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

Optimal. Leaf size=89 \[ \frac{(5 a+6 b) \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac{(5 a+6 b) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{1}{16} x (5 a+6 b)+\frac{a \sin (e+f x) \cos ^5(e+f x)}{6 f} \]

[Out]

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

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Rubi [A]  time = 0.052592, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4045, 2635, 8} \[ \frac{(5 a+6 b) \sin (e+f x) \cos ^3(e+f x)}{24 f}+\frac{(5 a+6 b) \sin (e+f x) \cos (e+f x)}{16 f}+\frac{1}{16} x (5 a+6 b)+\frac{a \sin (e+f x) \cos ^5(e+f x)}{6 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

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

Mathematica [A]  time = 0.102777, size = 68, normalized size = 0.76 \[ \frac{(45 a+48 b) \sin (2 (e+f x))+(9 a+6 b) \sin (4 (e+f x))+a \sin (6 (e+f x))+60 a e+60 a f x+72 b e+72 b f x}{192 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(60*a*e + 72*b*e + 60*a*f*x + 72*b*f*x + (45*a + 48*b)*Sin[2*(e + f*x)] + (9*a + 6*b)*Sin[4*(e + f*x)] + a*Sin
[6*(e + f*x)])/(192*f)

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Maple [A]  time = 0.053, size = 86, normalized size = 1. \begin{align*}{\frac{1}{f} \left ( a \left ({\frac{\sin \left ( fx+e \right ) }{6} \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( fx+e \right ) }{8}} \right ) }+{\frac{5\,fx}{16}}+{\frac{5\,e}{16}} \right ) +b \left ({\frac{\sin \left ( fx+e \right ) }{4} \left ( \left ( \cos \left ( fx+e \right ) \right ) ^{3}+{\frac{3\,\cos \left ( fx+e \right ) }{2}} \right ) }+{\frac{3\,fx}{8}}+{\frac{3\,e}{8}} \right ) \right ) } \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),x)

[Out]

1/f*(a*(1/6*(cos(f*x+e)^5+5/4*cos(f*x+e)^3+15/8*cos(f*x+e))*sin(f*x+e)+5/16*f*x+5/16*e)+b*(1/4*(cos(f*x+e)^3+3
/2*cos(f*x+e))*sin(f*x+e)+3/8*f*x+3/8*e))

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Maxima [A]  time = 1.4862, size = 139, normalized size = 1.56 \begin{align*} \frac{3 \,{\left (f x + e\right )}{\left (5 \, a + 6 \, b\right )} + \frac{3 \,{\left (5 \, a + 6 \, b\right )} \tan \left (f x + e\right )^{5} + 8 \,{\left (5 \, a + 6 \, b\right )} \tan \left (f x + e\right )^{3} + 3 \,{\left (11 \, a + 10 \, b\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} \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),x, algorithm="maxima")

[Out]

1/48*(3*(f*x + e)*(5*a + 6*b) + (3*(5*a + 6*b)*tan(f*x + e)^5 + 8*(5*a + 6*b)*tan(f*x + e)^3 + 3*(11*a + 10*b)
*tan(f*x + e))/(tan(f*x + e)^6 + 3*tan(f*x + e)^4 + 3*tan(f*x + e)^2 + 1))/f

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Fricas [A]  time = 0.495716, size = 167, normalized size = 1.88 \begin{align*} \frac{3 \,{\left (5 \, a + 6 \, b\right )} f x +{\left (8 \, a \cos \left (f x + e\right )^{5} + 2 \,{\left (5 \, a + 6 \, b\right )} \cos \left (f x + e\right )^{3} + 3 \,{\left (5 \, a + 6 \, b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{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),x, algorithm="fricas")

[Out]

1/48*(3*(5*a + 6*b)*f*x + (8*a*cos(f*x + e)^5 + 2*(5*a + 6*b)*cos(f*x + e)^3 + 3*(5*a + 6*b)*cos(f*x + e))*sin
(f*x + e))/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),x)

[Out]

Timed out

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Giac [A]  time = 1.34248, size = 140, normalized size = 1.57 \begin{align*} \frac{3 \,{\left (f x + e\right )}{\left (5 \, a + 6 \, b\right )} + \frac{15 \, a \tan \left (f x + e\right )^{5} + 18 \, b \tan \left (f x + e\right )^{5} + 40 \, a \tan \left (f x + e\right )^{3} + 48 \, b \tan \left (f x + e\right )^{3} + 33 \, a \tan \left (f x + e\right ) + 30 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3}}}{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),x, algorithm="giac")

[Out]

1/48*(3*(f*x + e)*(5*a + 6*b) + (15*a*tan(f*x + e)^5 + 18*b*tan(f*x + e)^5 + 40*a*tan(f*x + e)^3 + 48*b*tan(f*
x + e)^3 + 33*a*tan(f*x + e) + 30*b*tan(f*x + e))/(tan(f*x + e)^2 + 1)^3)/f