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

Optimal. Leaf size=192 \[ \frac {b \sin (c+d x) \left (-\left (a^2 (6 A-8 C)\right )+9 a b B+b^2 (3 A+2 C)\right )}{3 d}+\frac {a^2 (a B+3 A b) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {1}{2} x \left (2 a^3 C+6 a^2 b B+3 a b^2 (2 A+C)+b^3 B\right )-\frac {b^2 \sin (c+d x) \cos (c+d x) (6 a A-5 a C-3 b B)}{6 d}-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d} \]

[Out]

1/2*(6*a^2*b*B+b^3*B+2*a^3*C+3*a*b^2*(2*A+C))*x+a^2*(3*A*b+B*a)*arctanh(sin(d*x+c))/d+1/3*b*(9*a*b*B-a^2*(6*A-
8*C)+b^2*(3*A+2*C))*sin(d*x+c)/d-1/6*b^2*(6*A*a-3*B*b-5*C*a)*cos(d*x+c)*sin(d*x+c)/d-1/3*b*(3*A-C)*(a+b*cos(d*
x+c))^2*sin(d*x+c)/d+A*(a+b*cos(d*x+c))^3*tan(d*x+c)/d

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Rubi [A]  time = 0.59, antiderivative size = 192, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3047, 3049, 3033, 3023, 2735, 3770} \[ \frac {b \sin (c+d x) \left (a^2 (-(6 A-8 C))+9 a b B+b^2 (3 A+2 C)\right )}{3 d}+\frac {1}{2} x \left (6 a^2 b B+2 a^3 C+3 a b^2 (2 A+C)+b^3 B\right )+\frac {a^2 (a B+3 A b) \tanh ^{-1}(\sin (c+d x))}{d}-\frac {b^2 \sin (c+d x) \cos (c+d x) (6 a A-5 a C-3 b B)}{6 d}-\frac {b (3 A-C) \sin (c+d x) (a+b \cos (c+d x))^2}{3 d}+\frac {A \tan (c+d x) (a+b \cos (c+d x))^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((6*a^2*b*B + b^3*B + 2*a^3*C + 3*a*b^2*(2*A + C))*x)/2 + (a^2*(3*A*b + a*B)*ArcTanh[Sin[c + d*x]])/d + (b*(9*
a*b*B - a^2*(6*A - 8*C) + b^2*(3*A + 2*C))*Sin[c + d*x])/(3*d) - (b^2*(6*a*A - 3*b*B - 5*a*C)*Cos[c + d*x]*Sin
[c + d*x])/(6*d) - (b*(3*A - C)*(a + b*Cos[c + d*x])^2*Sin[c + d*x])/(3*d) + (A*(a + b*Cos[c + d*x])^3*Tan[c +
d*x])/d

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 3023

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

Rule 3033

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[(C*d*Cos[e + f*x]*Sin[e + f*x]*(a + b
*Sin[e + f*x])^(m + 1))/(b*f*(m + 3)), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*c*
(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e +
f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[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 3049

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] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e +
f*x])^m*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\int (a+b \cos (c+d x))^2 \left (3 A b+a B+(b B+a C) \cos (c+d x)-b (3 A-C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {1}{3} \int (a+b \cos (c+d x)) \left (3 a (3 A b+a B)+\left (3 A b^2+6 a b B+3 a^2 C+2 b^2 C\right ) \cos (c+d x)-b (6 a A-3 b B-5 a C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b^2 (6 a A-3 b B-5 a C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {1}{6} \int \left (6 a^2 (3 A b+a B)+3 \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 (2 A+C)\right ) \cos (c+d x)+2 b \left (9 a b B-a^2 (6 A-8 C)+b^2 (3 A+2 C)\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {b \left (9 a b B-a^2 (6 A-8 C)+b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}-\frac {b^2 (6 a A-3 b B-5 a C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\frac {1}{6} \int \left (6 a^2 (3 A b+a B)+3 \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 (2 A+C)\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{2} \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 (2 A+C)\right ) x+\frac {b \left (9 a b B-a^2 (6 A-8 C)+b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}-\frac {b^2 (6 a A-3 b B-5 a C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}+\left (a^2 (3 A b+a B)\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} \left (6 a^2 b B+b^3 B+2 a^3 C+3 a b^2 (2 A+C)\right ) x+\frac {a^2 (3 A b+a B) \tanh ^{-1}(\sin (c+d x))}{d}+\frac {b \left (9 a b B-a^2 (6 A-8 C)+b^2 (3 A+2 C)\right ) \sin (c+d x)}{3 d}-\frac {b^2 (6 a A-3 b B-5 a C) \cos (c+d x) \sin (c+d x)}{6 d}-\frac {b (3 A-C) (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {A (a+b \cos (c+d x))^3 \tan (c+d x)}{d}\\ \end {align*}

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Mathematica [A]  time = 1.30, size = 266, normalized size = 1.39 \[ \frac {\frac {12 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {12 a^3 A \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+3 b \sin (c+d x) \left (3 \left (4 a^2 C+4 a b B+b^2 C\right )+4 A b^2\right )-12 a^2 (a B+3 A b) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^2 (a B+3 A b) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+6 (c+d x) \left (2 a^3 C+6 a^2 b B+3 a b^2 (2 A+C)+b^3 B\right )+3 b^2 (3 a C+b B) \sin (2 (c+d x))+b^3 C \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(6*a^2*b*B + b^3*B + 2*a^3*C + 3*a*b^2*(2*A + C))*(c + d*x) - 12*a^2*(3*A*b + a*B)*Log[Cos[(c + d*x)/2] - S
in[(c + d*x)/2]] + 12*a^2*(3*A*b + a*B)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + (12*a^3*A*Sin[(c + d*x)/2])
/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + (12*a^3*A*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 3
*b*(4*A*b^2 + 3*(4*a*b*B + 4*a^2*C + b^2*C))*Sin[c + d*x] + 3*b^2*(b*B + 3*a*C)*Sin[2*(c + d*x)] + b^3*C*Sin[3
*(c + d*x)])/(12*d)

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fricas [A]  time = 0.46, size = 201, normalized size = 1.05 \[ \frac {3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 3 \, {\left (2 \, A + C\right )} a b^{2} + B b^{3}\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (2 \, C b^{3} \cos \left (d x + c\right )^{3} + 6 \, A a^{3} + 3 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (9 \, C a^{2} b + 9 \, B a b^{2} + {\left (3 \, A + 2 \, C\right )} b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )} \]

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)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="fricas")

[Out]

1/6*(3*(2*C*a^3 + 6*B*a^2*b + 3*(2*A + C)*a*b^2 + B*b^3)*d*x*cos(d*x + c) + 3*(B*a^3 + 3*A*a^2*b)*cos(d*x + c)
*log(sin(d*x + c) + 1) - 3*(B*a^3 + 3*A*a^2*b)*cos(d*x + c)*log(-sin(d*x + c) + 1) + (2*C*b^3*cos(d*x + c)^3 +
6*A*a^3 + 3*(3*C*a*b^2 + B*b^3)*cos(d*x + c)^2 + 2*(9*C*a^2*b + 9*B*a*b^2 + (3*A + 2*C)*b^3)*cos(d*x + c))*si
n(d*x + c))/(d*cos(d*x + c))

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giac [B]  time = 1.83, size = 418, normalized size = 2.18 \[ -\frac {\frac {12 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - 3 \, {\left (2 \, C a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2} + 3 \, C a b^{2} + B b^{3}\right )} {\left (d x + c\right )} - 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 6 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, C a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

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)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="giac")

[Out]

-1/6*(12*A*a^3*tan(1/2*d*x + 1/2*c)/(tan(1/2*d*x + 1/2*c)^2 - 1) - 3*(2*C*a^3 + 6*B*a^2*b + 6*A*a*b^2 + 3*C*a*
b^2 + B*b^3)*(d*x + c) - 6*(B*a^3 + 3*A*a^2*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) + 6*(B*a^3 + 3*A*a^2*b)*log(
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(18*C*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 9*
C*a*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*A*b^3*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^3*tan
(1/2*d*x + 1/2*c)^5 + 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 + 12*A*b^3*tan(1/2
*d*x + 1/2*c)^3 + 4*C*b^3*tan(1/2*d*x + 1/2*c)^3 + 18*C*a^2*b*tan(1/2*d*x + 1/2*c) + 18*B*a*b^2*tan(1/2*d*x +
1/2*c) + 9*C*a*b^2*tan(1/2*d*x + 1/2*c) + 6*A*b^3*tan(1/2*d*x + 1/2*c) + 3*B*b^3*tan(1/2*d*x + 1/2*c) + 6*C*b^
3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 0.30, size = 278, normalized size = 1.45 \[ \frac {A \,a^{3} \tan \left (d x +c \right )}{d}+\frac {a^{3} B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+a^{3} C x +\frac {C \,a^{3} c}{d}+\frac {3 A \,a^{2} b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+3 B x \,a^{2} b +\frac {3 B \,a^{2} b c}{d}+\frac {3 C \,a^{2} b \sin \left (d x +c \right )}{d}+3 A x a \,b^{2}+\frac {3 A a \,b^{2} c}{d}+\frac {3 B a \,b^{2} \sin \left (d x +c \right )}{d}+\frac {3 C a \,b^{2} \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {3 a \,b^{2} C x}{2}+\frac {3 C a \,b^{2} c}{2 d}+\frac {A \,b^{3} \sin \left (d x +c \right )}{d}+\frac {b^{3} B \cos \left (d x +c \right ) \sin \left (d x +c \right )}{2 d}+\frac {b^{3} B x}{2}+\frac {b^{3} B c}{2 d}+\frac {C \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right ) b^{3}}{3 d}+\frac {2 b^{3} C \sin \left (d x +c \right )}{3 d} \]

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)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x)

[Out]

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

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maxima [A]  time = 0.36, size = 216, normalized size = 1.12 \[ \frac {12 \, {\left (d x + c\right )} C a^{3} + 36 \, {\left (d x + c\right )} B a^{2} b + 36 \, {\left (d x + c\right )} A a b^{2} + 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b^{2} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{3} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{3} + 6 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, A a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, C a^{2} b \sin \left (d x + c\right ) + 36 \, B a b^{2} \sin \left (d x + c\right ) + 12 \, A b^{3} \sin \left (d x + c\right ) + 12 \, A a^{3} \tan \left (d x + c\right )}{12 \, d} \]

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)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x, algorithm="maxima")

[Out]

1/12*(12*(d*x + c)*C*a^3 + 36*(d*x + c)*B*a^2*b + 36*(d*x + c)*A*a*b^2 + 9*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*
a*b^2 + 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*b^3 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*b^3 + 6*B*a^3*(log(si
n(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 18*A*a^2*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 36*C*a
^2*b*sin(d*x + c) + 36*B*a*b^2*sin(d*x + c) + 12*A*b^3*sin(d*x + c) + 12*A*a^3*tan(d*x + c))/d

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mupad [B]  time = 4.00, size = 2470, normalized size = 12.86 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(2*A*a^3 + 2*A*b^3 + B*b^3 + 2*C*b^3 + 6*B*a*b^2 + 3*C*a*b^2 + 6*C*a^2*b) - tan(c/2 + (d*x
)/2)^7*(2*A*b^3 - 2*A*a^3 - B*b^3 + 2*C*b^3 + 6*B*a*b^2 - 3*C*a*b^2 + 6*C*a^2*b) + tan(c/2 + (d*x)/2)^3*(6*A*a
^3 + 2*A*b^3 - B*b^3 - (2*C*b^3)/3 + 6*B*a*b^2 - 3*C*a*b^2 + 6*C*a^2*b) - tan(c/2 + (d*x)/2)^5*(2*A*b^3 - 6*A*
a^3 + B*b^3 - (2*C*b^3)/3 + 6*B*a*b^2 + 3*C*a*b^2 + 6*C*a^2*b))/(d*(2*tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)
/2)^6 - tan(c/2 + (d*x)/2)^8 + 1)) - (atan((((B*a^3 + 3*A*a^2*b)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*A*a*b^2
+ 96*A*a^2*b + 96*B*a^2*b + 48*C*a*b^2) + tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^
2*b^4 + 288*A^2*a^4*b^2 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 96*A*B*a*b^5 +
192*A*B*a^5*b + 48*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a
^3*b^3))*(B*a^3 + 3*A*a^2*b)*1i - ((B*a^3 + 3*A*a^2*b)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*A*a*b^2 + 96*A*a^2
*b + 96*B*a^2*b + 48*C*a*b^2) - tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^2*b^4 + 28
8*A^2*a^4*b^2 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 96*A*B*a*b^5 + 192*A*B*a^
5*b + 48*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a^3*b^3))*(
B*a^3 + 3*A*a^2*b)*1i)/(((B*a^3 + 3*A*a^2*b)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*A*a*b^2 + 96*A*a^2*b + 96*B*
a^2*b + 48*C*a*b^2) + tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^2*b^4 + 288*A^2*a^4*
b^2 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 96*A*B*a*b^5 + 192*A*B*a^5*b + 48*B
*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a^3*b^3))*(B*a^3 + 3*
A*a^2*b) + ((B*a^3 + 3*A*a^2*b)*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*A*a*b^2 + 96*A*a^2*b + 96*B*a^2*b + 48*C*
a*b^2) - tan(c/2 + (d*x)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 96*B^2*
a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 96*A*B*a*b^5 + 192*A*B*a^5*b + 48*B*C*a*b^5 + 19
2*B*C*a^5*b + 576*A*B*a^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a^3*b^3))*(B*a^3 + 3*A*a^2*b) + 64
*B*C^2*a^9 - 64*B^2*C*a^9 - 192*B^3*a^8*b + 1728*A^3*a^4*b^5 - 1728*A^3*a^5*b^4 + 16*B^3*a^3*b^6 + 192*B^3*a^5
*b^4 - 32*B^3*a^6*b^3 + 576*B^3*a^7*b^2 + 192*A*C^2*a^8*b + 384*B^2*C*a^8*b + 48*A*B^2*a^2*b^7 + 768*A*B^2*a^4
*b^5 - 192*A*B^2*a^5*b^4 + 2880*A*B^2*a^6*b^3 - 1344*A*B^2*a^7*b^2 + 576*A^2*B*a^3*b^6 - 288*A^2*B*a^4*b^5 + 4
032*A^2*B*a^5*b^4 - 2880*A^2*B*a^6*b^3 + 432*A*C^2*a^4*b^5 + 576*A*C^2*a^6*b^3 + 1728*A^2*C*a^4*b^5 - 864*A^2*
C*a^5*b^4 + 1152*A^2*C*a^6*b^3 - 576*A^2*C*a^7*b^2 + 144*B*C^2*a^5*b^4 + 192*B*C^2*a^7*b^2 + 96*B^2*C*a^4*b^5
+ 640*B^2*C*a^6*b^3 - 96*B^2*C*a^7*b^2 - 384*A*B*C*a^8*b + 288*A*B*C*a^3*b^6 + 2496*A*B*C*a^5*b^4 - 576*A*B*C*
a^6*b^3 + 1536*A*B*C*a^7*b^2))*(B*a^3*2i + A*a^2*b*6i))/d + (atanh((2*tan(c/2 + (d*x)/2)*((B*b^3*1i)/2 + C*a^3
*1i + A*a*b^2*3i + B*a^2*b*3i + (C*a*b^2*3i)/2)*(32*B^2*a^6 + 8*B^2*b^6 + 32*C^2*a^6 + 288*A^2*a^2*b^4 + 288*A
^2*a^4*b^2 + 96*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 72*C^2*a^2*b^4 + 96*C^2*a^4*b^2 + 96*A*B*a*b^5 + 192*A*B*a^5*b
+ 48*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 288*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 320*B*C*a^3*b^3))/(2*(
(B*b^3*1i)/2 + C*a^3*1i + A*a*b^2*3i + B*a^2*b*3i + (C*a*b^2*3i)/2)^2*(32*B*a^3 + 16*B*b^3 + 32*C*a^3 + 96*A*a
*b^2 + 96*A*a^2*b + 96*B*a^2*b + 48*C*a*b^2) + 64*B*C^2*a^9 - 64*B^2*C*a^9 - 192*B^3*a^8*b + 1728*A^3*a^4*b^5
- 1728*A^3*a^5*b^4 + 16*B^3*a^3*b^6 + 192*B^3*a^5*b^4 - 32*B^3*a^6*b^3 + 576*B^3*a^7*b^2 + 192*A*C^2*a^8*b + 3
84*B^2*C*a^8*b + 48*A*B^2*a^2*b^7 + 768*A*B^2*a^4*b^5 - 192*A*B^2*a^5*b^4 + 2880*A*B^2*a^6*b^3 - 1344*A*B^2*a^
7*b^2 + 576*A^2*B*a^3*b^6 - 288*A^2*B*a^4*b^5 + 4032*A^2*B*a^5*b^4 - 2880*A^2*B*a^6*b^3 + 432*A*C^2*a^4*b^5 +
576*A*C^2*a^6*b^3 + 1728*A^2*C*a^4*b^5 - 864*A^2*C*a^5*b^4 + 1152*A^2*C*a^6*b^3 - 576*A^2*C*a^7*b^2 + 144*B*C^
2*a^5*b^4 + 192*B*C^2*a^7*b^2 + 96*B^2*C*a^4*b^5 + 640*B^2*C*a^6*b^3 - 96*B^2*C*a^7*b^2 - 384*A*B*C*a^8*b + 28
8*A*B*C*a^3*b^6 + 2496*A*B*C*a^5*b^4 - 576*A*B*C*a^6*b^3 + 1536*A*B*C*a^7*b^2))*(B*b^3*1i + C*a^3*2i + A*a*b^2
*6i + B*a^2*b*6i + C*a*b^2*3i))/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((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

Timed out

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