Integrand size = 23, antiderivative size = 119 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {1}{16} \left (5 a^2+12 a b+8 b^2\right ) x+\frac {\left (5 a^2+12 a b+8 b^2\right ) \cos (e+f x) \sin (e+f x)}{16 f}+\frac {a (5 a+8 b) \cos ^3(e+f x) \sin (e+f x)}{24 f}+\frac {a \cos ^5(e+f x) \sin (e+f x) \left (a+b+b \tan ^2(e+f x)\right )}{6 f} \]
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*sin(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
Time = 0.27 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.83 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {60 a^2 e+144 a b e+96 b^2 e+60 a^2 f x+144 a b f x+96 b^2 f x+\left (45 a^2+96 a b+48 b^2\right ) \sin (2 (e+f x))+3 a (3 a+4 b) \sin (4 (e+f x))+a^2 \sin (6 (e+f x))}{192 f} \]
(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)
Time = 0.30 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4634, 315, 298, 215, 216}
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.
\(\displaystyle \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^2}{\sec (e+f x)^6}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {\left (b \tan ^2(e+f x)+a+b\right )^2}{\left (\tan ^2(e+f x)+1\right )^4}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 315 |
\(\displaystyle \frac {\frac {1}{6} \int \frac {3 b (a+2 b) \tan ^2(e+f x)+(a+b) (5 a+6 b)}{\left (\tan ^2(e+f x)+1\right )^3}d\tan (e+f x)+\frac {a \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 298 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} \left (5 a^2+12 a b+8 b^2\right ) \int \frac {1}{\left (\tan ^2(e+f x)+1\right )^2}d\tan (e+f x)+\frac {a (5 a+8 b) \tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )+\frac {a \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} \left (5 a^2+12 a b+8 b^2\right ) \left (\frac {1}{2} \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)+\frac {\tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}\right )+\frac {a (5 a+8 b) \tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )+\frac {a \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {1}{6} \left (\frac {3}{4} \left (5 a^2+12 a b+8 b^2\right ) \left (\frac {1}{2} \arctan (\tan (e+f x))+\frac {\tan (e+f x)}{2 \left (\tan ^2(e+f x)+1\right )}\right )+\frac {a (5 a+8 b) \tan (e+f x)}{4 \left (\tan ^2(e+f x)+1\right )^2}\right )+\frac {a \tan (e+f x) \left (a+b \tan ^2(e+f x)+b\right )}{6 \left (\tan ^2(e+f x)+1\right )^3}}{f}\) |
((a*Tan[e + f*x]*(a + b + b*Tan[e + f*x]^2))/(6*(1 + Tan[e + f*x]^2)^3) + ((a*(5*a + 8*b)*Tan[e + f*x])/(4*(1 + Tan[e + f*x]^2)^2) + (3*(5*a^2 + 12* a*b + 8*b^2)*(ArcTan[Tan[e + f*x]]/2 + Tan[e + f*x]/(2*(1 + Tan[e + f*x]^2 ))))/4)/6)/f
3.2.77.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2), x_Symbol] :> Simp[(-( b*c - a*d))*x*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] - Simp[(a*d - b*c*( 2*p + 3))/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c, d, p}, x] && NeQ[b*c - a*d, 0] && (LtQ[p, -1] || ILtQ[1/2 + p, 0])
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(a*d - c*b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q - 1)/(2*a*b*(p + 1))), x] - Simp[1/(2*a*b*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^(q - 2)*S imp[c*(a*d - c*b*(2*p + 3)) + d*(a*d*(2*(q - 1) + 1) - b*c*(2*(p + q) + 1)) *x^2, x], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && LtQ[p, - 1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[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]
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 | \(\frac {5 a^{2} x}{16}+\frac {3 x a b}{4}+\frac {x \,b^{2}}{2}+\frac {a^{2} \sin \left (6 f x +6 e \right )}{192 f}+\frac {3 a^{2} \sin \left (4 f x +4 e \right )}{64 f}+\frac {\sin \left (4 f x +4 e \right ) a b}{16 f}+\frac {15 \sin \left (2 f x +2 e \right ) a^{2}}{64 f}+\frac {\sin \left (2 f x +2 e \right ) a b}{2 f}+\frac {\sin \left (2 f x +2 e \right ) b^{2}}{4 f}\) | \(119\) |
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
Time = 0.25 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.75 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (5 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} f x + {\left (8 \, a^{2} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{2} + 12 \, a b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (5 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, f} \]
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
Timed out. \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\text {Timed out} \]
Time = 0.27 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.13 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (5 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} + \frac {3 \, {\left (5 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (5 \, a^{2} + 12 \, a b + 6 \, b^{2}\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (11 \, a^{2} + 20 \, a b + 8 \, b^{2}\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{6} + 3 \, \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{2} + 1}}{48 \, f} \]
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)*t an(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
Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.26 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, {\left (5 \, a^{2} + 12 \, a b + 8 \, b^{2}\right )} {\left (f x + e\right )} + \frac {15 \, a^{2} \tan \left (f x + e\right )^{5} + 36 \, a b \tan \left (f x + e\right )^{5} + 24 \, b^{2} \tan \left (f x + e\right )^{5} + 40 \, a^{2} \tan \left (f x + e\right )^{3} + 96 \, a b \tan \left (f x + e\right )^{3} + 48 \, b^{2} \tan \left (f x + e\right )^{3} + 33 \, a^{2} \tan \left (f x + e\right ) + 60 \, a b \tan \left (f x + e\right ) + 24 \, b^{2} \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3}}}{48 \, f} \]
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*tan (f*x + e) + 24*b^2*tan(f*x + e))/(tan(f*x + e)^2 + 1)^3)/f
Time = 19.08 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.03 \[ \int \cos ^6(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=x\,\left (\frac {5\,a^2}{16}+\frac {3\,a\,b}{4}+\frac {b^2}{2}\right )+\frac {\left (\frac {5\,a^2}{16}+\frac {3\,a\,b}{4}+\frac {b^2}{2}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (\frac {5\,a^2}{6}+2\,a\,b+b^2\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {11\,a^2}{16}+\frac {5\,a\,b}{4}+\frac {b^2}{2}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^6+3\,{\mathrm {tan}\left (e+f\,x\right )}^4+3\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )} \]