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

Optimal. Leaf size=204 \[ -\frac {b \sin (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{2 d}+\frac {a \left (a^2 (A+2 C)+6 a b B+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {1}{2} b x \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right )-\frac {b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{2 d}+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d} \]

[Out]

1/2*b*(2*A*b^2+6*B*a*b+6*C*a^2+C*b^2)*x+1/2*a*(6*A*b^2+6*a*b*B+a^2*(A+2*C))*arctanh(sin(d*x+c))/d-1/2*b*(9*A*a
*b+4*B*a^2-2*B*b^2-6*C*a*b)*sin(d*x+c)/d-1/2*b^2*(4*A*b+2*B*a-C*b)*cos(d*x+c)*sin(d*x+c)/d+1/2*(3*A*b+2*B*a)*(
a+b*cos(d*x+c))^2*tan(d*x+c)/d+1/2*A*(a+b*cos(d*x+c))^3*sec(d*x+c)*tan(d*x+c)/d

________________________________________________________________________________________

Rubi [A]  time = 0.65, antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.122, Rules used = {3047, 3033, 3023, 2735, 3770} \[ -\frac {b \sin (c+d x) \left (4 a^2 B+9 a A b-6 a b C-2 b^2 B\right )}{2 d}+\frac {a \left (a^2 (A+2 C)+6 a b B+6 A b^2\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac {1}{2} b x \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right )-\frac {b^2 \sin (c+d x) \cos (c+d x) (2 a B+4 A b-b C)}{2 d}+\frac {(2 a B+3 A b) \tan (c+d x) (a+b \cos (c+d x))^2}{2 d}+\frac {A \tan (c+d x) \sec (c+d x) (a+b \cos (c+d x))^3}{2 d} \]

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]^3,x]

[Out]

(b*(2*A*b^2 + 6*a*b*B + 6*a^2*C + b^2*C)*x)/2 + (a*(6*A*b^2 + 6*a*b*B + a^2*(A + 2*C))*ArcTanh[Sin[c + d*x]])/
(2*d) - (b*(9*a*A*b + 4*a^2*B - 2*b^2*B - 6*a*b*C)*Sin[c + d*x])/(2*d) - (b^2*(4*A*b + 2*a*B - b*C)*Cos[c + d*
x]*Sin[c + d*x])/(2*d) + ((3*A*b + 2*a*B)*(a + b*Cos[c + d*x])^2*Tan[c + d*x])/(2*d) + (A*(a + b*Cos[c + d*x])
^3*Sec[c + d*x]*Tan[c + d*x])/(2*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 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 ^3(c+d x) \, dx &=\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+b \cos (c+d x))^2 \left (3 A b+2 a B+(2 b B+a (A+2 C)) \cos (c+d x)-2 b (A-C) \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx\\ &=\frac {(3 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \int (a+b \cos (c+d x)) \left (6 A b^2+6 a b B+a^2 (A+2 C)-b (a A-2 b B-4 a C) \cos (c+d x)-2 b (4 A b+2 a B-b C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b^2 (4 A b+2 a B-b C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {(3 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{4} \int \left (2 a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right )+2 b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) \cos (c+d x)-2 b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx\\ &=-\frac {b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \sin (c+d x)}{2 d}-\frac {b^2 (4 A b+2 a B-b C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {(3 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{4} \int \left (2 a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right )+2 b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx\\ &=\frac {1}{2} b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) x-\frac {b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \sin (c+d x)}{2 d}-\frac {b^2 (4 A b+2 a B-b C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {(3 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}+\frac {1}{2} \left (a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right )\right ) \int \sec (c+d x) \, dx\\ &=\frac {1}{2} b \left (2 A b^2+6 a b B+6 a^2 C+b^2 C\right ) x+\frac {a \left (6 A b^2+6 a b B+a^2 (A+2 C)\right ) \tanh ^{-1}(\sin (c+d x))}{2 d}-\frac {b \left (9 a A b+4 a^2 B-2 b^2 B-6 a b C\right ) \sin (c+d x)}{2 d}-\frac {b^2 (4 A b+2 a B-b C) \cos (c+d x) \sin (c+d x)}{2 d}+\frac {(3 A b+2 a B) (a+b \cos (c+d x))^2 \tan (c+d x)}{2 d}+\frac {A (a+b \cos (c+d x))^3 \sec (c+d x) \tan (c+d x)}{2 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 3.21, size = 318, normalized size = 1.56 \[ \frac {\frac {a^3 A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {a^3 A}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+2 b (c+d x) \left (6 a^2 C+6 a b B+2 A b^2+b^2 C\right )-2 a \left (a^2 (A+2 C)+6 a b B+6 A b^2\right ) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+2 a \left (a^2 (A+2 C)+6 a b B+6 A b^2\right ) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {4 a^2 (a B+3 A b) \sin \left (\frac {1}{2} (c+d x)\right )}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {4 a^2 (a B+3 A b) \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}+4 b^2 (3 a C+b B) \sin (c+d x)+b^3 C \sin (2 (c+d x))}{4 d} \]

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]^3,x]

[Out]

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

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 208, normalized size = 1.02 \[ \frac {2 \, {\left (6 \, C a^{2} b + 6 \, B a b^{2} + {\left (2 \, A + C\right )} b^{3}\right )} d x \cos \left (d x + c\right )^{2} + {\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + 2 \, C\right )} a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (C b^{3} \cos \left (d x + c\right )^{3} + A a^{3} + 2 \, {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{2} + 2 \, {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]

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)^3,x, algorithm="fricas")

[Out]

1/4*(2*(6*C*a^2*b + 6*B*a*b^2 + (2*A + C)*b^3)*d*x*cos(d*x + c)^2 + ((A + 2*C)*a^3 + 6*B*a^2*b + 6*A*a*b^2)*co
s(d*x + c)^2*log(sin(d*x + c) + 1) - ((A + 2*C)*a^3 + 6*B*a^2*b + 6*A*a*b^2)*cos(d*x + c)^2*log(-sin(d*x + c)
+ 1) + 2*(C*b^3*cos(d*x + c)^3 + A*a^3 + 2*(3*C*a*b^2 + B*b^3)*cos(d*x + c)^2 + 2*(B*a^3 + 3*A*a^2*b)*cos(d*x
+ c))*sin(d*x + c))/(d*cos(d*x + c)^2)

________________________________________________________________________________________

giac [B]  time = 0.31, size = 538, normalized size = 2.64 \[ \frac {{\left (6 \, C a^{2} b + 6 \, B a b^{2} + 2 \, A b^{3} + C b^{3}\right )} {\left (d x + c\right )} + {\left (A a^{3} + 2 \, C a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - {\left (A a^{3} + 2 \, C a^{3} + 6 \, B a^{2} b + 6 \, A a b^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 3 \, C b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, A a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 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 )^{4} - 1\right )}^{2}}}{2 \, 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)^3,x, algorithm="giac")

[Out]

1/2*((6*C*a^2*b + 6*B*a*b^2 + 2*A*b^3 + C*b^3)*(d*x + c) + (A*a^3 + 2*C*a^3 + 6*B*a^2*b + 6*A*a*b^2)*log(abs(t
an(1/2*d*x + 1/2*c) + 1)) - (A*a^3 + 2*C*a^3 + 6*B*a^2*b + 6*A*a*b^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(
A*a^3*tan(1/2*d*x + 1/2*c)^7 - 2*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^7 + 6*C*a*b^2*t
an(1/2*d*x + 1/2*c)^7 + 2*B*b^3*tan(1/2*d*x + 1/2*c)^7 - C*b^3*tan(1/2*d*x + 1/2*c)^7 + 3*A*a^3*tan(1/2*d*x +
1/2*c)^5 - 2*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^5 - 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^
5 - 2*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*C*b^3*tan(1/2*d*x + 1/2*c)^5 + 3*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 2*B*a^3
*tan(1/2*d*x + 1/2*c)^3 + 6*A*a^2*b*tan(1/2*d*x + 1/2*c)^3 - 6*C*a*b^2*tan(1/2*d*x + 1/2*c)^3 - 2*B*b^3*tan(1/
2*d*x + 1/2*c)^3 - 3*C*b^3*tan(1/2*d*x + 1/2*c)^3 + A*a^3*tan(1/2*d*x + 1/2*c) + 2*B*a^3*tan(1/2*d*x + 1/2*c)
+ 6*A*a^2*b*tan(1/2*d*x + 1/2*c) + 6*C*a*b^2*tan(1/2*d*x + 1/2*c) + 2*B*b^3*tan(1/2*d*x + 1/2*c) + C*b^3*tan(1
/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^4 - 1)^2)/d

________________________________________________________________________________________

maple [A]  time = 0.33, size = 267, normalized size = 1.31 \[ \frac {A \,a^{3} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a^{3} B \tan \left (d x +c \right )}{d}+\frac {C \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 A \,a^{2} b \tan \left (d x +c \right )}{d}+\frac {3 a^{2} b B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+3 C \,a^{2} b x +\frac {3 C \,a^{2} b c}{d}+\frac {3 A a \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+3 B x a \,b^{2}+\frac {3 B a \,b^{2} c}{d}+\frac {3 C \,b^{2} a \sin \left (d x +c \right )}{d}+A x \,b^{3}+\frac {A \,b^{3} c}{d}+\frac {b^{3} B \sin \left (d x +c \right )}{d}+\frac {b^{3} C \sin \left (d x +c \right ) \cos \left (d x +c \right )}{2 d}+\frac {b^{3} C x}{2}+\frac {b^{3} C c}{2 d} \]

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)^3,x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.37, size = 243, normalized size = 1.19 \[ \frac {12 \, {\left (d x + c\right )} C a^{2} b + 12 \, {\left (d x + c\right )} B a b^{2} + 4 \, {\left (d x + c\right )} A b^{3} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b^{3} - A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, C a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, B a^{2} b {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a b^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 12 \, C a b^{2} \sin \left (d x + c\right ) + 4 \, B b^{3} \sin \left (d x + c\right ) + 4 \, B a^{3} \tan \left (d x + c\right ) + 12 \, A a^{2} b \tan \left (d x + c\right )}{4 \, d} \]

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)^3,x, algorithm="maxima")

[Out]

1/4*(12*(d*x + c)*C*a^2*b + 12*(d*x + c)*B*a*b^2 + 4*(d*x + c)*A*b^3 + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*b^3
- A*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 2*C*a^3*(log(s
in(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*B*a^2*b*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 6*A*a*
b^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 12*C*a*b^2*sin(d*x + c) + 4*B*b^3*sin(d*x + c) + 4*B*a^3
*tan(d*x + c) + 12*A*a^2*b*tan(d*x + c))/d

________________________________________________________________________________________

mupad [B]  time = 4.65, size = 3879, normalized size = 19.01 \[ \text {result too large to display} \]

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)^3,x)

[Out]

(tan(c/2 + (d*x)/2)^7*(A*a^3 - 2*B*a^3 + 2*B*b^3 - C*b^3 - 6*A*a^2*b + 6*C*a*b^2) + tan(c/2 + (d*x)/2)^3*(3*A*
a^3 + 2*B*a^3 - 2*B*b^3 - 3*C*b^3 + 6*A*a^2*b - 6*C*a*b^2) - tan(c/2 + (d*x)/2)^5*(2*B*a^3 - 3*A*a^3 + 2*B*b^3
- 3*C*b^3 + 6*A*a^2*b + 6*C*a*b^2) + tan(c/2 + (d*x)/2)*(A*a^3 + 2*B*a^3 + 2*B*b^3 + C*b^3 + 6*A*a^2*b + 6*C*
a*b^2))/(d*(tan(c/2 + (d*x)/2)^8 - 2*tan(c/2 + (d*x)/2)^4 + 1)) - (atan(((((A*a^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B
*a^2*b)*(16*A*a^3 + 32*A*b^3 + 32*C*a^3 + 16*C*b^3 + 96*A*a*b^2 + 96*B*a*b^2 + 96*B*a^2*b + 96*C*a^2*b) + tan(
c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A^2*b^6 + 32*C^2*a^6 + 8*C^2*b^6 + 288*A^2*a^2*b^4 + 96*A^2*a^4*b^2 + 288*B^2*a
^2*b^4 + 288*B^2*a^4*b^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*a^6 + 32*A*C*b^6 + 192*A*B*a*b^5 + 96*A*B
*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 576*B*C*a^3*b^3)
)*((A*a^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2*b)*1i - (((A*a^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2*b)*(16*A*a^3 + 32
*A*b^3 + 32*C*a^3 + 16*C*b^3 + 96*A*a*b^2 + 96*B*a*b^2 + 96*B*a^2*b + 96*C*a^2*b) - tan(c/2 + (d*x)/2)*(8*A^2*
a^6 + 32*A^2*b^6 + 32*C^2*a^6 + 8*C^2*b^6 + 288*A^2*a^2*b^4 + 96*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 288*B^2*a^4*b
^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*a^6 + 32*A*C*b^6 + 192*A*B*a*b^5 + 96*A*B*a^5*b + 96*B*C*a*b^5
+ 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 576*B*C*a^3*b^3))*((A*a^3)/2 + C*a^3 +
3*A*a*b^2 + 3*B*a^2*b)*1i)/((((A*a^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2*b)*(16*A*a^3 + 32*A*b^3 + 32*C*a^3 + 16
*C*b^3 + 96*A*a*b^2 + 96*B*a*b^2 + 96*B*a^2*b + 96*C*a^2*b) + tan(c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A^2*b^6 + 32*
C^2*a^6 + 8*C^2*b^6 + 288*A^2*a^2*b^4 + 96*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 96*C^2*a^2*b^4 +
288*C^2*a^4*b^2 + 32*A*C*a^6 + 32*A*C*b^6 + 192*A*B*a*b^5 + 96*A*B*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^5*b + 576*
A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 576*B*C*a^3*b^3))*((A*a^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2*b
) + (((A*a^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2*b)*(16*A*a^3 + 32*A*b^3 + 32*C*a^3 + 16*C*b^3 + 96*A*a*b^2 + 96*
B*a*b^2 + 96*B*a^2*b + 96*C*a^2*b) - tan(c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A^2*b^6 + 32*C^2*a^6 + 8*C^2*b^6 + 288
*A^2*a^2*b^4 + 96*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*
a^6 + 32*A*C*b^6 + 192*A*B*a*b^5 + 96*A*B*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 192*A*C*a^2
*b^4 + 192*A*C*a^4*b^2 + 576*B*C*a^3*b^3))*((A*a^3)/2 + C*a^3 + 3*A*a*b^2 + 3*B*a^2*b) + 192*A^3*a*b^8 - 192*C
^3*a^8*b - 576*A^3*a^2*b^7 + 32*A^3*a^3*b^6 - 192*A^3*a^4*b^5 - 16*A^3*a^6*b^3 + 1728*B^3*a^4*b^5 - 1728*B^3*a
^5*b^4 + 16*C^3*a^3*b^6 + 192*C^3*a^5*b^4 - 32*C^3*a^6*b^3 + 576*C^3*a^7*b^2 + 48*A*C^2*a*b^8 - 192*A*C^2*a^8*
b + 192*A^2*C*a*b^8 - 48*A^2*C*a^8*b + 2880*A*B^2*a^3*b^6 - 4032*A*B^2*a^4*b^5 + 288*A*B^2*a^5*b^4 - 576*A*B^2
*a^6*b^3 + 1344*A^2*B*a^2*b^7 - 2880*A^2*B*a^3*b^6 + 192*A^2*B*a^4*b^5 - 768*A^2*B*a^5*b^4 - 48*A^2*B*a^7*b^2
+ 648*A*C^2*a^3*b^6 - 192*A*C^2*a^4*b^5 + 2208*A*C^2*a^5*b^4 - 1248*A*C^2*a^6*b^3 + 288*A*C^2*a^7*b^2 - 288*A^
2*C*a^2*b^7 + 1248*A^2*C*a^3*b^6 - 2208*A^2*C*a^4*b^5 + 192*A^2*C*a^5*b^4 - 648*A^2*C*a^6*b^3 + 48*B*C^2*a^2*b
^7 + 768*B*C^2*a^4*b^5 - 192*B*C^2*a^5*b^4 + 2880*B*C^2*a^6*b^3 - 1344*B*C^2*a^7*b^2 + 576*B^2*C*a^3*b^6 - 288
*B^2*C*a^4*b^5 + 4032*B^2*C*a^5*b^4 - 2880*B^2*C*a^6*b^3 + 768*A*B*C*a^2*b^7 - 576*A*B*C*a^3*b^6 + 5088*A*B*C*
a^4*b^5 - 5088*A*B*C*a^5*b^4 + 576*A*B*C*a^6*b^3 - 768*A*B*C*a^7*b^2))*(A*a^3*1i + C*a^3*2i + A*a*b^2*6i + B*a
^2*b*6i))/d - (b*atan(((b*(tan(c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A^2*b^6 + 32*C^2*a^6 + 8*C^2*b^6 + 288*A^2*a^2*b
^4 + 96*A^2*a^4*b^2 + 288*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*a^6 + 32*A
*C*b^6 + 192*A*B*a*b^5 + 96*A*B*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 192
*A*C*a^4*b^2 + 576*B*C*a^3*b^3) - (b*(2*A*b^2 + 6*C*a^2 + C*b^2 + 6*B*a*b)*(16*A*a^3 + 32*A*b^3 + 32*C*a^3 + 1
6*C*b^3 + 96*A*a*b^2 + 96*B*a*b^2 + 96*B*a^2*b + 96*C*a^2*b)*1i)/2)*(2*A*b^2 + 6*C*a^2 + C*b^2 + 6*B*a*b))/2 +
(b*(tan(c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A^2*b^6 + 32*C^2*a^6 + 8*C^2*b^6 + 288*A^2*a^2*b^4 + 96*A^2*a^4*b^2 +
288*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*a^6 + 32*A*C*b^6 + 192*A*B*a*b^5
+ 96*A*B*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 576*B*C
*a^3*b^3) + (b*(2*A*b^2 + 6*C*a^2 + C*b^2 + 6*B*a*b)*(16*A*a^3 + 32*A*b^3 + 32*C*a^3 + 16*C*b^3 + 96*A*a*b^2 +
96*B*a*b^2 + 96*B*a^2*b + 96*C*a^2*b)*1i)/2)*(2*A*b^2 + 6*C*a^2 + C*b^2 + 6*B*a*b))/2)/(192*A^3*a*b^8 - 192*C
^3*a^8*b - (b*(tan(c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A^2*b^6 + 32*C^2*a^6 + 8*C^2*b^6 + 288*A^2*a^2*b^4 + 96*A^2*
a^4*b^2 + 288*B^2*a^2*b^4 + 288*B^2*a^4*b^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*a^6 + 32*A*C*b^6 + 192
*A*B*a*b^5 + 96*A*B*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 192*A*C*a^4*b^2
+ 576*B*C*a^3*b^3) - (b*(2*A*b^2 + 6*C*a^2 + C*b^2 + 6*B*a*b)*(16*A*a^3 + 32*A*b^3 + 32*C*a^3 + 16*C*b^3 + 96
*A*a*b^2 + 96*B*a*b^2 + 96*B*a^2*b + 96*C*a^2*b)*1i)/2)*(2*A*b^2 + 6*C*a^2 + C*b^2 + 6*B*a*b)*1i)/2 + (b*(tan(
c/2 + (d*x)/2)*(8*A^2*a^6 + 32*A^2*b^6 + 32*C^2*a^6 + 8*C^2*b^6 + 288*A^2*a^2*b^4 + 96*A^2*a^4*b^2 + 288*B^2*a
^2*b^4 + 288*B^2*a^4*b^2 + 96*C^2*a^2*b^4 + 288*C^2*a^4*b^2 + 32*A*C*a^6 + 32*A*C*b^6 + 192*A*B*a*b^5 + 96*A*B
*a^5*b + 96*B*C*a*b^5 + 192*B*C*a^5*b + 576*A*B*a^3*b^3 + 192*A*C*a^2*b^4 + 192*A*C*a^4*b^2 + 576*B*C*a^3*b^3)
+ (b*(2*A*b^2 + 6*C*a^2 + C*b^2 + 6*B*a*b)*(16*A*a^3 + 32*A*b^3 + 32*C*a^3 + 16*C*b^3 + 96*A*a*b^2 + 96*B*a*b
^2 + 96*B*a^2*b + 96*C*a^2*b)*1i)/2)*(2*A*b^2 + 6*C*a^2 + C*b^2 + 6*B*a*b)*1i)/2 - 576*A^3*a^2*b^7 + 32*A^3*a^
3*b^6 - 192*A^3*a^4*b^5 - 16*A^3*a^6*b^3 + 1728*B^3*a^4*b^5 - 1728*B^3*a^5*b^4 + 16*C^3*a^3*b^6 + 192*C^3*a^5*
b^4 - 32*C^3*a^6*b^3 + 576*C^3*a^7*b^2 + 48*A*C^2*a*b^8 - 192*A*C^2*a^8*b + 192*A^2*C*a*b^8 - 48*A^2*C*a^8*b +
2880*A*B^2*a^3*b^6 - 4032*A*B^2*a^4*b^5 + 288*A*B^2*a^5*b^4 - 576*A*B^2*a^6*b^3 + 1344*A^2*B*a^2*b^7 - 2880*A
^2*B*a^3*b^6 + 192*A^2*B*a^4*b^5 - 768*A^2*B*a^5*b^4 - 48*A^2*B*a^7*b^2 + 648*A*C^2*a^3*b^6 - 192*A*C^2*a^4*b^
5 + 2208*A*C^2*a^5*b^4 - 1248*A*C^2*a^6*b^3 + 288*A*C^2*a^7*b^2 - 288*A^2*C*a^2*b^7 + 1248*A^2*C*a^3*b^6 - 220
8*A^2*C*a^4*b^5 + 192*A^2*C*a^5*b^4 - 648*A^2*C*a^6*b^3 + 48*B*C^2*a^2*b^7 + 768*B*C^2*a^4*b^5 - 192*B*C^2*a^5
*b^4 + 2880*B*C^2*a^6*b^3 - 1344*B*C^2*a^7*b^2 + 576*B^2*C*a^3*b^6 - 288*B^2*C*a^4*b^5 + 4032*B^2*C*a^5*b^4 -
2880*B^2*C*a^6*b^3 + 768*A*B*C*a^2*b^7 - 576*A*B*C*a^3*b^6 + 5088*A*B*C*a^4*b^5 - 5088*A*B*C*a^5*b^4 + 576*A*B
*C*a^6*b^3 - 768*A*B*C*a^7*b^2))*(2*A*b^2 + 6*C*a^2 + C*b^2 + 6*B*a*b))/d

________________________________________________________________________________________

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)**3,x)

[Out]

Timed out

________________________________________________________________________________________