\(\int \frac {\cos (e+f x)}{(a+b \sec ^2(e+f x))^3} \, dx\) [209]

Optimal result

Integrand size = 21, antiderivative size = 156 \[ \int \frac {\cos (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {3 b \left (8 a^2+12 a b+5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{7/2} (a+b)^{5/2} f}+\frac {\sin (e+f x)}{a^3 f}-\frac {b^3 \sin (e+f x)}{4 a^3 (a+b) f \left (a+b-a \sin ^2(e+f x)\right )^2}+\frac {3 b^2 (4 a+3 b) \sin (e+f x)}{8 a^3 (a+b)^2 f \left (a+b-a \sin ^2(e+f x)\right )} \] Output:

-3/8*b*(8*a^2+12*a*b+5*b^2)*arctanh(a^(1/2)*sin(f*x+e)/(a+b)^(1/2))/a^(7/2 
)/(a+b)^(5/2)/f+sin(f*x+e)/a^3/f-1/4*b^3*sin(f*x+e)/a^3/(a+b)/f/(a+b-a*sin 
(f*x+e)^2)^2+3/8*b^2*(4*a+3*b)*sin(f*x+e)/a^3/(a+b)^2/f/(a+b-a*sin(f*x+e)^ 
2)
 

Mathematica [A] (verified)

Time = 1.06 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.13 \[ \int \frac {\cos (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {-\frac {3 b \left (8 a^2+12 a b+5 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{(a+b)^{5/2}}+\frac {4 \sqrt {a} \sin (e+f x) \left (8 a^4+32 a^3 b+60 a^2 b^2+51 a b^3+15 b^4-a \left (16 a^3+48 a^2 b+60 a b^2+25 b^3\right ) \sin ^2(e+f x)+8 a^2 (a+b)^2 \sin ^4(e+f x)\right )}{(a+b)^2 (a+2 b+a \cos (2 (e+f x)))^2}}{8 a^{7/2} f} \] Input:

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

Output:

((-3*b*(8*a^2 + 12*a*b + 5*b^2)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b] 
])/(a + b)^(5/2) + (4*Sqrt[a]*Sin[e + f*x]*(8*a^4 + 32*a^3*b + 60*a^2*b^2 
+ 51*a*b^3 + 15*b^4 - a*(16*a^3 + 48*a^2*b + 60*a*b^2 + 25*b^3)*Sin[e + f* 
x]^2 + 8*a^2*(a + b)^2*Sin[e + f*x]^4))/((a + b)^2*(a + 2*b + a*Cos[2*(e + 
 f*x)])^2))/(8*a^(7/2)*f)
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 148, normalized size of antiderivative = 0.95, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3042, 4635, 300, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4635

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

\(\Big \downarrow \) 300

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

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {3 b \left (4 (a+b)^2+(2 a+b)^2\right ) \text {arctanh}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )}{8 a^{7/2} (a+b)^{5/2}}-\frac {b^3 \sin (e+f x)}{4 a^3 (a+b) \left (-a \sin ^2(e+f x)+a+b\right )^2}+\frac {3 b^2 (4 a+3 b) \sin (e+f x)}{8 a^3 (a+b)^2 \left (-a \sin ^2(e+f x)+a+b\right )}+\frac {\sin (e+f x)}{a^3}}{f}\)

Input:

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

Output:

((-3*b*(4*(a + b)^2 + (2*a + b)^2)*ArcTanh[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + 
 b]])/(8*a^(7/2)*(a + b)^(5/2)) + Sin[e + f*x]/a^3 - (b^3*Sin[e + f*x])/(4 
*a^3*(a + b)*(a + b - a*Sin[e + f*x]^2)^2) + (3*b^2*(4*a + 3*b)*Sin[e + f* 
x])/(8*a^3*(a + b)^2*(a + b - a*Sin[e + f*x]^2)))/f
 

Defintions of rubi rules used

rule 300

Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Int 
[PolynomialDivide[(a + b*x^2)^p, (c + d*x^2)^(-q), x], x] /; FreeQ[{a, b, c 
, d}, x] && NeQ[b*c - a*d, 0] && IGtQ[p, 0] && ILtQ[q, 0] && GeQ[p, -q]
 

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 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 = 2.22 (sec) , antiderivative size = 149, normalized size of antiderivative = 0.96

method	result	size
derivativedivides	\(\frac {\frac {\sin \left (f x +e \right )}{a^{3}}+\frac {b \left (\frac {-\frac {3 a b \left (4 a +3 b \right ) \sin \left (f x +e \right )^{3}}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (12 a +7 b \right ) b \sin \left (f x +e \right )}{8 a +8 b}}{\left (-a -b +a \sin \left (f x +e \right )^{2}\right )^{2}}-\frac {3 \left (8 a^{2}+12 a b +5 b^{2}\right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{a^{3}}}{f}\)	\(149\)
default	\(\frac {\frac {\sin \left (f x +e \right )}{a^{3}}+\frac {b \left (\frac {-\frac {3 a b \left (4 a +3 b \right ) \sin \left (f x +e \right )^{3}}{8 \left (a^{2}+2 a b +b^{2}\right )}+\frac {\left (12 a +7 b \right ) b \sin \left (f x +e \right )}{8 a +8 b}}{\left (-a -b +a \sin \left (f x +e \right )^{2}\right )^{2}}-\frac {3 \left (8 a^{2}+12 a b +5 b^{2}\right ) \operatorname {arctanh}\left (\frac {a \sin \left (f x +e \right )}{\sqrt {a \left (a +b \right )}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {a \left (a +b \right )}}\right )}{a^{3}}}{f}\)	\(149\)
risch	\(-\frac {i {\mathrm e}^{i \left (f x +e \right )}}{2 a^{3} f}+\frac {i {\mathrm e}^{-i \left (f x +e \right )}}{2 a^{3} f}-\frac {i b^{2} \left (12 a^{2} {\mathrm e}^{7 i \left (f x +e \right )}+9 a b \,{\mathrm e}^{7 i \left (f x +e \right )}+12 a^{2} {\mathrm e}^{5 i \left (f x +e \right )}+49 a b \,{\mathrm e}^{5 i \left (f x +e \right )}+28 b^{2} {\mathrm e}^{5 i \left (f x +e \right )}-12 a^{2} {\mathrm e}^{3 i \left (f x +e \right )}-49 a b \,{\mathrm e}^{3 i \left (f x +e \right )}-28 b^{2} {\mathrm e}^{3 i \left (f x +e \right )}-12 a^{2} {\mathrm e}^{i \left (f x +e \right )}-9 a b \,{\mathrm e}^{i \left (f x +e \right )}\right )}{4 a^{3} \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 {3 \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}{2 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f a}+\frac {9 \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^{2}}{4 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{2}}+\frac {15 b^{3} \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 )}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{3}}-\frac {3 \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}{2 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f a}-\frac {9 \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^{2}}{4 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{2}}-\frac {15 b^{3} \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 )}{16 \sqrt {a^{2}+a b}\, \left (a +b \right )^{2} f \,a^{3}}\)	\(593\)

Input:

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

Output:

1/f*(sin(f*x+e)/a^3+b/a^3*((-3/8*a*b*(4*a+3*b)/(a^2+2*a*b+b^2)*sin(f*x+e)^ 
3+1/8*(12*a+7*b)*b/(a+b)*sin(f*x+e))/(-a-b+a*sin(f*x+e)^2)^2-3/8*(8*a^2+12 
*a*b+5*b^2)/(a^2+2*a*b+b^2)/(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. 351 vs. \(2 (148) = 296\).

Time = 0.13 (sec) , antiderivative size = 727, normalized size of antiderivative = 4.66 \[ \int \frac {\cos (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[1/16*(3*(8*a^2*b^3 + 12*a*b^4 + 5*b^5 + (8*a^4*b + 12*a^3*b^2 + 5*a^2*b^3 
)*cos(f*x + e)^4 + 2*(8*a^3*b^2 + 12*a^2*b^3 + 5*a*b^4)*cos(f*x + e)^2)*sq 
rt(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*(8*a^4*b^2 + 34*a^3*b^3 + 41*a^2*b^4 + 
15*a*b^5 + 8*(a^6 + 3*a^5*b + 3*a^4*b^2 + a^3*b^3)*cos(f*x + e)^4 + (16*a^ 
5*b + 60*a^4*b^2 + 69*a^3*b^3 + 25*a^2*b^4)*cos(f*x + e)^2)*sin(f*x + e))/ 
((a^9 + 3*a^8*b + 3*a^7*b^2 + a^6*b^3)*f*cos(f*x + e)^4 + 2*(a^8*b + 3*a^7 
*b^2 + 3*a^6*b^3 + a^5*b^4)*f*cos(f*x + e)^2 + (a^7*b^2 + 3*a^6*b^3 + 3*a^ 
5*b^4 + a^4*b^5)*f), 1/8*(3*(8*a^2*b^3 + 12*a*b^4 + 5*b^5 + (8*a^4*b + 12* 
a^3*b^2 + 5*a^2*b^3)*cos(f*x + e)^4 + 2*(8*a^3*b^2 + 12*a^2*b^3 + 5*a*b^4) 
*cos(f*x + e)^2)*sqrt(-a^2 - a*b)*arctan(sqrt(-a^2 - a*b)*sin(f*x + e)/(a 
+ b)) + (8*a^4*b^2 + 34*a^3*b^3 + 41*a^2*b^4 + 15*a*b^5 + 8*(a^6 + 3*a^5*b 
 + 3*a^4*b^2 + a^3*b^3)*cos(f*x + e)^4 + (16*a^5*b + 60*a^4*b^2 + 69*a^3*b 
^3 + 25*a^2*b^4)*cos(f*x + e)^2)*sin(f*x + e))/((a^9 + 3*a^8*b + 3*a^7*b^2 
 + a^6*b^3)*f*cos(f*x + e)^4 + 2*(a^8*b + 3*a^7*b^2 + 3*a^6*b^3 + a^5*b^4) 
*f*cos(f*x + e)^2 + (a^7*b^2 + 3*a^6*b^3 + 3*a^5*b^4 + a^4*b^5)*f)]
 

Sympy [F(-1)]

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

integrate(cos(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 = 253, normalized size of antiderivative = 1.62 \[ \int \frac {\cos (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {3 \, {\left (8 \, a^{2} b + 12 \, a b^{2} + 5 \, b^{3}\right )} \log \left (\frac {a \sin \left (f x + e\right ) - \sqrt {{\left (a + b\right )} a}}{a \sin \left (f x + e\right ) + \sqrt {{\left (a + b\right )} a}}\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {{\left (a + b\right )} a}} - \frac {2 \, {\left (3 \, {\left (4 \, a^{2} b^{2} + 3 \, a b^{3}\right )} \sin \left (f x + e\right )^{3} - {\left (12 \, a^{2} b^{2} + 19 \, a b^{3} + 7 \, b^{4}\right )} \sin \left (f x + e\right )\right )}}{a^{7} + 4 \, a^{6} b + 6 \, a^{5} b^{2} + 4 \, a^{4} b^{3} + a^{3} b^{4} + {\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} \sin \left (f x + e\right )^{4} - 2 \, {\left (a^{7} + 3 \, a^{6} b + 3 \, a^{5} b^{2} + a^{4} b^{3}\right )} \sin \left (f x + e\right )^{2}} + \frac {16 \, \sin \left (f x + e\right )}{a^{3}}}{16 \, f} \] Input:

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

Output:

1/16*(3*(8*a^2*b + 12*a*b^2 + 5*b^3)*log((a*sin(f*x + e) - sqrt((a + b)*a) 
)/(a*sin(f*x + e) + sqrt((a + b)*a)))/((a^5 + 2*a^4*b + a^3*b^2)*sqrt((a + 
 b)*a)) - 2*(3*(4*a^2*b^2 + 3*a*b^3)*sin(f*x + e)^3 - (12*a^2*b^2 + 19*a*b 
^3 + 7*b^4)*sin(f*x + e))/(a^7 + 4*a^6*b + 6*a^5*b^2 + 4*a^4*b^3 + a^3*b^4 
 + (a^7 + 2*a^6*b + a^5*b^2)*sin(f*x + e)^4 - 2*(a^7 + 3*a^6*b + 3*a^5*b^2 
 + a^4*b^3)*sin(f*x + e)^2) + 16*sin(f*x + e)/a^3)/f
 

Giac [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 197, normalized size of antiderivative = 1.26 \[ \int \frac {\cos (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {3 \, {\left (8 \, a^{2} b + 12 \, a b^{2} + 5 \, b^{3}\right )} \arctan \left (\frac {a \sin \left (f x + e\right )}{\sqrt {-a^{2} - a b}}\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {-a^{2} - a b}} - \frac {12 \, a^{2} b^{2} \sin \left (f x + e\right )^{3} + 9 \, a b^{3} \sin \left (f x + e\right )^{3} - 12 \, a^{2} b^{2} \sin \left (f x + e\right ) - 19 \, a b^{3} \sin \left (f x + e\right ) - 7 \, b^{4} \sin \left (f x + e\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} {\left (a \sin \left (f x + e\right )^{2} - a - b\right )}^{2}} + \frac {8 \, \sin \left (f x + e\right )}{a^{3}}}{8 \, f} \] Input:

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

Output:

1/8*(3*(8*a^2*b + 12*a*b^2 + 5*b^3)*arctan(a*sin(f*x + e)/sqrt(-a^2 - a*b) 
)/((a^5 + 2*a^4*b + a^3*b^2)*sqrt(-a^2 - a*b)) - (12*a^2*b^2*sin(f*x + e)^ 
3 + 9*a*b^3*sin(f*x + e)^3 - 12*a^2*b^2*sin(f*x + e) - 19*a*b^3*sin(f*x + 
e) - 7*b^4*sin(f*x + e))/((a^5 + 2*a^4*b + a^3*b^2)*(a*sin(f*x + e)^2 - a 
- b)^2) + 8*sin(f*x + e)/a^3)/f
 

Mupad [B] (verification not implemented)

Time = 17.04 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.12 \[ \int \frac {\cos (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\sin \left (e+f\,x\right )}{a^3\,f}+\frac {\frac {\sin \left (e+f\,x\right )\,\left (7\,b^3+12\,a\,b^2\right )}{8\,\left (a+b\right )}-\frac {3\,{\sin \left (e+f\,x\right )}^3\,\left (4\,a^2\,b^2+3\,a\,b^3\right )}{8\,{\left (a+b\right )}^2}}{f\,\left (2\,a^4\,b-{\sin \left (e+f\,x\right )}^2\,\left (2\,a^5+2\,b\,a^4\right )+a^5+a^3\,b^2+a^5\,{\sin \left (e+f\,x\right )}^4\right )}-\frac {3\,b\,\mathrm {atanh}\left (\frac {\sqrt {a}\,\sin \left (e+f\,x\right )}{\sqrt {a+b}}\right )\,\left (8\,a^2+12\,a\,b+5\,b^2\right )}{8\,a^{7/2}\,f\,{\left (a+b\right )}^{5/2}} \] Input:

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

Output:

sin(e + f*x)/(a^3*f) + ((sin(e + f*x)*(12*a*b^2 + 7*b^3))/(8*(a + b)) - (3 
*sin(e + f*x)^3*(3*a*b^3 + 4*a^2*b^2))/(8*(a + b)^2))/(f*(2*a^4*b - sin(e 
+ f*x)^2*(2*a^4*b + 2*a^5) + a^5 + a^3*b^2 + a^5*sin(e + f*x)^4)) - (3*b*a 
tanh((a^(1/2)*sin(e + f*x))/(a + b)^(1/2))*(12*a*b + 8*a^2 + 5*b^2))/(8*a^ 
(7/2)*f*(a + b)^(5/2))
 

Reduce [B] (verification not implemented)

Time = 0.27 (sec) , antiderivative size = 1649, normalized size of antiderivative = 10.57 \[ \int \frac {\cos (e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(24*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**4*b + 36*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*b**2 + 15*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**3 - 48*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**4*b - 
 120*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*b**2 - 102*sqrt(a)*sqr 
t(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**3 - 30*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))*si 
n(e + f*x)**2*a*b**4 + 24*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**4*b + 84*sqrt(a)*sq 
rt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*ta 
n((e + f*x)/2))*a**3*b**2 + 111*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**3 + 66*s 
qrt(a)*sqrt(a + b)*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*s 
qrt(a)*tan((e + f*x)/2))*a*b**4 + 15*sqrt(a)*sqrt(a + b)*log(sqrt(a + b)*t 
an((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt(a)*tan((e + f*x)/2))*b**5 - 2...