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]

1/16*(5*a+6*b)*x+1/16*(5*a+6*b)*cos(f*x+e)*sin(f*x+e)/f+1/24*(5*a+6*b)*cos(f*x+e)^3*sin(f*x+e)/f+1/6*a*cos(f*x
+e)^5*sin(f*x+e)/f

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Rubi [A]  time = 0.05, antiderivative size = 89, normalized size of antiderivative = 1.00, 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 8

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

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 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]

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*}

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Mathematica [A]  time = 0.11, 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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fricas [A]  time = 0.57, size = 68, normalized size = 0.76 \[ \frac {3 \, {\left (5 \, a + 6 \, b\right )} f x + {\left (8 \, a \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a + 6 \, b\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (5 \, a + 6 \, b\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{48 \, f} \]

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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giac [A]  time = 0.23, size = 104, normalized size = 1.17 \[ \frac {3 \, {\left (f x + e\right )} {\left (5 \, a + 6 \, b\right )} + \frac {15 \, a \tan \left (f x + e\right )^{5} + 18 \, b \tan \left (f x + e\right )^{5} + 40 \, a \tan \left (f x + e\right )^{3} + 48 \, b \tan \left (f x + e\right )^{3} + 33 \, a \tan \left (f x + e\right ) + 30 \, b \tan \left (f x + e\right )}{{\left (\tan \left (f x + e\right )^{2} + 1\right )}^{3}}}{48 \, f} \]

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

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maple [A]  time = 1.68, size = 86, normalized size = 0.97 \[ \frac {a \left (\frac {\left (\cos ^{5}\left (f x +e \right )+\frac {5 \left (\cos ^{3}\left (f x +e \right )\right )}{4}+\frac {15 \cos \left (f x +e \right )}{8}\right ) \sin \left (f x +e \right )}{6}+\frac {5 f x}{16}+\frac {5 e}{16}\right )+b \left (\frac {\left (\cos ^{3}\left (f x +e \right )+\frac {3 \cos \left (f x +e \right )}{2}\right ) \sin \left (f x +e \right )}{4}+\frac {3 f x}{8}+\frac {3 e}{8}\right )}{f} \]

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 = 0.44, size = 103, normalized size = 1.16 \[ \frac {3 \, {\left (f x + e\right )} {\left (5 \, a + 6 \, b\right )} + \frac {3 \, {\left (5 \, a + 6 \, b\right )} \tan \left (f x + e\right )^{5} + 8 \, {\left (5 \, a + 6 \, b\right )} \tan \left (f x + e\right )^{3} + 3 \, {\left (11 \, a + 10 \, b\right )} \tan \left (f x + e\right )}{\tan \left (f x + e\right )^{6} + 3 \, \tan \left (f x + e\right )^{4} + 3 \, \tan \left (f x + e\right )^{2} + 1}}{48 \, f} \]

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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mupad [B]  time = 4.94, size = 91, normalized size = 1.02 \[ x\,\left (\frac {5\,a}{16}+\frac {3\,b}{8}\right )+\frac {\left (\frac {5\,a}{16}+\frac {3\,b}{8}\right )\,{\mathrm {tan}\left (e+f\,x\right )}^5+\left (\frac {5\,a}{6}+b\right )\,{\mathrm {tan}\left (e+f\,x\right )}^3+\left (\frac {11\,a}{16}+\frac {5\,b}{8}\right )\,\mathrm {tan}\left (e+f\,x\right )}{f\,\left ({\mathrm {tan}\left (e+f\,x\right )}^6+3\,{\mathrm {tan}\left (e+f\,x\right )}^4+3\,{\mathrm {tan}\left (e+f\,x\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

x*((5*a)/16 + (3*b)/8) + (tan(e + f*x)^5*((5*a)/16 + (3*b)/8) + tan(e + f*x)*((11*a)/16 + (5*b)/8) + tan(e + f
*x)^3*((5*a)/6 + b))/(f*(3*tan(e + f*x)^2 + 3*tan(e + f*x)^4 + tan(e + f*x)^6 + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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