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

Optimal. Leaf size=188 \[ \frac {a \left (3 a^2 B+12 a b C+10 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac {\left (2 a^3 C+6 a^2 b B+9 a b^2 C+3 b^3 B\right ) \tan (c+d x)}{3 d}+\frac {\left (3 a^3 B+12 a^2 b C+12 a b^2 B+8 b^3 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d} \]

[Out]

1/8*(3*B*a^3+12*B*a*b^2+12*C*a^2*b+8*C*b^3)*arctanh(sin(d*x+c))/d+1/3*(6*B*a^2*b+3*B*b^3+2*C*a^3+9*C*a*b^2)*ta
n(d*x+c)/d+1/8*a*(3*B*a^2+10*B*b^2+12*C*a*b)*sec(d*x+c)*tan(d*x+c)/d+1/6*a^2*(3*B*b+2*C*a)*sec(d*x+c)^2*tan(d*
x+c)/d+1/4*a*B*(a+b*cos(d*x+c))^2*sec(d*x+c)^3*tan(d*x+c)/d

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Rubi [A]  time = 0.55, antiderivative size = 188, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3029, 2989, 3031, 3021, 2748, 3767, 8, 3770} \[ \frac {\left (6 a^2 b B+2 a^3 C+9 a b^2 C+3 b^3 B\right ) \tan (c+d x)}{3 d}+\frac {\left (12 a^2 b C+3 a^3 B+12 a b^2 B+8 b^3 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (3 a^2 B+12 a b C+10 b^2 B\right ) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a^2 (2 a C+3 b B) \tan (c+d x) \sec ^2(c+d x)}{6 d}+\frac {a B \tan (c+d x) \sec ^3(c+d x) (a+b \cos (c+d x))^2}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^3*B + 12*a*b^2*B + 12*a^2*b*C + 8*b^3*C)*ArcTanh[Sin[c + d*x]])/(8*d) + ((6*a^2*b*B + 3*b^3*B + 2*a^3*C
+ 9*a*b^2*C)*Tan[c + d*x])/(3*d) + (a*(3*a^2*B + 10*b^2*B + 12*a*b*C)*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (a^2*
(3*b*B + 2*a*C)*Sec[c + d*x]^2*Tan[c + d*x])/(6*d) + (a*B*(a + b*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/
(4*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 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 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 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 (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^6(c+d x) \, dx &=\int (a+b \cos (c+d x))^3 (B+C \cos (c+d x)) \sec ^5(c+d x) \, dx\\ &=\frac {a B (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+b \cos (c+d x)) \left (2 a (3 b B+2 a C)+\left (3 a^2 B+4 b^2 B+8 a b C\right ) \cos (c+d x)+b (a B+4 b C) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx\\ &=\frac {a^2 (3 b B+2 a C) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{12} \int \left (-3 a \left (3 a^2 B+10 b^2 B+12 a b C\right )-4 \left (6 a^2 b B+3 b^3 B+2 a^3 C+9 a b^2 C\right ) \cos (c+d x)-3 b^2 (a B+4 b C) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx\\ &=\frac {a \left (3 a^2 B+10 b^2 B+12 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (3 b B+2 a C) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{24} \int \left (-8 \left (6 a^2 b B+3 b^3 B+2 a^3 C+9 a b^2 C\right )-3 \left (3 a^3 B+12 a b^2 B+12 a^2 b C+8 b^3 C\right ) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {a \left (3 a^2 B+10 b^2 B+12 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (3 b B+2 a C) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {1}{3} \left (-6 a^2 b B-3 b^3 B-2 a^3 C-9 a b^2 C\right ) \int \sec ^2(c+d x) \, dx-\frac {1}{8} \left (-3 a^3 B-12 a b^2 B-12 a^2 b C-8 b^3 C\right ) \int \sec (c+d x) \, dx\\ &=\frac {\left (3 a^3 B+12 a b^2 B+12 a^2 b C+8 b^3 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a \left (3 a^2 B+10 b^2 B+12 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (3 b B+2 a C) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {\left (6 a^2 b B+3 b^3 B+2 a^3 C+9 a b^2 C\right ) \operatorname {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d}\\ &=\frac {\left (3 a^3 B+12 a b^2 B+12 a^2 b C+8 b^3 C\right ) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {\left (6 a^2 b B+3 b^3 B+2 a^3 C+9 a b^2 C\right ) \tan (c+d x)}{3 d}+\frac {a \left (3 a^2 B+10 b^2 B+12 a b C\right ) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (3 b B+2 a C) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a B (a+b \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}\\ \end {align*}

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Mathematica [A]  time = 0.81, size = 140, normalized size = 0.74 \[ \frac {3 \left (3 a^3 B+12 a^2 b C+12 a b^2 B+8 b^3 C\right ) \tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \left (6 a^3 B \sec ^3(c+d x)+9 a \left (a^2 B+4 a b C+4 b^2 B\right ) \sec (c+d x)+8 a^2 (a C+3 b B) \tan ^2(c+d x)+24 \left (a^3 C+3 a^2 b B+3 a b^2 C+b^3 B\right )\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(3*a^3*B + 12*a*b^2*B + 12*a^2*b*C + 8*b^3*C)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(24*(3*a^2*b*B + b^3*B +
a^3*C + 3*a*b^2*C) + 9*a*(a^2*B + 4*b^2*B + 4*a*b*C)*Sec[c + d*x] + 6*a^3*B*Sec[c + d*x]^3 + 8*a^2*(3*b*B + a
*C)*Tan[c + d*x]^2))/(24*d)

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fricas [A]  time = 0.47, size = 211, normalized size = 1.12 \[ \frac {3 \, {\left (3 \, B a^{3} + 12 \, C a^{2} b + 12 \, B a b^{2} + 8 \, C b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, B a^{3} + 12 \, C a^{2} b + 12 \, B a b^{2} + 8 \, C b^{3}\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, B a^{3} + 8 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 9 \, C a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{3} + 9 \, {\left (B a^{3} + 4 \, C a^{2} b + 4 \, B a b^{2}\right )} \cos \left (d x + c\right )^{2} + 8 \, {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(3*(3*B*a^3 + 12*C*a^2*b + 12*B*a*b^2 + 8*C*b^3)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*B*a^3 + 12*C
*a^2*b + 12*B*a*b^2 + 8*C*b^3)*cos(d*x + c)^4*log(-sin(d*x + c) + 1) + 2*(6*B*a^3 + 8*(2*C*a^3 + 6*B*a^2*b + 9
*C*a*b^2 + 3*B*b^3)*cos(d*x + c)^3 + 9*(B*a^3 + 4*C*a^2*b + 4*B*a*b^2)*cos(d*x + c)^2 + 8*(C*a^3 + 3*B*a^2*b)*
cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

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giac [B]  time = 0.43, size = 586, normalized size = 3.12 \[ \frac {3 \, {\left (3 \, B a^{3} + 12 \, C a^{2} b + 12 \, B a b^{2} + 8 \, C b^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, B a^{3} + 12 \, C a^{2} b + 12 \, B a b^{2} + 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 (15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 72 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 120 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 216 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 72 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 120 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 216 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 72 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, C a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 72 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, 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 )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*B*a^3 + 12*C*a^2*b + 12*B*a*b^2 + 8*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*B*a^3 + 12*C*a
^2*b + 12*B*a*b^2 + 8*C*b^3)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 24*C*a^
3*tan(1/2*d*x + 1/2*c)^7 - 72*B*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 36*B*a*b^2*
tan(1/2*d*x + 1/2*c)^7 - 72*C*a*b^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*b^3*tan(1/2*d*x + 1/2*c)^7 + 9*B*a^3*tan(1/2
*d*x + 1/2*c)^5 + 40*C*a^3*tan(1/2*d*x + 1/2*c)^5 + 120*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 36*C*a^2*b*tan(1/2*d*
x + 1/2*c)^5 - 36*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 216*C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 72*B*b^3*tan(1/2*d*x +
1/2*c)^5 + 9*B*a^3*tan(1/2*d*x + 1/2*c)^3 - 40*C*a^3*tan(1/2*d*x + 1/2*c)^3 - 120*B*a^2*b*tan(1/2*d*x + 1/2*c
)^3 - 36*C*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 36*B*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 216*C*a*b^2*tan(1/2*d*x + 1/2*c)
^3 - 72*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 15*B*a^3*tan(1/2*d*x + 1/2*c) + 24*C*a^3*tan(1/2*d*x + 1/2*c) + 72*B*a^
2*b*tan(1/2*d*x + 1/2*c) + 36*C*a^2*b*tan(1/2*d*x + 1/2*c) + 36*B*a*b^2*tan(1/2*d*x + 1/2*c) + 72*C*a*b^2*tan(
1/2*d*x + 1/2*c) + 24*B*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

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maple [A]  time = 0.41, size = 290, normalized size = 1.54 \[ \frac {2 C \,a^{3} \tan \left (d x +c \right )}{3 d}+\frac {C \,a^{3} \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{3 d}+\frac {a^{3} B \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{3} B \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {3 C \,a^{2} b \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {3 C \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {2 a^{2} b B \tan \left (d x +c \right )}{d}+\frac {a^{2} b B \tan \left (d x +c \right ) \left (\sec ^{2}\left (d x +c \right )\right )}{d}+\frac {3 C a \,b^{2} \tan \left (d x +c \right )}{d}+\frac {3 B a \,b^{2} \tan \left (d x +c \right ) \sec \left (d x +c \right )}{2 d}+\frac {3 B a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {b^{3} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {b^{3} B \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3/d*C*a^3*tan(d*x+c)+1/3/d*C*a^3*tan(d*x+c)*sec(d*x+c)^2+1/4/d*a^3*B*tan(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))+3/2/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^2*b*B*tan(d*x+c)+1/d*a^2*b*B*tan(d*x+c)*sec(d*x+c)^2+3/d*C*a*b^2*tan(d*x+c)+3/2/d
*B*a*b^2*tan(d*x+c)*sec(d*x+c)+3/2/d*B*a*b^2*ln(sec(d*x+c)+tan(d*x+c))+1/d*b^3*C*ln(sec(d*x+c)+tan(d*x+c))+1/d
*b^3*B*tan(d*x+c)

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maxima [A]  time = 0.34, size = 273, normalized size = 1.45 \[ \frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} C a^{3} + 48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} b - 3 \, B a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, C a^{2} b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a b^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, C b^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 144 \, C a b^{2} \tan \left (d x + c\right ) + 48 \, B b^{3} \tan \left (d x + c\right )}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*C*a^3 + 48*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2*b - 3*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(s
in(d*x + c) - 1)) - 36*C*a^2*b*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c)
- 1)) - 36*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)) + 24
*C*b^3*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 144*C*a*b^2*tan(d*x + c) + 48*B*b^3*tan(d*x + c))/d

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mupad [B]  time = 5.39, size = 395, normalized size = 2.10 \[ \frac {\mathrm {atanh}\left (\frac {4\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,B\,a^3}{8}+\frac {3\,C\,a^2\,b}{2}+\frac {3\,B\,a\,b^2}{2}+C\,b^3\right )}{\frac {3\,B\,a^3}{2}+6\,C\,a^2\,b+6\,B\,a\,b^2+4\,C\,b^3}\right )\,\left (\frac {3\,B\,a^3}{4}+3\,C\,a^2\,b+3\,B\,a\,b^2+2\,C\,b^3\right )}{d}-\frac {\left (2\,B\,b^3-\frac {5\,B\,a^3}{4}+2\,C\,a^3-3\,B\,a\,b^2+6\,B\,a^2\,b+6\,C\,a\,b^2-3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (3\,B\,a\,b^2-6\,B\,b^3-\frac {10\,C\,a^3}{3}-\frac {3\,B\,a^3}{4}-10\,B\,a^2\,b-18\,C\,a\,b^2+3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (6\,B\,b^3-\frac {3\,B\,a^3}{4}+\frac {10\,C\,a^3}{3}+3\,B\,a\,b^2+10\,B\,a^2\,b+18\,C\,a\,b^2+3\,C\,a^2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (-\frac {5\,B\,a^3}{4}-2\,B\,b^3-2\,C\,a^3-3\,B\,a\,b^2-6\,B\,a^2\,b-6\,C\,a\,b^2-3\,C\,a^2\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atanh((4*tan(c/2 + (d*x)/2)*((3*B*a^3)/8 + C*b^3 + (3*B*a*b^2)/2 + (3*C*a^2*b)/2))/((3*B*a^3)/2 + 4*C*b^3 + 6
*B*a*b^2 + 6*C*a^2*b))*((3*B*a^3)/4 + 2*C*b^3 + 3*B*a*b^2 + 3*C*a^2*b))/d - (tan(c/2 + (d*x)/2)^7*(2*B*b^3 - (
5*B*a^3)/4 + 2*C*a^3 - 3*B*a*b^2 + 6*B*a^2*b + 6*C*a*b^2 - 3*C*a^2*b) + tan(c/2 + (d*x)/2)^3*(6*B*b^3 - (3*B*a
^3)/4 + (10*C*a^3)/3 + 3*B*a*b^2 + 10*B*a^2*b + 18*C*a*b^2 + 3*C*a^2*b) - tan(c/2 + (d*x)/2)^5*((3*B*a^3)/4 +
6*B*b^3 + (10*C*a^3)/3 - 3*B*a*b^2 + 10*B*a^2*b + 18*C*a*b^2 - 3*C*a^2*b) - tan(c/2 + (d*x)/2)*((5*B*a^3)/4 +
2*B*b^3 + 2*C*a^3 + 3*B*a*b^2 + 6*B*a^2*b + 6*C*a*b^2 + 3*C*a^2*b))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (
d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1))

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

[Out]

Timed out

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