\(\int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx\) [1164]

Optimal result

Integrand size = 32, antiderivative size = 194 \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=-\frac {i \text {arctanh}\left (\frac {\sqrt {2} \sqrt {a} \sqrt {c+d \tan (e+f x)}}{\sqrt {c-i d} \sqrt {a+i a \tan (e+f x)}}\right )}{\sqrt {2} \sqrt {a} (c-i d)^{3/2} f}-\frac {1}{(i c-d) f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}}+\frac {(c-3 i d) d \sqrt {a+i a \tan (e+f x)}}{a (c-i d) (c+i d)^2 f \sqrt {c+d \tan (e+f x)}} \] Output:

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

Mathematica [A] (verified)

Time = 0.83 (sec) , antiderivative size = 171, normalized size of antiderivative = 0.88 \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\frac {i a \arctan \left (\frac {\sqrt {-a (c-i d)} \sqrt {a+i a \tan (e+f x)}}{\sqrt {2} a \sqrt {c+d \tan (e+f x)}}\right )}{\sqrt {2} (-a (c-i d))^{3/2} f}+\frac {i c^2+c d-2 i d^2+d (i c+3 d) \tan (e+f x)}{(c-i d) (c+i d)^2 f \sqrt {a+i a \tan (e+f x)} \sqrt {c+d \tan (e+f x)}} \] Input:

Integrate[1/(Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2)),x]
 

Output:

(I*a*ArcTan[(Sqrt[-(a*(c - I*d))]*Sqrt[a + I*a*Tan[e + f*x]])/(Sqrt[2]*a*S 
qrt[c + d*Tan[e + f*x]])])/(Sqrt[2]*(-(a*(c - I*d)))^(3/2)*f) + (I*c^2 + c 
*d - (2*I)*d^2 + d*(I*c + 3*d)*Tan[e + f*x])/((c - I*d)*(c + I*d)^2*f*Sqrt 
[a + I*a*Tan[e + f*x]]*Sqrt[c + d*Tan[e + f*x]])
 

Rubi [A] (verified)

Time = 0.86 (sec) , antiderivative size = 220, normalized size of antiderivative = 1.13, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.281, Rules used = {3042, 4042, 27, 3042, 4081, 27, 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.

\(\displaystyle \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4042

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4081

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4027

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

\(\Big \downarrow \) 221

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

Input:

Int[1/(Sqrt[a + I*a*Tan[e + f*x]]*(c + d*Tan[e + f*x])^(3/2)),x]
 

Output:

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

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

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 4042

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

rule 4081

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 3195 vs. \(2 (160 ) = 320\).

Time = 0.66 (sec) , antiderivative size = 3196, normalized size of antiderivative = 16.47

method	result	size
derivativedivides	\(\text {Expression too large to display}\)	\(3196\)
default	\(\text {Expression too large to display}\)	\(3196\)

Input:

int(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x,method=_RETURNVERB 
OSE)
 

Output:

-1/4/f*(-7*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) 
))*2^(1/2)*c^4*d*(-a*(I*d-c))^(1/2)*tan(f*x+e)+2*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)))*2^(1/2)*c^2*d^3*(-a*(I*d-c))^(1/2 
)*tan(f*x+e)+2*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)))*2^(1/2)*d^5*(-a*(I*d-c))^(1/2)*tan(f*x+e)^2+2*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)))*2^(1/2)*c^5*(-a*(I*d-c))^ 
(1/2)*tan(f*x+e)+4*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)))*2^(1/2)*c^4*d*(-a*(I*d-c))^(1/2)-4*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)))*2^(1/2)*c^2*d^3*(-a*(I*d-c))^(1/ 
2)-4*(a*(c+d*tan(f*x+e))*(1+I*tan(f*x+e)))^(1/2)*c^5+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)))*2^(1/2)*d^5*(-a*(I*d-c))^(1/2 
)*tan(f*x+e)-6*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...
 

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. 673 vs. \(2 (150) = 300\).

Time = 0.17 (sec) , antiderivative size = 673, normalized size of antiderivative = 3.47 \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm=" 
fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\int \frac {1}{\sqrt {i a \left (\tan {\left (e + f x \right )} - i\right )} \left (c + d \tan {\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/(a+I*a*tan(f*x+e))**(1/2)/(c+d*tan(f*x+e))**(3/2),x)
 

Output:

Integral(1/(sqrt(I*a*(tan(e + f*x) - I))*(c + d*tan(e + f*x))**(3/2)), x)
 

Maxima [F]

\[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\int { \frac {1}{\sqrt {i \, a \tan \left (f x + e\right ) + a} {\left (d \tan \left (f x + e\right ) + c\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm=" 
maxima")
 

Output:

integrate(1/(sqrt(I*a*tan(f*x + e) + a)*(d*tan(f*x + e) + c)^(3/2)), x)
 

Giac [F(-2)]

Exception generated. \[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\text {Exception raised: TypeError} \] Input:

integrate(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x, algorithm=" 
giac")
 

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:Error: Bad Argument TypeError: Bad 
Argument TypeError: Bad Argument TypeError: Bad Argument TypeRecursive ass 
umption s
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \frac {1}{\sqrt {a+i a \tan (e+f x)} (c+d \tan (e+f x))^{3/2}} \, dx=\text {too large to display} \] Input:

int(1/(a+I*a*tan(f*x+e))^(1/2)/(c+d*tan(f*x+e))^(3/2),x)
 

Output:

(2*sqrt(a)*(8*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f* 
x)**2*c**3*d**2*i + 56*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*t 
an(e + f*x)**2*c**2*d**3 - 88*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d 
 + c)*tan(e + f*x)**2*c*d**4*i - 40*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + 
f*x)*d + c)*tan(e + f*x)**2*d**5 + 4*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + 
 f*x)*d + c)*tan(e + f*x)*c**4*d*i + 24*sqrt(tan(e + f*x)*i + 1)*sqrt(tan( 
e + f*x)*d + c)*tan(e + f*x)*c**3*d**2 - 16*sqrt(tan(e + f*x)*i + 1)*sqrt( 
tan(e + f*x)*d + c)*tan(e + f*x)*c**2*d**3*i + 24*sqrt(tan(e + f*x)*i + 1) 
*sqrt(tan(e + f*x)*d + c)*tan(e + f*x)*c*d**4 - 20*sqrt(tan(e + f*x)*i + 1 
)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x)*d**5*i - sqrt(tan(e + f*x)*i + 1)* 
sqrt(tan(e + f*x)*d + c)*c**5*i - 9*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + 
f*x)*d + c)*c**4*d + 30*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)* 
c**3*d**2*i + 62*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c**2*d* 
*3 - 65*sqrt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*c*d**4*i - 25*sq 
rt(tan(e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*d**5 + 1250*int((sqrt(tan( 
e + f*x)*i + 1)*sqrt(tan(e + f*x)*d + c)*tan(e + f*x))/(625*tan(e + f*x)** 
4*c**15*d**2*i - 3000*tan(e + f*x)**4*c**14*d**3 + 4725*tan(e + f*x)**4*c* 
*13*d**4*i - 16130*tan(e + f*x)**4*c**12*d**5 + 40681*tan(e + f*x)**4*c**1 
1*d**6*i - 58084*tan(e + f*x)**4*c**10*d**7 + 350821*tan(e + f*x)**4*c**9* 
d**8*i - 1094078*tan(e + f*x)**4*c**8*d**9 - 1380053*tan(e + f*x)**4*c*...