3.239 \(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx\)

Optimal. Leaf size=236 \[ \frac {a \left (4 a^2 A+15 a b B+12 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac {a^2 (5 a B+7 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {\left (8 a^3 A+30 a^2 b B+30 a A b^2+15 b^3 B\right ) \tan (c+d x)}{15 d}+\frac {\left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

[Out]

1/8*(9*A*a^2*b+4*A*b^3+3*B*a^3+12*B*a*b^2)*arctanh(sin(d*x+c))/d+1/15*(8*A*a^3+30*A*a*b^2+30*B*a^2*b+15*B*b^3)
*tan(d*x+c)/d+1/8*(9*A*a^2*b+4*A*b^3+3*B*a^3+12*B*a*b^2)*sec(d*x+c)*tan(d*x+c)/d+1/15*a*(4*A*a^2+12*A*b^2+15*B
*a*b)*sec(d*x+c)^2*tan(d*x+c)/d+1/20*a^2*(7*A*b+5*B*a)*sec(d*x+c)^3*tan(d*x+c)/d+1/5*a*A*(a+b*cos(d*x+c))^2*se
c(d*x+c)^4*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.49, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {2989, 3031, 3021, 2748, 3768, 3770, 3767, 8} \[ \frac {\left (8 a^3 A+30 a^2 b B+30 a A b^2+15 b^3 B\right ) \tan (c+d x)}{15 d}+\frac {\left (9 a^2 A b+3 a^3 B+12 a b^2 B+4 A b^3\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (4 a^2 A+15 a b B+12 A b^2\right ) \tan (c+d x) \sec ^2(c+d x)}{15 d}+\frac {\left (9 a^2 A b+3 a^3 B+12 a b^2 B+4 A b^3\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a^2 (5 a B+7 A b) \tan (c+d x) \sec ^3(c+d x)}{20 d}+\frac {a A \tan (c+d x) \sec ^4(c+d x) (a+b \cos (c+d x))^2}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*ArcTanh[Sin[c + d*x]])/(8*d) + ((8*a^3*A + 30*a*A*b^2 + 30*a^2*b
*B + 15*b^3*B)*Tan[c + d*x])/(15*d) + ((9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*Sec[c + d*x]*Tan[c + d*x])
/(8*d) + (a*(4*a^2*A + 12*A*b^2 + 15*a*b*B)*Sec[c + d*x]^2*Tan[c + d*x])/(15*d) + (a^2*(7*A*b + 5*a*B)*Sec[c +
d*x]^3*Tan[c + d*x])/(20*d) + (a*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 2989

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

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 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))^3 (A+B \cos (c+d x)) \sec ^6(c+d x) \, dx &=\frac {a 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 (a (7 A b+5 a B)+\left (4 a^2 A+5 A b^2+10 a b B\right ) \cos (c+d x)+b (2 a A+5 b B) \cos ^2(c+d x)\right ) \sec ^5(c+d x) \, dx\\ &=\frac {a^2 (7 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a 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 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 ) \cos (c+d x)-4 b^2 (2 a A+5 b B) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a \left (4 a^2 A+12 A b^2+15 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (7 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{60} \int \left (-15 \left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right )-4 \left (8 a^3 A+30 a A b^2+30 a^2 b B+15 b^3 B\right ) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a \left (4 a^2 A+12 A b^2+15 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (7 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{4} \left (-9 a^2 A b-4 A b^3-3 a^3 B-12 a b^2 B\right ) \int \sec ^3(c+d x) \, dx-\frac {1}{15} \left (-8 a^3 A-30 a A b^2-30 a^2 b B-15 b^3 B\right ) \int \sec ^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 ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \left (4 a^2 A+12 A b^2+15 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (7 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}-\frac {1}{8} \left (-9 a^2 A b-4 A b^3-3 a^3 B-12 a b^2 B\right ) \int \sec (c+d x) \, dx-\frac {\left (8 a^3 A+30 a A b^2+30 a^2 b B+15 b^3 B\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{15 d}\\ &=\frac {\left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (8 a^3 A+30 a A b^2+30 a^2 b B+15 b^3 B\right ) \tan (c+d x)}{15 d}+\frac {\left (9 a^2 A b+4 A b^3+3 a^3 B+12 a b^2 B\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a \left (4 a^2 A+12 A b^2+15 a b B\right ) \sec ^2(c+d x) \tan (c+d x)}{15 d}+\frac {a^2 (7 A b+5 a B) \sec ^3(c+d x) \tan (c+d x)}{20 d}+\frac {a A (a+b \cos (c+d x))^2 \sec ^4(c+d x) \tan (c+d x)}{5 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 3.26, size = 181, normalized size = 0.77 \[ \frac {15 \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (30 a^2 (a B+3 A b) \sec ^3(c+d x)+8 \left (3 a^3 A \tan ^4(c+d x)+5 a \left (2 a^2 A+3 a b B+3 A b^2\right ) \tan ^2(c+d x)+15 \left (a^3 A+3 a^2 b B+3 a A b^2+b^3 B\right )\right )+15 \left (3 a^3 B+9 a^2 A b+12 a b^2 B+4 A b^3\right ) \sec (c+d x)\right )}{120 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(15*(9*a^2*A*b + 4*A*b^3 + 3*a^3*B + 12*a*b^2*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(15*(9*a^2*A*b + 4*A*b^3
+ 3*a^3*B + 12*a*b^2*B)*Sec[c + d*x] + 30*a^2*(3*A*b + a*B)*Sec[c + d*x]^3 + 8*(15*(a^3*A + 3*a*A*b^2 + 3*a^2
*b*B + b^3*B) + 5*a*(2*a^2*A + 3*A*b^2 + 3*a*b*B)*Tan[c + d*x]^2 + 3*a^3*A*Tan[c + d*x]^4)))/(120*d)

________________________________________________________________________________________

fricas [A]  time = 0.86, size = 249, normalized size = 1.06 \[ \frac {15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (8 \, A a^{3} + 30 \, B a^{2} b + 30 \, A a b^{2} + 15 \, B b^{3}\right )} \cos \left (d x + c\right )^{4} + 24 \, A a^{3} + 15 \, {\left (3 \, B a^{3} + 9 \, A a^{2} b + 12 \, B a b^{2} + 4 \, A b^{3}\right )} \cos \left (d x + c\right )^{3} + 8 \, {\left (4 \, A a^{3} + 15 \, B a^{2} b + 15 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} + 30 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{240 \, d \cos \left (d x + c\right )^{5}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/240*(15*(3*B*a^3 + 9*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 15*(3*B*a^3 + 9*
A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*cos(d*x + c)^5*log(-sin(d*x + c) + 1) + 2*(8*(8*A*a^3 + 30*B*a^2*b + 30*A*a*b^
2 + 15*B*b^3)*cos(d*x + c)^4 + 24*A*a^3 + 15*(3*B*a^3 + 9*A*a^2*b + 12*B*a*b^2 + 4*A*b^3)*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 + 30*(B*a^3 + 3*A*a^2*b)*cos(d*x + c))*sin(d*x + c))/(d*cos(
d*x + c)^5)

________________________________________________________________________________________

giac [B]  time = 0.89, size = 722, normalized size = 3.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^6,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)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*B*a^3 + 9*A*
a^2*b + 12*B*a*b^2 + 4*A*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 - 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

________________________________________________________________________________________

maple [A]  time = 0.15, size = 382, normalized size = 1.62 \[ \frac {8 A \,a^{3} \tan \left (d x +c \right )}{15 d}+\frac {A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{4}\left (d x +c \right )\right )}{5 d}+\frac {4 A \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{15 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 A \,a^{2} b \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {9 A \,a^{2} b \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {9 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {2 a^{2} b B \tan \left (d x +c \right )}{d}+\frac {a^{2} b B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {2 A \,b^{2} a \tan \left (d x +c \right )}{d}+\frac {A a \,b^{2} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 B a \,b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {3 B \,b^{2} a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {A \,b^{3} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {A \,b^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {b^{3} B \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*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+1/4/d*a^3*B*t
an(d*x+c)*sec(d*x+c)^3+3/8/d*a^3*B*sec(d*x+c)*tan(d*x+c)+3/8/d*a^3*B*ln(sec(d*x+c)+tan(d*x+c))+3/4/d*A*a^2*b*t
an(d*x+c)*sec(d*x+c)^3+9/8/d*A*a^2*b*sec(d*x+c)*tan(d*x+c)+9/8/d*A*a^2*b*ln(sec(d*x+c)+tan(d*x+c))+2/d*a^2*b*B
*tan(d*x+c)+1/d*a^2*b*B*tan(d*x+c)*sec(d*x+c)^2+2/d*A*b^2*a*tan(d*x+c)+1/d*A*a*b^2*tan(d*x+c)*sec(d*x+c)^2+3/2
/d*B*a*b^2*tan(d*x+c)*sec(d*x+c)+3/2/d*B*b^2*a*ln(sec(d*x+c)+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*B*tan(d*x+c)

________________________________________________________________________________________

maxima [A]  time = 0.51, size = 341, normalized size = 1.44 \[ \frac {16 \, {\left (3 \, \tan \left (d x + c\right )^{5} + 10 \, \tan \left (d x + c\right )^{3} + 15 \, \tan \left (d x + c\right )\right )} A a^{3} + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b + 240 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a b^{2} - 15 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 45 \, A a^{2} b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 180 \, B a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 60 \, A b^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 240 \, B b^{3} \tan \left (d x + c\right )}{240 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(A+B*cos(d*x+c))*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 + 240*(tan(d*x + c)^3 + 3*tan(d*x + c
))*B*a^2*b + 240*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a*b^2 - 15*B*a^3*(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)) - 45*A*a^2*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(si
n(d*x + c) - 1)) - 180*B*a*b^2*(2*sin(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)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 240*
B*b^3*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 3.89, size = 470, normalized size = 1.99 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,B\,a^3}{8}+\frac {9\,A\,a^2\,b}{8}+\frac {3\,B\,a\,b^2}{2}+\frac {A\,b^3}{2}\right )}{\frac {3\,B\,a^3}{2}+\frac {9\,A\,a^2\,b}{2}+6\,B\,a\,b^2+2\,A\,b^3}\right )\,\left (\frac {3\,B\,a^3}{4}+\frac {9\,A\,a^2\,b}{4}+3\,B\,a\,b^2+A\,b^3\right )}{d}-\frac {\left (2\,A\,a^3-A\,b^3-\frac {5\,B\,a^3}{4}+2\,B\,b^3+6\,A\,a\,b^2-\frac {15\,A\,a^2\,b}{4}-3\,B\,a\,b^2+6\,B\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+\left (2\,A\,b^3-\frac {8\,A\,a^3}{3}+\frac {B\,a^3}{2}-8\,B\,b^3-16\,A\,a\,b^2+\frac {3\,A\,a^2\,b}{2}+6\,B\,a\,b^2-16\,B\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {116\,A\,a^3}{15}+20\,B\,a^2\,b+20\,A\,a\,b^2+12\,B\,b^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {8\,A\,a^3}{3}-2\,A\,b^3-\frac {B\,a^3}{2}-8\,B\,b^3-16\,A\,a\,b^2-\frac {3\,A\,a^2\,b}{2}-6\,B\,a\,b^2-16\,B\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,A\,a^3+A\,b^3+\frac {5\,B\,a^3}{4}+2\,B\,b^3+6\,A\,a\,b^2+\frac {15\,A\,a^2\,b}{4}+3\,B\,a\,b^2+6\,B\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*cos(c + d*x))*(a + b*cos(c + d*x))^3)/cos(c + d*x)^6,x)

[Out]

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

[Out]

Timed out

________________________________________________________________________________________