\(\int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx\) [234]

Optimal result

Integrand size = 29, antiderivative size = 137 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {1}{2} \left (6 a^2 A b+A b^3+2 a^3 B+3 a b^2 B\right ) x+\frac {a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {b \left (9 a A b+8 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 d}+\frac {b^2 (3 A b+5 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {b B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.36 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3069, 3112, 3102, 2814, 3855} \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {b \left (8 a^2 B+9 a A b+2 b^2 B\right ) \sin (c+d x)}{3 d}+\frac {1}{2} x \left (2 a^3 B+6 a^2 A b+3 a b^2 B+A b^3\right )+\frac {b^2 (5 a B+3 A b) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {b B \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

[In]

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

[Out]

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

Rule 2814

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 3069

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*
x])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f
*x])^n*Simp[a^2*A*d*(m + n + 1) + b*B*(b*c*(m - 1) + a*d*(n + 1)) + (a*d*(2*A*b + a*B)*(m + n + 1) - b*B*(a*c
- b*d*(m + n)))*Sin[e + f*x] + b*(A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))*Sin[e + f*x]^2, x], x], x] /;
FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m
, 1] &&  !(IGtQ[n, 1] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3102

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 3112

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 3855

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = \frac {b B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{3} \int (a+b \cos (c+d x)) \left (3 a^2 A+\left (6 a A b+3 a^2 B+2 b^2 B\right ) \cos (c+d x)+b (3 A b+5 a B) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = \frac {b^2 (3 A b+5 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {b B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{6} \int \left (6 a^3 A+3 \left (6 a^2 A b+A b^3+2 a^3 B+3 a b^2 B\right ) \cos (c+d x)+2 b \left (9 a A b+8 a^2 B+2 b^2 B\right ) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx \\ & = \frac {b \left (9 a A b+8 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 d}+\frac {b^2 (3 A b+5 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {b B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{6} \int \left (6 a^3 A+3 \left (6 a^2 A b+A b^3+2 a^3 B+3 a b^2 B\right ) \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = \frac {1}{2} \left (6 a^2 A b+A b^3+2 a^3 B+3 a b^2 B\right ) x+\frac {b \left (9 a A b+8 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 d}+\frac {b^2 (3 A b+5 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {b B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d}+\left (a^3 A\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} \left (6 a^2 A b+A b^3+2 a^3 B+3 a b^2 B\right ) x+\frac {a^3 A \text {arctanh}(\sin (c+d x))}{d}+\frac {b \left (9 a A b+8 a^2 B+2 b^2 B\right ) \sin (c+d x)}{3 d}+\frac {b^2 (3 A b+5 a B) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {b B (a+b \cos (c+d x))^2 \sin (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.70 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.16 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {6 \left (6 a^2 A b+A b^3+2 a^3 B+3 a b^2 B\right ) (c+d x)-12 a^3 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+12 a^3 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+9 b \left (4 a A b+4 a^2 B+b^2 B\right ) \sin (c+d x)+3 b^2 (A b+3 a B) \sin (2 (c+d x))+b^3 B \sin (3 (c+d x))}{12 d} \]

[In]

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

[Out]

(6*(6*a^2*A*b + A*b^3 + 2*a^3*B + 3*a*b^2*B)*(c + d*x) - 12*a^3*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 1
2*a^3*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 9*b*(4*a*A*b + 4*a^2*B + b^2*B)*Sin[c + d*x] + 3*b^2*(A*b +
3*a*B)*Sin[2*(c + d*x)] + b^3*B*Sin[3*(c + d*x)])/(12*d)

Maple [A] (verified)

Time = 2.92 (sec) , antiderivative size = 135, normalized size of antiderivative = 0.99

method	result	size
parts	\(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (A \,b^{3}+3 B a \,b^{2}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (3 A a \,b^{2}+3 B \,a^{2} b \right ) \sin \left (d x +c \right )}{d}+\frac {\left (3 A \,a^{2} b +B \,a^{3}\right ) \left (d x +c \right )}{d}+\frac {B \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}\)	\(135\)
parallelrisch	\(\frac {-12 A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+12 A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+3 \left (A \,b^{3}+3 B a \,b^{2}\right ) \sin \left (2 d x +2 c \right )+B \sin \left (3 d x +3 c \right ) b^{3}+9 \left (4 A a \,b^{2}+4 B \,a^{2} b +B \,b^{3}\right ) \sin \left (d x +c \right )+36 x \left (A \,a^{2} b +\frac {1}{6} A \,b^{3}+\frac {1}{3} B \,a^{3}+\frac {1}{2} B a \,b^{2}\right ) d}{12 d}\)	\(139\)
derivativedivides	\(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \left (d x +c \right )+3 A \,a^{2} b \left (d x +c \right )+3 B \sin \left (d x +c \right ) a^{2} b +3 A \sin \left (d x +c \right ) a \,b^{2}+3 B a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(151\)
default	\(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \left (d x +c \right )+3 A \,a^{2} b \left (d x +c \right )+3 B \sin \left (d x +c \right ) a^{2} b +3 A \sin \left (d x +c \right ) a \,b^{2}+3 B a \,b^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+A \,b^{3} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+\frac {B \,b^{3} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}}{d}\)	\(151\)
risch	\(3 x A \,a^{2} b +\frac {x A \,b^{3}}{2}+a^{3} B x +\frac {3 x B a \,b^{2}}{2}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} A a \,b^{2}}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2} b}{2 d}-\frac {3 i {\mathrm e}^{i \left (d x +c \right )} B \,b^{3}}{8 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} A a \,b^{2}}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2} b}{2 d}+\frac {3 i {\mathrm e}^{-i \left (d x +c \right )} B \,b^{3}}{8 d}+\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (3 d x +3 c \right ) B \,b^{3}}{12 d}+\frac {\sin \left (2 d x +2 c \right ) A \,b^{3}}{4 d}+\frac {3 \sin \left (2 d x +2 c \right ) B a \,b^{2}}{4 d}\)	\(247\)
norman	\(\frac {\left (3 A \,a^{2} b +\frac {1}{2} A \,b^{3}+B \,a^{3}+\frac {3}{2} B a \,b^{2}\right ) x +\left (3 A \,a^{2} b +\frac {1}{2} A \,b^{3}+B \,a^{3}+\frac {3}{2} B a \,b^{2}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 A \,a^{2} b +2 A \,b^{3}+4 B \,a^{3}+6 B a \,b^{2}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (12 A \,a^{2} b +2 A \,b^{3}+4 B \,a^{3}+6 B a \,b^{2}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (18 A \,a^{2} b +3 A \,b^{3}+6 B \,a^{3}+9 B a \,b^{2}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {b \left (6 A a b -A \,b^{2}+6 B \,a^{2}-3 B a b +2 B \,b^{2}\right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {b \left (6 A a b +A \,b^{2}+6 B \,a^{2}+3 B a b +2 B \,b^{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {b \left (54 A a b -3 A \,b^{2}+54 B \,a^{2}-9 B a b +10 B \,b^{2}\right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {b \left (54 A a b +3 A \,b^{2}+54 B \,a^{2}+9 B a b +10 B \,b^{2}\right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4}}+\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {A \,a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(426\)

[In]

int((a+cos(d*x+c)*b)^3*(A+B*cos(d*x+c))*sec(d*x+c),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 131, normalized size of antiderivative = 0.96 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {3 \, A a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, A a^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (2 \, B a^{3} + 6 \, A a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} d x + {\left (2 \, B b^{3} \cos \left (d x + c\right )^{2} + 18 \, B a^{2} b + 18 \, A a b^{2} + 4 \, B b^{3} + 3 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \, d} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\int \left (A + B \cos {\left (c + d x \right )}\right ) \left (a + b \cos {\left (c + d x \right )}\right )^{3} \sec {\left (c + d x \right )}\, dx \]

[In]

integrate((a+b*cos(d*x+c))**3*(A+B*cos(d*x+c))*sec(d*x+c),x)

[Out]

Integral((A + B*cos(c + d*x))*(a + b*cos(c + d*x))**3*sec(c + d*x), x)

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.06 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {12 \, {\left (d x + c\right )} B a^{3} + 36 \, {\left (d x + c\right )} A a^{2} b + 9 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b^{2} + 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{3} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B b^{3} + 12 \, A a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) + 36 \, B a^{2} b \sin \left (d x + c\right ) + 36 \, A a b^{2} \sin \left (d x + c\right )}{12 \, d} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (129) = 258\).

Time = 0.33 (sec) , antiderivative size = 314, normalized size of antiderivative = 2.29 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\frac {6 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 6 \, A a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 3 \, {\left (2 \, B a^{3} + 6 \, A a^{2} b + 3 \, B a b^{2} + A b^{3}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (18 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 18 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 36 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 36 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, B a^{2} b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, A a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 9 \, B a b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, B b^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

[In]

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

[Out]

1/6*(6*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 6*A*a^3*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 3*(2*B*a^3 + 6*
A*a^2*b + 3*B*a*b^2 + A*b^3)*(d*x + c) + 2*(18*B*a^2*b*tan(1/2*d*x + 1/2*c)^5 + 18*A*a*b^2*tan(1/2*d*x + 1/2*c
)^5 - 9*B*a*b^2*tan(1/2*d*x + 1/2*c)^5 - 3*A*b^3*tan(1/2*d*x + 1/2*c)^5 + 6*B*b^3*tan(1/2*d*x + 1/2*c)^5 + 36*
B*a^2*b*tan(1/2*d*x + 1/2*c)^3 + 36*A*a*b^2*tan(1/2*d*x + 1/2*c)^3 + 4*B*b^3*tan(1/2*d*x + 1/2*c)^3 + 18*B*a^2
*b*tan(1/2*d*x + 1/2*c) + 18*A*a*b^2*tan(1/2*d*x + 1/2*c) + 9*B*a*b^2*tan(1/2*d*x + 1/2*c) + 3*A*b^3*tan(1/2*d
*x + 1/2*c) + 6*B*b^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

Mupad [B] (verification not implemented)

Time = 2.04 (sec) , antiderivative size = 1924, normalized size of antiderivative = 14.04 \[ \int (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \sec (c+d x) \, dx=\text {Too large to display} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*(A*b^3 + 2*B*b^3 + 6*A*a*b^2 + 3*B*a*b^2 + 6*B*a^2*b) + tan(c/2 + (d*x)/2)^3*((4*B*b^3)/3
+ 12*A*a*b^2 + 12*B*a^2*b) + tan(c/2 + (d*x)/2)^5*(2*B*b^3 - A*b^3 + 6*A*a*b^2 - 3*B*a*b^2 + 6*B*a^2*b))/(d*(3
*tan(c/2 + (d*x)/2)^2 + 3*tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^6 + 1)) + (atan(((((A*b^3*1i)/2 + B*a^3*1i
+ A*a^2*b*3i + (B*a*b^2*3i)/2)*(32*A*a^3 + 16*A*b^3 + 32*B*a^3 + 96*A*a^2*b + 48*B*a*b^2) + tan(c/2 + (d*x)/2
)*(32*A^2*a^6 + 8*A^2*b^6 + 32*B^2*a^6 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 72*B^2*a^2*b^4 + 96*B^2*a^4*b^2 +
48*A*B*a*b^5 + 192*A*B*a^5*b + 320*A*B*a^3*b^3))*((A*b^3*1i)/2 + B*a^3*1i + A*a^2*b*3i + (B*a*b^2*3i)/2)*1i -
(((A*b^3*1i)/2 + B*a^3*1i + A*a^2*b*3i + (B*a*b^2*3i)/2)*(32*A*a^3 + 16*A*b^3 + 32*B*a^3 + 96*A*a^2*b + 48*B*a
*b^2) - tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + 32*B^2*a^6 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 72*B^2*a^
2*b^4 + 96*B^2*a^4*b^2 + 48*A*B*a*b^5 + 192*A*B*a^5*b + 320*A*B*a^3*b^3))*((A*b^3*1i)/2 + B*a^3*1i + A*a^2*b*3
i + (B*a*b^2*3i)/2)*1i)/((((A*b^3*1i)/2 + B*a^3*1i + A*a^2*b*3i + (B*a*b^2*3i)/2)*(32*A*a^3 + 16*A*b^3 + 32*B*
a^3 + 96*A*a^2*b + 48*B*a*b^2) + tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + 32*B^2*a^6 + 96*A^2*a^2*b^4 + 28
8*A^2*a^4*b^2 + 72*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 48*A*B*a*b^5 + 192*A*B*a^5*b + 320*A*B*a^3*b^3))*((A*b^3*1i)
/2 + B*a^3*1i + A*a^2*b*3i + (B*a*b^2*3i)/2) + (((A*b^3*1i)/2 + B*a^3*1i + A*a^2*b*3i + (B*a*b^2*3i)/2)*(32*A*
a^3 + 16*A*b^3 + 32*B*a^3 + 96*A*a^2*b + 48*B*a*b^2) - tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + 32*B^2*a^6
+ 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 72*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 48*A*B*a*b^5 + 192*A*B*a^5*b + 320*A*B
*a^3*b^3))*((A*b^3*1i)/2 + B*a^3*1i + A*a^2*b*3i + (B*a*b^2*3i)/2) + 64*A*B^2*a^9 - 64*A^2*B*a^9 - 192*A^3*a^8
*b + 16*A^3*a^3*b^6 + 192*A^3*a^5*b^4 - 32*A^3*a^6*b^3 + 576*A^3*a^7*b^2 + 384*A^2*B*a^8*b + 144*A*B^2*a^5*b^4
+ 192*A*B^2*a^7*b^2 + 96*A^2*B*a^4*b^5 + 640*A^2*B*a^6*b^3 - 96*A^2*B*a^7*b^2))*(A*b^3 + 2*B*a^3 + 6*A*a^2*b
+ 3*B*a*b^2))/d - (A*a^3*atan((A*a^3*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + 32*B^2*a^6 + 96*A^2*a^2*b^4
+ 288*A^2*a^4*b^2 + 72*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 48*A*B*a*b^5 + 192*A*B*a^5*b + 320*A*B*a^3*b^3) + A*a^3
*(32*A*a^3 + 16*A*b^3 + 32*B*a^3 + 96*A*a^2*b + 48*B*a*b^2))*1i + A*a^3*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^
2*b^6 + 32*B^2*a^6 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 72*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 48*A*B*a*b^5 + 192*A
*B*a^5*b + 320*A*B*a^3*b^3) - A*a^3*(32*A*a^3 + 16*A*b^3 + 32*B*a^3 + 96*A*a^2*b + 48*B*a*b^2))*1i)/(64*A*B^2*
a^9 - 64*A^2*B*a^9 - 192*A^3*a^8*b + A*a^3*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8*A^2*b^6 + 32*B^2*a^6 + 96*A^2*a
^2*b^4 + 288*A^2*a^4*b^2 + 72*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 48*A*B*a*b^5 + 192*A*B*a^5*b + 320*A*B*a^3*b^3) +
A*a^3*(32*A*a^3 + 16*A*b^3 + 32*B*a^3 + 96*A*a^2*b + 48*B*a*b^2)) - A*a^3*(tan(c/2 + (d*x)/2)*(32*A^2*a^6 + 8
*A^2*b^6 + 32*B^2*a^6 + 96*A^2*a^2*b^4 + 288*A^2*a^4*b^2 + 72*B^2*a^2*b^4 + 96*B^2*a^4*b^2 + 48*A*B*a*b^5 + 19
2*A*B*a^5*b + 320*A*B*a^3*b^3) - A*a^3*(32*A*a^3 + 16*A*b^3 + 32*B*a^3 + 96*A*a^2*b + 48*B*a*b^2)) + 16*A^3*a^
3*b^6 + 192*A^3*a^5*b^4 - 32*A^3*a^6*b^3 + 576*A^3*a^7*b^2 + 384*A^2*B*a^8*b + 144*A*B^2*a^5*b^4 + 192*A*B^2*a
^7*b^2 + 96*A^2*B*a^4*b^5 + 640*A^2*B*a^6*b^3 - 96*A^2*B*a^7*b^2))*2i)/d