\(\int (a+a \cos (c+d x))^4 (A+C \cos ^2(c+d x)) \sec (c+d x) \, dx\) [31]

Optimal result

Integrand size = 31, antiderivative size = 177 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {1}{2} a^4 (12 A+7 C) x+\frac {a^4 A \text {arctanh}(\sin (c+d x))}{d}+\frac {a^4 (10 A+7 C) \sin (c+d x)}{2 d}+\frac {a C (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(8 A+7 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d} \]

[Out]

1/2*a^4*(12*A+7*C)*x+a^4*A*arctanh(sin(d*x+c))/d+1/2*a^4*(10*A+7*C)*sin(d*x+c)/d+1/5*a*C*(a+a*cos(d*x+c))^3*si
n(d*x+c)/d+1/5*C*(a+a*cos(d*x+c))^4*sin(d*x+c)/d+1/15*(5*A+7*C)*(a^2+a^2*cos(d*x+c))^2*sin(d*x+c)/d+1/6*(8*A+7
*C)*(a^4+a^4*cos(d*x+c))*sin(d*x+c)/d

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3125, 3055, 3047, 3102, 2814, 3855} \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {a^4 A \text {arctanh}(\sin (c+d x))}{d}+\frac {a^4 (10 A+7 C) \sin (c+d x)}{2 d}+\frac {(8 A+7 C) \sin (c+d x) \left (a^4 \cos (c+d x)+a^4\right )}{6 d}+\frac {1}{2} a^4 x (12 A+7 C)+\frac {(5 A+7 C) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )^2}{15 d}+\frac {a C \sin (c+d x) (a \cos (c+d x)+a)^3}{5 d}+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^4}{5 d} \]

[In]

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

[Out]

(a^4*(12*A + 7*C)*x)/2 + (a^4*A*ArcTanh[Sin[c + d*x]])/d + (a^4*(10*A + 7*C)*Sin[c + d*x])/(2*d) + (a*C*(a + a
*Cos[c + d*x])^3*Sin[c + d*x])/(5*d) + (C*(a + a*Cos[c + d*x])^4*Sin[c + d*x])/(5*d) + ((5*A + 7*C)*(a^2 + a^2
*Cos[c + d*x])^2*Sin[c + d*x])/(15*d) + ((8*A + 7*C)*(a^4 + a^4*Cos[c + d*x])*Sin[c + d*x])/(6*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 3125

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)*
sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(
n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(b*d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n*Si
mp[A*b*d*(m + n + 2) + C*(a*c*m + b*d*(n + 1)) + C*(a*d*m - b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a,
b, c, d, e, f, A, C, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] &&  !LtQ[m, -2^
(-1)] && NeQ[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 {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x))^4 (5 a A+4 a C \cos (c+d x)) \sec (c+d x) \, dx}{5 a} \\ & = \frac {a C (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {\int (a+a \cos (c+d x))^3 \left (20 a^2 A+4 a^2 (5 A+7 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{20 a} \\ & = \frac {a C (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {\int (a+a \cos (c+d x))^2 \left (60 a^3 A+20 a^3 (8 A+7 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{60 a} \\ & = \frac {a C (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(8 A+7 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {\int (a+a \cos (c+d x)) \left (120 a^4 A+60 a^4 (10 A+7 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{120 a} \\ & = \frac {a C (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(8 A+7 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {\int \left (120 a^5 A+\left (120 a^5 A+60 a^5 (10 A+7 C)\right ) \cos (c+d x)+60 a^5 (10 A+7 C) \cos ^2(c+d x)\right ) \sec (c+d x) \, dx}{120 a} \\ & = \frac {a^4 (10 A+7 C) \sin (c+d x)}{2 d}+\frac {a C (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(8 A+7 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\frac {\int \left (120 a^5 A+60 a^5 (12 A+7 C) \cos (c+d x)\right ) \sec (c+d x) \, dx}{120 a} \\ & = \frac {1}{2} a^4 (12 A+7 C) x+\frac {a^4 (10 A+7 C) \sin (c+d x)}{2 d}+\frac {a C (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(8 A+7 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d}+\left (a^4 A\right ) \int \sec (c+d x) \, dx \\ & = \frac {1}{2} a^4 (12 A+7 C) x+\frac {a^4 A \text {arctanh}(\sin (c+d x))}{d}+\frac {a^4 (10 A+7 C) \sin (c+d x)}{2 d}+\frac {a C (a+a \cos (c+d x))^3 \sin (c+d x)}{5 d}+\frac {C (a+a \cos (c+d x))^4 \sin (c+d x)}{5 d}+\frac {(5 A+7 C) \left (a^2+a^2 \cos (c+d x)\right )^2 \sin (c+d x)}{15 d}+\frac {(8 A+7 C) \left (a^4+a^4 \cos (c+d x)\right ) \sin (c+d x)}{6 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.56 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.83 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {a^4 \left (1440 A d x+840 C d x-240 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+240 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+30 (54 A+49 C) \sin (c+d x)+240 (A+2 C) \sin (2 (c+d x))+20 A \sin (3 (c+d x))+145 C \sin (3 (c+d x))+30 C \sin (4 (c+d x))+3 C \sin (5 (c+d x))\right )}{240 d} \]

[In]

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

[Out]

(a^4*(1440*A*d*x + 840*C*d*x - 240*A*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 240*A*Log[Cos[(c + d*x)/2] + S
in[(c + d*x)/2]] + 30*(54*A + 49*C)*Sin[c + d*x] + 240*(A + 2*C)*Sin[2*(c + d*x)] + 20*A*Sin[3*(c + d*x)] + 14
5*C*Sin[3*(c + d*x)] + 30*C*Sin[4*(c + d*x)] + 3*C*Sin[5*(c + d*x)]))/(240*d)

Maple [A] (verified)

Time = 5.74 (sec) , antiderivative size = 120, normalized size of antiderivative = 0.68

method	result	size
parallelrisch	\(-\frac {a^{4} \left (A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-A -2 C \right ) \sin \left (2 d x +2 c \right )+\left (-\frac {A}{12}-\frac {29 C}{48}\right ) \sin \left (3 d x +3 c \right )-\frac {\sin \left (4 d x +4 c \right ) C}{8}-\frac {\sin \left (5 d x +5 c \right ) C}{80}+\left (-\frac {27 A}{4}-\frac {49 C}{8}\right ) \sin \left (d x +c \right )-6 d x \left (A +\frac {7 C}{12}\right )\right )}{d}\)	\(120\)
parts	\(\frac {a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {\left (a^{4} A +6 C \,a^{4}\right ) \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3 d}+\frac {\left (4 a^{4} A +4 C \,a^{4}\right ) \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {\left (6 a^{4} A +C \,a^{4}\right ) \sin \left (d x +c \right )}{d}+\frac {C \,a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5 d}+\frac {4 C \,a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )}{d}+\frac {4 a^{4} A \left (d x +c \right )}{d}\)	\(209\)
derivativedivides	\(\frac {\frac {a^{4} A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {C \,a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 C \,a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+6 a^{4} A \sin \left (d x +c \right )+2 C \,a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 a^{4} A \left (d x +c \right )+4 C \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{4} \sin \left (d x +c \right )}{d}\)	\(230\)
default	\(\frac {\frac {a^{4} A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {C \,a^{4} \left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{5}+4 a^{4} A \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+4 C \,a^{4} \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+6 a^{4} A \sin \left (d x +c \right )+2 C \,a^{4} \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+4 a^{4} A \left (d x +c \right )+4 C \,a^{4} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+a^{4} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+C \,a^{4} \sin \left (d x +c \right )}{d}\)	\(230\)
risch	\(6 a^{4} x A +\frac {7 a^{4} C x}{2}-\frac {27 i {\mathrm e}^{i \left (d x +c \right )} a^{4} A}{8 d}-\frac {49 i {\mathrm e}^{i \left (d x +c \right )} C \,a^{4}}{16 d}+\frac {27 i {\mathrm e}^{-i \left (d x +c \right )} a^{4} A}{8 d}+\frac {49 i {\mathrm e}^{-i \left (d x +c \right )} C \,a^{4}}{16 d}+\frac {a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a^{4} A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}+\frac {\sin \left (5 d x +5 c \right ) C \,a^{4}}{80 d}+\frac {\sin \left (4 d x +4 c \right ) C \,a^{4}}{8 d}+\frac {\sin \left (3 d x +3 c \right ) a^{4} A}{12 d}+\frac {29 \sin \left (3 d x +3 c \right ) C \,a^{4}}{48 d}+\frac {\sin \left (2 d x +2 c \right ) a^{4} A}{d}+\frac {2 \sin \left (2 d x +2 c \right ) C \,a^{4}}{d}\)	\(242\)
norman	\(\frac {\left (6 a^{4} A +\frac {7}{2} C \,a^{4}\right ) x +\left (6 a^{4} A +\frac {7}{2} C \,a^{4}\right ) x \left (\tan ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (90 a^{4} A +\frac {105}{2} C \,a^{4}\right ) x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (90 a^{4} A +\frac {105}{2} C \,a^{4}\right ) x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (36 a^{4} A +21 C \,a^{4}\right ) x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (36 a^{4} A +21 C \,a^{4}\right ) x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (120 a^{4} A +70 C \,a^{4}\right ) x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\frac {a^{4} \left (10 A +7 C \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{4} \left (18 A +25 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}+\frac {a^{4} \left (166 A +119 C \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {a^{4} \left (238 A +233 C \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d}+\frac {2 a^{4} \left (310 A +231 C \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}+\frac {2 a^{4} \left (350 A +281 C \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{6}}+\frac {a^{4} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a^{4} A \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\)	\(383\)

[In]

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

[Out]

-a^4*(A*ln(tan(1/2*d*x+1/2*c)-1)-A*ln(tan(1/2*d*x+1/2*c)+1)+(-A-2*C)*sin(2*d*x+2*c)+(-1/12*A-29/48*C)*sin(3*d*
x+3*c)-1/8*sin(4*d*x+4*c)*C-1/80*sin(5*d*x+5*c)*C+(-27/4*A-49/8*C)*sin(d*x+c)-6*d*x*(A+7/12*C))/d

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 138, normalized size of antiderivative = 0.78 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {15 \, {\left (12 \, A + 7 \, C\right )} a^{4} d x + 15 \, A a^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, A a^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + {\left (6 \, C a^{4} \cos \left (d x + c\right )^{4} + 30 \, C a^{4} \cos \left (d x + c\right )^{3} + 2 \, {\left (5 \, A + 34 \, C\right )} a^{4} \cos \left (d x + c\right )^{2} + 15 \, {\left (4 \, A + 7 \, C\right )} a^{4} \cos \left (d x + c\right ) + 2 \, {\left (100 \, A + 83 \, C\right )} a^{4}\right )} \sin \left (d x + c\right )}{30 \, d} \]

[In]

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

[Out]

1/30*(15*(12*A + 7*C)*a^4*d*x + 15*A*a^4*log(sin(d*x + c) + 1) - 15*A*a^4*log(-sin(d*x + c) + 1) + (6*C*a^4*co
s(d*x + c)^4 + 30*C*a^4*cos(d*x + c)^3 + 2*(5*A + 34*C)*a^4*cos(d*x + c)^2 + 15*(4*A + 7*C)*a^4*cos(d*x + c) +
2*(100*A + 83*C)*a^4)*sin(d*x + c))/d

Sympy [F]

\[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=a^{4} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 4 A \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 A \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 A \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 C \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 6 C \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 4 C \cos ^{5}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{6}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \]

[In]

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

[Out]

a**4*(Integral(A*sec(c + d*x), x) + Integral(4*A*cos(c + d*x)*sec(c + d*x), x) + Integral(6*A*cos(c + d*x)**2*
sec(c + d*x), x) + Integral(4*A*cos(c + d*x)**3*sec(c + d*x), x) + Integral(A*cos(c + d*x)**4*sec(c + d*x), x)
+ Integral(C*cos(c + d*x)**2*sec(c + d*x), x) + Integral(4*C*cos(c + d*x)**3*sec(c + d*x), x) + Integral(6*C*
cos(c + d*x)**4*sec(c + d*x), x) + Integral(4*C*cos(c + d*x)**5*sec(c + d*x), x) + Integral(C*cos(c + d*x)**6*
sec(c + d*x), x))

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 222, normalized size of antiderivative = 1.25 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=-\frac {40 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a^{4} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A a^{4} - 480 \, {\left (d x + c\right )} A a^{4} - 8 \, {\left (3 \, \sin \left (d x + c\right )^{5} - 10 \, \sin \left (d x + c\right )^{3} + 15 \, \sin \left (d x + c\right )\right )} C a^{4} + 240 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{4} - 15 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 120 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{4} - 120 \, A a^{4} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right ) - 720 \, A a^{4} \sin \left (d x + c\right ) - 120 \, C a^{4} \sin \left (d x + c\right )}{120 \, d} \]

[In]

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

[Out]

-1/120*(40*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a^4 - 120*(2*d*x + 2*c + sin(2*d*x + 2*c))*A*a^4 - 480*(d*x + c
)*A*a^4 - 8*(3*sin(d*x + c)^5 - 10*sin(d*x + c)^3 + 15*sin(d*x + c))*C*a^4 + 240*(sin(d*x + c)^3 - 3*sin(d*x +
c))*C*a^4 - 15*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*C*a^4 - 120*(2*d*x + 2*c + sin(2*d*x +
2*c))*C*a^4 - 120*A*a^4*log(sec(d*x + c) + tan(d*x + c)) - 720*A*a^4*sin(d*x + c) - 120*C*a^4*sin(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.40 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {30 \, A a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 30 \, A a^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 15 \, {\left (12 \, A a^{4} + 7 \, C a^{4}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (150 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 105 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} + 680 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 490 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 1180 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 896 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 920 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 790 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 270 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 375 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{5}}}{30 \, d} \]

[In]

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

[Out]

1/30*(30*A*a^4*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 30*A*a^4*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 15*(12*A*a^4
+ 7*C*a^4)*(d*x + c) + 2*(150*A*a^4*tan(1/2*d*x + 1/2*c)^9 + 105*C*a^4*tan(1/2*d*x + 1/2*c)^9 + 680*A*a^4*tan
(1/2*d*x + 1/2*c)^7 + 490*C*a^4*tan(1/2*d*x + 1/2*c)^7 + 1180*A*a^4*tan(1/2*d*x + 1/2*c)^5 + 896*C*a^4*tan(1/2
*d*x + 1/2*c)^5 + 920*A*a^4*tan(1/2*d*x + 1/2*c)^3 + 790*C*a^4*tan(1/2*d*x + 1/2*c)^3 + 270*A*a^4*tan(1/2*d*x
+ 1/2*c) + 375*C*a^4*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^5)/d

Mupad [B] (verification not implemented)

Time = 0.96 (sec) , antiderivative size = 202, normalized size of antiderivative = 1.14 \[ \int (a+a \cos (c+d x))^4 \left (A+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx=\frac {12\,A\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+2\,A\,a^4\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+7\,C\,a^4\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )+A\,a^4\,\sin \left (2\,c+2\,d\,x\right )+\frac {A\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{12}+2\,C\,a^4\,\sin \left (2\,c+2\,d\,x\right )+\frac {29\,C\,a^4\,\sin \left (3\,c+3\,d\,x\right )}{48}+\frac {C\,a^4\,\sin \left (4\,c+4\,d\,x\right )}{8}+\frac {C\,a^4\,\sin \left (5\,c+5\,d\,x\right )}{80}+\frac {27\,A\,a^4\,\sin \left (c+d\,x\right )}{4}+\frac {49\,C\,a^4\,\sin \left (c+d\,x\right )}{8}}{d} \]

[In]

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

[Out]

(12*A*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + 2*A*a^4*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) +
7*C*a^4*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)) + A*a^4*sin(2*c + 2*d*x) + (A*a^4*sin(3*c + 3*d*x))/12 +
2*C*a^4*sin(2*c + 2*d*x) + (29*C*a^4*sin(3*c + 3*d*x))/48 + (C*a^4*sin(4*c + 4*d*x))/8 + (C*a^4*sin(5*c + 5*d*
x))/80 + (27*A*a^4*sin(c + d*x))/4 + (49*C*a^4*sin(c + d*x))/8)/d