\(\int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^2(c+d x) \, dx\) [313]

Optimal result

Integrand size = 41, antiderivative size = 121 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {1}{2} a^2 (2 A+4 B+3 C) x+\frac {a^2 (2 A+B) \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 (2 A-2 B-3 C) \sin (c+d x)}{2 d}-\frac {(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {A (a+a \cos (c+d x))^2 \tan (c+d x)}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {3122, 3055, 3047, 3102, 2814, 3855} \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a^2 (2 A+B) \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 (2 A-2 B-3 C) \sin (c+d x)}{2 d}+\frac {1}{2} a^2 x (2 A+4 B+3 C)-\frac {(2 A-C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{2 d}+\frac {A \tan (c+d x) (a \cos (c+d x)+a)^2}{d} \]

[In]

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

[Out]

(a^2*(2*A + 4*B + 3*C)*x)/2 + (a^2*(2*A + B)*ArcTanh[Sin[c + d*x]])/d - (a^2*(2*A - 2*B - 3*C)*Sin[c + d*x])/(
2*d) - ((2*A - C)*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(2*d) + (A*(a + a*Cos[c + d*x])^2*Tan[c + d*x])/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 3047

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

Rule 3055

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 - 1)*(c + d*Sin[e + f*
x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))
*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &
& NeQ[c^2 - d^2, 0] && GtQ[m, 1/2] &&  !LtQ[n, -1] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[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 3122

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

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 {A (a+a \cos (c+d x))^2 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x))^2 (a (2 A+B)-a (2 A-C) \cos (c+d x)) \sec (c+d x) \, dx}{a} \\ & = -\frac {(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {A (a+a \cos (c+d x))^2 \tan (c+d x)}{d}+\frac {\int (a+a \cos (c+d x)) \left (2 a^2 (2 A+B)-a^2 (2 A-2 B-3 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a} \\ & = -\frac {(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {A (a+a \cos (c+d x))^2 \tan (c+d x)}{d}+\frac {\int \left (2 a^3 (2 A+B)+\left (2 a^3 (2 A+B)-a^3 (2 A-2 B-3 C)\right ) \cos (c+d x)-a^3 (2 A-2 B-3 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{2 a} \\ & = -\frac {a^2 (2 A-2 B-3 C) \sin (c+d x)}{2 d}-\frac {(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {A (a+a \cos (c+d x))^2 \tan (c+d x)}{d}+\frac {\int \left (2 a^3 (2 A+B)+a^3 (2 A+4 B+3 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{2 a} \\ & = \frac {1}{2} a^2 (2 A+4 B+3 C) x-\frac {a^2 (2 A-2 B-3 C) \sin (c+d x)}{2 d}-\frac {(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {A (a+a \cos (c+d x))^2 \tan (c+d x)}{d}+\left (a^2 (2 A+B)\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a^2 (2 A+4 B+3 C) x+\frac {a^2 (2 A+B) \text {arctanh}(\sin (c+d x))}{d}-\frac {a^2 (2 A-2 B-3 C) \sin (c+d x)}{2 d}-\frac {(2 A-C) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{2 d}+\frac {A (a+a \cos (c+d x))^2 \tan (c+d x)}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.38 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.44 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {a^2 \left (4 A c+8 B c+6 c C+4 A d x+8 B d x+6 C d x-8 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+8 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+4 (B+2 C) \sin (c+d x)+C \sin (2 (c+d x))+4 A \tan (c+d x)\right )}{4 d} \]

[In]

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

[Out]

(a^2*(4*A*c + 8*B*c + 6*c*C + 4*A*d*x + 8*B*d*x + 6*C*d*x - 8*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - 4*B
*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 8*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] + 4*B*Log[Cos[(c + d*
x)/2] + Sin[(c + d*x)/2]] + 4*(B + 2*C)*Sin[c + d*x] + C*Sin[2*(c + d*x)] + 4*A*Tan[c + d*x]))/(4*d)

Maple [A] (verified)

Time = 5.51 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.02

method	result	size
parts	\(\frac {a^{2} A \tan \left (d x +c \right )}{d}+\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (B \,a^{2}+2 a^{2} C \right ) \sin \left (d x +c \right )}{d}+\frac {\left (A \,a^{2}+2 B \,a^{2}+a^{2} C \right ) \left (d x +c \right )}{d}+\frac {a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}\)	\(124\)
parallelrisch	\(-\frac {2 \left (\cos \left (d x +c \right ) \left (A +\frac {B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-\cos \left (d x +c \right ) \left (A +\frac {B}{2}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-\frac {B}{4}-\frac {C}{2}\right ) \sin \left (2 d x +2 c \right )-\frac {\sin \left (3 d x +3 c \right ) C}{16}-\frac {x d \left (A +2 B +\frac {3 C}{2}\right ) \cos \left (d x +c \right )}{2}-\frac {\sin \left (d x +c \right ) \left (A +\frac {C}{8}\right )}{2}\right ) a^{2}}{d \cos \left (d x +c \right )}\)	\(127\)
derivativedivides	\(\frac {A \,a^{2} \left (d x +c \right )+B \sin \left (d x +c \right ) a^{2}+a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B \,a^{2} \left (d x +c \right )+2 a^{2} C \sin \left (d x +c \right )+A \,a^{2} \tan \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \left (d x +c \right )}{d}\)	\(137\)
default	\(\frac {A \,a^{2} \left (d x +c \right )+B \sin \left (d x +c \right ) a^{2}+a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+2 B \,a^{2} \left (d x +c \right )+2 a^{2} C \sin \left (d x +c \right )+A \,a^{2} \tan \left (d x +c \right )+B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{2} C \left (d x +c \right )}{d}\)	\(137\)
risch	\(a^{2} x A +2 a^{2} B x +\frac {3 a^{2} C x}{2}-\frac {i a^{2} C \,{\mathrm e}^{2 i \left (d x +c \right )}}{8 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} B \,a^{2}}{2 d}-\frac {i {\mathrm e}^{i \left (d x +c \right )} a^{2} C}{d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} B \,a^{2}}{2 d}+\frac {i {\mathrm e}^{-i \left (d x +c \right )} a^{2} C}{d}+\frac {i a^{2} C \,{\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}+\frac {2 i A \,a^{2}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}-\frac {2 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}+\frac {2 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}\)	\(246\)
norman	\(\frac {\left (-A \,a^{2}-2 B \,a^{2}-\frac {3}{2} a^{2} C \right ) x +\left (-3 A \,a^{2}-6 B \,a^{2}-\frac {9}{2} a^{2} C \right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (A \,a^{2}+2 B \,a^{2}+\frac {3}{2} a^{2} C \right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (3 A \,a^{2}+6 B \,a^{2}+\frac {9}{2} a^{2} C \right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-2 A \,a^{2}-4 B \,a^{2}-3 a^{2} C \right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (2 A \,a^{2}+4 B \,a^{2}+3 a^{2} C \right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 a^{2} \left (6 A -C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (2 A -2 B -3 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (2 A -B -2 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (2 A +B +2 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {a^{2} \left (2 A +2 B +5 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}+\frac {a^{2} \left (2 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} \left (2 A +B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(401\)

[In]

int((a+cos(d*x+c)*a)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.07 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {{\left (2 \, A + 4 \, B + 3 \, C\right )} a^{2} d x \cos \left (d x + c\right ) + {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (C a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, A a^{2}\right )} \sin \left (d x + c\right )}{2 \, d \cos \left (d x + c\right )} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=a^{2} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 A \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int A \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \cos {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 B \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 2 C \cos ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

[In]

integrate((a+a*cos(d*x+c))**2*(A+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**2,x)

[Out]

a**2*(Integral(A*sec(c + d*x)**2, x) + Integral(2*A*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(A*cos(c + d*x)
**2*sec(c + d*x)**2, x) + Integral(B*cos(c + d*x)*sec(c + d*x)**2, x) + Integral(2*B*cos(c + d*x)**2*sec(c + d
*x)**2, x) + Integral(B*cos(c + d*x)**3*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**2*sec(c + d*x)**2, x) +
Integral(2*C*cos(c + d*x)**3*sec(c + d*x)**2, x) + Integral(C*cos(c + d*x)**4*sec(c + d*x)**2, x))

Maxima [A] (verification not implemented)

none

Time = 0.21 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.25 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {4 \, {\left (d x + c\right )} A a^{2} + 8 \, {\left (d x + c\right )} B a^{2} + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 4 \, {\left (d x + c\right )} C a^{2} + 4 \, A a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 2 \, B a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 4 \, B a^{2} \sin \left (d x + c\right ) + 8 \, C a^{2} \sin \left (d x + c\right ) + 4 \, A a^{2} \tan \left (d x + c\right )}{4 \, d} \]

[In]

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

[Out]

1/4*(4*(d*x + c)*A*a^2 + 8*(d*x + c)*B*a^2 + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 + 4*(d*x + c)*C*a^2 + 4*A*
a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 2*B*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1))
+ 4*B*a^2*sin(d*x + c) + 8*C*a^2*sin(d*x + c) + 4*A*a^2*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.37 (sec) , antiderivative size = 198, normalized size of antiderivative = 1.64 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=-\frac {\frac {4 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} - {\left (2 \, A a^{2} + 4 \, B a^{2} + 3 \, C a^{2}\right )} {\left (d x + c\right )} - 2 \, {\left (2 \, A a^{2} + B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) + 2 \, {\left (2 \, A a^{2} + B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 1.95 (sec) , antiderivative size = 232, normalized size of antiderivative = 1.92 \[ \int (a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx=\frac {2\,A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+4\,B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+3\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )-A\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,4{}\mathrm {i}-B\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,1{}\mathrm {i}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )\,2{}\mathrm {i}}{d}+\frac {\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2}+C\,a^2\,\sin \left (2\,c+2\,d\,x\right )+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{8}+A\,a^2\,\sin \left (c+d\,x\right )+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{8}}{d\,\cos \left (c+d\,x\right )} \]

[In]

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

[Out]

(2*A*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - A*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/2))*
4i + 4*B*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) - B*a^2*atan((sin(c/2 + (d*x)/2)*1i)/cos(c/2 + (d*x)/
2))*2i + 3*C*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + ((B*a^2*sin(2*c + 2*d*x))/2 + C*a^2*sin(2*c
+ 2*d*x) + (C*a^2*sin(3*c + 3*d*x))/8 + A*a^2*sin(c + d*x) + (C*a^2*sin(c + d*x))/8)/(d*cos(c + d*x))