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

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

[Out]

1/16*(5*a^4+36*a^2*b^2+8*b^4)*x+4/5*a*b*(4*a^2+5*b^2)*sin(d*x+c)/d+1/16*(5*a^4+36*a^2*b^2+8*b^4)*cos(d*x+c)*si
n(d*x+c)/d+1/24*a^2*(5*a^2+32*b^2)*cos(d*x+c)^3*sin(d*x+c)/d+7/15*a^3*b*cos(d*x+c)^4*sin(d*x+c)/d+1/6*a^2*cos(
d*x+c)^5*(a+b*sec(d*x+c))^2*sin(d*x+c)/d-4/15*a*b*(4*a^2+5*b^2)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.38, antiderivative size = 213, normalized size of antiderivative = 1.00, 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 8

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

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*}

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Mathematica [A]  time = 0.46, size = 156, normalized size = 0.73 \[ \frac {5 a^4 \sin (6 (c+d x))+48 a^3 b \sin (5 (c+d x))+45 a^2 \left (a^2+4 b^2\right ) \sin (4 (c+d x))+480 a b \left (5 a^2+6 b^2\right ) \sin (c+d x)+80 a b \left (5 a^2+4 b^2\right ) \sin (3 (c+d x))+60 \left (5 a^4+36 a^2 b^2+8 b^4\right ) (c+d x)+15 \left (15 a^4+96 a^2 b^2+16 b^4\right ) \sin (2 (c+d x))}{960 d} \]

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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fricas [A]  time = 0.54, size = 150, normalized size = 0.70 \[ \frac {15 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} d x + {\left (40 \, a^{4} \cos \left (d x + c\right )^{5} + 192 \, a^{3} b \cos \left (d x + c\right )^{4} + 512 \, a^{3} b + 640 \, a b^{3} + 10 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2}\right )} \cos \left (d x + c\right )^{3} + 64 \, {\left (4 \, a^{3} b + 5 \, a b^{3}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d} \]

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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giac [B]  time = 0.25, size = 550, normalized size = 2.58 \[ \frac {15 \, {\left (5 \, a^{4} + 36 \, a^{2} b^{2} + 8 \, b^{4}\right )} {\left (d x + c\right )} - \frac {2 \, {\left (165 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 960 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 900 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 960 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} + 120 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{11} - 25 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 2240 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 1260 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 3520 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 450 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 4992 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 360 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 5760 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 240 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 450 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 4992 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 360 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 5760 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 240 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 25 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2240 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 1260 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3520 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 360 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 165 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 960 \, a^{3} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 900 \, a^{2} b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 960 \, a b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 120 \, b^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{6}}}{240 \, d} \]

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

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maple [A]  time = 1.60, size = 174, normalized size = 0.82 \[ \frac {a^{4} \left (\frac {\left (\cos ^{5}\left (d x +c \right )+\frac {5 \left (\cos ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \cos \left (d x +c \right )}{8}\right ) \sin \left (d x +c \right )}{6}+\frac {5 d x}{16}+\frac {5 c}{16}\right )+\frac {4 a^{3} b \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+6 a^{2} b^{2} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {4 a \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+b^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

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*(2+cos(d*x+c)^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 = 0.37, size = 170, normalized size = 0.80 \[ -\frac {5 \, {\left (4 \, \sin \left (2 \, d x + 2 \, c\right )^{3} - 60 \, d x - 60 \, c - 9 \, \sin \left (4 \, d x + 4 \, c\right ) - 48 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{4} - 256 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} a^{3} b - 180 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} a^{2} b^{2} + 1280 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a b^{3} - 240 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} b^{4}}{960 \, d} \]

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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mupad [B]  time = 1.09, size = 214, normalized size = 1.00 \[ \frac {5\,a^4\,x}{16}+\frac {b^4\,x}{2}+\frac {9\,a^2\,b^2\,x}{4}+\frac {15\,a^4\,\sin \left (2\,c+2\,d\,x\right )}{64\,d}+\frac {3\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{64\,d}+\frac {a^4\,\sin \left (6\,c+6\,d\,x\right )}{192\,d}+\frac {b^4\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {a\,b^3\,\sin \left (3\,c+3\,d\,x\right )}{3\,d}+\frac {5\,a^3\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {a^3\,b\,\sin \left (5\,c+5\,d\,x\right )}{20\,d}+\frac {3\,a^2\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {3\,a^2\,b^2\,\sin \left (4\,c+4\,d\,x\right )}{16\,d}+\frac {3\,a\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,a^3\,b\,\sin \left (c+d\,x\right )}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(5*a^4*x)/16 + (b^4*x)/2 + (9*a^2*b^2*x)/4 + (15*a^4*sin(2*c + 2*d*x))/(64*d) + (3*a^4*sin(4*c + 4*d*x))/(64*d
) + (a^4*sin(6*c + 6*d*x))/(192*d) + (b^4*sin(2*c + 2*d*x))/(4*d) + (a*b^3*sin(3*c + 3*d*x))/(3*d) + (5*a^3*b*
sin(3*c + 3*d*x))/(12*d) + (a^3*b*sin(5*c + 5*d*x))/(20*d) + (3*a^2*b^2*sin(2*c + 2*d*x))/(2*d) + (3*a^2*b^2*s
in(4*c + 4*d*x))/(16*d) + (3*a*b^3*sin(c + d*x))/d + (5*a^3*b*sin(c + d*x))/(2*d)

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

[Out]

Timed out

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