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

Optimal. Leaf size=278 \[ \frac{b^{7/2} (9 a+8 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^5 f (a+b)^{3/2}}+\frac{b \left (-7 a^2 b+5 a^3+12 a b^2+32 b^3\right ) \tan (e+f x)}{16 a^4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\left (15 a^2-26 a b+48 b^2\right ) \sin (e+f x) \cos (e+f x)}{48 a^3 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac{x \left (-12 a^2 b+5 a^3+24 a b^2-64 b^3\right )}{16 a^5}+\frac{(5 a-8 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\sin (e+f x) \cos ^5(e+f x)}{6 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

[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))

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Rubi [A]  time = 0.346716, antiderivative size = 278, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ \frac{b^{7/2} (9 a+8 b) \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a+b}}\right )}{2 a^5 f (a+b)^{3/2}}+\frac{b \left (-7 a^2 b+5 a^3+12 a b^2+32 b^3\right ) \tan (e+f x)}{16 a^4 f (a+b) \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\left (15 a^2-26 a b+48 b^2\right ) \sin (e+f x) \cos (e+f x)}{48 a^3 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac{x \left (-12 a^2 b+5 a^3+24 a b^2-64 b^3\right )}{16 a^5}+\frac{(5 a-8 b) \sin (e+f x) \cos ^3(e+f x)}{24 a^2 f \left (a+b \tan ^2(e+f x)+b\right )}+\frac{\sin (e+f x) \cos ^5(e+f x)}{6 a f \left (a+b \tan ^2(e+f x)+b\right )} \]

Antiderivative was successfully verified.

[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 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]

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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^6(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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{\operatorname{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 ) \operatorname{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 ) \operatorname{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) \tan ^{-1}\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]  time = 4.49534, size = 499, normalized size = 1.79 \[ \frac{\sec ^4(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (12 x \left (-12 a^2 b+5 a^3+24 a b^2-64 b^3\right ) (a \cos (2 (e+f x))+a+2 b)+\frac{3 a \left (15 a^2-32 a b+48 b^2\right ) \sin (2 e) \cos (2 f x) (a \cos (2 (e+f x))+a+2 b)}{f}+\frac{3 a \left (15 a^2-32 a b+48 b^2\right ) \cos (2 e) \sin (2 f x) (a \cos (2 (e+f x))+a+2 b)}{f}+\frac{3 a^2 (3 a-4 b) \sin (4 e) \cos (4 f x) (a \cos (2 (e+f x))+a+2 b)}{f}+\frac{a^3 \sin (6 e) \cos (6 f x) (a \cos (2 (e+f x))+a+2 b)}{f}+\frac{3 a^2 (3 a-4 b) \cos (4 e) \sin (4 f x) (a \cos (2 (e+f x))+a+2 b)}{f}+\frac{a^3 \cos (6 e) \sin (6 f x) (a \cos (2 (e+f x))+a+2 b)}{f}-\frac{96 b^4 ((a+2 b) \sin (2 e)-a \sin (2 f x))}{f (a+b) (\cos (e)-\sin (e)) (\sin (e)+\cos (e))}-\frac{96 b^4 (9 a+8 b) (\cos (2 e)-i \sin (2 e)) (a \cos (2 (e+f x))+a+2 b) \tan ^{-1}\left (\frac{(\cos (2 e)-i \sin (2 e)) \sec (f x) (a \sin (2 e+f x)-(a+2 b) \sin (f x))}{2 \sqrt{a+b} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{f (a+b)^{3/2} \sqrt{b (\cos (e)-i \sin (e))^4}}\right )}{768 a^5 \left (a+b \sec ^2(e+f x)\right )^2} \]

Warning: Unable to verify antiderivative.

[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)

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Maple [A]  time = 0.109, size = 442, normalized size = 1.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 0.772644, size = 1801, normalized size = 6.48 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.28722, size = 385, normalized size = 1.38 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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