3.485 \(\int \cos ^6(c+d x) (a+b \sec (c+d x))^4 \, dx\)

Optimal. Leaf size=213 \[ -\frac{4 a b \left (4 a^2+5 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{4 a b \left (4 a^2+5 b^2\right ) \sin (c+d x)}{5 d}+\frac{a^2 \left (5 a^2+32 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{\left (36 a^2 b^2+5 a^4+8 b^4\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (36 a^2 b^2+5 a^4+8 b^4\right )+\frac{7 a^3 b \sin (c+d x) \cos ^4(c+d x)}{15 d}+\frac{a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d} \]

[Out]

((5*a^4 + 36*a^2*b^2 + 8*b^4)*x)/16 + (4*a*b*(4*a^2 + 5*b^2)*Sin[c + d*x])/(5*d) + ((5*a^4 + 36*a^2*b^2 + 8*b^
4)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a^2*(5*a^2 + 32*b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) + (7*a^3*b*Co
s[c + d*x]^4*Sin[c + d*x])/(15*d) + (a^2*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(6*d) - (4*a*b*(4
*a^2 + 5*b^2)*Sin[c + d*x]^3)/(15*d)

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Rubi [A]  time = 0.380011, antiderivative size = 213, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {3841, 4074, 4047, 2633, 4045, 2635, 8} \[ -\frac{4 a b \left (4 a^2+5 b^2\right ) \sin ^3(c+d x)}{15 d}+\frac{4 a b \left (4 a^2+5 b^2\right ) \sin (c+d x)}{5 d}+\frac{a^2 \left (5 a^2+32 b^2\right ) \sin (c+d x) \cos ^3(c+d x)}{24 d}+\frac{\left (36 a^2 b^2+5 a^4+8 b^4\right ) \sin (c+d x) \cos (c+d x)}{16 d}+\frac{1}{16} x \left (36 a^2 b^2+5 a^4+8 b^4\right )+\frac{7 a^3 b \sin (c+d x) \cos ^4(c+d x)}{15 d}+\frac{a^2 \sin (c+d x) \cos ^5(c+d x) (a+b \sec (c+d x))^2}{6 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4,x]

[Out]

((5*a^4 + 36*a^2*b^2 + 8*b^4)*x)/16 + (4*a*b*(4*a^2 + 5*b^2)*Sin[c + d*x])/(5*d) + ((5*a^4 + 36*a^2*b^2 + 8*b^
4)*Cos[c + d*x]*Sin[c + d*x])/(16*d) + (a^2*(5*a^2 + 32*b^2)*Cos[c + d*x]^3*Sin[c + d*x])/(24*d) + (7*a^3*b*Co
s[c + d*x]^4*Sin[c + d*x])/(15*d) + (a^2*Cos[c + d*x]^5*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(6*d) - (4*a*b*(4
*a^2 + 5*b^2)*Sin[c + d*x]^3)/(15*d)

Rule 3841

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a^2*C
ot[e + f*x]*(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^n)/(f*n), x] - Dist[1/(d*n), Int[(a + b*Csc[e + f*x]
)^(m - 3)*(d*Csc[e + f*x])^(n + 1)*Simp[a^2*b*(m - 2*n - 2) - a*(3*b^2*n + a^2*(n + 1))*Csc[e + f*x] - b*(b^2*
n + a^2*(m + n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && GtQ[m, 2]
&& ((IntegerQ[m] && LtQ[n, -1]) || (IntegersQ[m + 1/2, 2*n] && LeQ[n, -1]))

Rule 4074

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(d*Csc[e + f*x])^(n + 1)*Simp[n*(B*a + A*b) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rule 4047

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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(c+d x) (a+b \sec (c+d x))^4 \, dx &=\frac{a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}+\frac{1}{6} \int \cos ^5(c+d x) (a+b \sec (c+d x)) \left (14 a^2 b+a \left (5 a^2+18 b^2\right ) \sec (c+d x)+3 b \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{7 a^3 b \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac{a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}-\frac{1}{30} \int \cos ^4(c+d x) \left (-5 a^2 \left (5 a^2+32 b^2\right )-24 a b \left (4 a^2+5 b^2\right ) \sec (c+d x)-15 b^2 \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx\\ &=\frac{7 a^3 b \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac{a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}-\frac{1}{30} \int \cos ^4(c+d x) \left (-5 a^2 \left (5 a^2+32 b^2\right )-15 b^2 \left (a^2+2 b^2\right ) \sec ^2(c+d x)\right ) \, dx+\frac{1}{5} \left (4 a b \left (4 a^2+5 b^2\right )\right ) \int \cos ^3(c+d x) \, dx\\ &=\frac{a^2 \left (5 a^2+32 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{7 a^3 b \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac{a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}-\frac{1}{8} \left (-5 a^4-36 a^2 b^2-8 b^4\right ) \int \cos ^2(c+d x) \, dx-\frac{\left (4 a b \left (4 a^2+5 b^2\right )\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{5 d}\\ &=\frac{4 a b \left (4 a^2+5 b^2\right ) \sin (c+d x)}{5 d}+\frac{\left (5 a^4+36 a^2 b^2+8 b^4\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^2 \left (5 a^2+32 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{7 a^3 b \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac{a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}-\frac{4 a b \left (4 a^2+5 b^2\right ) \sin ^3(c+d x)}{15 d}-\frac{1}{16} \left (-5 a^4-36 a^2 b^2-8 b^4\right ) \int 1 \, dx\\ &=\frac{1}{16} \left (5 a^4+36 a^2 b^2+8 b^4\right ) x+\frac{4 a b \left (4 a^2+5 b^2\right ) \sin (c+d x)}{5 d}+\frac{\left (5 a^4+36 a^2 b^2+8 b^4\right ) \cos (c+d x) \sin (c+d x)}{16 d}+\frac{a^2 \left (5 a^2+32 b^2\right ) \cos ^3(c+d x) \sin (c+d x)}{24 d}+\frac{7 a^3 b \cos ^4(c+d x) \sin (c+d x)}{15 d}+\frac{a^2 \cos ^5(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{6 d}-\frac{4 a b \left (4 a^2+5 b^2\right ) \sin ^3(c+d x)}{15 d}\\ \end{align*}

Mathematica [A]  time = 0.470141, size = 156, normalized size = 0.73 \[ \frac{60 \left (36 a^2 b^2+5 a^4+8 b^4\right ) (c+d x)+45 a^2 \left (a^2+4 b^2\right ) \sin (4 (c+d x))+480 a b \left (5 a^2+6 b^2\right ) \sin (c+d x)+80 a b \left (5 a^2+4 b^2\right ) \sin (3 (c+d x))+15 \left (96 a^2 b^2+15 a^4+16 b^4\right ) \sin (2 (c+d x))+48 a^3 b \sin (5 (c+d x))+5 a^4 \sin (6 (c+d x))}{960 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + b*Sec[c + d*x])^4,x]

[Out]

(60*(5*a^4 + 36*a^2*b^2 + 8*b^4)*(c + d*x) + 480*a*b*(5*a^2 + 6*b^2)*Sin[c + d*x] + 15*(15*a^4 + 96*a^2*b^2 +
16*b^4)*Sin[2*(c + d*x)] + 80*a*b*(5*a^2 + 4*b^2)*Sin[3*(c + d*x)] + 45*a^2*(a^2 + 4*b^2)*Sin[4*(c + d*x)] + 4
8*a^3*b*Sin[5*(c + d*x)] + 5*a^4*Sin[6*(c + d*x)])/(960*d)

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Maple [A]  time = 0.064, size = 174, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ({a}^{4} \left ({\frac{\sin \left ( dx+c \right ) }{6} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\cos \left ( dx+c \right ) }{8}} \right ) }+{\frac{5\,dx}{16}}+{\frac{5\,c}{16}} \right ) +{\frac{4\,{a}^{3}b\sin \left ( dx+c \right ) }{5} \left ({\frac{8}{3}}+ \left ( \cos \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) }+6\,{a}^{2}{b}^{2} \left ( 1/4\, \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3/2\,\cos \left ( dx+c \right ) \right ) \sin \left ( dx+c \right ) +3/8\,dx+3/8\,c \right ) +{\frac{4\,a{b}^{3} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{2}+2 \right ) \sin \left ( dx+c \right ) }{3}}+{b}^{4} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+b*sec(d*x+c))^4,x)

[Out]

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

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Maxima [A]  time = 1.22446, size = 230, normalized size = 1.08 \begin{align*} -\frac{5 \,{\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 256 \,{\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} b - 180 \,{\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b^{2} + 1280 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{3} - 240 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{4}}{960 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/960*(5*(4*sin(2*d*x + 2*c)^3 - 60*d*x - 60*c - 9*sin(4*d*x + 4*c) - 48*sin(2*d*x + 2*c))*a^4 - 256*(3*sin(d
*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*a^3*b - 180*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x +
2*c))*a^2*b^2 + 1280*(sin(d*x + c)^3 - 3*sin(d*x + c))*a*b^3 - 240*(2*d*x + 2*c + sin(2*d*x + 2*c))*b^4)/d

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Fricas [A]  time = 1.68636, size = 358, normalized size = 1.68 \begin{align*} \frac{15 \,{\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} d x +{\left (40 \, a^{4} \cos \left (d x + c\right )^{5} + 192 \, a^{3} b \cos \left (d x + c\right )^{4} + 512 \, a^{3} b + 640 \, a b^{3} + 10 \,{\left (5 \, a^{4} + 36 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 64 \,{\left (4 \, a^{3} b + 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \,{\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^4,x, algorithm="fricas")

[Out]

1/240*(15*(5*a^4 + 36*a^2*b^2 + 8*b^4)*d*x + (40*a^4*cos(d*x + c)^5 + 192*a^3*b*cos(d*x + c)^4 + 512*a^3*b + 6
40*a*b^3 + 10*(5*a^4 + 36*a^2*b^2)*cos(d*x + c)^3 + 64*(4*a^3*b + 5*a*b^3)*cos(d*x + c)^2 + 15*(5*a^4 + 36*a^2
*b^2 + 8*b^4)*cos(d*x + c))*sin(d*x + c))/d

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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(d*x+c)**6*(a+b*sec(d*x+c))**4,x)

[Out]

Timed out

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Giac [B]  time = 1.28985, size = 743, normalized size = 3.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+b*sec(d*x+c))^4,x, algorithm="giac")

[Out]

1/240*(15*(5*a^4 + 36*a^2*b^2 + 8*b^4)*(d*x + c) - 2*(165*a^4*tan(1/2*d*x + 1/2*c)^11 - 960*a^3*b*tan(1/2*d*x
+ 1/2*c)^11 + 900*a^2*b^2*tan(1/2*d*x + 1/2*c)^11 - 960*a*b^3*tan(1/2*d*x + 1/2*c)^11 + 120*b^4*tan(1/2*d*x +
1/2*c)^11 - 25*a^4*tan(1/2*d*x + 1/2*c)^9 - 2240*a^3*b*tan(1/2*d*x + 1/2*c)^9 + 1260*a^2*b^2*tan(1/2*d*x + 1/2
*c)^9 - 3520*a*b^3*tan(1/2*d*x + 1/2*c)^9 + 360*b^4*tan(1/2*d*x + 1/2*c)^9 + 450*a^4*tan(1/2*d*x + 1/2*c)^7 -
4992*a^3*b*tan(1/2*d*x + 1/2*c)^7 + 360*a^2*b^2*tan(1/2*d*x + 1/2*c)^7 - 5760*a*b^3*tan(1/2*d*x + 1/2*c)^7 + 2
40*b^4*tan(1/2*d*x + 1/2*c)^7 - 450*a^4*tan(1/2*d*x + 1/2*c)^5 - 4992*a^3*b*tan(1/2*d*x + 1/2*c)^5 - 360*a^2*b
^2*tan(1/2*d*x + 1/2*c)^5 - 5760*a*b^3*tan(1/2*d*x + 1/2*c)^5 - 240*b^4*tan(1/2*d*x + 1/2*c)^5 + 25*a^4*tan(1/
2*d*x + 1/2*c)^3 - 2240*a^3*b*tan(1/2*d*x + 1/2*c)^3 - 1260*a^2*b^2*tan(1/2*d*x + 1/2*c)^3 - 3520*a*b^3*tan(1/
2*d*x + 1/2*c)^3 - 360*b^4*tan(1/2*d*x + 1/2*c)^3 - 165*a^4*tan(1/2*d*x + 1/2*c) - 960*a^3*b*tan(1/2*d*x + 1/2
*c) - 900*a^2*b^2*tan(1/2*d*x + 1/2*c) - 960*a*b^3*tan(1/2*d*x + 1/2*c) - 120*b^4*tan(1/2*d*x + 1/2*c))/(tan(1
/2*d*x + 1/2*c)^2 + 1)^6)/d