\(\int (a+a \cos (c+d x))^2 (A+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [15]

Optimal result

Integrand size = 33, antiderivative size = 110 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=a^2 C x+\frac {a^2 (A+2 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.39 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {3123, 3054, 3047, 3100, 2814, 3855} \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a^2 (A+2 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {A \tan (c+d x) \sec (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{3 d}+a^2 C x+\frac {A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[In]

Int[(a + a*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

a^2*C*x + (a^2*(A + 2*C)*ArcTanh[Sin[c + d*x]])/d + (a^2*(A + C)*Tan[c + d*x])/d + (A*(a^2 + a^2*Cos[c + d*x])
*Sec[c + d*x]*Tan[c + d*x])/(3*d) + (A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^2*Tan[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 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 3054

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^2)*(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)*(b*c + a*d))), x] - Dist[b/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x
])^(m - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[a*A*d*(m - n - 2) - B*(a*c*(m - 1) + b*d*(n + 1)) - (A*b*d*(m + n
+ 1) - B*(b*c*m - a*d*(n + 1)))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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 3100

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 3123

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (C_.)*s
in[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(c^2*C + A*d^2))*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Si
n[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*(a*c*m + b*d*(n + 1)) - b*(A*d^2*(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, 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 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x))^2 (2 a A+3 a C \cos (c+d x)) \sec ^3(c+d x) \, dx}{3 a} \\ & = \frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int (a+a \cos (c+d x)) \left (6 a^2 (A+C)+6 a^2 C \cos (c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a} \\ & = \frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \left (6 a^3 (A+C)+\left (6 a^3 C+6 a^3 (A+C)\right ) \cos (c+d x)+6 a^3 C \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx}{6 a} \\ & = \frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {\int \left (6 a^3 (A+2 C)+6 a^3 C \cos (c+d x)\right ) \sec (c+d x) \, dx}{6 a} \\ & = a^2 C x+\frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\left (a^2 (A+2 C)\right ) \int \sec (c+d x) \, dx \\ & = a^2 C x+\frac {a^2 (A+2 C) \text {arctanh}(\sin (c+d x))}{d}+\frac {a^2 (A+C) \tan (c+d x)}{d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{3 d}+\frac {A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.55 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {a^2 \left (3 C d x+3 (A+2 C) \text {arctanh}(\sin (c+d x))+3 (2 A+C+A \sec (c+d x)) \tan (c+d x)+A \tan ^3(c+d x)\right )}{3 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^2*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]

[Out]

(a^2*(3*C*d*x + 3*(A + 2*C)*ArcTanh[Sin[c + d*x]] + 3*(2*A + C + A*Sec[c + d*x])*Tan[c + d*x] + A*Tan[c + d*x]
^3))/(3*d)

Maple [A] (verified)

Time = 7.72 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {A \,a^{2} \tan \left (d x +c \right )+a^{2} C \left (d x +c \right )+2 A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{2} C \tan \left (d x +c \right )}{d}\)	\(119\)
default	\(\frac {A \,a^{2} \tan \left (d x +c \right )+a^{2} C \left (d x +c \right )+2 A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )-A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+a^{2} C \tan \left (d x +c \right )}{d}\)	\(119\)
parts	\(-\frac {A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,a^{2}+a^{2} C \right ) \tan \left (d x +c \right )}{d}+\frac {a^{2} C \left (d x +c \right )}{d}+\frac {a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{d}+\frac {A \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {2 a^{2} C \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}\)	\(128\)
parallelrisch	\(\frac {2 \left (-\frac {3 \left (A +2 C \right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {3 \left (A +2 C \right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {d x C \cos \left (3 d x +3 c \right )}{2}+\left (\frac {5 A}{6}+\frac {C}{2}\right ) \sin \left (3 d x +3 c \right )+\sin \left (2 d x +2 c \right ) A +\frac {3 d x C \cos \left (d x +c \right )}{2}+\frac {3 \left (A +\frac {C}{3}\right ) \sin \left (d x +c \right )}{2}\right ) a^{2}}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(170\)
risch	\(a^{2} C x -\frac {2 i a^{2} \left (3 A \,{\mathrm e}^{5 i \left (d x +c \right )}-3 A \,{\mathrm e}^{4 i \left (d x +c \right )}-3 C \,{\mathrm e}^{4 i \left (d x +c \right )}-12 A \,{\mathrm e}^{2 i \left (d x +c \right )}-6 C \,{\mathrm e}^{2 i \left (d x +c \right )}-3 A \,{\mathrm e}^{i \left (d x +c \right )}-5 A -3 C \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}+\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}+\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) C}{d}-\frac {A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}-\frac {2 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) C}{d}\)	\(196\)
norman	\(\frac {a^{2} C x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{2} C x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a^{2} C x +\frac {8 a^{2} C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-a^{2} C x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{2} C x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{2} C x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{2} C x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{2} C x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {2 a^{2} \left (A +C \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{2} \left (2 A +3 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (3 A +C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {2 a^{2} \left (5 A +3 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {4 a^{2} \left (14 A +3 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {2 a^{2} \left (25 A -3 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}+\frac {a^{2} \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{2} \left (A +2 C \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(382\)

[In]

int((a+cos(d*x+c)*a)^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(A*a^2*tan(d*x+c)+a^2*C*(d*x+c)+2*A*a^2*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+2*a^2*C*
ln(sec(d*x+c)+tan(d*x+c))-A*a^2*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+a^2*C*tan(d*x+c))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 131, normalized size of antiderivative = 1.19 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {6 \, C a^{2} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A + 2 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left ({\left (5 \, A + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 3 \, A a^{2} \cos \left (d x + c\right ) + A a^{2}\right )} \sin \left (d x + c\right )}{6 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/6*(6*C*a^2*d*x*cos(d*x + c)^3 + 3*(A + 2*C)*a^2*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(A + 2*C)*a^2*cos(d
*x + c)^3*log(-sin(d*x + c) + 1) + 2*((5*A + 3*C)*a^2*cos(d*x + c)^2 + 3*A*a^2*cos(d*x + c) + A*a^2)*sin(d*x +
c))/(d*cos(d*x + c)^3)

Sympy [F(-1)]

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

[In]

integrate((a+a*cos(d*x+c))**2*(A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.25 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {2 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 6 \, {\left (d x + c\right )} C a^{2} - 3 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, C a^{2} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a^{2} \tan \left (d x + c\right ) + 6 \, C a^{2} \tan \left (d x + c\right )}{6 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

1/6*(2*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 6*(d*x + c)*C*a^2 - 3*A*a^2*(2*sin(d*x + c)/(sin(d*x + c)^2 -
1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1)) + 6*C*a^2*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1))
+ 6*A*a^2*tan(d*x + c) + 6*C*a^2*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 187, normalized size of antiderivative = 1.70 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {3 \, {\left (d x + c\right )} C a^{2} + 3 \, {\left (A a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (A a^{2} + 2 \, C a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 8 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{3 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^2*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)*C*a^2 + 3*(A*a^2 + 2*C*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(A*a^2 + 2*C*a^2)*log(abs(
tan(1/2*d*x + 1/2*c) - 1)) - 2*(3*A*a^2*tan(1/2*d*x + 1/2*c)^5 + 3*C*a^2*tan(1/2*d*x + 1/2*c)^5 - 8*A*a^2*tan(
1/2*d*x + 1/2*c)^3 - 6*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 9*A*a^2*tan(1/2*d*x + 1/2*c) + 3*C*a^2*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 = 1.22 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.67 \[ \int (a+a \cos (c+d x))^2 \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {2\,A\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,C\,a^2\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {4\,C\,a^2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {5\,A\,a^2\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^2\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {C\,a^2\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )} \]

[In]

int(((A + C*cos(c + d*x)^2)*(a + a*cos(c + d*x))^2)/cos(c + d*x)^4,x)

[Out]

(2*A*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*C*a^2*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (4*C*a^2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (5*A*a^2*sin(c + d*x))/(3*d*cos(c + d*x)) +
(A*a^2*sin(c + d*x))/(d*cos(c + d*x)^2) + (A*a^2*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (C*a^2*sin(c + d*x))/(d*
cos(c + d*x))