\(\int \frac {\sec ^4(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [213]

Optimal result

Integrand size = 23, antiderivative size = 123 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+4 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 b^{3/2} (a+b)^{5/2} f}-\frac {a \tan (e+f x)}{4 b (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {(a+4 b) \tan (e+f x)}{8 b (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )} \] Output:

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

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 4.17 (sec) , antiderivative size = 283, normalized size of antiderivative = 2.30 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^6(e+f x) \left (-\frac {(a+4 b) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x)))^2 (\cos (2 e)-i \sin (2 e))}{b \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}-\frac {4 (a+b) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{a (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}+\frac {(a+2 b+a \cos (2 (e+f x))) ((a+4 b) \sin (2 e)-(a-2 b) \sin (2 f x))}{b (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right )}{64 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^3} \] Input:

Integrate[Sec[e + f*x]^4/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*(-(((a + 4*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 
*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x)] 
)^2*(Cos[2*e] - I*Sin[2*e]))/(b*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) 
) - (4*(a + b)*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/(a*(Cos[e] - Sin[e])*( 
Cos[e] + Sin[e])) + ((a + 2*b + a*Cos[2*(e + f*x)])*((a + 4*b)*Sin[2*e] - 
(a - 2*b)*Sin[2*f*x]))/(b*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(64*(a + 
b)^2*f*(a + b*Sec[e + f*x]^2)^3)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4634, 298, 215, 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.

\(\displaystyle \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\sec (e+f x)^4}{\left (a+b \sec (e+f x)^2\right )^3}dx\)

\(\Big \downarrow \) 4634

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

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 218

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

Input:

Int[Sec[e + f*x]^4/(a + b*Sec[e + f*x]^2)^3,x]
 

Output:

(-1/4*(a*Tan[e + f*x])/(b*(a + b)*(a + b + b*Tan[e + f*x]^2)^2) + ((a + 4* 
b)*(ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]]/(2*Sqrt[b]*(a + b)^(3/2)) + 
 Tan[e + f*x]/(2*(a + b)*(a + b + b*Tan[e + f*x]^2))))/(4*b*(a + b)))/f
 

Defintions of rubi rules used

rule 215

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) 
/(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1))   Int[(a + b*x^2)^(p + 1 
), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 
*p])
 

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 298

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( 
b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 
2*p + 3))/(2*a*b*(p + 1))   Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, 
 c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
 

rule 3042

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

rule 4634

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) 
)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f 
Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), 
x]^p, 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 = 1.12 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {\frac {\left (a +4 b \right ) \tan \left (f x +e \right )^{3}}{8 a^{2}+16 a b +8 b^{2}}-\frac {\left (a -4 b \right ) \tan \left (f x +e \right )}{8 \left (a +b \right ) b}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (a +4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) b \sqrt {\left (a +b \right ) b}}}{f}\)	\(118\)
default	\(\frac {\frac {\frac {\left (a +4 b \right ) \tan \left (f x +e \right )^{3}}{8 a^{2}+16 a b +8 b^{2}}-\frac {\left (a -4 b \right ) \tan \left (f x +e \right )}{8 \left (a +b \right ) b}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (a +4 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) b \sqrt {\left (a +b \right ) b}}}{f}\)	\(118\)
risch	\(\frac {i \left (-a^{3} {\mathrm e}^{6 i \left (f x +e \right )}-4 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}-3 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}-2 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+8 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+16 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}-3 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+4 \,{\mathrm e}^{2 i \left (f x +e \right )} a^{2} b +16 \,{\mathrm e}^{2 i \left (f x +e \right )} a \,b^{2}-a^{3}+2 a^{2} b \right )}{4 a \left (a +b \right )^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2} b}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right ) a}{16 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f b}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right ) a}{16 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f b}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {-2 i b a -2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f}\)	\(569\)

Input:

int(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x,method=_RETURNVERBOSE)
 

Output:

1/f*((1/8*(a+4*b)/(a^2+2*a*b+b^2)*tan(f*x+e)^3-1/8*(a-4*b)/(a+b)/b*tan(f*x 
+e))/(a+b+b*tan(f*x+e)^2)^2+1/8*(a+4*b)/(a^2+2*a*b+b^2)/b/((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. 287 vs. \(2 (109) = 218\).

Time = 0.13 (sec) , antiderivative size = 654, normalized size of antiderivative = 5.32 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="fricas")
 

Output:

[-1/32*(((a^3 + 4*a^2*b)*cos(f*x + e)^4 + a*b^2 + 4*b^3 + 2*(a^2*b + 4*a*b 
^2)*cos(f*x + e)^2)*sqrt(-a*b - b^2)*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*b)*cos(f*x + e)^3 - b* 
cos(f*x + e))*sqrt(-a*b - b^2)*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2 
*a*b*cos(f*x + e)^2 + b^2)) + 4*((a^3*b - a^2*b^2 - 2*a*b^3)*cos(f*x + e)^ 
3 - (a^2*b^2 + 5*a*b^3 + 4*b^4)*cos(f*x + e))*sin(f*x + e))/((a^5*b^2 + 3* 
a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 + 2*(a^4*b^3 + 3*a^3*b^4 + 
 3*a^2*b^5 + a*b^6)*f*cos(f*x + e)^2 + (a^3*b^4 + 3*a^2*b^5 + 3*a*b^6 + b^ 
7)*f), -1/16*(((a^3 + 4*a^2*b)*cos(f*x + e)^4 + a*b^2 + 4*b^3 + 2*(a^2*b + 
 4*a*b^2)*cos(f*x + e)^2)*sqrt(a*b + b^2)*arctan(1/2*((a + 2*b)*cos(f*x + 
e)^2 - b)/(sqrt(a*b + b^2)*cos(f*x + e)*sin(f*x + e))) + 2*((a^3*b - a^2*b 
^2 - 2*a*b^3)*cos(f*x + e)^3 - (a^2*b^2 + 5*a*b^3 + 4*b^4)*cos(f*x + e))*s 
in(f*x + e))/((a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f*cos(f*x + e)^4 
 + 2*(a^4*b^3 + 3*a^3*b^4 + 3*a^2*b^5 + a*b^6)*f*cos(f*x + e)^2 + (a^3*b^4 
 + 3*a^2*b^5 + 3*a*b^6 + b^7)*f)]
 

Sympy [F]

\[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\int \frac {\sec ^{4}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \] Input:

integrate(sec(f*x+e)**4/(a+b*sec(f*x+e)**2)**3,x)
 

Output:

Integral(sec(e + f*x)**4/(a + b*sec(e + f*x)**2)**3, x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.52 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {{\left (a + 4 \, b\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {{\left (a b + 4 \, b^{2}\right )} \tan \left (f x + e\right )^{3} - {\left (a^{2} - 3 \, a b - 4 \, b^{2}\right )} \tan \left (f x + e\right )}{a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5} + {\left (a^{2} b^{3} + 2 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{3} b^{2} + 3 \, a^{2} b^{3} + 3 \, a b^{4} + b^{5}\right )} \tan \left (f x + e\right )^{2}}}{8 \, f} \] Input:

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="maxima")
 

Output:

1/8*((a + 4*b)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^2*b + 2*a*b^2 + 
b^3)*sqrt((a + b)*b)) + ((a*b + 4*b^2)*tan(f*x + e)^3 - (a^2 - 3*a*b - 4*b 
^2)*tan(f*x + e))/(a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5 + (a^2*b^ 
3 + 2*a*b^4 + b^5)*tan(f*x + e)^4 + 2*(a^3*b^2 + 3*a^2*b^3 + 3*a*b^4 + b^5 
)*tan(f*x + e)^2))/f
 

Giac [A] (verification not implemented)

Time = 0.18 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.33 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {{\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} {\left (a + 4 \, b\right )}}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sqrt {a b + b^{2}}} + \frac {a b \tan \left (f x + e\right )^{3} + 4 \, b^{2} \tan \left (f x + e\right )^{3} - a^{2} \tan \left (f x + e\right ) + 3 \, a b \tan \left (f x + e\right ) + 4 \, b^{2} \tan \left (f x + e\right )}{{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}}}{8 \, f} \] Input:

integrate(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x, algorithm="giac")
 

Output:

1/8*((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b 
 + b^2)))*(a + 4*b)/((a^2*b + 2*a*b^2 + b^3)*sqrt(a*b + b^2)) + (a*b*tan(f 
*x + e)^3 + 4*b^2*tan(f*x + e)^3 - a^2*tan(f*x + e) + 3*a*b*tan(f*x + e) + 
 4*b^2*tan(f*x + e))/((a^2*b + 2*a*b^2 + b^3)*(b*tan(f*x + e)^2 + a + b)^2 
))/f
 

Mupad [B] (verification not implemented)

Time = 17.66 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {{\mathrm {tan}\left (e+f\,x\right )}^3\,\left (a+4\,b\right )}{8\,{\left (a+b\right )}^2}-\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (a-4\,b\right )}{8\,b\,\left (a+b\right )}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (a+4\,b\right )}{8\,b^{3/2}\,f\,{\left (a+b\right )}^{5/2}} \] Input:

int(1/(cos(e + f*x)^4*(a + b/cos(e + f*x)^2)^3),x)
 

Output:

((tan(e + f*x)^3*(a + 4*b))/(8*(a + b)^2) - (tan(e + f*x)*(a - 4*b))/(8*b* 
(a + b)))/(f*(2*a*b + a^2 + b^2 + tan(e + f*x)^2*(2*a*b + 2*b^2) + b^2*tan 
(e + f*x)^4)) + (atan((b^(1/2)*tan(e + f*x))/(a + b)^(1/2))*(a + 4*b))/(8* 
b^(3/2)*f*(a + b)^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 1026, normalized size of antiderivative = 8.34 \[ \int \frac {\sec ^4(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

int(sec(f*x+e)^4/(a+b*sec(f*x+e)^2)^3,x)
 

Output:

(sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b) 
)*sin(e + f*x)**4*a**3 + 4*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + 
f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**2*b - 2*sqrt(b)*sqrt(a + b) 
*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a* 
*3 - 10*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/ 
sqrt(b))*sin(e + f*x)**2*a**2*b - 8*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)* 
tan((e + f*x)/2) - sqrt(a))/sqrt(b))*sin(e + f*x)**2*a*b**2 + sqrt(b)*sqrt 
(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a**3 + 6*sq 
rt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a))/sqrt(b))*a 
**2*b + 9*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) - sqrt(a) 
)/sqrt(b))*a*b**2 + 4*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/ 
2) - sqrt(a))/sqrt(b))*b**3 + sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e 
 + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a**3 + 4*sqrt(b)*sqrt(a + b 
)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**4*a 
**2*b - 2*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a) 
)/sqrt(b))*sin(e + f*x)**2*a**3 - 10*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b) 
*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + f*x)**2*a**2*b - 8*sqrt(b)*s 
qrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + sqrt(a))/sqrt(b))*sin(e + 
f*x)**2*a*b**2 + sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e + f*x)/2) + 
sqrt(a))/sqrt(b))*a**3 + 6*sqrt(b)*sqrt(a + b)*atan((sqrt(a + b)*tan((e...