\(\int \cos ^6(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\) [177]

Optimal result

Integrand size = 23, antiderivative size = 119 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {1}{16} \left (5 a^2+12 a b+8 b^2\right ) x+\frac {\left (5 a^2+12 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {a (5 a+8 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a \cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )}{6 f} \]

[Out]

1/16*(5*a^2+12*a*b+8*b^2)*x+1/16*(5*a^2+12*a*b+8*b^2)*cos(f*x+e)*sin(f*x+e)/f+1/24*a*(5*a+8*b)*cos(f*x+e)^3*si
n(f*x+e)/f+1/6*a*cos(f*x+e)^5*sin(f*x+e)*(a+b+b*tan(f*x+e)^2)/f

Rubi [A] (verified)

Time = 0.17 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {4231, 424, 393, 205, 209} \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\left (5 a^2+12 a b+8 b^2\right ) \sin (e+f x) \cos (e+f x)}{16 f}+\frac {1}{16} x \left (5 a^2+12 a b+8 b^2\right )+\frac {a (5 a+8 b) \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac {a \sin (e+f x) \cos ^5(e+f x) \left (a+b \tan ^2(e+f x)+b\right )}{6 f} \]

[In]

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

[Out]

((5*a^2 + 12*a*b + 8*b^2)*x)/16 + ((5*a^2 + 12*a*b + 8*b^2)*Cos[e + f*x]*Sin[e + f*x])/(16*f) + (a*(5*a + 8*b)
*Cos[e + f*x]^3*Sin[e + f*x])/(24*f) + (a*Cos[e + f*x]^5*Sin[e + f*x]*(a + b + b*Tan[e + f*x]^2))/(6*f)

Rule 205

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^n)^(p + 1)/(a*n*(p + 1))), x] + Dist[(n*(p
+ 1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (
IntegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[
p])

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 393

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

Rule 424

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

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 {\left (a+b+b x^2\right )^2}{\left (1+x^2\right )^4} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {a \cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )}{6 f}+\frac {\text {Subst}\left (\int \frac {(a+b) (5 a+6 b)+3 b (a+2 b) x^2}{\left (1+x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{6 f} \\ & = \frac {a (5 a+8 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a \cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )}{6 f}+\frac {\left (5 a^2+12 a b+8 b^2\right ) \text {Subst}\left (\int \frac {1}{\left (1+x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{8 f} \\ & = \frac {\left (5 a^2+12 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {a (5 a+8 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a \cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )}{6 f}+\frac {\left (5 a^2+12 a b+8 b^2\right ) \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{16 f} \\ & = \frac {1}{16} \left (5 a^2+12 a b+8 b^2\right ) x+\frac {\left (5 a^2+12 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {a (5 a+8 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a \cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )}{6 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.83 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {60 a^2 e+144 a b e+96 b^2 e+60 a^2 f x+144 a b f x+96 b^2 f x+\left (45 a^2+96 a b+48 b^2\right ) \sin (2 (e+f x))+3 a (3 a+4 b) \sin (4 (e+f x))+a^2 \sin (6 (e+f x))}{192 f} \]

[In]

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

[Out]

(60*a^2*e + 144*a*b*e + 96*b^2*e + 60*a^2*f*x + 144*a*b*f*x + 96*b^2*f*x + (45*a^2 + 96*a*b + 48*b^2)*Sin[2*(e
+ f*x)] + 3*a*(3*a + 4*b)*Sin[4*(e + f*x)] + a^2*Sin[6*(e + f*x)])/(192*f)

Maple [A] (verified)

Time = 0.49 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.69

method	result	size
parallelrisch	\(\frac {\left (45 a^{2}+96 a b +48 b^{2}\right ) \sin \left (2 f x +2 e \right )+\left (9 a^{2}+12 a b \right ) \sin \left (4 f x +4 e \right )+a^{2} \sin \left (6 f x +6 e \right )+60 x \left (a^{2}+\frac {12}{5} a b +\frac {8}{5} b^{2}\right ) f}{192 f}\)	\(82\)
derivativedivides	\(\frac {a^{2} \left (\frac {\left (\cos \left (f x +e \right )^{5}+\frac {5 \cos \left (f x +e \right )^{3}}{4}+\frac {15 \cos \left (f x +e \right )}{8}\right ) \sin \left (f x +e \right )}{6}+\frac {5 f x}{16}+\frac {5 e}{16}\right )+2 a b \left (\frac {\left (\cos \left (f x +e \right )^{3}+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+b^{2} \left (\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\)	\(116\)
default	\(\frac {a^{2} \left (\frac {\left (\cos \left (f x +e \right )^{5}+\frac {5 \cos \left (f x +e \right )^{3}}{4}+\frac {15 \cos \left (f x +e \right )}{8}\right ) \sin \left (f x +e \right )}{6}+\frac {5 f x}{16}+\frac {5 e}{16}\right )+2 a b \left (\frac {\left (\cos \left (f x +e \right )^{3}+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )+b^{2} \left (\frac {\sin \left (f x +e \right ) \cos \left (f x +e \right )}{2}+\frac {f x}{2}+\frac {e}{2}\right )}{f}\)	\(116\)
risch	\(\frac {5 a^{2} x}{16}+\frac {3 x a b}{4}+\frac {x \,b^{2}}{2}+\frac {a^{2} \sin \left (6 f x +6 e \right )}{192 f}+\frac {3 a^{2} \sin \left (4 f x +4 e \right )}{64 f}+\frac {\sin \left (4 f x +4 e \right ) a b}{16 f}+\frac {15 \sin \left (2 f x +2 e \right ) a^{2}}{64 f}+\frac {\sin \left (2 f x +2 e \right ) a b}{2 f}+\frac {\sin \left (2 f x +2 e \right ) b^{2}}{4 f}\)	\(119\)

[In]

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

[Out]

1/192*((45*a^2+96*a*b+48*b^2)*sin(2*f*x+2*e)+(9*a^2+12*a*b)*sin(4*f*x+4*e)+a^2*sin(6*f*x+6*e)+60*x*(a^2+12/5*a
*b+8/5*b^2)*f)/f

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (5 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} f x + {\left (8 \, a^{2} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{2} + 12 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, f} \]

[In]

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

[Out]

1/48*(3*(5*a^2 + 12*a*b + 8*b^2)*f*x + (8*a^2*cos(f*x + e)^5 + 2*(5*a^2 + 12*a*b)*cos(f*x + e)^3 + 3*(5*a^2 +
12*a*b + 8*b^2)*cos(f*x + e))*sin(f*x + e))/f

Sympy [F(-1)]

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

[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 = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.13 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (5 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} + \frac {3 \, {\left (5 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (5 \, a^{2} + 12 \, a b + 6 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{2} + 20 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{6} + 3 \, \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{2} + 1}}{48 \, f} \]

[In]

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

[Out]

1/48*(3*(5*a^2 + 12*a*b + 8*b^2)*(f*x + e) + (3*(5*a^2 + 12*a*b + 8*b^2)*tan(f*x + e)^5 + 8*(5*a^2 + 12*a*b +
6*b^2)*tan(f*x + e)^3 + 3*(11*a^2 + 20*a*b + 8*b^2)*tan(f*x + e))/(tan(f*x + e)^6 + 3*tan(f*x + e)^4 + 3*tan(f
*x + e)^2 + 1))/f

Giac [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.26 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (5 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{5} + 36 \, a b \tan \left (f x + e\right )^{5} + 24 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} + 96 \, a b \tan \left (f x + e\right )^{3} + 48 \, b^{2} \tan \left (f x + e\right )^{3} + 33 \, a^{2} \tan \left (f x + e\right ) + 60 \, a b \tan \left (f x + e\right ) + 24 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3}}}{48 \, f} \]

[In]

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

[Out]

1/48*(3*(5*a^2 + 12*a*b + 8*b^2)*(f*x + e) + (15*a^2*tan(f*x + e)^5 + 36*a*b*tan(f*x + e)^5 + 24*b^2*tan(f*x +
e)^5 + 40*a^2*tan(f*x + e)^3 + 96*a*b*tan(f*x + e)^3 + 48*b^2*tan(f*x + e)^3 + 33*a^2*tan(f*x + e) + 60*a*b*t
an(f*x + e) + 24*b^2*tan(f*x + e))/(tan(f*x + e)^2 + 1)^3)/f

Mupad [B] (verification not implemented)

Time = 19.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.03 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=x\,\left (\frac {5\,a^2}{16}+\frac {3\,a\,b}{4}+\frac {b^2}{2}\right )+\frac {\left (\frac {5\,a^2}{16}+\frac {3\,a\,b}{4}+\frac {b^2}{2}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (\frac {5\,a^2}{6}+2\,a\,b+b^2\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {11\,a^2}{16}+\frac {5\,a\,b}{4}+\frac {b^2}{2}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^6+3\,{\mathrm {tan}\left (e+f\,x\right )}^4+3\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )} \]

[In]

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

[Out]

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