\(\int \frac {\sec ^3(e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [207]

Optimal result

Integrand size = 23, antiderivative size = 125 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(4 a+b) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{3/2} (a+b)^{5/2} f}-\frac {b \sin (e+f x)}{4 a (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {(4 a+b) \sin (e+f x)}{8 a (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )} \] Output:

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

Mathematica [A] (verified)

Time = 0.67 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.30 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x)))^3 \sec ^6(e+f x) \left (\frac {8 \sin (e+f x)}{\left (a+b-a \sin ^2(e+f x)\right )^2}-(4 a+b) \left (\frac {3 \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {a} (a+b)^{5/2}}+\frac {4 \sin (e+f x) \left (5 (a+b)-3 a \sin ^2(e+f x)\right )}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))^2}\right )\right )}{192 a f \left (a+b \sec ^2(e+f x)\right )^3} \] Input:

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

Output:

-1/192*((a + 2*b + a*Cos[2*(e + f*x)])^3*Sec[e + f*x]^6*((8*Sin[e + f*x])/ 
(a + b - a*Sin[e + f*x]^2)^2 - (4*a + b)*((3*ArcTanh[(Sqrt[a]*Sin[e + f*x] 
)/Sqrt[a + b]])/(Sqrt[a]*(a + b)^(5/2)) + (4*Sin[e + f*x]*(5*(a + b) - 3*a 
*Sin[e + f*x]^2))/((a + b)^2*(a + 2*b + a*Cos[2*(e + f*x)])^2))))/(a*f*(a 
+ b*Sec[e + f*x]^2)^3)
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 125, 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, 4635, 298, 215, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4635

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

\(\Big \downarrow \) 298

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

\(\Big \downarrow \) 215

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

(-1/4*(b*Sin[e + f*x])/(a*(a + b)*(a + b - a*Sin[e + f*x]^2)^2) + ((4*a + 
b)*(ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]]/(2*Sqrt[a]*(a + b)^(3/2)) 
+ Sin[e + f*x]/(2*(a + b)*(a + b - a*Sin[e + f*x]^2))))/(4*a*(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 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 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 4635

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

Maple [A] (verified)

Time = 1.27 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.99

method	result	size
derivativedivides	\(\frac {\frac {-\frac {\left (4 a +b \right ) \sin \left (f x +e \right )^{3}}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 a -b \right ) \sin \left (f x +e \right )}{8 a \left (a +b \right )}}{\left (-a -b +a \sin \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (4 a +b \right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) a \sqrt {a \left (a +b \right )}}}{f}\)	\(124\)
default	\(\frac {\frac {-\frac {\left (4 a +b \right ) \sin \left (f x +e \right )^{3}}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (4 a -b \right ) \sin \left (f x +e \right )}{8 a \left (a +b \right )}}{\left (-a -b +a \sin \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (4 a +b \right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) a \sqrt {a \left (a +b \right )}}}{f}\)	\(124\)
risch	\(-\frac {i \left (4 a^{2} {\mathrm e}^{7 i \left (f x +e \right )}+a b \,{\mathrm e}^{7 i \left (f x +e \right )}+4 a^{2} {\mathrm e}^{5 i \left (f x +e \right )}+9 a b \,{\mathrm e}^{5 i \left (f x +e \right )}-4 b^{2} {\mathrm e}^{5 i \left (f x +e \right )}-4 a^{2} {\mathrm e}^{3 i \left (f x +e \right )}-9 a b \,{\mathrm e}^{3 i \left (f x +e \right )}+4 b^{2} {\mathrm e}^{3 i \left (f x +e \right )}-4 a^{2} {\mathrm e}^{i \left (f x +e \right )}-a b \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 a \left (a +b \right )^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{4 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right ) b}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f a}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right )}{4 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \left (a +b \right ) {\mathrm e}^{i \left (f x +e \right )}}{\sqrt {a^{2}+a b}}-1\right ) b}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f a}\)	\(419\)

Input:

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

Output:

1/f*((-1/8*(4*a+b)/(a^2+2*a*b+b^2)*sin(f*x+e)^3+1/8*(4*a-b)/a/(a+b)*sin(f* 
x+e))/(-a-b+a*sin(f*x+e)^2)^2+1/8*(4*a+b)/(a^2+2*a*b+b^2)/a/(a*(a+b))^(1/2 
)*arctanh(a*sin(f*x+e)/(a*(a+b))^(1/2)))
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 260 vs. \(2 (117) = 234\).

Time = 0.11 (sec) , antiderivative size = 544, normalized size of antiderivative = 4.35 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [\frac {{\left ({\left (4 \, a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{4} + 4 \, a b^{2} + b^{3} + 2 \, {\left (4 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {a^{2} + a b} \log \left (-\frac {a \cos \left (f x + e\right )^{2} - 2 \, \sqrt {a^{2} + a b} \sin \left (f x + e\right ) - 2 \, a - b}{a \cos \left (f x + e\right )^{2} + b}\right ) + 2 \, {\left (2 \, a^{3} b + a^{2} b^{2} - a b^{3} + {\left (4 \, a^{4} + 5 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}, -\frac {{\left ({\left (4 \, a^{3} + a^{2} b\right )} \cos \left (f x + e\right )^{4} + 4 \, a b^{2} + b^{3} + 2 \, {\left (4 \, a^{2} b + a b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-a^{2} - a b} \arctan \left (\frac {\sqrt {-a^{2} - a b} \sin \left (f x + e\right )}{a + b}\right ) - {\left (2 \, a^{3} b + a^{2} b^{2} - a b^{3} + {\left (4 \, a^{4} + 5 \, a^{3} b + a^{2} b^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}{8 \, {\left ({\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 3 \, a^{5} b^{2} + 3 \, a^{4} b^{3} + a^{3} b^{4}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 3 \, a^{4} b^{3} + 3 \, a^{3} b^{4} + a^{2} b^{5}\right )} f\right )}}\right ] \] Input:

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

Output:

[1/16*(((4*a^3 + a^2*b)*cos(f*x + e)^4 + 4*a*b^2 + b^3 + 2*(4*a^2*b + a*b^ 
2)*cos(f*x + e)^2)*sqrt(a^2 + a*b)*log(-(a*cos(f*x + e)^2 - 2*sqrt(a^2 + a 
*b)*sin(f*x + e) - 2*a - b)/(a*cos(f*x + e)^2 + b)) + 2*(2*a^3*b + a^2*b^2 
 - a*b^3 + (4*a^4 + 5*a^3*b + a^2*b^2)*cos(f*x + e)^2)*sin(f*x + e))/((a^7 
 + 3*a^6*b + 3*a^5*b^2 + a^4*b^3)*f*cos(f*x + e)^4 + 2*(a^6*b + 3*a^5*b^2 
+ 3*a^4*b^3 + a^3*b^4)*f*cos(f*x + e)^2 + (a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 
 + a^2*b^5)*f), -1/8*(((4*a^3 + a^2*b)*cos(f*x + e)^4 + 4*a*b^2 + b^3 + 2* 
(4*a^2*b + a*b^2)*cos(f*x + e)^2)*sqrt(-a^2 - a*b)*arctan(sqrt(-a^2 - a*b) 
*sin(f*x + e)/(a + b)) - (2*a^3*b + a^2*b^2 - a*b^3 + (4*a^4 + 5*a^3*b + a 
^2*b^2)*cos(f*x + e)^2)*sin(f*x + e))/((a^7 + 3*a^6*b + 3*a^5*b^2 + a^4*b^ 
3)*f*cos(f*x + e)^4 + 2*(a^6*b + 3*a^5*b^2 + 3*a^4*b^3 + a^3*b^4)*f*cos(f* 
x + e)^2 + (a^5*b^2 + 3*a^4*b^3 + 3*a^3*b^4 + a^2*b^5)*f)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.70 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (4 \, a + b\right )} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {{\left (a + b\right )} a}} + \frac {2 \, {\left ({\left (4 \, a^{2} + a b\right )} \sin \left (f x + e\right )^{3} - {\left (4 \, a^{2} + 3 \, a b - b^{2}\right )} \sin \left (f x + e\right )\right )}}{a^{5} + 4 \, a^{4} b + 6 \, a^{3} b^{2} + 4 \, a^{2} b^{3} + a b^{4} + {\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sin \left (f x + e\right )^{4} - 2 \, {\left (a^{5} + 3 \, a^{4} b + 3 \, a^{3} b^{2} + a^{2} b^{3}\right )} \sin \left (f x + e\right )^{2}}}{16 \, f} \] Input:

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

Output:

-1/16*((4*a + b)*log((a*sin(f*x + e) - sqrt((a + b)*a))/(a*sin(f*x + e) + 
sqrt((a + b)*a)))/((a^3 + 2*a^2*b + a*b^2)*sqrt((a + b)*a)) + 2*((4*a^2 + 
a*b)*sin(f*x + e)^3 - (4*a^2 + 3*a*b - b^2)*sin(f*x + e))/(a^5 + 4*a^4*b + 
 6*a^3*b^2 + 4*a^2*b^3 + a*b^4 + (a^5 + 2*a^4*b + a^3*b^2)*sin(f*x + e)^4 
- 2*(a^5 + 3*a^4*b + 3*a^3*b^2 + a^2*b^3)*sin(f*x + e)^2))/f
 

Giac [A] (verification not implemented)

Time = 0.23 (sec) , antiderivative size = 155, normalized size of antiderivative = 1.24 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (4 \, a + b\right )} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} \sqrt {-a^{2} - a b}} + \frac {4 \, a^{2} \sin \left (f x + e\right )^{3} + a b \sin \left (f x + e\right )^{3} - 4 \, a^{2} \sin \left (f x + e\right ) - 3 \, a b \sin \left (f x + e\right ) + b^{2} \sin \left (f x + e\right )}{{\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} {\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2}}}{8 \, f} \] Input:

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

Output:

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

Mupad [B] (verification not implemented)

Time = 16.75 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.03 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (4\,a+b\right )}{8\,a^{3/2}\,f\,{\left (a+b\right )}^{5/2}}-\frac {\frac {{\sin \left (e+f\,x\right )}^3\,\left (4\,a+b\right )}{8\,{\left (a+b\right )}^2}-\frac {\sin \left (e+f\,x\right )\,\left (4\,a-b\right )}{8\,a\,\left (a+b\right )}}{f\,\left (2\,a\,b+a^2+b^2-{\sin \left (e+f\,x\right )}^2\,\left (2\,a^2+2\,b\,a\right )+a^2\,{\sin \left (e+f\,x\right )}^4\right )} \] Input:

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 1184, normalized size of antiderivative = 9.47 \[ \int \frac {\sec ^3(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 4*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b 
) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**4*a**3 - sqrt(a)*sqrt(a + b) 
*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f* 
x)/2))*sin(e + f*x)**4*a**2*b + 8*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan( 
(e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)** 
2*a**3 + 10*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt 
(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**2*a**2*b + 2*sqrt(a)*s 
qrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*t 
an((e + f*x)/2))*sin(e + f*x)**2*a*b**2 - 4*sqrt(a)*sqrt(a + b)*log(sqrt(a 
 + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*a**3 
 - 9*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) 
 - 2*sqrt(a)*tan((e + f*x)/2))*a**2*b - 6*sqrt(a)*sqrt(a + b)*log(sqrt(a + 
 b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*a*b**2 
 - sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 
 2*sqrt(a)*tan((e + f*x)/2))*b**3 + 4*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)* 
tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f* 
x)**4*a**3 + sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqr 
t(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**4*a**2*b - 8*sqrt(a)* 
sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) + 2*sqrt(a)* 
tan((e + f*x)/2))*sin(e + f*x)**2*a**3 - 10*sqrt(a)*sqrt(a + b)*log(sqr...