\(\int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [40]

Optimal result

Integrand size = 23, antiderivative size = 105 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {a^2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{7/2} f}-\frac {a^2 \cot (e+f x)}{(a+b)^3 f}-\frac {(2 a+b) \cot ^3(e+f x)}{3 (a+b)^2 f}-\frac {\cot ^5(e+f x)}{5 (a+b) f} \] Output:

-a^2*b^(1/2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/(a+b)^(7/2)/f-a^2*cot( 
f*x+e)/(a+b)^3/f-1/3*(2*a+b)*cot(f*x+e)^3/(a+b)^2/f-1/5*cot(f*x+e)^5/(a+b) 
/f
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 1.67 (sec) , antiderivative size = 318, normalized size of antiderivative = 3.03 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^2(e+f x) \left (240 a^2 b \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (\cos (2 e)-i \sin (2 e))+\sqrt {a+b} \csc (e) \csc ^5(e+f x) \sqrt {b (\cos (e)-i \sin (e))^4} \left (10 \left (8 a^2+b^2\right ) \sin (f x)-30 b (3 a+b) \sin (2 e+f x)-40 a^2 \sin (2 e+3 f x)+30 a b \sin (2 e+3 f x)+10 b^2 \sin (2 e+3 f x)+15 a b \sin (4 e+3 f x)+8 a^2 \sin (4 e+5 f x)-9 a b \sin (4 e+5 f x)-2 b^2 \sin (4 e+5 f x)\right )\right )}{480 (a+b)^{7/2} f \left (a+b \sec ^2(e+f x)\right ) \sqrt {b (\cos (e)-i \sin (e))^4}} \] Input:

Integrate[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^2*(240*a^2*b*ArcTan[(Sec[f*x] 
*(Cos[2*e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sq 
rt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(Cos[2*e] - I*Sin[2*e]) + Sqrt[a 
 + b]*Csc[e]*Csc[e + f*x]^5*Sqrt[b*(Cos[e] - I*Sin[e])^4]*(10*(8*a^2 + b^2 
)*Sin[f*x] - 30*b*(3*a + b)*Sin[2*e + f*x] - 40*a^2*Sin[2*e + 3*f*x] + 30* 
a*b*Sin[2*e + 3*f*x] + 10*b^2*Sin[2*e + 3*f*x] + 15*a*b*Sin[4*e + 3*f*x] + 
 8*a^2*Sin[4*e + 5*f*x] - 9*a*b*Sin[4*e + 5*f*x] - 2*b^2*Sin[4*e + 5*f*x]) 
))/(480*(a + b)^(7/2)*f*(a + b*Sec[e + f*x]^2)*Sqrt[b*(Cos[e] - I*Sin[e])^ 
4])
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.92, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4620, 364, 2009}

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.

\(\displaystyle \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {1}{\sin (e+f x)^6 \left (a+b \sec (e+f x)^2\right )}dx\)

\(\Big \downarrow \) 4620

\(\displaystyle \frac {\int \frac {\cot ^6(e+f x) \left (\tan ^2(e+f x)+1\right )^2}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 364

\(\displaystyle \frac {\int \left (\frac {\cot ^6(e+f x)}{a+b}+\frac {(2 a+b) \cot ^4(e+f x)}{(a+b)^2}+\frac {a^2 \cot ^2(e+f x)}{(a+b)^3}-\frac {a^2 b}{(a+b)^3 \left (b \tan ^2(e+f x)+a+b\right )}\right )d\tan (e+f x)}{f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {a^2 \sqrt {b} \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{7/2}}-\frac {a^2 \cot (e+f x)}{(a+b)^3}-\frac {\cot ^5(e+f x)}{5 (a+b)}-\frac {(2 a+b) \cot ^3(e+f x)}{3 (a+b)^2}}{f}\)

Input:

Int[Csc[e + f*x]^6/(a + b*Sec[e + f*x]^2),x]
 

Output:

(-((a^2*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a + b)^(7/2)) 
 - (a^2*Cot[e + f*x])/(a + b)^3 - ((2*a + b)*Cot[e + f*x]^3)/(3*(a + b)^2) 
 - Cot[e + f*x]^5/(5*(a + b)))/f
 

Defintions of rubi rules used

rule 364

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_))/((c_) + (d_.)*(x_)^2), 
x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^2)^p/(c + d*x^2)), x], x 
] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && (In 
tegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

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

rule 4620

Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_ 
)]^(m_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m 
+ 1)/f   Subst[Int[x^m*(ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p/(1 + f 
f^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, 
 x] && IntegerQ[m/2] && IntegerQ[n/2]
 

Maple [A] (verified)

Time = 0.76 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.89

method	result	size
derivativedivides	\(\frac {-\frac {1}{5 \left (a +b \right ) \tan \left (f x +e \right )^{5}}-\frac {a^{2}}{\left (a +b \right )^{3} \tan \left (f x +e \right )}-\frac {2 a +b}{3 \left (a +b \right )^{2} \tan \left (f x +e \right )^{3}}-\frac {a^{2} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{3} \sqrt {\left (a +b \right ) b}}}{f}\)	\(93\)
default	\(\frac {-\frac {1}{5 \left (a +b \right ) \tan \left (f x +e \right )^{5}}-\frac {a^{2}}{\left (a +b \right )^{3} \tan \left (f x +e \right )}-\frac {2 a +b}{3 \left (a +b \right )^{2} \tan \left (f x +e \right )^{3}}-\frac {a^{2} b \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{\left (a +b \right )^{3} \sqrt {\left (a +b \right ) b}}}{f}\)	\(93\)
risch	\(\frac {2 i \left (15 a b \,{\mathrm e}^{8 i \left (f x +e \right )}-90 a b \,{\mathrm e}^{6 i \left (f x +e \right )}-30 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}-80 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}-10 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+40 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-30 a b \,{\mathrm e}^{2 i \left (f x +e \right )}-10 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}-8 a^{2}+9 a b +2 b^{2}\right )}{15 f \left (a +b \right )^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5}}-\frac {\sqrt {-\left (a +b \right ) b}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{2 \left (a +b \right )^{4} f}+\frac {\sqrt {-\left (a +b \right ) b}\, a^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{2 \left (a +b \right )^{4} f}\)	\(257\)

Input:

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
 

Output:

1/f*(-1/5/(a+b)/tan(f*x+e)^5-a^2/(a+b)^3/tan(f*x+e)-1/3*(2*a+b)/(a+b)^2/ta 
n(f*x+e)^3-a^2*b/(a+b)^3/((a+b)*b)^(1/2)*arctan(b*tan(f*x+e)/((a+b)*b)^(1/ 
2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (93) = 186\).

Time = 0.11 (sec) , antiderivative size = 587, normalized size of antiderivative = 5.59 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\left [-\frac {4 \, {\left (8 \, a^{2} - 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 20 \, {\left (4 \, a^{2} - 3 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) + 60 \, a^{2} \cos \left (f x + e\right )}{60 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )} \sin \left (f x + e\right )}, -\frac {2 \, {\left (8 \, a^{2} - 9 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 10 \, {\left (4 \, a^{2} - 3 \, a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) + 30 \, a^{2} \cos \left (f x + e\right )}{30 \, {\left ({\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{4} - 2 \, {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} f\right )} \sin \left (f x + e\right )}\right ] \] Input:

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
 

Output:

[-1/60*(4*(8*a^2 - 9*a*b - 2*b^2)*cos(f*x + e)^5 - 20*(4*a^2 - 3*a*b - b^2 
)*cos(f*x + e)^3 - 15*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 + a^2)*sq 
rt(-b/(a + b))*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^ 
2)*cos(f*x + e)^2 + 4*((a^2 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)* 
cos(f*x + e))*sqrt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2 
*a*b*cos(f*x + e)^2 + b^2))*sin(f*x + e) + 60*a^2*cos(f*x + e))/(((a^3 + 3 
*a^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^ 
3)*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f)*sin(f*x + e)), -1 
/30*(2*(8*a^2 - 9*a*b - 2*b^2)*cos(f*x + e)^5 - 10*(4*a^2 - 3*a*b - b^2)*c 
os(f*x + e)^3 - 15*(a^2*cos(f*x + e)^4 - 2*a^2*cos(f*x + e)^2 + a^2)*sqrt( 
b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/(b*co 
s(f*x + e)*sin(f*x + e)))*sin(f*x + e) + 30*a^2*cos(f*x + e))/(((a^3 + 3*a 
^2*b + 3*a*b^2 + b^3)*f*cos(f*x + e)^4 - 2*(a^3 + 3*a^2*b + 3*a*b^2 + b^3) 
*f*cos(f*x + e)^2 + (a^3 + 3*a^2*b + 3*a*b^2 + b^3)*f)*sin(f*x + e))]
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\text {Timed out} \] Input:

integrate(csc(f*x+e)**6/(a+b*sec(f*x+e)**2),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.30 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {15 \, a^{2} b \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{4} + 5 \, {\left (2 \, a^{2} + 3 \, a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
 

Output:

-1/15*(15*a^2*b*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^3 + 3*a^2*b + 3 
*a*b^2 + b^3)*sqrt((a + b)*b)) + (15*a^2*tan(f*x + e)^4 + 5*(2*a^2 + 3*a*b 
 + b^2)*tan(f*x + e)^2 + 3*a^2 + 6*a*b + 3*b^2)/((a^3 + 3*a^2*b + 3*a*b^2 
+ b^3)*tan(f*x + e)^5))/f
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.65 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {15 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} a^{2} b}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a b + b^{2}}} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{4} + 10 \, a^{2} \tan \left (f x + e\right )^{2} + 15 \, a b \tan \left (f x + e\right )^{2} + 5 \, b^{2} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \tan \left (f x + e\right )^{5}}}{15 \, f} \] Input:

integrate(csc(f*x+e)^6/(a+b*sec(f*x+e)^2),x, algorithm="giac")
 

Output:

-1/15*(15*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqr 
t(a*b + b^2)))*a^2*b/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a*b + b^2)) + ( 
15*a^2*tan(f*x + e)^4 + 10*a^2*tan(f*x + e)^2 + 15*a*b*tan(f*x + e)^2 + 5* 
b^2*tan(f*x + e)^2 + 3*a^2 + 6*a*b + 3*b^2)/((a^3 + 3*a^2*b + 3*a*b^2 + b^ 
3)*tan(f*x + e)^5))/f
 

Mupad [B] (verification not implemented)

Time = 12.94 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.07 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {1}{5\,\left (a+b\right )}+\frac {{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,a+b\right )}{3\,{\left (a+b\right )}^2}+\frac {a^2\,{\mathrm {tan}\left (e+f\,x\right )}^4}{{\left (a+b\right )}^3}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^5}-\frac {a^2\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (a^3+3\,a^2\,b+3\,a\,b^2+b^3\right )}{{\left (a+b\right )}^{7/2}}\right )}{f\,{\left (a+b\right )}^{7/2}} \] Input:

int(1/(sin(e + f*x)^6*(a + b/cos(e + f*x)^2)),x)
 

Output:

- (1/(5*(a + b)) + (tan(e + f*x)^2*(2*a + b))/(3*(a + b)^2) + (a^2*tan(e + 
 f*x)^4)/(a + b)^3)/(f*tan(e + f*x)^5) - (a^2*b^(1/2)*atan((b^(1/2)*tan(e 
+ f*x)*(3*a*b^2 + 3*a^2*b + a^3 + b^3))/(a + b)^(7/2)))/(f*(a + b)^(7/2))
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 329, normalized size of antiderivative = 3.13 \[ \int \frac {\csc ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {-15 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {a}}{\sqrt {b}}\right ) \sin \left (f x +e \right )^{5} a^{2}-15 \sqrt {b}\, \sqrt {a +b}\, \mathit {atan} \left (\frac {\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\sqrt {a}}{\sqrt {b}}\right ) \sin \left (f x +e \right )^{5} a^{2}-8 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a^{3}+\cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a^{2} b +11 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} a \,b^{2}+2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{4} b^{3}-4 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{3}-7 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a^{2} b -2 \cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} a \,b^{2}+\cos \left (f x +e \right ) \sin \left (f x +e \right )^{2} b^{3}-3 \cos \left (f x +e \right ) a^{3}-9 \cos \left (f x +e \right ) a^{2} b -9 \cos \left (f x +e \right ) a \,b^{2}-3 \cos \left (f x +e \right ) b^{3}}{15 \sin \left (f x +e \right )^{5} f \left (a^{4}+4 a^{3} b +6 a^{2} b^{2}+4 a \,b^{3}+b^{4}\right )} \] Input:

int(csc(f*x+e)^6/(a+b*sec(f*x+e)^2),x)
 

Output:

( - 15*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/s 
qrt(b))*sin(e + f*x)**5*a**2 - 15*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*ta 
n((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**5*a**2 - 8*cos(e + f*x)*s 
in(e + f*x)**4*a**3 + cos(e + f*x)*sin(e + f*x)**4*a**2*b + 11*cos(e + f*x 
)*sin(e + f*x)**4*a*b**2 + 2*cos(e + f*x)*sin(e + f*x)**4*b**3 - 4*cos(e + 
 f*x)*sin(e + f*x)**2*a**3 - 7*cos(e + f*x)*sin(e + f*x)**2*a**2*b - 2*cos 
(e + f*x)*sin(e + f*x)**2*a*b**2 + cos(e + f*x)*sin(e + f*x)**2*b**3 - 3*c 
os(e + f*x)*a**3 - 9*cos(e + f*x)*a**2*b - 9*cos(e + f*x)*a*b**2 - 3*cos(e 
 + f*x)*b**3)/(15*sin(e + f*x)**5*f*(a**4 + 4*a**3*b + 6*a**2*b**2 + 4*a*b 
**3 + b**4))