3.547 \(\int (a+b \cos (c+d x))^3 (A+C \cos ^2(c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=227 \[ \frac {a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac {b \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{30 d}+\frac {3 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{20 d} \]

[Out]

1/8*b*(4*b^2*(A+2*C)+3*a^2*(3*A+4*C))*arctanh(sin(d*x+c))/d+1/15*a*(15*b^2*(2*A+3*C)+2*a^2*(4*A+5*C))*tan(d*x+
c)/d+3/40*b*(2*A*b^2+5*a^2*(3*A+4*C))*sec(d*x+c)*tan(d*x+c)/d+1/30*a*(3*A*b^2+2*a^2*(4*A+5*C))*sec(d*x+c)^2*ta
n(d*x+c)/d+3/20*A*b*(a+b*cos(d*x+c))^2*sec(d*x+c)^3*tan(d*x+c)/d+1/5*A*(a+b*cos(d*x+c))^3*sec(d*x+c)^4*tan(d*x
+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.71, antiderivative size = 227, 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, 3047, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac {a \left (2 a^2 (4 A+5 C)+15 b^2 (2 A+3 C)\right ) \tan (c+d x)}{15 d}+\frac {b \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (2 a^2 (4 A+5 C)+3 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{30 d}+\frac {3 b \left (5 a^2 (3 A+4 C)+2 A b^2\right ) \tan (c+d x) \sec (c+d x)}{40 d}+\frac {A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^3}{5 d}+\frac {3 A b \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{20 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(b*(4*b^2*(A + 2*C) + 3*a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]])/(8*d) + (a*(15*b^2*(2*A + 3*C) + 2*a^2*(4*A +
5*C))*Tan[c + d*x])/(15*d) + (3*b*(2*A*b^2 + 5*a^2*(3*A + 4*C))*Sec[c + d*x]*Tan[c + d*x])/(40*d) + (a*(3*A*b^
2 + 2*a^2*(4*A + 5*C))*Sec[c + d*x]^2*Tan[c + d*x])/(30*d) + (3*A*b*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[
c + d*x])/(20*d) + (A*(a + b*Cos[c + d*x])^3*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 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + 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*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(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, B, 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 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 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))^3 \left (A+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{5} \int (a+b \cos (c+d x))^2 \left (3 A b+a (4 A+5 C) \cos (c+d x)+b (A+5 C) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {3 A b (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{20} \int (a+b \cos (c+d x)) \left (2 \left (3 A b^2+2 a^2 (4 A+5 C)\right )+a b (29 A+40 C) \cos (c+d x)+b^2 (7 A+20 C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A b (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{60} \int \left (-9 b \left (2 A b^2+5 a^2 (3 A+4 C)\right )-4 a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \cos (c+d x)-3 b^3 (7 A+20 C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {3 b \left (2 A b^2+5 a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A b (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{120} \int \left (-8 a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )-15 b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {3 b \left (2 A b^2+5 a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A b (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}+\frac {1}{8} \left (b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right )\right ) \int \sec (c+d x) \, dx+\frac {1}{15} \left (a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )\right ) \int \sec ^2(c+d x) \, dx\\ &=\frac {b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {3 b \left (2 A b^2+5 a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A b (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {\left (a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right )\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {b \left (4 b^2 (A+2 C)+3 a^2 (3 A+4 C)\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (15 b^2 (2 A+3 C)+2 a^2 (4 A+5 C)\right ) \tan (c+d x)}{15 d}+\frac {3 b \left (2 A b^2+5 a^2 (3 A+4 C)\right ) \sec (c+d x) \tan (c+d x)}{40 d}+\frac {a \left (3 A b^2+2 a^2 (4 A+5 C)\right ) \sec ^2(c+d x) \tan (c+d x)}{30 d}+\frac {3 A b (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {A (a+b \cos (c+d x))^3 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 2.26, size = 150, normalized size = 0.66 \[ \frac {15 b \left (3 a^2 (3 A+4 C)+4 b^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (8 a \left (5 \left (a^2 (2 A+C)+3 A b^2\right ) \tan ^2(c+d x)+15 \left (a^2+3 b^2\right ) (A+C)+3 a^2 A \tan ^4(c+d x)\right )+15 b \left (3 a^2 (3 A+4 C)+4 A b^2\right ) \sec (c+d x)+90 a^2 A b \sec ^3(c+d x)\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Cos[c + d*x])^3*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^6,x]

[Out]

(15*b*(4*b^2*(A + 2*C) + 3*a^2*(3*A + 4*C))*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*b*(4*A*b^2 + 3*a^2*(3*A +
4*C))*Sec[c + d*x] + 90*a^2*A*b*Sec[c + d*x]^3 + 8*a*(15*(a^2 + 3*b^2)*(A + C) + 5*(3*A*b^2 + a^2*(2*A + C))*
Tan[c + d*x]^2 + 3*a^2*A*Tan[c + d*x]^4)))/(120*d)

________________________________________________________________________________________

fricas [A]  time = 0.82, size = 225, normalized size = 0.99 \[ \frac {15 \, {\left (3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \, {\left (A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (90 \, A a^{2} b \cos \left (d x + c\right ) + 8 \, {\left (2 \, {\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, {\left (2 \, A + 3 \, C\right )} a b^{2}\right )} \cos \left (d x + c\right )^{4} + 24 \, A a^{3} + 15 \, {\left (3 \, {\left (3 \, A + 4 \, C\right )} a^{2} b + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left ({\left (4 \, A + 5 \, C\right )} a^{3} + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2}\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="fricas")

[Out]

1/240*(15*(3*(3*A + 4*C)*a^2*b + 4*(A + 2*C)*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*(3*A + 4*C)*a^2
*b + 4*(A + 2*C)*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(90*A*a^2*b*cos(d*x + c) + 8*(2*(4*A + 5*C)*a^
3 + 15*(2*A + 3*C)*a*b^2)*cos(d*x + c)^4 + 24*A*a^3 + 15*(3*(3*A + 4*C)*a^2*b + 4*A*b^3)*cos(d*x + c)^3 + 8*((
4*A + 5*C)*a^3 + 15*A*a*b^2)*cos(d*x + c)^2)*sin(d*x + c))/(d*cos(d*x + c)^5)

________________________________________________________________________________________

giac [B]  time = 0.52, size = 656, normalized size = 2.89 \[ \frac {15 \, {\left (9 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 8 \, C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (9 \, A a^{2} b + 12 \, C a^{2} b + 4 \, A b^{3} + 8 \, C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (120 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 120 \, C 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} - 180 \, C 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} + 360 \, C 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} - 160 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 320 \, C 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} + 360 \, C 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} - 1440 \, C 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} + 464 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 400 \, C a^{3} \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} + 2160 \, C a b^{2} \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} - 320 \, C 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} - 360 \, C 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} - 1440 \, C 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} + 120 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 120 \, C 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 ) + 180 \, C 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 ) + 360 \, C 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 )\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((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="giac")

[Out]

1/120*(15*(9*A*a^2*b + 12*C*a^2*b + 4*A*b^3 + 8*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(9*A*a^2*b + 12
*C*a^2*b + 4*A*b^3 + 8*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(120*A*a^3*tan(1/2*d*x + 1/2*c)^9 + 120*C
*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 - 180*C*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 + 360*C*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 - 160*A*a^
3*tan(1/2*d*x + 1/2*c)^7 - 320*C*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 + 360*C*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 - 1440*C*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 + 464*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 400*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 1200*A*a*b^2*ta
n(1/2*d*x + 1/2*c)^5 + 2160*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 160*A*a^3*tan(1/2*d*x + 1/2*c)^3 - 320*C*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 - 360*C*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 - 1440*C*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 + 120*A*a^3*tan(1/
2*d*x + 1/2*c) + 120*C*a^3*tan(1/2*d*x + 1/2*c) + 225*A*a^2*b*tan(1/2*d*x + 1/2*c) + 180*C*a^2*b*tan(1/2*d*x +
1/2*c) + 360*A*a*b^2*tan(1/2*d*x + 1/2*c) + 360*C*a*b^2*tan(1/2*d*x + 1/2*c) + 60*A*b^3*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 = 338, normalized size = 1.49 \[ \frac {8 A \,a^{3} \tan \left (d x +c \right )}{15 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {2 C \,a^{3} \tan \left (d x +c \right )}{3 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {3 A \,a^{2} b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 A \,a^{2} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {9 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 C \,a^{2} b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {3 C \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 A a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {A a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 C a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {A \,b^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {b^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x)

[Out]

8/15/d*A*a^3*tan(d*x+c)+1/5/d*A*a^3*tan(d*x+c)*sec(d*x+c)^4+4/15/d*A*a^3*tan(d*x+c)*sec(d*x+c)^2+2/3/d*C*a^3*t
an(d*x+c)+1/3/d*C*a^3*tan(d*x+c)*sec(d*x+c)^2+3/4/d*A*a^2*b*tan(d*x+c)*sec(d*x+c)^3+9/8/d*A*a^2*b*sec(d*x+c)*t
an(d*x+c)+9/8/d*A*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+3/2/d*C*a^2*b*tan(d*x+c)*sec(d*x+c)+3/2/d*C*a^2*b*ln(sec(d*x
+c)+tan(d*x+c))+2/d*A*a*b^2*tan(d*x+c)+1/d*A*a*b^2*tan(d*x+c)*sec(d*x+c)^2+3/d*C*a*b^2*tan(d*x+c)+1/2/d*A*b^3*
tan(d*x+c)*sec(d*x+c)+1/2/d*A*b^3*ln(sec(d*x+c)+tan(d*x+c))+1/d*b^3*C*ln(sec(d*x+c)+tan(d*x+c))

________________________________________________________________________________________

maxima [A]  time = 0.62, size = 296, normalized size = 1.30 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 80 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} - 45 \, A a^{2} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, C a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 120 \, C b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 720 \, C a b^{2} \tan \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+C*cos(d*x+c)^2)*sec(d*x+c)^6,x, algorithm="maxima")

[Out]

1/240*(16*(3*tan(d*x + c)^5 + 10*tan(d*x + c)^3 + 15*tan(d*x + c))*A*a^3 + 80*(tan(d*x + c)^3 + 3*tan(d*x + c)
)*C*a^3 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b^2 - 45*A*a^2*b*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(s
in(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)) - 180*C*a^2*b*(2*si
n(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) - 60*A*b^3*(2*sin(d*x + c)/(s
in(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 120*C*b^3*(log(sin(d*x + c) + 1) - log(s
in(d*x + c) - 1)) + 720*C*a*b^2*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 4.82, size = 445, normalized size = 1.96 \[ \frac {b\,\mathrm {atanh}\left (\frac {b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (9\,A\,a^2+4\,A\,b^2+12\,C\,a^2+8\,C\,b^2\right )}{2\,\left (2\,A\,b^3+4\,C\,b^3+\frac {9\,A\,a^2\,b}{2}+6\,C\,a^2\,b\right )}\right )\,\left (9\,A\,a^2+4\,A\,b^2+12\,C\,a^2+8\,C\,b^2\right )}{4\,d}-\frac {\left (2\,A\,a^3-A\,b^3+2\,C\,a^3+6\,A\,a\,b^2-\frac {15\,A\,a^2\,b}{4}+6\,C\,a\,b^2-3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,A\,b^3-\frac {8\,A\,a^3}{3}-\frac {16\,C\,a^3}{3}-16\,A\,a\,b^2+\frac {3\,A\,a^2\,b}{2}-24\,C\,a\,b^2+6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^3}{15}+\frac {20\,C\,a^3}{3}+20\,A\,a\,b^2+36\,C\,a\,b^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^3}{3}-2\,A\,b^3-\frac {16\,C\,a^3}{3}-16\,A\,a\,b^2-\frac {3\,A\,a^2\,b}{2}-24\,C\,a\,b^2-6\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^3+A\,b^3+2\,C\,a^3+6\,A\,a\,b^2+\frac {15\,A\,a^2\,b}{4}+6\,C\,a\,b^2+3\,C\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

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))^3)/cos(c + d*x)^6,x)

[Out]

(b*atanh((b*tan(c/2 + (d*x)/2)*(9*A*a^2 + 4*A*b^2 + 12*C*a^2 + 8*C*b^2))/(2*(2*A*b^3 + 4*C*b^3 + (9*A*a^2*b)/2
+ 6*C*a^2*b)))*(9*A*a^2 + 4*A*b^2 + 12*C*a^2 + 8*C*b^2))/(4*d) - (tan(c/2 + (d*x)/2)^9*(2*A*a^3 - A*b^3 + 2*C
*a^3 + 6*A*a*b^2 - (15*A*a^2*b)/4 + 6*C*a*b^2 - 3*C*a^2*b) - tan(c/2 + (d*x)/2)^3*((8*A*a^3)/3 + 2*A*b^3 + (16
*C*a^3)/3 + 16*A*a*b^2 + (3*A*a^2*b)/2 + 24*C*a*b^2 + 6*C*a^2*b) - tan(c/2 + (d*x)/2)^7*((8*A*a^3)/3 - 2*A*b^3
+ (16*C*a^3)/3 + 16*A*a*b^2 - (3*A*a^2*b)/2 + 24*C*a*b^2 - 6*C*a^2*b) + tan(c/2 + (d*x)/2)^5*((116*A*a^3)/15
+ (20*C*a^3)/3 + 20*A*a*b^2 + 36*C*a*b^2) + tan(c/2 + (d*x)/2)*(2*A*a^3 + A*b^3 + 2*C*a^3 + 6*A*a*b^2 + (15*A*
a^2*b)/4 + 6*C*a*b^2 + 3*C*a^2*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))**3*(A+C*cos(d*x+c)**2)*sec(d*x+c)**6,x)

[Out]

Timed out

________________________________________________________________________________________