\(\int \frac {\sin (e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [56]

Optimal result

Integrand size = 21, antiderivative size = 116 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {15 \sqrt {b} \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{8 a^{7/2} f}-\frac {15 \cos (e+f x)}{8 a^3 f}+\frac {\cos ^5(e+f x)}{4 a f \left (b+a \cos ^2(e+f x)\right )^2}+\frac {5 \cos ^3(e+f x)}{8 a^2 f \left (b+a \cos ^2(e+f x)\right )} \] Output:

15/8*b^(1/2)*arctan(a^(1/2)*cos(f*x+e)/b^(1/2))/a^(7/2)/f-15/8*cos(f*x+e)/ 
a^3/f+1/4*cos(f*x+e)^5/a/f/(b+a*cos(f*x+e)^2)^2+5/8*cos(f*x+e)^3/a^2/f/(b+ 
a*cos(f*x+e)^2)
 

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 4.57 (sec) , antiderivative size = 656, normalized size of antiderivative = 5.66 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x)))^3 \sec ^6(e+f x) \left (15 \left (a^3+64 b^3\right ) \arctan \left (\frac {\left (-\sqrt {a}-i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}-\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )+15 \left (a^3+64 b^3\right ) \arctan \left (\frac {\left (-\sqrt {a}+i \sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2}\right ) \sin (e) \tan \left (\frac {f x}{2}\right )+\cos (e) \left (\sqrt {a}+\sqrt {a+b} \sqrt {(\cos (e)-i \sin (e))^2} \tan \left (\frac {f x}{2}\right )\right )}{\sqrt {b}}\right )+\frac {\sqrt {a} \left (24 a^4 \sqrt {b} \cos (e+f x)-24 a^3 b^{3/2} \cos (e+f x)-144 a^2 b^{5/2} \cos (e+f x)+512 b^{9/2} \cos (e+f x)-72 a^3 b^{3/2} \cos (e+f x) \cos (2 (e+f x))-24 a^3 \sqrt {b} \cos (e+f x) (a+2 b+a \cos (2 (e+f x)))+72 a^2 b^{3/2} \cos (e+f x) (a+2 b+a \cos (2 (e+f x)))-1152 b^{7/2} \cos (e+f x) (a+2 b+a \cos (2 (e+f x)))-15 a^{5/2} \arctan \left (\frac {\sqrt {a}-\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 (e+f x)))^2-15 a^{5/2} \arctan \left (\frac {\sqrt {a}+\sqrt {a+b} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {b}}\right ) (a+2 b+a \cos (2 (e+f x)))^2-512 b^{5/2} \cos (e) \cos (f x) (a+2 b+a \cos (2 (e+f x)))^2+512 b^{5/2} (a+2 b+a \cos (2 (e+f x)))^2 \sin (e) \sin (f x)+6 a^4 \sqrt {b} \csc (e+f x) \sin (4 (e+f x))\right )}{(a+2 b+a \cos (2 (e+f x)))^2}\right )}{4096 a^{7/2} b^{5/2} f \left (a+b \sec ^2(e+f x)\right )^3} \] Input:

Integrate[Sin[e + f*x]/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*(15*(a^3 + 64*b^3)*ArcTan 
[((-Sqrt[a] - I*Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/ 
2] + Cos[e]*(Sqrt[a] - Sqrt[a + b]*Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2 
]))/Sqrt[b]] + 15*(a^3 + 64*b^3)*ArcTan[((-Sqrt[a] + I*Sqrt[a + b]*Sqrt[(C 
os[e] - I*Sin[e])^2])*Sin[e]*Tan[(f*x)/2] + Cos[e]*(Sqrt[a] + Sqrt[a + b]* 
Sqrt[(Cos[e] - I*Sin[e])^2]*Tan[(f*x)/2]))/Sqrt[b]] + (Sqrt[a]*(24*a^4*Sqr 
t[b]*Cos[e + f*x] - 24*a^3*b^(3/2)*Cos[e + f*x] - 144*a^2*b^(5/2)*Cos[e + 
f*x] + 512*b^(9/2)*Cos[e + f*x] - 72*a^3*b^(3/2)*Cos[e + f*x]*Cos[2*(e + f 
*x)] - 24*a^3*Sqrt[b]*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) + 72*a^2 
*b^(3/2)*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) - 1152*b^(7/2)*Cos[e 
+ f*x]*(a + 2*b + a*Cos[2*(e + f*x)]) - 15*a^(5/2)*ArcTan[(Sqrt[a] - Sqrt[ 
a + b]*Tan[(e + f*x)/2])/Sqrt[b]]*(a + 2*b + a*Cos[2*(e + f*x)])^2 - 15*a^ 
(5/2)*ArcTan[(Sqrt[a] + Sqrt[a + b]*Tan[(e + f*x)/2])/Sqrt[b]]*(a + 2*b + 
a*Cos[2*(e + f*x)])^2 - 512*b^(5/2)*Cos[e]*Cos[f*x]*(a + 2*b + a*Cos[2*(e 
+ f*x)])^2 + 512*b^(5/2)*(a + 2*b + a*Cos[2*(e + f*x)])^2*Sin[e]*Sin[f*x] 
+ 6*a^4*Sqrt[b]*Csc[e + f*x]*Sin[4*(e + f*x)]))/(a + 2*b + a*Cos[2*(e + f* 
x)])^2))/(4096*a^(7/2)*b^(5/2)*f*(a + b*Sec[e + f*x]^2)^3)
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {3042, 4621, 252, 252, 262, 218}

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.

\(\displaystyle \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sin (e+f x)}{\left (a+b \sec (e+f x)^2\right )^3}dx\)

\(\Big \downarrow \) 4621

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

\(\Big \downarrow \) 252

\(\displaystyle -\frac {\frac {5 \int \frac {\cos ^4(e+f x)}{\left (a \cos ^2(e+f x)+b\right )^2}d\cos (e+f x)}{4 a}-\frac {\cos ^5(e+f x)}{4 a \left (a \cos ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 252

\(\displaystyle -\frac {\frac {5 \left (\frac {3 \int \frac {\cos ^2(e+f x)}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{2 a}-\frac {\cos ^3(e+f x)}{2 a \left (a \cos ^2(e+f x)+b\right )}\right )}{4 a}-\frac {\cos ^5(e+f x)}{4 a \left (a \cos ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 262

\(\displaystyle -\frac {\frac {5 \left (\frac {3 \left (\frac {\cos (e+f x)}{a}-\frac {b \int \frac {1}{a \cos ^2(e+f x)+b}d\cos (e+f x)}{a}\right )}{2 a}-\frac {\cos ^3(e+f x)}{2 a \left (a \cos ^2(e+f x)+b\right )}\right )}{4 a}-\frac {\cos ^5(e+f x)}{4 a \left (a \cos ^2(e+f x)+b\right )^2}}{f}\)

\(\Big \downarrow \) 218

\(\displaystyle -\frac {\frac {5 \left (\frac {3 \left (\frac {\cos (e+f x)}{a}-\frac {\sqrt {b} \arctan \left (\frac {\sqrt {a} \cos (e+f x)}{\sqrt {b}}\right )}{a^{3/2}}\right )}{2 a}-\frac {\cos ^3(e+f x)}{2 a \left (a \cos ^2(e+f x)+b\right )}\right )}{4 a}-\frac {\cos ^5(e+f x)}{4 a \left (a \cos ^2(e+f x)+b\right )^2}}{f}\)

Input:

Int[Sin[e + f*x]/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

-((-1/4*Cos[e + f*x]^5/(a*(b + a*Cos[e + f*x]^2)^2) + (5*((3*(-((Sqrt[b]*A 
rcTan[(Sqrt[a]*Cos[e + f*x])/Sqrt[b]])/a^(3/2)) + Cos[e + f*x]/a))/(2*a) - 
 Cos[e + f*x]^3/(2*a*(b + a*Cos[e + f*x]^2))))/(4*a))/f)
 

Defintions of rubi rules used

rule 218

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R 
t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
 

rule 252

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x 
)^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* 
(p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c 
}, x] && LtQ[p, -1] && GtQ[m, 1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi 
alQ[a, b, c, 2, m, p, x]
 

rule 262

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) 
^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ 
(b*(m + 2*p + 1)))   Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b 
, c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c 
, 2, m, p, x]
 

rule 3042

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

rule 4621

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

Maple [A] (verified)

Time = 2.15 (sec) , antiderivative size = 83, normalized size of antiderivative = 0.72

method	result	size
derivativedivides	\(\frac {-\frac {1}{a^{3} \sec \left (f x +e \right )}-\frac {b \left (\frac {\frac {7 b \sec \left (f x +e \right )^{3}}{8}+\frac {9 a \sec \left (f x +e \right )}{8}}{\left (a +b \sec \left (f x +e \right )^{2}\right )^{2}}+\frac {15 \arctan \left (\frac {b \sec \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}}{f}\)	\(83\)
default	\(\frac {-\frac {1}{a^{3} \sec \left (f x +e \right )}-\frac {b \left (\frac {\frac {7 b \sec \left (f x +e \right )^{3}}{8}+\frac {9 a \sec \left (f x +e \right )}{8}}{\left (a +b \sec \left (f x +e \right )^{2}\right )^{2}}+\frac {15 \arctan \left (\frac {b \sec \left (f x +e \right )}{\sqrt {a b}}\right )}{8 \sqrt {a b}}\right )}{a^{3}}}{f}\)	\(83\)
risch	\(-\frac {{\mathrm e}^{i \left (f x +e \right )}}{2 a^{3} f}-\frac {{\mathrm e}^{-i \left (f x +e \right )}}{2 a^{3} f}-\frac {b \left (9 a \,{\mathrm e}^{7 i \left (f x +e \right )}+27 a \,{\mathrm e}^{5 i \left (f x +e \right )}+28 b \,{\mathrm e}^{5 i \left (f x +e \right )}+27 a \,{\mathrm e}^{3 i \left (f x +e \right )}+28 b \,{\mathrm e}^{3 i \left (f x +e \right )}+9 a \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 a^{3} \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2} f}-\frac {15 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{16 a^{4} f}+\frac {15 i \sqrt {a b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {a b}\, {\mathrm e}^{i \left (f x +e \right )}}{a}+1\right )}{16 a^{4} f}\)	\(249\)

Input:

int(sin(f*x+e)/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/f*(-1/a^3/sec(f*x+e)-1/a^3*b*((7/8*b*sec(f*x+e)^3+9/8*a*sec(f*x+e))/(a+b 
*sec(f*x+e)^2)^2+15/8/(a*b)^(1/2)*arctan(b*sec(f*x+e)/(a*b)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 299, normalized size of antiderivative = 2.58 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [-\frac {16 \, a^{2} \cos \left (f x + e\right )^{5} + 50 \, a b \cos \left (f x + e\right )^{3} + 30 \, b^{2} \cos \left (f x + e\right ) - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {-\frac {b}{a}} \log \left (-\frac {a \cos \left (f x + e\right )^{2} + 2 \, a \sqrt {-\frac {b}{a}} \cos \left (f x + e\right ) - b}{a \cos \left (f x + e\right )^{2} + b}\right )}{16 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}, -\frac {8 \, a^{2} \cos \left (f x + e\right )^{5} + 25 \, a b \cos \left (f x + e\right )^{3} + 15 \, b^{2} \cos \left (f x + e\right ) - 15 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\right )} \sqrt {\frac {b}{a}} \arctan \left (\frac {a \sqrt {\frac {b}{a}} \cos \left (f x + e\right )}{b}\right )}{8 \, {\left (a^{5} f \cos \left (f x + e\right )^{4} + 2 \, a^{4} b f \cos \left (f x + e\right )^{2} + a^{3} b^{2} f\right )}}\right ] \] Input:

integrate(sin(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
 

Output:

[-1/16*(16*a^2*cos(f*x + e)^5 + 50*a*b*cos(f*x + e)^3 + 30*b^2*cos(f*x + e 
) - 15*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(-b/a)*log(-( 
a*cos(f*x + e)^2 + 2*a*sqrt(-b/a)*cos(f*x + e) - b)/(a*cos(f*x + e)^2 + b) 
))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b^2*f), -1/8*(8* 
a^2*cos(f*x + e)^5 + 25*a*b*cos(f*x + e)^3 + 15*b^2*cos(f*x + e) - 15*(a^2 
*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(b/a)*arctan(a*sqrt(b/a) 
*cos(f*x + e)/b))/(a^5*f*cos(f*x + e)^4 + 2*a^4*b*f*cos(f*x + e)^2 + a^3*b 
^2*f)]
                                                                                    
                                                                                    
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Timed out} \] Input:

integrate(sin(f*x+e)/(a+b*sec(f*x+e)**2)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 103, normalized size of antiderivative = 0.89 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {9 \, a b \cos \left (f x + e\right )^{3} + 7 \, b^{2} \cos \left (f x + e\right )}{a^{5} \cos \left (f x + e\right )^{4} + 2 \, a^{4} b \cos \left (f x + e\right )^{2} + a^{3} b^{2}} - \frac {15 \, b \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {a b} a^{3}} + \frac {8 \, \cos \left (f x + e\right )}{a^{3}}}{8 \, f} \] Input:

integrate(sin(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
 

Output:

-1/8*((9*a*b*cos(f*x + e)^3 + 7*b^2*cos(f*x + e))/(a^5*cos(f*x + e)^4 + 2* 
a^4*b*cos(f*x + e)^2 + a^3*b^2) - 15*b*arctan(a*cos(f*x + e)/sqrt(a*b))/(s 
qrt(a*b)*a^3) + 8*cos(f*x + e)/a^3)/f
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.77 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {15 \, b \arctan \left (\frac {a \cos \left (f x + e\right )}{\sqrt {a b}}\right )}{8 \, \sqrt {a b} a^{3} f} - \frac {\cos \left (f x + e\right )}{a^{3} f} - \frac {9 \, a b \cos \left (f x + e\right )^{3} + 7 \, b^{2} \cos \left (f x + e\right )}{8 \, {\left (a \cos \left (f x + e\right )^{2} + b\right )}^{2} a^{3} f} \] Input:

integrate(sin(f*x+e)/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
 

Output:

15/8*b*arctan(a*cos(f*x + e)/sqrt(a*b))/(sqrt(a*b)*a^3*f) - cos(f*x + e)/( 
a^3*f) - 1/8*(9*a*b*cos(f*x + e)^3 + 7*b^2*cos(f*x + e))/((a*cos(f*x + e)^ 
2 + b)^2*a^3*f)
 

Mupad [B] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 105, normalized size of antiderivative = 0.91 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {15\,\sqrt {b}\,\mathrm {atan}\left (\frac {\sqrt {a}\,\cos \left (e+f\,x\right )}{\sqrt {b}}\right )}{8\,a^{7/2}\,f}-\frac {\frac {7\,b^2\,\cos \left (e+f\,x\right )}{8}+\frac {9\,a\,b\,{\cos \left (e+f\,x\right )}^3}{8}}{f\,\left (a^5\,{\cos \left (e+f\,x\right )}^4+2\,a^4\,b\,{\cos \left (e+f\,x\right )}^2+a^3\,b^2\right )}-\frac {\cos \left (e+f\,x\right )}{a^3\,f} \] Input:

int(sin(e + f*x)/(a + b/cos(e + f*x)^2)^3,x)
 

Output:

(15*b^(1/2)*atan((a^(1/2)*cos(e + f*x))/b^(1/2)))/(8*a^(7/2)*f) - ((7*b^2* 
cos(e + f*x))/8 + (9*a*b*cos(e + f*x)^3)/8)/(f*(a^3*b^2 + a^5*cos(e + f*x) 
^4 + 2*a^4*b*cos(e + f*x)^2)) - cos(e + f*x)/(a^3*f)
 

Reduce [B] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 1410, normalized size of antiderivative = 12.16 \[ \int \frac {\sin (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(sin(f*x+e)/(a+b*sec(f*x+e)^2)^3,x)
 

Output:

( - 15*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt( 
b))*sin(e + f*x)**4*a**4 - 30*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f 
*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b - 15*sqrt(b)*sqrt(a)*ata 
n((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**2*b 
**2 + 30*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqr 
t(b))*sin(e + f*x)**2*a**4 + 90*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + 
 f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**3*b + 90*sqrt(b)*sqrt(a)*a 
tan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**2 
*b**2 + 30*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/s 
qrt(b))*sin(e + f*x)**2*a*b**3 - 15*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan( 
(e + f*x)/2) - sqrt(a))/sqrt(b))*a**4 - 60*sqrt(b)*sqrt(a)*atan((sqrt(a + 
b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**3*b - 90*sqrt(b)*sqrt(a)*atan(( 
sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**2*b**2 - 60*sqrt(b)*sq 
rt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a*b**3 - 15*s 
qrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*b**4 
 + 15*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b 
))*sin(e + f*x)**4*a**4 + 30*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f* 
x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3*b + 15*sqrt(b)*sqrt(a)*atan 
((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**2*b* 
*2 - 30*sqrt(b)*sqrt(a)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/s...