\(\int \frac {\cos ^6(e+f x)}{(a+b \sec ^2(e+f x))^2} \, dx\) [205]

Optimal result

Integrand size = 23, antiderivative size = 278 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\left (5 a^3-12 a^2 b+24 a b^2-64 b^3\right ) x}{16 a^5}+\frac {b^{7/2} (9 a+8 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^5 (a+b)^{3/2} f}+\frac {\left (15 a^2-26 a b+48 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {(5 a-8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) \tan (e+f x)}{16 a^4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \]

[Out]

1/16*(5*a^3-12*a^2*b+24*a*b^2-64*b^3)*x/a^5+1/2*b^(7/2)*(9*a+8*b)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a^5/(
a+b)^(3/2)/f+1/48*(15*a^2-26*a*b+48*b^2)*cos(f*x+e)*sin(f*x+e)/a^3/f/(a+b+b*tan(f*x+e)^2)+1/24*(5*a-8*b)*cos(f
*x+e)^3*sin(f*x+e)/a^2/f/(a+b+b*tan(f*x+e)^2)+1/6*cos(f*x+e)^5*sin(f*x+e)/a/f/(a+b+b*tan(f*x+e)^2)+1/16*b*(5*a
^3-7*a^2*b+12*a*b^2+32*b^3)*tan(f*x+e)/a^4/(a+b)/f/(a+b+b*tan(f*x+e)^2)

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4231, 425, 541, 536, 209, 211} \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {b^{7/2} (9 a+8 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^5 f (a+b)^{3/2}}+\frac {(5 a-8 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\left (15 a^2-26 a b+48 b^2\right ) \sin (e+f x) \cos (e+f x)}{48 a^3 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac {x \left (5 a^3-12 a^2 b+24 a b^2-64 b^3\right )}{16 a^5}+\frac {b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) \tan (e+f x)}{16 a^4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

[In]

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

[Out]

((5*a^3 - 12*a^2*b + 24*a*b^2 - 64*b^3)*x)/(16*a^5) + (b^(7/2)*(9*a + 8*b)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[
a + b]])/(2*a^5*(a + b)^(3/2)*f) + ((15*a^2 - 26*a*b + 48*b^2)*Cos[e + f*x]*Sin[e + f*x])/(48*a^3*f*(a + b + b
*Tan[e + f*x]^2)) + ((5*a - 8*b)*Cos[e + f*x]^3*Sin[e + f*x])/(24*a^2*f*(a + b + b*Tan[e + f*x]^2)) + (Cos[e +
f*x]^5*Sin[e + f*x])/(6*a*f*(a + b + b*Tan[e + f*x]^2)) + (b*(5*a^3 - 7*a^2*b + 12*a*b^2 + 32*b^3)*Tan[e + f*
x])/(16*a^4*(a + b)*f*(a + b + b*Tan[e + f*x]^2))

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 211

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

Rule 425

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[(-b)*x*(a + b*x^n)^(p + 1)*
((c + d*x^n)^(q + 1)/(a*n*(p + 1)*(b*c - a*d))), x] + Dist[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1
)*(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c,
d, n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 536

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 541

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[(
-(b*e - a*f))*x*(a + b*x^n)^(p + 1)*((c + d*x^n)^(q + 1)/(a*n*(b*c - a*d)*(p + 1))), x] + Dist[1/(a*n*(b*c - a
*d)*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[c*(b*e - a*f) + e*n*(b*c - a*d)*(p + 1) + d*(b*e - a*
f)*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, q}, x] && LtQ[p, -1]

Rule 4231

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fre
eFactors[Tan[e + f*x], x]}, Dist[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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^4 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {-5 a+b-7 b x^2}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{6 a f} \\ & = \frac {(5 a-8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\text {Subst}\left (\int \frac {15 a^2-a b+8 b^2+5 (5 a-8 b) b x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{24 a^2 f} \\ & = \frac {\left (15 a^2-26 a b+48 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {(5 a-8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {-3 \left (5 a^3+3 a^2 b-2 a b^2-16 b^3\right )-3 b \left (15 a^2-26 a b+48 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{48 a^3 f} \\ & = \frac {\left (15 a^2-26 a b+48 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {(5 a-8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) \tan (e+f x)}{16 a^4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}-\frac {\text {Subst}\left (\int \frac {-6 \left (5 a^4-2 a^3 b+5 a^2 b^2-28 a b^3-32 b^4\right )-6 b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{96 a^4 (a+b) f} \\ & = \frac {\left (15 a^2-26 a b+48 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {(5 a-8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) \tan (e+f x)}{16 a^4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\left (b^4 (9 a+8 b)\right ) \text {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^5 (a+b) f}+\frac {\left (5 a^3-12 a^2 b+24 a b^2-64 b^3\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 a^5 f} \\ & = \frac {\left (5 a^3-12 a^2 b+24 a b^2-64 b^3\right ) x}{16 a^5}+\frac {b^{7/2} (9 a+8 b) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 a^5 (a+b)^{3/2} f}+\frac {\left (15 a^2-26 a b+48 b^2\right ) \cos (e+f x) \sin (e+f x)}{48 a^3 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {(5 a-8 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f \left (a+b+b \tan ^2(e+f x)\right )}+\frac {b \left (5 a^3-7 a^2 b+12 a b^2+32 b^3\right ) \tan (e+f x)}{16 a^4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 4.75 (sec) , antiderivative size = 499, normalized size of antiderivative = 1.79 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^4(e+f x) \left (12 \left (5 a^3-12 a^2 b+24 a b^2-64 b^3\right ) x (a+2 b+a \cos (2 (e+f x)))-\frac {96 b^4 (9 a+8 b) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x))) (\cos (2 e)-i \sin (2 e))}{(a+b)^{3/2} f \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {3 a \left (15 a^2-32 a b+48 b^2\right ) \cos (2 f x) (a+2 b+a \cos (2 (e+f x))) \sin (2 e)}{f}+\frac {3 a^2 (3 a-4 b) \cos (4 f x) (a+2 b+a \cos (2 (e+f x))) \sin (4 e)}{f}+\frac {a^3 \cos (6 f x) (a+2 b+a \cos (2 (e+f x))) \sin (6 e)}{f}+\frac {3 a \left (15 a^2-32 a b+48 b^2\right ) \cos (2 e) (a+2 b+a \cos (2 (e+f x))) \sin (2 f x)}{f}-\frac {96 b^4 ((a+2 b) \sin (2 e)-a \sin (2 f x))}{(a+b) f (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}+\frac {3 a^2 (3 a-4 b) \cos (4 e) (a+2 b+a \cos (2 (e+f x))) \sin (4 f x)}{f}+\frac {a^3 \cos (6 e) (a+2 b+a \cos (2 (e+f x))) \sin (6 f x)}{f}\right )}{768 a^5 \left (a+b \sec ^2(e+f x)\right )^2} \]

[In]

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

[Out]

((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*(12*(5*a^3 - 12*a^2*b + 24*a*b^2 - 64*b^3)*x*(a + 2*b + a*Cos[2
*(e + f*x)]) - (96*b^4*(9*a + 8*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)])*(Cos[2*e] - I*Sin[2*e]
))/((a + b)^(3/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (3*a*(15*a^2 - 32*a*b + 48*b^2)*Cos[2*f*x]*(a + 2*b + a*C
os[2*(e + f*x)])*Sin[2*e])/f + (3*a^2*(3*a - 4*b)*Cos[4*f*x]*(a + 2*b + a*Cos[2*(e + f*x)])*Sin[4*e])/f + (a^3
*Cos[6*f*x]*(a + 2*b + a*Cos[2*(e + f*x)])*Sin[6*e])/f + (3*a*(15*a^2 - 32*a*b + 48*b^2)*Cos[2*e]*(a + 2*b + a
*Cos[2*(e + f*x)])*Sin[2*f*x])/f - (96*b^4*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/((a + b)*f*(Cos[e] - Sin[e])*(
Cos[e] + Sin[e])) + (3*a^2*(3*a - 4*b)*Cos[4*e]*(a + 2*b + a*Cos[2*(e + f*x)])*Sin[4*f*x])/f + (a^3*Cos[6*e]*(
a + 2*b + a*Cos[2*(e + f*x)])*Sin[6*f*x])/f))/(768*a^5*(a + b*Sec[e + f*x]^2)^2)

Maple [A] (verified)

Time = 6.04 (sec) , antiderivative size = 210, normalized size of antiderivative = 0.76

method	result	size
derivativedivides	\(\frac {\frac {b^{4} \left (\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (9 a +8 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{5}}+\frac {\frac {\left (\frac {5}{16} a^{3}-\frac {3}{4} a^{2} b +\frac {3}{2} a \,b^{2}\right ) \tan \left (f x +e \right )^{5}+\left (3 a \,b^{2}+\frac {5}{6} a^{3}-2 a^{2} b \right ) \tan \left (f x +e \right )^{3}+\left (-\frac {5}{4} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {11}{16} a^{3}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3}}+\frac {\left (5 a^{3}-12 a^{2} b +24 a \,b^{2}-64 b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{a^{5}}}{f}\)	\(210\)
default	\(\frac {\frac {b^{4} \left (\frac {a \tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\left (9 a +8 b \right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{5}}+\frac {\frac {\left (\frac {5}{16} a^{3}-\frac {3}{4} a^{2} b +\frac {3}{2} a \,b^{2}\right ) \tan \left (f x +e \right )^{5}+\left (3 a \,b^{2}+\frac {5}{6} a^{3}-2 a^{2} b \right ) \tan \left (f x +e \right )^{3}+\left (-\frac {5}{4} a^{2} b +\frac {3}{2} a \,b^{2}+\frac {11}{16} a^{3}\right ) \tan \left (f x +e \right )}{\left (1+\tan \left (f x +e \right )^{2}\right )^{3}}+\frac {\left (5 a^{3}-12 a^{2} b +24 a \,b^{2}-64 b^{3}\right ) \arctan \left (\tan \left (f x +e \right )\right )}{16}}{a^{5}}}{f}\)	\(210\)
risch	\(\frac {5 x}{16 a^{2}}-\frac {3 x b}{4 a^{3}}+\frac {3 x \,b^{2}}{2 a^{4}}-\frac {4 x \,b^{3}}{a^{5}}+\frac {i {\mathrm e}^{4 i \left (f x +e \right )} b}{32 a^{3} f}+\frac {i b^{4} \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}{a^{5} \left (a +b \right ) f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}-\frac {15 i {\mathrm e}^{2 i \left (f x +e \right )}}{128 a^{2} f}-\frac {i {\mathrm e}^{-2 i \left (f x +e \right )} b}{4 a^{3} f}-\frac {i {\mathrm e}^{-4 i \left (f x +e \right )} b}{32 a^{3} f}+\frac {i {\mathrm e}^{2 i \left (f x +e \right )} b}{4 a^{3} f}+\frac {3 i {\mathrm e}^{-4 i \left (f x +e \right )}}{128 a^{2} f}+\frac {15 i {\mathrm e}^{-2 i \left (f x +e \right )}}{128 a^{2} f}+\frac {3 i {\mathrm e}^{-2 i \left (f x +e \right )} b^{2}}{8 a^{4} f}-\frac {3 i {\mathrm e}^{2 i \left (f x +e \right )} b^{2}}{8 a^{4} f}-\frac {3 i {\mathrm e}^{4 i \left (f x +e \right )}}{128 a^{2} f}+\frac {9 \sqrt {-\left (a +b \right ) b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{4 \left (a +b \right )^{2} f \,a^{4}}+\frac {2 \sqrt {-\left (a +b \right ) b}\, b^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{\left (a +b \right )^{2} f \,a^{5}}-\frac {9 \sqrt {-\left (a +b \right ) b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{4 \left (a +b \right )^{2} f \,a^{4}}-\frac {2 \sqrt {-\left (a +b \right ) b}\, b^{4} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{\left (a +b \right )^{2} f \,a^{5}}+\frac {\sin \left (6 f x +6 e \right )}{192 a^{2} f}\)	\(547\)

[In]

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

[Out]

1/f*(b^4/a^5*(1/2*a/(a+b)*tan(f*x+e)/(a+b+b*tan(f*x+e)^2)+1/2*(9*a+8*b)/(a+b)/((a+b)*b)^(1/2)*arctan(b*tan(f*x
+e)/((a+b)*b)^(1/2)))+1/a^5*(((5/16*a^3-3/4*a^2*b+3/2*a*b^2)*tan(f*x+e)^5+(3*a*b^2+5/6*a^3-2*a^2*b)*tan(f*x+e)
^3+(-5/4*a^2*b+3/2*a*b^2+11/16*a^3)*tan(f*x+e))/(1+tan(f*x+e)^2)^3+1/16*(5*a^3-12*a^2*b+24*a*b^2-64*b^3)*arcta
n(tan(f*x+e))))

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 789, normalized size of antiderivative = 2.84 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {3 \, {\left (5 \, a^{5} - 7 \, a^{4} b + 12 \, a^{3} b^{2} - 40 \, a^{2} b^{3} - 64 \, a b^{4}\right )} f x \cos \left (f x + e\right )^{2} + 3 \, {\left (5 \, a^{4} b - 7 \, a^{3} b^{2} + 12 \, a^{2} b^{3} - 40 \, a b^{4} - 64 \, b^{5}\right )} f x + 6 \, {\left (9 \, a b^{4} + 8 \, b^{5} + {\left (9 \, a^{2} b^{3} + 8 \, a b^{4}\right )} \cos \left (f x + e\right )^{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 ) + {\left (8 \, {\left (a^{5} + a^{4} b\right )} \cos \left (f x + e\right )^{7} + 2 \, {\left (5 \, a^{5} - 3 \, a^{4} b - 8 \, a^{3} b^{2}\right )} \cos \left (f x + e\right )^{5} + {\left (15 \, a^{5} - 11 \, a^{4} b + 22 \, a^{3} b^{2} + 48 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{4} b - 7 \, a^{3} b^{2} + 12 \, a^{2} b^{3} + 32 \, a b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, {\left ({\left (a^{7} + a^{6} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{6} b + a^{5} b^{2}\right )} f\right )}}, \frac {3 \, {\left (5 \, a^{5} - 7 \, a^{4} b + 12 \, a^{3} b^{2} - 40 \, a^{2} b^{3} - 64 \, a b^{4}\right )} f x \cos \left (f x + e\right )^{2} + 3 \, {\left (5 \, a^{4} b - 7 \, a^{3} b^{2} + 12 \, a^{2} b^{3} - 40 \, a b^{4} - 64 \, b^{5}\right )} f x - 12 \, {\left (9 \, a b^{4} + 8 \, b^{5} + {\left (9 \, a^{2} b^{3} + 8 \, a b^{4}\right )} \cos \left (f x + e\right )^{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 ) + {\left (8 \, {\left (a^{5} + a^{4} b\right )} \cos \left (f x + e\right )^{7} + 2 \, {\left (5 \, a^{5} - 3 \, a^{4} b - 8 \, a^{3} b^{2}\right )} \cos \left (f x + e\right )^{5} + {\left (15 \, a^{5} - 11 \, a^{4} b + 22 \, a^{3} b^{2} + 48 \, a^{2} b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{4} b - 7 \, a^{3} b^{2} + 12 \, a^{2} b^{3} + 32 \, a b^{4}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, {\left ({\left (a^{7} + a^{6} b\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{6} b + a^{5} b^{2}\right )} f\right )}}\right ] \]

[In]

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

[Out]

[1/48*(3*(5*a^5 - 7*a^4*b + 12*a^3*b^2 - 40*a^2*b^3 - 64*a*b^4)*f*x*cos(f*x + e)^2 + 3*(5*a^4*b - 7*a^3*b^2 +
12*a^2*b^3 - 40*a*b^4 - 64*b^5)*f*x + 6*(9*a*b^4 + 8*b^5 + (9*a^2*b^3 + 8*a*b^4)*cos(f*x + e)^2)*sqrt(-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)*co
s(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)) + (8*(a^5 + a^4*b)*cos(f*x + e)^7 + 2*(5*a^5 - 3*a^4*b - 8*a^3*b^2)*cos(f*x + e)^5 + (15*a^
5 - 11*a^4*b + 22*a^3*b^2 + 48*a^2*b^3)*cos(f*x + e)^3 + 3*(5*a^4*b - 7*a^3*b^2 + 12*a^2*b^3 + 32*a*b^4)*cos(f
*x + e))*sin(f*x + e))/((a^7 + a^6*b)*f*cos(f*x + e)^2 + (a^6*b + a^5*b^2)*f), 1/48*(3*(5*a^5 - 7*a^4*b + 12*a
^3*b^2 - 40*a^2*b^3 - 64*a*b^4)*f*x*cos(f*x + e)^2 + 3*(5*a^4*b - 7*a^3*b^2 + 12*a^2*b^3 - 40*a*b^4 - 64*b^5)*
f*x - 12*(9*a*b^4 + 8*b^5 + (9*a^2*b^3 + 8*a*b^4)*cos(f*x + e)^2)*sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*
x + e)^2 - b)*sqrt(b/(a + b))/(b*cos(f*x + e)*sin(f*x + e))) + (8*(a^5 + a^4*b)*cos(f*x + e)^7 + 2*(5*a^5 - 3*
a^4*b - 8*a^3*b^2)*cos(f*x + e)^5 + (15*a^5 - 11*a^4*b + 22*a^3*b^2 + 48*a^2*b^3)*cos(f*x + e)^3 + 3*(5*a^4*b
- 7*a^3*b^2 + 12*a^2*b^3 + 32*a*b^4)*cos(f*x + e))*sin(f*x + e))/((a^7 + a^6*b)*f*cos(f*x + e)^2 + (a^6*b + a^
5*b^2)*f)]

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 369, normalized size of antiderivative = 1.33 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {24 \, {\left (9 \, a b^{4} + 8 \, b^{5}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{6} + a^{5} b\right )} \sqrt {{\left (a + b\right )} b}} + \frac {3 \, {\left (5 \, a^{3} b - 7 \, a^{2} b^{2} + 12 \, a b^{3} + 32 \, b^{4}\right )} \tan \left (f x + e\right )^{7} + {\left (15 \, a^{4} + 34 \, a^{3} b - 41 \, a^{2} b^{2} + 156 \, a b^{3} + 288 \, b^{4}\right )} \tan \left (f x + e\right )^{5} + {\left (40 \, a^{4} + 17 \, a^{3} b - 35 \, a^{2} b^{2} + 204 \, a b^{3} + 288 \, b^{4}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{4} + 2 \, a^{3} b - 5 \, a^{2} b^{2} + 28 \, a b^{3} + 32 \, b^{4}\right )} \tan \left (f x + e\right )}{{\left (a^{5} b + a^{4} b^{2}\right )} \tan \left (f x + e\right )^{8} + {\left (a^{6} + 5 \, a^{5} b + 4 \, a^{4} b^{2}\right )} \tan \left (f x + e\right )^{6} + a^{6} + 2 \, a^{5} b + a^{4} b^{2} + 3 \, {\left (a^{6} + 3 \, a^{5} b + 2 \, a^{4} b^{2}\right )} \tan \left (f x + e\right )^{4} + {\left (3 \, a^{6} + 7 \, a^{5} b + 4 \, a^{4} b^{2}\right )} \tan \left (f x + e\right )^{2}} + \frac {3 \, {\left (5 \, a^{3} - 12 \, a^{2} b + 24 \, a b^{2} - 64 \, b^{3}\right )} {\left (f x + e\right )}}{a^{5}}}{48 \, f} \]

[In]

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

[Out]

1/48*(24*(9*a*b^4 + 8*b^5)*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/((a^6 + a^5*b)*sqrt((a + b)*b)) + (3*(5*a^3*
b - 7*a^2*b^2 + 12*a*b^3 + 32*b^4)*tan(f*x + e)^7 + (15*a^4 + 34*a^3*b - 41*a^2*b^2 + 156*a*b^3 + 288*b^4)*tan
(f*x + e)^5 + (40*a^4 + 17*a^3*b - 35*a^2*b^2 + 204*a*b^3 + 288*b^4)*tan(f*x + e)^3 + 3*(11*a^4 + 2*a^3*b - 5*
a^2*b^2 + 28*a*b^3 + 32*b^4)*tan(f*x + e))/((a^5*b + a^4*b^2)*tan(f*x + e)^8 + (a^6 + 5*a^5*b + 4*a^4*b^2)*tan
(f*x + e)^6 + a^6 + 2*a^5*b + a^4*b^2 + 3*(a^6 + 3*a^5*b + 2*a^4*b^2)*tan(f*x + e)^4 + (3*a^6 + 7*a^5*b + 4*a^
4*b^2)*tan(f*x + e)^2) + 3*(5*a^3 - 12*a^2*b + 24*a*b^2 - 64*b^3)*(f*x + e)/a^5)/f

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 270, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {24 \, b^{4} \tan \left (f x + e\right )}{{\left (a^{5} + a^{4} b\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}} + \frac {24 \, {\left (9 \, a b^{4} + 8 \, b^{5}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{6} + a^{5} b\right )} \sqrt {a b + b^{2}}} + \frac {3 \, {\left (5 \, a^{3} - 12 \, a^{2} b + 24 \, a b^{2} - 64 \, b^{3}\right )} {\left (f x + e\right )}}{a^{5}} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{5} - 36 \, a b \tan \left (f x + e\right )^{5} + 72 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} - 96 \, a b \tan \left (f x + e\right )^{3} + 144 \, b^{2} \tan \left (f x + e\right )^{3} + 33 \, a^{2} \tan \left (f x + e\right ) - 60 \, a b \tan \left (f x + e\right ) + 72 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{4}}}{48 \, f} \]

[In]

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

[Out]

1/48*(24*b^4*tan(f*x + e)/((a^5 + a^4*b)*(b*tan(f*x + e)^2 + a + b)) + 24*(9*a*b^4 + 8*b^5)*(pi*floor((f*x + e
)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^6 + a^5*b)*sqrt(a*b + b^2)) + 3*(5*a^3 - 12*a
^2*b + 24*a*b^2 - 64*b^3)*(f*x + e)/a^5 + (15*a^2*tan(f*x + e)^5 - 36*a*b*tan(f*x + e)^5 + 72*b^2*tan(f*x + e)
^5 + 40*a^2*tan(f*x + e)^3 - 96*a*b*tan(f*x + e)^3 + 144*b^2*tan(f*x + e)^3 + 33*a^2*tan(f*x + e) - 60*a*b*tan
(f*x + e) + 72*b^2*tan(f*x + e))/((tan(f*x + e)^2 + 1)^3*a^4))/f

Mupad [B] (verification not implemented)

Time = 23.68 (sec) , antiderivative size = 3310, normalized size of antiderivative = 11.91 \[ \int \frac {\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

((tan(e + f*x)*(28*a*b^3 + 2*a^3*b + 11*a^4 + 32*b^4 - 5*a^2*b^2))/(16*a^4*(a + b)) + (tan(e + f*x)^5*(156*a*b
^3 + 34*a^3*b + 15*a^4 + 288*b^4 - 41*a^2*b^2))/(48*a^4*(a + b)) + (tan(e + f*x)^3*(204*a*b^3 + 17*a^3*b + 40*
a^4 + 288*b^4 - 35*a^2*b^2))/(48*a^4*(a + b)) + (b*tan(e + f*x)^7*(12*a*b^2 - 7*a^2*b + 5*a^3 + 32*b^3))/(16*a
^4*(a + b)))/(f*(a + b + tan(e + f*x)^2*(3*a + 4*b) + tan(e + f*x)^4*(3*a + 6*b) + b*tan(e + f*x)^8 + tan(e +
f*x)^6*(a + 4*b))) - (atan(-((((((8*a^10*b^7 + 15*a^11*b^6 + (23*a^12*b^5)/4 - (3*a^13*b^4)/4 - (3*a^14*b^3)/4
- (5*a^15*b^2)/4)/(2*a^13*b + a^14 + a^12*b^2) - (tan(e + f*x)*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64i)*(20
48*a^10*b^5 + 5120*a^11*b^4 + 4096*a^12*b^3 + 1024*a^13*b^2))/(4096*a^5*(2*a^9*b + a^10 + a^8*b^2)))*(a*b^2*24
i - a^2*b*12i + a^3*5i - b^3*64i))/(32*a^5) - (tan(e + f*x)*(14336*a*b^10 + 8192*b^11 + 5248*a^2*b^9 - 64*a^3*
b^8 + 64*a^4*b^7 - 568*a^5*b^6 + 169*a^6*b^5 - 70*a^7*b^4 + 25*a^8*b^3))/(128*(2*a^9*b + a^10 + a^8*b^2)))*(a*
b^2*24i - a^2*b*12i + a^3*5i - b^3*64i)*1i)/(32*a^5) - (((((8*a^10*b^7 + 15*a^11*b^6 + (23*a^12*b^5)/4 - (3*a^
13*b^4)/4 - (3*a^14*b^3)/4 - (5*a^15*b^2)/4)/(2*a^13*b + a^14 + a^12*b^2) + (tan(e + f*x)*(a*b^2*24i - a^2*b*1
2i + a^3*5i - b^3*64i)*(2048*a^10*b^5 + 5120*a^11*b^4 + 4096*a^12*b^3 + 1024*a^13*b^2))/(4096*a^5*(2*a^9*b + a
^10 + a^8*b^2)))*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64i))/(32*a^5) + (tan(e + f*x)*(14336*a*b^10 + 8192*b^1
1 + 5248*a^2*b^9 - 64*a^3*b^8 + 64*a^4*b^7 - 568*a^5*b^6 + 169*a^6*b^5 - 70*a^7*b^4 + 25*a^8*b^3))/(128*(2*a^9
*b + a^10 + a^8*b^2)))*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64i)*1i)/(32*a^5))/((72*a*b^12 + 64*b^13 - 11*a^2
*b^11 + (19*a^3*b^10)/8 + (267*a^4*b^9)/32 - (101*a^5*b^8)/16 + (655*a^6*b^7)/256 - (225*a^7*b^6)/256)/(2*a^13
*b + a^14 + a^12*b^2) + (((((8*a^10*b^7 + 15*a^11*b^6 + (23*a^12*b^5)/4 - (3*a^13*b^4)/4 - (3*a^14*b^3)/4 - (5
*a^15*b^2)/4)/(2*a^13*b + a^14 + a^12*b^2) - (tan(e + f*x)*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64i)*(2048*a^
10*b^5 + 5120*a^11*b^4 + 4096*a^12*b^3 + 1024*a^13*b^2))/(4096*a^5*(2*a^9*b + a^10 + a^8*b^2)))*(a*b^2*24i - a
^2*b*12i + a^3*5i - b^3*64i))/(32*a^5) - (tan(e + f*x)*(14336*a*b^10 + 8192*b^11 + 5248*a^2*b^9 - 64*a^3*b^8 +
64*a^4*b^7 - 568*a^5*b^6 + 169*a^6*b^5 - 70*a^7*b^4 + 25*a^8*b^3))/(128*(2*a^9*b + a^10 + a^8*b^2)))*(a*b^2*2
4i - a^2*b*12i + a^3*5i - b^3*64i))/(32*a^5) + (((((8*a^10*b^7 + 15*a^11*b^6 + (23*a^12*b^5)/4 - (3*a^13*b^4)/
4 - (3*a^14*b^3)/4 - (5*a^15*b^2)/4)/(2*a^13*b + a^14 + a^12*b^2) + (tan(e + f*x)*(a*b^2*24i - a^2*b*12i + a^3
*5i - b^3*64i)*(2048*a^10*b^5 + 5120*a^11*b^4 + 4096*a^12*b^3 + 1024*a^13*b^2))/(4096*a^5*(2*a^9*b + a^10 + a^
8*b^2)))*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64i))/(32*a^5) + (tan(e + f*x)*(14336*a*b^10 + 8192*b^11 + 5248
*a^2*b^9 - 64*a^3*b^8 + 64*a^4*b^7 - 568*a^5*b^6 + 169*a^6*b^5 - 70*a^7*b^4 + 25*a^8*b^3))/(128*(2*a^9*b + a^1
0 + a^8*b^2)))*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64i))/(32*a^5)))*(a*b^2*24i - a^2*b*12i + a^3*5i - b^3*64
i)*1i)/(16*a^5*f) - (atan((((-b^7*(a + b)^3)^(1/2)*((tan(e + f*x)*(14336*a*b^10 + 8192*b^11 + 5248*a^2*b^9 - 6
4*a^3*b^8 + 64*a^4*b^7 - 568*a^5*b^6 + 169*a^6*b^5 - 70*a^7*b^4 + 25*a^8*b^3))/(128*(2*a^9*b + a^10 + a^8*b^2)
) - (((8*a^10*b^7 + 15*a^11*b^6 + (23*a^12*b^5)/4 - (3*a^13*b^4)/4 - (3*a^14*b^3)/4 - (5*a^15*b^2)/4)/(2*a^13*
b + a^14 + a^12*b^2) - (tan(e + f*x)*(-b^7*(a + b)^3)^(1/2)*(9*a + 8*b)*(2048*a^10*b^5 + 5120*a^11*b^4 + 4096*
a^12*b^3 + 1024*a^13*b^2))/(512*(2*a^9*b + a^10 + a^8*b^2)*(3*a^7*b + a^8 + a^5*b^3 + 3*a^6*b^2)))*(-b^7*(a +
b)^3)^(1/2)*(9*a + 8*b))/(4*(3*a^7*b + a^8 + a^5*b^3 + 3*a^6*b^2)))*(9*a + 8*b)*1i)/(4*(3*a^7*b + a^8 + a^5*b^
3 + 3*a^6*b^2)) + ((-b^7*(a + b)^3)^(1/2)*((tan(e + f*x)*(14336*a*b^10 + 8192*b^11 + 5248*a^2*b^9 - 64*a^3*b^8
+ 64*a^4*b^7 - 568*a^5*b^6 + 169*a^6*b^5 - 70*a^7*b^4 + 25*a^8*b^3))/(128*(2*a^9*b + a^10 + a^8*b^2)) + (((8*
a^10*b^7 + 15*a^11*b^6 + (23*a^12*b^5)/4 - (3*a^13*b^4)/4 - (3*a^14*b^3)/4 - (5*a^15*b^2)/4)/(2*a^13*b + a^14
+ a^12*b^2) + (tan(e + f*x)*(-b^7*(a + b)^3)^(1/2)*(9*a + 8*b)*(2048*a^10*b^5 + 5120*a^11*b^4 + 4096*a^12*b^3
+ 1024*a^13*b^2))/(512*(2*a^9*b + a^10 + a^8*b^2)*(3*a^7*b + a^8 + a^5*b^3 + 3*a^6*b^2)))*(-b^7*(a + b)^3)^(1/
2)*(9*a + 8*b))/(4*(3*a^7*b + a^8 + a^5*b^3 + 3*a^6*b^2)))*(9*a + 8*b)*1i)/(4*(3*a^7*b + a^8 + a^5*b^3 + 3*a^6
*b^2)))/((72*a*b^12 + 64*b^13 - 11*a^2*b^11 + (19*a^3*b^10)/8 + (267*a^4*b^9)/32 - (101*a^5*b^8)/16 + (655*a^6
*b^7)/256 - (225*a^7*b^6)/256)/(2*a^13*b + a^14 + a^12*b^2) - ((-b^7*(a + b)^3)^(1/2)*((tan(e + f*x)*(14336*a*
b^10 + 8192*b^11 + 5248*a^2*b^9 - 64*a^3*b^8 + 64*a^4*b^7 - 568*a^5*b^6 + 169*a^6*b^5 - 70*a^7*b^4 + 25*a^8*b^
3))/(128*(2*a^9*b + a^10 + a^8*b^2)) - (((8*a^10*b^7 + 15*a^11*b^6 + (23*a^12*b^5)/4 - (3*a^13*b^4)/4 - (3*a^1
4*b^3)/4 - (5*a^15*b^2)/4)/(2*a^13*b + a^14 + a^12*b^2) - (tan(e + f*x)*(-b^7*(a + b)^3)^(1/2)*(9*a + 8*b)*(20
48*a^10*b^5 + 5120*a^11*b^4 + 4096*a^12*b^3 + 1024*a^13*b^2))/(512*(2*a^9*b + a^10 + a^8*b^2)*(3*a^7*b + a^8 +
a^5*b^3 + 3*a^6*b^2)))*(-b^7*(a + b)^3)^(1/2)*(9*a + 8*b))/(4*(3*a^7*b + a^8 + a^5*b^3 + 3*a^6*b^2)))*(9*a +
8*b))/(4*(3*a^7*b + a^8 + a^5*b^3 + 3*a^6*b^2)) + ((-b^7*(a + b)^3)^(1/2)*((tan(e + f*x)*(14336*a*b^10 + 8192*
b^11 + 5248*a^2*b^9 - 64*a^3*b^8 + 64*a^4*b^7 - 568*a^5*b^6 + 169*a^6*b^5 - 70*a^7*b^4 + 25*a^8*b^3))/(128*(2*
a^9*b + a^10 + a^8*b^2)) + (((8*a^10*b^7 + 15*a^11*b^6 + (23*a^12*b^5)/4 - (3*a^13*b^4)/4 - (3*a^14*b^3)/4 - (
5*a^15*b^2)/4)/(2*a^13*b + a^14 + a^12*b^2) + (tan(e + f*x)*(-b^7*(a + b)^3)^(1/2)*(9*a + 8*b)*(2048*a^10*b^5
+ 5120*a^11*b^4 + 4096*a^12*b^3 + 1024*a^13*b^2))/(512*(2*a^9*b + a^10 + a^8*b^2)*(3*a^7*b + a^8 + a^5*b^3 + 3
*a^6*b^2)))*(-b^7*(a + b)^3)^(1/2)*(9*a + 8*b))/(4*(3*a^7*b + a^8 + a^5*b^3 + 3*a^6*b^2)))*(9*a + 8*b))/(4*(3*
a^7*b + a^8 + a^5*b^3 + 3*a^6*b^2))))*(-b^7*(a + b)^3)^(1/2)*(9*a + 8*b)*1i)/(2*f*(3*a^7*b + a^8 + a^5*b^3 + 3
*a^6*b^2))