3.301 \(\int \cos ^5(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=221 \[ -\frac {a \left (4 a^2 A+15 a b B+12 A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {\left (4 a^3 A+15 a^2 b B+14 a A b^2+5 b^3 B\right ) \sin (c+d x)}{5 d}+\frac {\left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right )+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d} \]

[Out]

1/8*(9*A*a^2*b+4*A*b^3+3*B*a^3+12*B*a*b^2)*x+1/5*(4*A*a^3+14*A*a*b^2+15*B*a^2*b+5*B*b^3)*sin(d*x+c)/d+1/8*(9*A
*a^2*b+4*A*b^3+3*B*a^3+12*B*a*b^2)*cos(d*x+c)*sin(d*x+c)/d+1/20*a^2*(7*A*b+5*B*a)*cos(d*x+c)^3*sin(d*x+c)/d+1/
5*a*A*cos(d*x+c)^4*(a+b*sec(d*x+c))^2*sin(d*x+c)/d-1/15*a*(4*A*a^2+12*A*b^2+15*B*a*b)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.49, antiderivative size = 221, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {4025, 4074, 4047, 2635, 8, 4044, 3013} \[ -\frac {a \left (4 a^2 A+15 a b B+12 A b^2\right ) \sin ^3(c+d x)}{15 d}+\frac {\left (4 a^3 A+15 a^2 b B+14 a A b^2+5 b^3 B\right ) \sin (c+d x)}{5 d}+\frac {\left (9 a^2 A b+3 a^3 B+12 a b^2 B+4 A b^3\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x \left (9 a^2 A b+3 a^3 B+12 a b^2 B+4 A b^3\right )+\frac {a^2 (5 a B+7 A b) \sin (c+d x) \cos ^3(c+d x)}{20 d}+\frac {a A \sin (c+d x) \cos ^4(c+d x) (a+b \sec (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

((9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*x)/8 + ((4*a^3*A + 14*a*A*b^2 + 15*a^2*b*B + 5*b^3*B)*Sin[c + d*
x])/(5*d) + ((9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*(7*A*b + 5*a
*B)*Cos[c + d*x]^3*Sin[c + d*x])/(20*d) + (a*A*Cos[c + d*x]^4*(a + b*Sec[c + d*x])^2*Sin[c + d*x])/(5*d) - (a*
(4*a^2*A + 12*A*b^2 + 15*a*b*B)*Sin[c + d*x]^3)/(15*d)

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 3013

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 4025

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(a*A*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m - 1)*(d*Csc[e + f*x])^n)/(f*n), x]
+ Dist[1/(d*n), Int[(a + b*Csc[e + f*x])^(m - 2)*(d*Csc[e + f*x])^(n + 1)*Simp[a*(a*B*n - A*b*(m - n - 1)) + (
2*a*b*B*n + A*(b^2*n + a^2*(1 + n)))*Csc[e + f*x] + b*(b*B*n + a*A*(m + n))*Csc[e + f*x]^2, x], x], x] /; Free
Q[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 1] && LeQ[n, -1]

Rule 4044

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]

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 ^5(c+d x) (a+b \sec (c+d x))^3 (A+B \sec (c+d x)) \, dx &=\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {1}{5} \int \cos ^4(c+d x) (a+b \sec (c+d x)) \left (-a (7 A b+5 a B)-\left (4 a^2 A+5 A b^2+10 a b B\right ) \sec (c+d x)-b (2 a A+5 b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 (7 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos ^3(c+d x) \left (4 a \left (4 a^2 A+12 A b^2+15 a b B\right )+5 \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \sec (c+d x)+4 b^2 (2 a A+5 b B) \sec ^2(c+d x)\right ) \, dx\\ &=\frac {a^2 (7 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos ^3(c+d x) \left (4 a \left (4 a^2 A+12 A b^2+15 a b B\right )+4 b^2 (2 a A+5 b B) \sec ^2(c+d x)\right ) \, dx+\frac {1}{4} \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac {\left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (7 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}+\frac {1}{20} \int \cos (c+d x) \left (4 b^2 (2 a A+5 b B)+4 a \left (4 a^2 A+12 A b^2+15 a b B\right ) \cos ^2(c+d x)\right ) \, dx+\frac {1}{8} \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \int 1 \, dx\\ &=\frac {1}{8} \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) x+\frac {\left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (7 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {\operatorname {Subst}\left (\int \left (4 b^2 (2 a A+5 b B)+4 a \left (4 a^2 A+12 A b^2+15 a b B\right )-4 a \left (4 a^2 A+12 A b^2+15 a b B\right ) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{20 d}\\ &=\frac {1}{8} \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) x+\frac {\left (4 a^3 A+14 a A b^2+15 a^2 b B+5 b^3 B\right ) \sin (c+d x)}{5 d}+\frac {\left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac {a^2 (7 A b+5 a B) \cos ^3(c+d x) \sin (c+d x)}{20 d}+\frac {a A \cos ^4(c+d x) (a+b \sec (c+d x))^2 \sin (c+d x)}{5 d}-\frac {a \left (4 a^2 A+12 A b^2+15 a b B\right ) \sin ^3(c+d x)}{15 d}\\ \end {align*}

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Mathematica [A]  time = 0.73, size = 176, normalized size = 0.80 \[ \frac {6 a^3 A \sin (5 (c+d x))+10 a \left (5 a^2 A+12 a b B+12 A b^2\right ) \sin (3 (c+d x))+15 a^2 (a B+3 A b) \sin (4 (c+d x))+60 (c+d x) \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right )+60 \left (5 a^3 A+18 a^2 b B+18 a A b^2+8 b^3 B\right ) \sin (c+d x)+120 \left (a^3 B+3 a^2 A b+3 a b^2 B+A b^3\right ) \sin (2 (c+d x))}{480 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + b*Sec[c + d*x])^3*(A + B*Sec[c + d*x]),x]

[Out]

(60*(9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*(c + d*x) + 60*(5*a^3*A + 18*a*A*b^2 + 18*a^2*b*B + 8*b^3*B)*
Sin[c + d*x] + 120*(3*a^2*A*b + A*b^3 + a^3*B + 3*a*b^2*B)*Sin[2*(c + d*x)] + 10*a*(5*a^2*A + 12*A*b^2 + 12*a*
b*B)*Sin[3*(c + d*x)] + 15*a^2*(3*A*b + a*B)*Sin[4*(c + d*x)] + 6*a^3*A*Sin[5*(c + d*x)])/(480*d)

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fricas [A]  time = 0.45, size = 174, normalized size = 0.79 \[ \frac {15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} d x + {\left (24 \, A a^{3} \cos \left (d x + c\right )^{4} + 64 \, A a^{3} + 240 \, B a^{2} b + 240 \, A a b^{2} + 120 \, B b^{3} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A a^{3} + 15 \, B a^{2} b + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/120*(15*(3*B*a^3 + 9*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*d*x + (24*A*a^3*cos(d*x + c)^4 + 64*A*a^3 + 240*B*a^2*b
+ 240*A*a*b^2 + 120*B*b^3 + 30*(B*a^3 + 3*A*a^2*b)*cos(d*x + c)^3 + 8*(4*A*a^3 + 15*B*a^2*b + 15*A*a*b^2)*cos
(d*x + c)^2 + 15*(3*B*a^3 + 9*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*cos(d*x + c))*sin(d*x + c))/d

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giac [B]  time = 0.34, size = 672, normalized size = 3.04 \[ \frac {15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (120 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 225 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 360 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 180 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 160 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 30 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 90 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 960 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 360 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 480 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 464 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1200 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 720 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 160 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 30 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 90 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 960 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 960 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 360 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 480 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 120 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 225 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 360 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 180 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, B b^{3} \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 )}^{5}}}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/120*(15*(3*B*a^3 + 9*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*(d*x + c) + 2*(120*A*a^3*tan(1/2*d*x + 1/2*c)^9 - 75*B*
a^3*tan(1/2*d*x + 1/2*c)^9 - 225*A*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*B*a^2*b*tan(1/2*d*x + 1/2*c)^9 + 360*A*a
*b^2*tan(1/2*d*x + 1/2*c)^9 - 180*B*a*b^2*tan(1/2*d*x + 1/2*c)^9 - 60*A*b^3*tan(1/2*d*x + 1/2*c)^9 + 120*B*b^3
*tan(1/2*d*x + 1/2*c)^9 + 160*A*a^3*tan(1/2*d*x + 1/2*c)^7 - 30*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 90*A*a^2*b*tan(
1/2*d*x + 1/2*c)^7 + 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 360*B*a*b^2*tan
(1/2*d*x + 1/2*c)^7 - 120*A*b^3*tan(1/2*d*x + 1/2*c)^7 + 480*B*b^3*tan(1/2*d*x + 1/2*c)^7 + 464*A*a^3*tan(1/2*
d*x + 1/2*c)^5 + 1200*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 1200*A*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 720*B*b^3*tan(1/2
*d*x + 1/2*c)^5 + 160*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 30*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 90*A*a^2*b*tan(1/2*d*x
+ 1/2*c)^3 + 960*B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 960*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 360*B*a*b^2*tan(1/2*d*x
+ 1/2*c)^3 + 120*A*b^3*tan(1/2*d*x + 1/2*c)^3 + 480*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 120*A*a^3*tan(1/2*d*x + 1/
2*c) + 75*B*a^3*tan(1/2*d*x + 1/2*c) + 225*A*a^2*b*tan(1/2*d*x + 1/2*c) + 360*B*a^2*b*tan(1/2*d*x + 1/2*c) + 3
60*A*a*b^2*tan(1/2*d*x + 1/2*c) + 180*B*a*b^2*tan(1/2*d*x + 1/2*c) + 60*A*b^3*tan(1/2*d*x + 1/2*c) + 120*B*b^3
*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

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maple [A]  time = 2.11, size = 227, normalized size = 1.03 \[ \frac {\frac {A \,a^{3} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+a^{3} B \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+3 A \,a^{2} b \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+a^{2} b B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+A a \,b^{2} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+3 B a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+b^{3} B \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x)

[Out]

1/d*(1/5*A*a^3*(8/3+cos(d*x+c)^4+4/3*cos(d*x+c)^2)*sin(d*x+c)+a^3*B*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x
+c)+3/8*d*x+3/8*c)+3*A*a^2*b*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+a^2*b*B*(2+cos(d*x+c
)^2)*sin(d*x+c)+A*a*b^2*(2+cos(d*x+c)^2)*sin(d*x+c)+3*B*a*b^2*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+A*b^3*
(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+b^3*B*sin(d*x+c))

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maxima [A]  time = 0.68, size = 217, normalized size = 0.98 \[ \frac {32 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} A a^{3} + 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{3} + 45 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} b - 480 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b^{2} + 360 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} + 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} + 480 \, B b^{3} \sin \left (d x + c\right )}{480 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+b*sec(d*x+c))^3*(A+B*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/480*(32*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*A*a^3 + 15*(12*d*x + 12*c + sin(4*d*x + 4*c
) + 8*sin(2*d*x + 2*c))*B*a^3 + 45*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a^2*b - 480*(sin(
d*x + c)^3 - 3*sin(d*x + c))*B*a^2*b - 480*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a*b^2 + 360*(2*d*x + 2*c + sin(
2*d*x + 2*c))*B*a*b^2 + 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*b^3 + 480*B*b^3*sin(d*x + c))/d

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mupad [B]  time = 2.73, size = 277, normalized size = 1.25 \[ \frac {A\,b^3\,x}{2}+\frac {3\,B\,a^3\,x}{8}+\frac {9\,A\,a^2\,b\,x}{8}+\frac {3\,B\,a\,b^2\,x}{2}+\frac {5\,A\,a^3\,\sin \left (c+d\,x\right )}{8\,d}+\frac {B\,b^3\,\sin \left (c+d\,x\right )}{d}+\frac {5\,A\,a^3\,\sin \left (3\,c+3\,d\,x\right )}{48\,d}+\frac {A\,a^3\,\sin \left (5\,c+5\,d\,x\right )}{80\,d}+\frac {A\,b^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^3\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,A\,a^2\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,b^2\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {3\,A\,a^2\,b\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {3\,B\,a\,b^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {B\,a^2\,b\,\sin \left (3\,c+3\,d\,x\right )}{4\,d}+\frac {9\,A\,a\,b^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {9\,B\,a^2\,b\,\sin \left (c+d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^5*(A + B/cos(c + d*x))*(a + b/cos(c + d*x))^3,x)

[Out]

(A*b^3*x)/2 + (3*B*a^3*x)/8 + (9*A*a^2*b*x)/8 + (3*B*a*b^2*x)/2 + (5*A*a^3*sin(c + d*x))/(8*d) + (B*b^3*sin(c
+ d*x))/d + (5*A*a^3*sin(3*c + 3*d*x))/(48*d) + (A*a^3*sin(5*c + 5*d*x))/(80*d) + (A*b^3*sin(2*c + 2*d*x))/(4*
d) + (B*a^3*sin(2*c + 2*d*x))/(4*d) + (B*a^3*sin(4*c + 4*d*x))/(32*d) + (3*A*a^2*b*sin(2*c + 2*d*x))/(4*d) + (
A*a*b^2*sin(3*c + 3*d*x))/(4*d) + (3*A*a^2*b*sin(4*c + 4*d*x))/(32*d) + (3*B*a*b^2*sin(2*c + 2*d*x))/(4*d) + (
B*a^2*b*sin(3*c + 3*d*x))/(4*d) + (9*A*a*b^2*sin(c + d*x))/(4*d) + (9*B*a^2*b*sin(c + d*x))/(4*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)**5*(a+b*sec(d*x+c))**3*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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