3.539 \(\int (a+b \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=187 \[ \frac {\left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac {a b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a b (3 A+4 C) \tan (c+d x) \sec (c+d x)}{4 d}+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{10 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

[Out]

1/4*a*b*(3*A+4*C)*arctanh(sin(d*x+c))/d+1/15*(5*b^2*(2*A+3*C)+2*a^2*(4*A+5*C))*tan(d*x+c)/d+1/4*a*b*(3*A+4*C)*
sec(d*x+c)*tan(d*x+c)/d+1/15*(2*A*b^2+a^2*(4*A+5*C))*sec(d*x+c)^2*tan(d*x+c)/d+1/10*a*A*b*sec(d*x+c)^3*tan(d*x
+c)/d+1/5*A*(a+b*cos(d*x+c))^2*sec(d*x+c)^4*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.45, antiderivative size = 187, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {3048, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {\left (2 a^2 (4 A+5 C)+5 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac {\left (a^2 (4 A+5 C)+2 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac {a b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {a b (3 A+4 C) \tan (c+d x) \sec (c+d x)}{4 d}+\frac {a A b \tan (c+d x) \sec ^3(c+d x)}{10 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(a*b*(3*A + 4*C)*ArcTanh[Sin[c + d*x]])/(4*d) + ((5*b^2*(2*A + 3*C) + 2*a^2*(4*A + 5*C))*Tan[c + d*x])/(15*d)
+ (a*b*(3*A + 4*C)*Sec[c + d*x]*Tan[c + d*x])/(4*d) + ((2*A*b^2 + a^2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x]
)/(15*d) + (a*A*b*Sec[c + d*x]^3*Tan[c + d*x])/(10*d) + (A*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^4*Tan[c + d*x])
/(5*d)

Rule 8

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3021

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m +
1)*(a^2 - b^2)), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B + a*
C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e,
f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3031

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rule 3048

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C + A*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[
e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m
- 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + c*C*(b*c*m + a*d*(n + 1)) - (A*d*(a*d*(n +
2) - b*c*(n + 1)) - C*(b*c*d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b*(A*d^2*(m + n + 2) + C*(c^2*(
m + 1) + d^2*(n + 1)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, C}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3767

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

Rule 3768

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

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin {align*} \int (a+b \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+b \cos (c+d x)) \left (2 A b+a (4 A+5 C) \cos (c+d x)+b (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{20} \int \left (-4 \left (2 A b^2+a^2 (4 A+5 C)\right )-10 a b (3 A+4 C) \cos (c+d x)-4 b^2 (2 A+5 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{60} \int \left (-30 a b (3 A+4 C)-4 \left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{2} (a b (3 A+4 C)) \int \sec ^3(c+d x) \, dx-\frac {1}{15} \left (-5 b^2 (2 A+3 C)-2 a^2 (4 A+5 C)\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {a b (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{4} (a b (3 A+4 C)) \int \sec (c+d x) \, dx-\frac {\left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {a b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{4 d}+\frac {\left (5 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {a b (3 A+4 C) \sec (c+d x) \tan (c+d x)}{4 d}+\frac {\left (2 A b^2+a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a A b \sec ^3(c+d x) \tan (c+d x)}{10 d}+\frac {A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.10, size = 115, normalized size = 0.61 \[ \frac {\tan (c+d x) \left (20 \left (a^2 (2 A+C)+A b^2\right ) \tan ^2(c+d x)+60 \left (a^2+b^2\right ) (A+C)+12 a^2 A \tan ^4(c+d x)+15 a b (3 A+4 C) \sec (c+d x)+30 a A b \sec ^3(c+d x)\right )+15 a b (3 A+4 C) \tanh ^{-1}(\sin (c+d x))}{60 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(15*a*b*(3*A + 4*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(60*(a^2 + b^2)*(A + C) + 15*a*b*(3*A + 4*C)*Sec[c +
d*x] + 30*a*A*b*Sec[c + d*x]^3 + 20*(A*b^2 + a^2*(2*A + C))*Tan[c + d*x]^2 + 12*a^2*A*Tan[c + d*x]^4))/(60*d)

________________________________________________________________________________________

fricas [A]  time = 1.48, size = 180, normalized size = 0.96 \[ \frac {15 \, {\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (15 \, {\left (3 \, A + 4 \, C\right )} a b \cos \left (d x + c\right )^{3} + 4 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{2} + 5 \, {\left (2 \, A + 3 \, C\right )} b^{2}\right )} \cos \left (d x + c\right )^{4} + 30 \, A a b \cos \left (d x + c\right ) + 12 \, A a^{2} + 4 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{2} + 5 \, A b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/120*(15*(3*A + 4*C)*a*b*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*A + 4*C)*a*b*cos(d*x + c)^5*log(-sin(d*
x + c) + 1) + 2*(15*(3*A + 4*C)*a*b*cos(d*x + c)^3 + 4*(2*(4*A + 5*C)*a^2 + 5*(2*A + 3*C)*b^2)*cos(d*x + c)^4
+ 30*A*a*b*cos(d*x + c) + 12*A*a^2 + 4*((4*A + 5*C)*a^2 + 5*A*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x +
c)^5)

________________________________________________________________________________________

giac [B]  time = 0.56, size = 532, normalized size = 2.84 \[ \frac {15 \, {\left (3 \, A a b + 4 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (3 \, A a b + 4 \, C a b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 75 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 60 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 60 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 80 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 30 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 120 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 240 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 232 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 200 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 360 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 80 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 30 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 160 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 240 \, C b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 60 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 75 \, A a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, A b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 60 \, C b^{2} \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}}}{60 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/60*(15*(3*A*a*b + 4*C*a*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*A*a*b + 4*C*a*b)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(60*A*a^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*a^2*tan(1/2*d*x + 1/2*c)^9 - 75*A*a*b*tan(1/2*d*x +
1/2*c)^9 - 60*C*a*b*tan(1/2*d*x + 1/2*c)^9 + 60*A*b^2*tan(1/2*d*x + 1/2*c)^9 + 60*C*b^2*tan(1/2*d*x + 1/2*c)^9
- 80*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 160*C*a^2*tan(1/2*d*x + 1/2*c)^7 + 30*A*a*b*tan(1/2*d*x + 1/2*c)^7 + 120*
C*a*b*tan(1/2*d*x + 1/2*c)^7 - 160*A*b^2*tan(1/2*d*x + 1/2*c)^7 - 240*C*b^2*tan(1/2*d*x + 1/2*c)^7 + 232*A*a^2
*tan(1/2*d*x + 1/2*c)^5 + 200*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 200*A*b^2*tan(1/2*d*x + 1/2*c)^5 + 360*C*b^2*tan(
1/2*d*x + 1/2*c)^5 - 80*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 160*C*a^2*tan(1/2*d*x + 1/2*c)^3 - 30*A*a*b*tan(1/2*d*x
+ 1/2*c)^3 - 120*C*a*b*tan(1/2*d*x + 1/2*c)^3 - 160*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 240*C*b^2*tan(1/2*d*x + 1/
2*c)^3 + 60*A*a^2*tan(1/2*d*x + 1/2*c) + 60*C*a^2*tan(1/2*d*x + 1/2*c) + 75*A*a*b*tan(1/2*d*x + 1/2*c) + 60*C*
a*b*tan(1/2*d*x + 1/2*c) + 60*A*b^2*tan(1/2*d*x + 1/2*c) + 60*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c
)^2 - 1)^5)/d

________________________________________________________________________________________

maple [A]  time = 0.43, size = 257, normalized size = 1.37 \[ \frac {8 a^{2} A \tan \left (d x +c \right )}{15 d}+\frac {a^{2} A \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 a^{2} A \left (\sec ^{2}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{15 d}+\frac {2 a^{2} C \tan \left (d x +c \right )}{3 d}+\frac {a^{2} C \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a A b \left (\sec ^{3}\left (d x +c \right )\right ) \tan \left (d x +c \right )}{2 d}+\frac {3 a A b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{4 d}+\frac {3 A a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {C a b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{d}+\frac {C a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 A \,b^{2} \tan \left (d x +c \right )}{3 d}+\frac {A \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {b^{2} C \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)

[Out]

8/15*a^2*A*tan(d*x+c)/d+1/5/d*a^2*A*tan(d*x+c)*sec(d*x+c)^4+4/15*a^2*A*sec(d*x+c)^2*tan(d*x+c)/d+2/3/d*a^2*C*t
an(d*x+c)+1/3/d*a^2*C*tan(d*x+c)*sec(d*x+c)^2+1/2*a*A*b*sec(d*x+c)^3*tan(d*x+c)/d+3/4*a*A*b*sec(d*x+c)*tan(d*x
+c)/d+3/4/d*A*a*b*ln(sec(d*x+c)+tan(d*x+c))+1/d*C*a*b*tan(d*x+c)*sec(d*x+c)+1/d*C*a*b*ln(sec(d*x+c)+tan(d*x+c)
)+2/3/d*A*b^2*tan(d*x+c)+1/3/d*A*b^2*tan(d*x+c)*sec(d*x+c)^2+1/d*b^2*C*tan(d*x+c)

________________________________________________________________________________________

maxima [A]  time = 0.73, size = 216, normalized size = 1.16 \[ \frac {8 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{2} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b^{2} - 15 \, A a b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, C a b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C b^{2} \tan \left (d x + c\right )}{120 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/120*(8*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))
*C*a^2 + 40*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b^2 - 15*A*a*b*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x
+ c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) - 60*C*a*b*(2*sin(d*x + c
)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*C*b^2*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 4.81, size = 322, normalized size = 1.72 \[ \frac {a\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A+4\,C\right )}{2\,d}-\frac {\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2-\frac {5\,A\,a\,b}{2}-2\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (A\,a\,b-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-8\,C\,b^2-\frac {8\,A\,a^2}{3}+4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^2}{15}+\frac {20\,A\,b^2}{3}+\frac {20\,C\,a^2}{3}+12\,C\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^2}{3}-\frac {16\,A\,b^2}{3}-\frac {16\,C\,a^2}{3}-8\,C\,b^2-A\,a\,b-4\,C\,a\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^2+2\,A\,b^2+2\,C\,a^2+2\,C\,b^2+\frac {5\,A\,a\,b}{2}+2\,C\,a\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^2)/cos(c + d*x)^6,x)

[Out]

(a*b*atanh(tan(c/2 + (d*x)/2))*(3*A + 4*C))/(2*d) - (tan(c/2 + (d*x)/2)^9*(2*A*a^2 + 2*A*b^2 + 2*C*a^2 + 2*C*b
^2 - (5*A*a*b)/2 - 2*C*a*b) - tan(c/2 + (d*x)/2)^3*((8*A*a^2)/3 + (16*A*b^2)/3 + (16*C*a^2)/3 + 8*C*b^2 + A*a*
b + 4*C*a*b) - tan(c/2 + (d*x)/2)^7*((8*A*a^2)/3 + (16*A*b^2)/3 + (16*C*a^2)/3 + 8*C*b^2 - A*a*b - 4*C*a*b) +
tan(c/2 + (d*x)/2)^5*((116*A*a^2)/15 + (20*A*b^2)/3 + (20*C*a^2)/3 + 12*C*b^2) + tan(c/2 + (d*x)/2)*(2*A*a^2 +
2*A*b^2 + 2*C*a^2 + 2*C*b^2 + (5*A*a*b)/2 + 2*C*a*b))/(d*(5*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 +
10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))

________________________________________________________________________________________

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((a+b*cos(d*x+c))**2*(A+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

________________________________________________________________________________________