\(\int \cos ^5(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\) [170]

Optimal result

Integrand size = 23, antiderivative size = 53 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {(a+b)^2 \sin (e+f x)}{f}-\frac {2 a (a+b) \sin ^3(e+f x)}{3 f}+\frac {a^2 \sin ^5(e+f x)}{5 f} \]

[Out]

(a+b)^2*sin(f*x+e)/f-2/3*a*(a+b)*sin(f*x+e)^3/f+1/5*a^2*sin(f*x+e)^5/f

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 53, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {4232, 200} \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {a^2 \sin ^5(e+f x)}{5 f}-\frac {2 a (a+b) \sin ^3(e+f x)}{3 f}+\frac {(a+b)^2 \sin (e+f x)}{f} \]

[In]

Int[Cos[e + f*x]^5*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

((a + b)^2*Sin[e + f*x])/f - (2*a*(a + b)*Sin[e + f*x]^3)/(3*f) + (a^2*Sin[e + f*x]^5)/(5*f)

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 4232

Int[sec[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = Fr
eeFactors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[ExpandToSum[b + a*(1 - ff^2*x^2)^(n/2), x]^p/(1 - ff^2*x^2)^
((m + n*p + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[(m - 1)/2] && IntegerQ[n
/2] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \left (a+b-a x^2\right )^2 \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (a^2 \left (1+\frac {b (2 a+b)}{a^2}\right )-2 a^2 \left (1+\frac {b}{a}\right ) x^2+a^2 x^4\right ) \, dx,x,\sin (e+f x)\right )}{f} \\ & = \frac {(a+b)^2 \sin (e+f x)}{f}-\frac {2 a (a+b) \sin ^3(e+f x)}{3 f}+\frac {a^2 \sin ^5(e+f x)}{5 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 2.00 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {b^2 \cos (f x) \sin (e)}{f}+\frac {b^2 \cos (e) \sin (f x)}{f}+\frac {a^2 \sin (e+f x)}{f}+\frac {2 a b \sin (e+f x)}{f}-\frac {2 a^2 \sin ^3(e+f x)}{3 f}-\frac {2 a b \sin ^3(e+f x)}{3 f}+\frac {a^2 \sin ^5(e+f x)}{5 f} \]

[In]

Integrate[Cos[e + f*x]^5*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

(b^2*Cos[f*x]*Sin[e])/f + (b^2*Cos[e]*Sin[f*x])/f + (a^2*Sin[e + f*x])/f + (2*a*b*Sin[e + f*x])/f - (2*a^2*Sin
[e + f*x]^3)/(3*f) - (2*a*b*Sin[e + f*x]^3)/(3*f) + (a^2*Sin[e + f*x]^5)/(5*f)

Maple [A] (verified)

Time = 0.47 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.26

method	result	size
derivativedivides	\(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos \left (f x +e \right )^{4}+\frac {4 \cos \left (f x +e \right )^{2}}{3}\right ) \sin \left (f x +e \right )}{5}+\frac {2 a b \left (\cos \left (f x +e \right )^{2}+2\right ) \sin \left (f x +e \right )}{3}+\sin \left (f x +e \right ) b^{2}}{f}\)	\(67\)
default	\(\frac {\frac {a^{2} \left (\frac {8}{3}+\cos \left (f x +e \right )^{4}+\frac {4 \cos \left (f x +e \right )^{2}}{3}\right ) \sin \left (f x +e \right )}{5}+\frac {2 a b \left (\cos \left (f x +e \right )^{2}+2\right ) \sin \left (f x +e \right )}{3}+\sin \left (f x +e \right ) b^{2}}{f}\)	\(67\)
parallelrisch	\(\frac {150 \sin \left (f x +e \right ) a^{2}+360 \sin \left (f x +e \right ) a b +240 \sin \left (f x +e \right ) b^{2}+3 \sin \left (5 f x +5 e \right ) a^{2}+25 \sin \left (3 f x +3 e \right ) a^{2}+40 \sin \left (3 f x +3 e \right ) a b}{240 f}\)	\(80\)
risch	\(\frac {5 a^{2} \sin \left (f x +e \right )}{8 f}+\frac {3 \sin \left (f x +e \right ) a b}{2 f}+\frac {\sin \left (f x +e \right ) b^{2}}{f}+\frac {a^{2} \sin \left (5 f x +5 e \right )}{80 f}+\frac {5 a^{2} \sin \left (3 f x +3 e \right )}{48 f}+\frac {\sin \left (3 f x +3 e \right ) a b}{6 f}\)	\(92\)
norman	\(\frac {-\frac {2 \left (a^{2}+2 a b +b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {2 \left (a^{2}+2 a b +b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{15}}{f}+\frac {2 \left (5 a^{2}+2 a b -3 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f}-\frac {2 \left (5 a^{2}+2 a b -3 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{13}}{3 f}-\frac {2 \left (43 a^{2}-50 a b -45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 f}+\frac {2 \left (43 a^{2}-50 a b -45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{15 f}+\frac {2 \left (109 a^{2}+10 a b +45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{15 f}-\frac {2 \left (109 a^{2}+10 a b +45 b^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{15 f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3}}\)	\(271\)

[In]

int(cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)

[Out]

1/f*(1/5*a^2*(8/3+cos(f*x+e)^4+4/3*cos(f*x+e)^2)*sin(f*x+e)+2/3*a*b*(cos(f*x+e)^2+2)*sin(f*x+e)+sin(f*x+e)*b^2
)

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.11 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {{\left (3 \, a^{2} \cos \left (f x + e\right )^{4} + 2 \, {\left (2 \, a^{2} + 5 \, a b\right )} \cos \left (f x + e\right )^{2} + 8 \, a^{2} + 20 \, a b + 15 \, b^{2}\right )} \sin \left (f x + e\right )}{15 \, f} \]

[In]

integrate(cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/15*(3*a^2*cos(f*x + e)^4 + 2*(2*a^2 + 5*a*b)*cos(f*x + e)^2 + 8*a^2 + 20*a*b + 15*b^2)*sin(f*x + e)/f

Sympy [F(-1)]

Timed out. \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\text {Timed out} \]

[In]

integrate(cos(f*x+e)**5*(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.04 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, a^{2} \sin \left (f x + e\right )^{5} - 10 \, {\left (a^{2} + a b\right )} \sin \left (f x + e\right )^{3} + 15 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} \sin \left (f x + e\right )}{15 \, f} \]

[In]

integrate(cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

1/15*(3*a^2*sin(f*x + e)^5 - 10*(a^2 + a*b)*sin(f*x + e)^3 + 15*(a^2 + 2*a*b + b^2)*sin(f*x + e))/f

Giac [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.43 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {3 \, a^{2} \sin \left (f x + e\right )^{5} - 10 \, a^{2} \sin \left (f x + e\right )^{3} - 10 \, a b \sin \left (f x + e\right )^{3} + 15 \, a^{2} \sin \left (f x + e\right ) + 30 \, a b \sin \left (f x + e\right ) + 15 \, b^{2} \sin \left (f x + e\right )}{15 \, f} \]

[In]

integrate(cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/15*(3*a^2*sin(f*x + e)^5 - 10*a^2*sin(f*x + e)^3 - 10*a*b*sin(f*x + e)^3 + 15*a^2*sin(f*x + e) + 30*a*b*sin(
f*x + e) + 15*b^2*sin(f*x + e))/f

Mupad [B] (verification not implemented)

Time = 18.54 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.83 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\sin \left (e+f\,x\right )\,{\left (a+b\right )}^2+\frac {a^2\,{\sin \left (e+f\,x\right )}^5}{5}-\frac {2\,a\,{\sin \left (e+f\,x\right )}^3\,\left (a+b\right )}{3}}{f} \]

[In]

int(cos(e + f*x)^5*(a + b/cos(e + f*x)^2)^2,x)

[Out]

(sin(e + f*x)*(a + b)^2 + (a^2*sin(e + f*x)^5)/5 - (2*a*sin(e + f*x)^3*(a + b))/3)/f