x)*c**9*d**2 - 3024*tan(e + f*x)*c**8*d**3*i + 4104*tan(e + f*x)*c* 
*7*d**4 + 3606*tan(e + f*x)*c**6*d**5*i - 2112*tan(e + f*x)*c**5*d**6 - 79 
2*tan(e + f*x)*c**4*d**7*i + 168*tan(e + f*x)*c**3*d**8 + 15*tan(e + f*...