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

Optimal. Leaf size=163 \[ \frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^4 f \sqrt {a+b}}+\frac {(5 a-6 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f}+\frac {\left (5 a^2-6 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 a^3 f}+\frac {x \left (5 a^3-6 a^2 b+8 a b^2-16 b^3\right )}{16 a^4}+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 a f} \]

[Out]

1/16*(5*a^3-6*a^2*b+8*a*b^2-16*b^3)*x/a^4+1/16*(5*a^2-6*a*b+8*b^2)*cos(f*x+e)*sin(f*x+e)/a^3/f+1/24*(5*a-6*b)*
cos(f*x+e)^3*sin(f*x+e)/a^2/f+1/6*cos(f*x+e)^5*sin(f*x+e)/a/f+b^(7/2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/a
^4/f/(a+b)^(1/2)

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Rubi [A]  time = 0.24, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {4146, 414, 527, 522, 203, 205} \[ \frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^4 f \sqrt {a+b}}+\frac {\left (5 a^2-6 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 a^3 f}+\frac {x \left (-6 a^2 b+5 a^3+8 a b^2-16 b^3\right )}{16 a^4}+\frac {(5 a-6 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f}+\frac {\sin (e+f x) \cos ^5(e+f x)}{6 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 203

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

Rule 205

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

Rule 414

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]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 522

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 527

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 4146

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*} \int \frac {\cos ^6(e+f x)}{a+b \sec ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {1}{\left (1+x^2\right )^4 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f}-\frac {\operatorname {Subst}\left (\int \frac {-5 a+b-5 b x^2}{\left (1+x^2\right )^3 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{6 a f}\\ &=\frac {(5 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f}+\frac {\operatorname {Subst}\left (\int \frac {3 \left (5 a^2-a b+2 b^2\right )+3 (5 a-6 b) b x^2}{\left (1+x^2\right )^2 \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{24 a^2 f}\\ &=\frac {\left (5 a^2-6 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac {(5 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f}-\frac {\operatorname {Subst}\left (\int \frac {-3 \left (5 a^3-a^2 b+2 a b^2-8 b^3\right )-3 b \left (5 a^2-6 a b+8 b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{48 a^3 f}\\ &=\frac {\left (5 a^2-6 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac {(5 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f}+\frac {b^4 \operatorname {Subst}\left (\int \frac {1}{a+b+b x^2} \, dx,x,\tan (e+f x)\right )}{a^4 f}+\frac {\left (5 a^3-6 a^2 b+8 a b^2-16 b^3\right ) \operatorname {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 a^4 f}\\ &=\frac {\left (5 a^3-6 a^2 b+8 a b^2-16 b^3\right ) x}{16 a^4}+\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a^4 \sqrt {a+b} f}+\frac {\left (5 a^2-6 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 a^3 f}+\frac {(5 a-6 b) \cos ^3(e+f x) \sin (e+f x)}{24 a^2 f}+\frac {\cos ^5(e+f x) \sin (e+f x)}{6 a f}\\ \end {align*}

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Mathematica [A]  time = 0.85, size = 133, normalized size = 0.82 \[ \frac {a^3 \sin (6 (e+f x))+3 a \left (15 a^2-16 a b+16 b^2\right ) \sin (2 (e+f x))+3 a^2 (3 a-2 b) \sin (4 (e+f x))+12 \left (5 a^3-6 a^2 b+8 a b^2-16 b^3\right ) (e+f x)+\frac {192 b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{\sqrt {a+b}}}{192 a^4 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(5*a^3 - 6*a^2*b + 8*a*b^2 - 16*b^3)*(e + f*x) + (192*b^(7/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/
Sqrt[a + b] + 3*a*(15*a^2 - 16*a*b + 16*b^2)*Sin[2*(e + f*x)] + 3*a^2*(3*a - 2*b)*Sin[4*(e + f*x)] + a^3*Sin[6
*(e + f*x)])/(192*a^4*f)

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fricas [A]  time = 0.78, size = 424, normalized size = 2.60 \[ \left [\frac {12 \, b^{3} \sqrt {-\frac {b}{a + b}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right ) + 3 \, {\left (5 \, a^{3} - 6 \, a^{2} b + 8 \, a b^{2} - 16 \, b^{3}\right )} f x + {\left (8 \, a^{3} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{3} - 6 \, a^{2} b + 8 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, a^{4} f}, -\frac {24 \, b^{3} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - 3 \, {\left (5 \, a^{3} - 6 \, a^{2} b + 8 \, a b^{2} - 16 \, b^{3}\right )} f x - {\left (8 \, a^{3} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 6 \, a^{2} b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{3} - 6 \, a^{2} b + 8 \, a b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, a^{4} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(12*b^3*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)*cos(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)) + 3*(5*a^3 - 6*a^2*b + 8*a*b^2 - 16*b^3)*f*x + (8*a^3*cos(f*x +
e)^5 + 2*(5*a^3 - 6*a^2*b)*cos(f*x + e)^3 + 3*(5*a^3 - 6*a^2*b + 8*a*b^2)*cos(f*x + e))*sin(f*x + e))/(a^4*f),
-1/48*(24*b^3*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))) - 3*(5*a^3 - 6*a^2*b + 8*a*b^2 - 16*b^3)*f*x - (8*a^3*cos(f*x + e)^5 + 2*(5*a^3 - 6*a^2*b)*cos(f*x +
e)^3 + 3*(5*a^3 - 6*a^2*b + 8*a*b^2)*cos(f*x + e))*sin(f*x + e))/(a^4*f)]

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giac [A]  time = 0.24, size = 229, normalized size = 1.40 \[ \frac {\frac {48 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\relax (b) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )} b^{4}}{\sqrt {a b + b^{2}} a^{4}} + \frac {3 \, {\left (5 \, a^{3} - 6 \, a^{2} b + 8 \, a b^{2} - 16 \, b^{3}\right )} {\left (f x + e\right )}}{a^{4}} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{5} - 18 \, a b \tan \left (f x + e\right )^{5} + 24 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} - 48 \, a b \tan \left (f x + e\right )^{3} + 48 \, b^{2} \tan \left (f x + e\right )^{3} + 33 \, a^{2} \tan \left (f x + e\right ) - 30 \, a b \tan \left (f x + e\right ) + 24 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3} a^{3}}}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))*b^4/(sqrt(a*b + b^2)*a
^4) + 3*(5*a^3 - 6*a^2*b + 8*a*b^2 - 16*b^3)*(f*x + e)/a^4 + (15*a^2*tan(f*x + e)^5 - 18*a*b*tan(f*x + e)^5 +
24*b^2*tan(f*x + e)^5 + 40*a^2*tan(f*x + e)^3 - 48*a*b*tan(f*x + e)^3 + 48*b^2*tan(f*x + e)^3 + 33*a^2*tan(f*x
+ e) - 30*a*b*tan(f*x + e) + 24*b^2*tan(f*x + e))/((tan(f*x + e)^2 + 1)^3*a^3))/f

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maple [B]  time = 1.65, size = 359, normalized size = 2.20 \[ \frac {b^{4} \arctan \left (\frac {\tan \left (f x +e \right ) b}{\sqrt {\left (a +b \right ) b}}\right )}{f \,a^{4} \sqrt {\left (a +b \right ) b}}+\frac {5 \left (\tan ^{5}\left (f x +e \right )\right )}{16 f a \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {3 \left (\tan ^{5}\left (f x +e \right )\right ) b}{8 f \,a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}+\frac {\left (\tan ^{5}\left (f x +e \right )\right ) b^{2}}{2 f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}+\frac {\left (\tan ^{3}\left (f x +e \right )\right ) b^{2}}{f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}+\frac {5 \left (\tan ^{3}\left (f x +e \right )\right )}{6 f a \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {\left (\tan ^{3}\left (f x +e \right )\right ) b}{f \,a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {5 \tan \left (f x +e \right ) b}{8 f \,a^{2} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}+\frac {\tan \left (f x +e \right ) b^{2}}{2 f \,a^{3} \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}+\frac {11 \tan \left (f x +e \right )}{16 f a \left (1+\tan ^{2}\left (f x +e \right )\right )^{3}}-\frac {\arctan \left (\tan \left (f x +e \right )\right ) b^{3}}{f \,a^{4}}+\frac {5 \arctan \left (\tan \left (f x +e \right )\right )}{16 f a}-\frac {3 \arctan \left (\tan \left (f x +e \right )\right ) b}{8 f \,a^{2}}+\frac {\arctan \left (\tan \left (f x +e \right )\right ) b^{2}}{2 f \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/f/a^4*b^4/((a+b)*b)^(1/2)*arctan(tan(f*x+e)*b/((a+b)*b)^(1/2))+5/16/f/a/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5-3/8/
f/a^2/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5*b+1/2/f/a^3/(tan(f*x+e)^2+1)^3*tan(f*x+e)^5*b^2+1/f/a^3/(tan(f*x+e)^2+1)
^3*tan(f*x+e)^3*b^2+5/6/f/a/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3-1/f/a^2/(tan(f*x+e)^2+1)^3*tan(f*x+e)^3*b-5/8/f/a^
2/(tan(f*x+e)^2+1)^3*tan(f*x+e)*b+1/2/f/a^3/(tan(f*x+e)^2+1)^3*tan(f*x+e)*b^2+11/16/f/a/(tan(f*x+e)^2+1)^3*tan
(f*x+e)-1/f/a^4*arctan(tan(f*x+e))*b^3+5/16/f/a*arctan(tan(f*x+e))-3/8/f/a^2*arctan(tan(f*x+e))*b+1/2/f/a^3*ar
ctan(tan(f*x+e))*b^2

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maxima [A]  time = 0.45, size = 189, normalized size = 1.16 \[ \frac {\frac {48 \, b^{4} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} a^{4}} + \frac {3 \, {\left (5 \, a^{2} - 6 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (5 \, a^{2} - 6 \, a b + 6 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{2} - 10 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )}{a^{3} \tan \left (f x + e\right )^{6} + 3 \, a^{3} \tan \left (f x + e\right )^{4} + 3 \, a^{3} \tan \left (f x + e\right )^{2} + a^{3}} + \frac {3 \, {\left (5 \, a^{3} - 6 \, a^{2} b + 8 \, a b^{2} - 16 \, b^{3}\right )} {\left (f x + e\right )}}{a^{4}}}{48 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(48*b^4*arctan(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*a^4) + (3*(5*a^2 - 6*a*b + 8*b^2)*tan(f*x
+ e)^5 + 8*(5*a^2 - 6*a*b + 6*b^2)*tan(f*x + e)^3 + 3*(11*a^2 - 10*a*b + 8*b^2)*tan(f*x + e))/(a^3*tan(f*x +
e)^6 + 3*a^3*tan(f*x + e)^4 + 3*a^3*tan(f*x + e)^2 + a^3) + 3*(5*a^3 - 6*a^2*b + 8*a*b^2 - 16*b^3)*(f*x + e)/a
^4)/f

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mupad [B]  time = 6.15, size = 1979, normalized size = 12.14 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((tan(e + f*x)*(11*a^2 - 10*a*b + 8*b^2))/(16*a^3) + (tan(e + f*x)^3*(5*a^2 - 6*a*b + 6*b^2))/(6*a^3) + (tan(e
+ f*x)^5*(5*a^2 - 6*a*b + 8*b^2))/(16*a^3))/(f*(3*tan(e + f*x)^2 + 3*tan(e + f*x)^4 + tan(e + f*x)^6 + 1)) +
(atan(((((((2*a^8*b^5 - (a^9*b^4)/2 + (a^10*b^3)/4 - (5*a^11*b^2)/4)/a^9 - (tan(e + f*x)*(2048*a^8*b^3 + 1024*
a^9*b^2)*(a*b^2*8i - a^2*b*6i + a^3*5i - b^3*16i))/(4096*a^10))*(a*b^2*8i - a^2*b*6i + a^3*5i - b^3*16i))/(32*
a^4) - (tan(e + f*x)*(512*b^9 - 256*a*b^8 + 256*a^2*b^7 - 256*a^3*b^6 + 116*a^4*b^5 - 60*a^5*b^4 + 25*a^6*b^3)
)/(128*a^6))*(a*b^2*8i - a^2*b*6i + a^3*5i - b^3*16i)*1i)/(32*a^4) - (((((2*a^8*b^5 - (a^9*b^4)/2 + (a^10*b^3)
/4 - (5*a^11*b^2)/4)/a^9 + (tan(e + f*x)*(2048*a^8*b^3 + 1024*a^9*b^2)*(a*b^2*8i - a^2*b*6i + a^3*5i - b^3*16i
))/(4096*a^10))*(a*b^2*8i - a^2*b*6i + a^3*5i - b^3*16i))/(32*a^4) + (tan(e + f*x)*(512*b^9 - 256*a*b^8 + 256*
a^2*b^7 - 256*a^3*b^6 + 116*a^4*b^5 - 60*a^5*b^4 + 25*a^6*b^3))/(128*a^6))*(a*b^2*8i - a^2*b*6i + a^3*5i - b^3
*16i)*1i)/(32*a^4))/((((((2*a^8*b^5 - (a^9*b^4)/2 + (a^10*b^3)/4 - (5*a^11*b^2)/4)/a^9 - (tan(e + f*x)*(2048*a
^8*b^3 + 1024*a^9*b^2)*(a*b^2*8i - a^2*b*6i + a^3*5i - b^3*16i))/(4096*a^10))*(a*b^2*8i - a^2*b*6i + a^3*5i -
b^3*16i))/(32*a^4) - (tan(e + f*x)*(512*b^9 - 256*a*b^8 + 256*a^2*b^7 - 256*a^3*b^6 + 116*a^4*b^5 - 60*a^5*b^4
+ 25*a^6*b^3))/(128*a^6))*(a*b^2*8i - a^2*b*6i + a^3*5i - b^3*16i))/(32*a^4) - ((5*a*b^10)/4 - b^11 - (11*a^2
*b^9)/8 + (29*a^3*b^8)/32 - (15*a^4*b^7)/32 + (25*a^5*b^6)/128)/a^9 + (((((2*a^8*b^5 - (a^9*b^4)/2 + (a^10*b^3
)/4 - (5*a^11*b^2)/4)/a^9 + (tan(e + f*x)*(2048*a^8*b^3 + 1024*a^9*b^2)*(a*b^2*8i - a^2*b*6i + a^3*5i - b^3*16
i))/(4096*a^10))*(a*b^2*8i - a^2*b*6i + a^3*5i - b^3*16i))/(32*a^4) + (tan(e + f*x)*(512*b^9 - 256*a*b^8 + 256
*a^2*b^7 - 256*a^3*b^6 + 116*a^4*b^5 - 60*a^5*b^4 + 25*a^6*b^3))/(128*a^6))*(a*b^2*8i - a^2*b*6i + a^3*5i - b^
3*16i))/(32*a^4)))*(a*b^2*8i - a^2*b*6i + a^3*5i - b^3*16i)*1i)/(16*a^4*f) + (atan((((((-b^7*(a + b))^(1/2)*((
2*a^8*b^5 - (a^9*b^4)/2 + (a^10*b^3)/4 - (5*a^11*b^2)/4)/(2*a^9) - (tan(e + f*x)*(2048*a^8*b^3 + 1024*a^9*b^2)
*(-b^7*(a + b))^(1/2))/(512*a^6*(a^4*b + a^5))))/(2*(a^4*b + a^5)) - (tan(e + f*x)*(512*b^9 - 256*a*b^8 + 256*
a^2*b^7 - 256*a^3*b^6 + 116*a^4*b^5 - 60*a^5*b^4 + 25*a^6*b^3))/(256*a^6))*(-b^7*(a + b))^(1/2)*1i)/(a^4*b + a
^5) - ((((-b^7*(a + b))^(1/2)*((2*a^8*b^5 - (a^9*b^4)/2 + (a^10*b^3)/4 - (5*a^11*b^2)/4)/(2*a^9) + (tan(e + f*
x)*(2048*a^8*b^3 + 1024*a^9*b^2)*(-b^7*(a + b))^(1/2))/(512*a^6*(a^4*b + a^5))))/(2*(a^4*b + a^5)) + (tan(e +
f*x)*(512*b^9 - 256*a*b^8 + 256*a^2*b^7 - 256*a^3*b^6 + 116*a^4*b^5 - 60*a^5*b^4 + 25*a^6*b^3))/(256*a^6))*(-b
^7*(a + b))^(1/2)*1i)/(a^4*b + a^5))/(((((-b^7*(a + b))^(1/2)*((2*a^8*b^5 - (a^9*b^4)/2 + (a^10*b^3)/4 - (5*a^
11*b^2)/4)/(2*a^9) - (tan(e + f*x)*(2048*a^8*b^3 + 1024*a^9*b^2)*(-b^7*(a + b))^(1/2))/(512*a^6*(a^4*b + a^5))
))/(2*(a^4*b + a^5)) - (tan(e + f*x)*(512*b^9 - 256*a*b^8 + 256*a^2*b^7 - 256*a^3*b^6 + 116*a^4*b^5 - 60*a^5*b
^4 + 25*a^6*b^3))/(256*a^6))*(-b^7*(a + b))^(1/2))/(a^4*b + a^5) - ((5*a*b^10)/4 - b^11 - (11*a^2*b^9)/8 + (29
*a^3*b^8)/32 - (15*a^4*b^7)/32 + (25*a^5*b^6)/128)/a^9 + ((((-b^7*(a + b))^(1/2)*((2*a^8*b^5 - (a^9*b^4)/2 + (
a^10*b^3)/4 - (5*a^11*b^2)/4)/(2*a^9) + (tan(e + f*x)*(2048*a^8*b^3 + 1024*a^9*b^2)*(-b^7*(a + b))^(1/2))/(512
*a^6*(a^4*b + a^5))))/(2*(a^4*b + a^5)) + (tan(e + f*x)*(512*b^9 - 256*a*b^8 + 256*a^2*b^7 - 256*a^3*b^6 + 116
*a^4*b^5 - 60*a^5*b^4 + 25*a^6*b^3))/(256*a^6))*(-b^7*(a + b))^(1/2))/(a^4*b + a^5)))*(-b^7*(a + b))^(1/2)*1i)
/(f*(a^4*b + a^5))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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