\(\int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx\) [25]

Optimal result

Integrand size = 31, antiderivative size = 125 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=a^3 B x+\frac {a^3 (5 A+7 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac {(5 A+3 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d} \]

[Out]

a^3*B*x+1/2*a^3*(5*A+7*B)*arctanh(sin(d*x+c))/d+5/2*a^3*(A+B)*tan(d*x+c)/d+1/6*(5*A+3*B)*(a^3+a^3*cos(d*x+c))*
sec(d*x+c)*tan(d*x+c)/d+1/3*a*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 = 125, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {3054, 3047, 3100, 2814, 3855} \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {a^3 (5 A+7 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac {(5 A+3 B) \tan (c+d x) \sec (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+a^3 B x+\frac {a A \tan (c+d x) \sec ^2(c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[In]

Int[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]

[Out]

a^3*B*x + (a^3*(5*A + 7*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (5*a^3*(A + B)*Tan[c + d*x])/(2*d) + ((5*A + 3*B)*(a
^3 + a^3*Cos[c + d*x])*Sec[c + d*x]*Tan[c + d*x])/(6*d) + (a*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 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+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{3} \int (a+a \cos (c+d x))^2 (a (5 A+3 B)+3 a B \cos (c+d x)) \sec ^3(c+d x) \, dx \\ & = \frac {(5 A+3 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int (a+a \cos (c+d x)) \left (15 a^2 (A+B)+6 a^2 B \cos (c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {(5 A+3 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (15 a^3 (A+B)+\left (6 a^3 B+15 a^3 (A+B)\right ) \cos (c+d x)+6 a^3 B \cos ^2(c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac {(5 A+3 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{6} \int \left (3 a^3 (5 A+7 B)+6 a^3 B \cos (c+d x)\right ) \sec (c+d x) \, dx \\ & = a^3 B x+\frac {5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac {(5 A+3 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^2(c+d x) \tan (c+d x)}{3 d}+\frac {1}{2} \left (a^3 (5 A+7 B)\right ) \int \sec (c+d x) \, dx \\ & = a^3 B x+\frac {a^3 (5 A+7 B) \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 (A+B) \tan (c+d x)}{2 d}+\frac {(5 A+3 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec (c+d x) \tan (c+d x)}{6 d}+\frac {a 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 = 1.10 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.56 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {a^3 \left (6 B d x+3 (5 A+7 B) \text {arctanh}(\sin (c+d x))+3 (8 A+6 B+(3 A+B) \sec (c+d x)) \tan (c+d x)+2 A \tan ^3(c+d x)\right )}{6 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^4,x]

[Out]

(a^3*(6*B*d*x + 3*(5*A + 7*B)*ArcTanh[Sin[c + d*x]] + 3*(8*A + 6*B + (3*A + B)*Sec[c + d*x])*Tan[c + d*x] + 2*
A*Tan[c + d*x]^3))/(6*d)

Maple [A] (verified)

Time = 4.02 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.14

method	result	size
parts	\(-\frac {A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (A \,a^{3}+3 B \,a^{3}\right ) \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (3 A \,a^{3}+B \,a^{3}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {\left (3 A \,a^{3}+3 B \,a^{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{3} \left (d x +c \right )}{d}\)	\(142\)
parallelrisch	\(\frac {3 \left (-\frac {5 \left (A +\frac {7 B}{5}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2}+\frac {5 \left (A +\frac {7 B}{5}\right ) \left (\frac {\cos \left (3 d x +3 c \right )}{3}+\cos \left (d x +c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2}+\frac {d x B \cos \left (3 d x +3 c \right )}{3}+\left (A +\frac {B}{3}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {11 A}{9}+B \right ) \sin \left (3 d x +3 c \right )+d x B \cos \left (d x +c \right )+\frac {5 \left (A +\frac {3 B}{5}\right ) \sin \left (d x +c \right )}{3}\right ) a^{3}}{d \left (\cos \left (3 d x +3 c \right )+3 \cos \left (d x +c \right )\right )}\)	\(171\)
derivativedivides	\(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \left (d x +c \right )+3 A \,a^{3} \tan \left (d x +c \right )+3 B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 B \,a^{3} \tan \left (d x +c \right )-A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(176\)
default	\(\frac {A \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+B \,a^{3} \left (d x +c \right )+3 A \,a^{3} \tan \left (d x +c \right )+3 B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 B \,a^{3} \tan \left (d x +c \right )-A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(176\)
risch	\(a^{3} B x -\frac {i a^{3} \left (9 A \,{\mathrm e}^{5 i \left (d x +c \right )}+3 B \,{\mathrm e}^{5 i \left (d x +c \right )}-18 A \,{\mathrm e}^{4 i \left (d x +c \right )}-18 B \,{\mathrm e}^{4 i \left (d x +c \right )}-48 A \,{\mathrm e}^{2 i \left (d x +c \right )}-36 B \,{\mathrm e}^{2 i \left (d x +c \right )}-9 A \,{\mathrm e}^{i \left (d x +c \right )}-3 B \,{\mathrm e}^{i \left (d x +c \right )}-22 A -18 B \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {5 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}+\frac {5 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}\)	\(221\)
norman	\(\frac {a^{3} B x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a^{3} B x \left (\tan ^{14}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a^{3} B x -a^{3} B x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{3} B x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+3 a^{3} B x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{3} B x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-3 a^{3} B x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {5 a^{3} \left (A +B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {8 a^{3} \left (2 A +3 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{3} \left (5 A +6 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{3} \left (11 A +7 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {4 a^{3} \left (23 A +12 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a^{3} \left (37 A +33 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}-\frac {a^{3} \left (53 A -3 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{3}}-\frac {a^{3} \left (5 A +7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {a^{3} \left (5 A +7 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\)	\(395\)

[In]

int((a+cos(d*x+c)*a)^3*(A+B*cos(d*x+c))*sec(d*x+c)^4,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 141, normalized size of antiderivative = 1.13 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {12 \, B a^{3} d x \cos \left (d x + c\right )^{3} + 3 \, {\left (5 \, A + 7 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A + 7 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (2 \, {\left (11 \, A + 9 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 2 \, A a^{3}\right )} \sin \left (d x + c\right )}{12 \, d \cos \left (d x + c\right )^{3}} \]

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="fricas")

[Out]

1/12*(12*B*a^3*d*x*cos(d*x + c)^3 + 3*(5*A + 7*B)*a^3*cos(d*x + c)^3*log(sin(d*x + c) + 1) - 3*(5*A + 7*B)*a^3
*cos(d*x + c)^3*log(-sin(d*x + c) + 1) + 2*(2*(11*A + 9*B)*a^3*cos(d*x + c)^2 + 3*(3*A + B)*a^3*cos(d*x + c) +
2*A*a^3)*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**3*(A+B*cos(d*x+c))*sec(d*x+c)**4,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.70 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {4 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 12 \, {\left (d x + c\right )} B a^{3} - 9 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 3 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 6 \, A a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 18 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 36 \, A a^{3} \tan \left (d x + c\right ) + 36 \, B a^{3} \tan \left (d x + c\right )}{12 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="maxima")

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 189, normalized size of antiderivative = 1.51 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {6 \, {\left (d x + c\right )} B a^{3} + 3 \, {\left (5 \, A a^{3} + 7 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (5 \, A a^{3} + 7 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (15 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 15 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 36 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 33 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3}}}{6 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^4,x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*B*a^3 + 3*(5*A*a^3 + 7*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(5*A*a^3 + 7*B*a^3)*log(
abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(15*A*a^3*tan(1/2*d*x + 1/2*c)^5 + 15*B*a^3*tan(1/2*d*x + 1/2*c)^5 - 40*A*a
^3*tan(1/2*d*x + 1/2*c)^3 - 36*B*a^3*tan(1/2*d*x + 1/2*c)^3 + 33*A*a^3*tan(1/2*d*x + 1/2*c) + 21*B*a^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 = 0.36 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.67 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^4(c+d x) \, dx=\frac {5\,A\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {2\,B\,a^3\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {7\,B\,a^3\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{d}+\frac {11\,A\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,\cos \left (c+d\,x\right )}+\frac {3\,A\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2}+\frac {A\,a^3\,\sin \left (c+d\,x\right )}{3\,d\,{\cos \left (c+d\,x\right )}^3}+\frac {3\,B\,a^3\,\sin \left (c+d\,x\right )}{d\,\cos \left (c+d\,x\right )}+\frac {B\,a^3\,\sin \left (c+d\,x\right )}{2\,d\,{\cos \left (c+d\,x\right )}^2} \]

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^3)/cos(c + d*x)^4,x)

[Out]

(5*A*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (2*B*a^3*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)
))/d + (7*B*a^3*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/d + (11*A*a^3*sin(c + d*x))/(3*d*cos(c + d*x)) +
(3*A*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2) + (A*a^3*sin(c + d*x))/(3*d*cos(c + d*x)^3) + (3*B*a^3*sin(c + d*
x))/(d*cos(c + d*x)) + (B*a^3*sin(c + d*x))/(2*d*cos(c + d*x)^2)