\(\int \cos ^5(e+f x) (a+b \sec ^2(e+f x)) \, dx\) [157]

Optimal result

Integrand size = 21, antiderivative size = 50 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {(a+b) \sin (e+f x)}{f}-\frac {(2 a+b) \sin ^3(e+f x)}{3 f}+\frac {a \sin ^5(e+f x)}{5 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4129, 3092, 380} \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {(2 a+b) \sin ^3(e+f x)}{3 f}+\frac {(a+b) \sin (e+f x)}{f}+\frac {a \sin ^5(e+f x)}{5 f} \]

[In]

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

[Out]

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

Rule 380

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

Rule 3092

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[-f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rule 4129

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

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

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.42 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {a \sin (e+f x)}{f}+\frac {b \sin (e+f x)}{f}-\frac {2 a \sin ^3(e+f x)}{3 f}-\frac {b \sin ^3(e+f x)}{3 f}+\frac {a \sin ^5(e+f x)}{5 f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.43 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.98

method	result	size
parallelrisch	\(\frac {\left (25 a +20 b \right ) \sin \left (3 f x +3 e \right )+3 \sin \left (5 f x +5 e \right ) a +150 \sin \left (f x +e \right ) \left (a +\frac {6 b}{5}\right )}{240 f}\)	\(49\)
derivativedivides	\(\frac {\frac {a \left (\frac {8}{3}+\cos \left (f x +e \right )^{4}+\frac {4 \cos \left (f x +e \right )^{2}}{3}\right ) \sin \left (f x +e \right )}{5}+\frac {b \left (\cos \left (f x +e \right )^{2}+2\right ) \sin \left (f x +e \right )}{3}}{f}\)	\(54\)
default	\(\frac {\frac {a \left (\frac {8}{3}+\cos \left (f x +e \right )^{4}+\frac {4 \cos \left (f x +e \right )^{2}}{3}\right ) \sin \left (f x +e \right )}{5}+\frac {b \left (\cos \left (f x +e \right )^{2}+2\right ) \sin \left (f x +e \right )}{3}}{f}\)	\(54\)
risch	\(\frac {5 a \sin \left (f x +e \right )}{8 f}+\frac {3 \sin \left (f x +e \right ) b}{4 f}+\frac {a \sin \left (5 f x +5 e \right )}{80 f}+\frac {5 a \sin \left (3 f x +3 e \right )}{48 f}+\frac {\sin \left (3 f x +3 e \right ) b}{12 f}\)	\(71\)
norman	\(\frac {-\frac {2 \left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{f}+\frac {2 \left (a +b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{f}-\frac {2 \left (a +5 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 f}+\frac {2 \left (a +5 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{3 f}-\frac {4 \left (19 a +5 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{15 f}+\frac {4 \left (19 a +5 b \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{15 f}}{\left (1+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}\right )^{5} \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )}\)	\(157\)

[In]

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

[Out]

1/240*((25*a+20*b)*sin(3*f*x+3*e)+3*sin(5*f*x+5*e)*a+150*sin(f*x+e)*(a+6/5*b))/f

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.90 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {{\left (3 \, a \cos \left (f x + e\right )^{4} + {\left (4 \, a + 5 \, b\right )} \cos \left (f x + e\right )^{2} + 8 \, a + 10 \, b\right )} \sin \left (f x + e\right )}{15 \, f} \]

[In]

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

[Out]

1/15*(3*a*cos(f*x + e)^4 + (4*a + 5*b)*cos(f*x + e)^2 + 8*a + 10*b)*sin(f*x + e)/f

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, a \sin \left (f x + e\right )^{5} - 5 \, {\left (2 \, a + b\right )} \sin \left (f x + e\right )^{3} + 15 \, {\left (a + b\right )} \sin \left (f x + e\right )}{15 \, f} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 57, normalized size of antiderivative = 1.14 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, a \sin \left (f x + e\right )^{5} - 10 \, a \sin \left (f x + e\right )^{3} - 5 \, b \sin \left (f x + e\right )^{3} + 15 \, a \sin \left (f x + e\right ) + 15 \, b \sin \left (f x + e\right )}{15 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 18.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.86 \[ \int \cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {\frac {a\,{\sin \left (e+f\,x\right )}^5}{5}+\left (-\frac {2\,a}{3}-\frac {b}{3}\right )\,{\sin \left (e+f\,x\right )}^3+\left (a+b\right )\,\sin \left (e+f\,x\right )}{f} \]

[In]

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

[Out]

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