\(\int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx\) [26]

Optimal result

Integrand size = 31, antiderivative size = 154 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {5 a^3 (3 A+4 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (9 A+11 B) \tan (c+d x)}{3 d}+\frac {a^3 (27 A+28 B) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[Out]

5/8*a^3*(3*A+4*B)*arctanh(sin(d*x+c))/d+1/3*a^3*(9*A+11*B)*tan(d*x+c)/d+1/24*a^3*(27*A+28*B)*sec(d*x+c)*tan(d*
x+c)/d+1/6*(3*A+2*B)*(a^3+a^3*cos(d*x+c))*sec(d*x+c)^2*tan(d*x+c)/d+1/4*a*A*(a+a*cos(d*x+c))^2*sec(d*x+c)^3*ta
n(d*x+c)/d

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.226, Rules used = {3054, 3047, 3100, 2827, 3852, 8, 3855} \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {5 a^3 (3 A+4 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (9 A+11 B) \tan (c+d x)}{3 d}+\frac {a^3 (27 A+28 B) \tan (c+d x) \sec (c+d x)}{24 d}+\frac {(3 A+2 B) \tan (c+d x) \sec ^2(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{6 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^2}{4 d} \]

[In]

Int[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]

[Out]

(5*a^3*(3*A + 4*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^3*(9*A + 11*B)*Tan[c + d*x])/(3*d) + (a^3*(27*A + 28*B)*S
ec[c + d*x]*Tan[c + d*x])/(24*d) + ((3*A + 2*B)*(a^3 + a^3*Cos[c + d*x])*Sec[c + d*x]^2*Tan[c + d*x])/(6*d) +
(a*A*(a + a*Cos[c + d*x])^2*Sec[c + d*x]^3*Tan[c + d*x])/(4*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

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 3852

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[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+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+a \cos (c+d x))^2 (2 a (3 A+2 B)+a (A+4 B) \cos (c+d x)) \sec ^4(c+d x) \, dx \\ & = \frac {(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{12} \int (a+a \cos (c+d x)) \left (a^2 (27 A+28 B)+a^2 (9 A+16 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{12} \int \left (a^3 (27 A+28 B)+\left (a^3 (9 A+16 B)+a^3 (27 A+28 B)\right ) \cos (c+d x)+a^3 (9 A+16 B) \cos ^2(c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {a^3 (27 A+28 B) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{24} \int \left (8 a^3 (9 A+11 B)+15 a^3 (3 A+4 B) \cos (c+d x)\right ) \sec ^2(c+d x) \, dx \\ & = \frac {a^3 (27 A+28 B) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} \left (5 a^3 (3 A+4 B)\right ) \int \sec (c+d x) \, dx+\frac {1}{3} \left (a^3 (9 A+11 B)\right ) \int \sec ^2(c+d x) \, dx \\ & = \frac {5 a^3 (3 A+4 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (27 A+28 B) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d}-\frac {\left (a^3 (9 A+11 B)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d} \\ & = \frac {5 a^3 (3 A+4 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^3 (9 A+11 B) \tan (c+d x)}{3 d}+\frac {a^3 (27 A+28 B) \sec (c+d x) \tan (c+d x)}{24 d}+\frac {(3 A+2 B) \left (a^3+a^3 \cos (c+d x)\right ) \sec ^2(c+d x) \tan (c+d x)}{6 d}+\frac {a A (a+a \cos (c+d x))^2 \sec ^3(c+d x) \tan (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.98 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.12 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {15 a^3 A \text {arctanh}(\sin (c+d x))}{8 d}+\frac {5 a^3 B \text {arctanh}(\sin (c+d x))}{2 d}+\frac {4 a^3 A \tan (c+d x)}{d}+\frac {4 a^3 B \tan (c+d x)}{d}+\frac {15 a^3 A \sec (c+d x) \tan (c+d x)}{8 d}+\frac {3 a^3 B \sec (c+d x) \tan (c+d x)}{2 d}+\frac {a^3 A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a^3 A \tan ^3(c+d x)}{d}+\frac {a^3 B \tan ^3(c+d x)}{3 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]

[Out]

(15*a^3*A*ArcTanh[Sin[c + d*x]])/(8*d) + (5*a^3*B*ArcTanh[Sin[c + d*x]])/(2*d) + (4*a^3*A*Tan[c + d*x])/d + (4
*a^3*B*Tan[c + d*x])/d + (15*a^3*A*Sec[c + d*x]*Tan[c + d*x])/(8*d) + (3*a^3*B*Sec[c + d*x]*Tan[c + d*x])/(2*d
) + (a^3*A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (a^3*A*Tan[c + d*x]^3)/d + (a^3*B*Tan[c + d*x]^3)/(3*d)

Maple [A] (verified)

Time = 4.71 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.16

method	result	size
parallelrisch	\(\frac {10 \left (-\frac {3 \left (A +\frac {4 B}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{4}+\frac {3 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (A +\frac {4 B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4}+\left (A +\frac {13 B}{15}\right ) \sin \left (2 d x +2 c \right )+\left (\frac {3 A}{8}+\frac {3 B}{10}\right ) \sin \left (3 d x +3 c \right )+\left (\frac {3 A}{10}+\frac {11 B}{30}\right ) \sin \left (4 d x +4 c \right )+\frac {23 \left (A +\frac {12 B}{23}\right ) \sin \left (d x +c \right )}{40}\right ) a^{3}}{d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(178\)
parts	\(\frac {A \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (A \,a^{3}+3 B \,a^{3}\right ) \tan \left (d x +c \right )}{d}-\frac {\left (3 A \,a^{3}+B \,a^{3}\right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {\left (3 A \,a^{3}+3 B \,a^{3}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right ) a^{3}}{d}\)	\(181\)
derivativedivides	\(\frac {A \,a^{3} \tan \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 B \,a^{3} \tan \left (d x +c \right )-3 A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(219\)
default	\(\frac {A \,a^{3} \tan \left (d x +c \right )+B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+3 A \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 B \,a^{3} \tan \left (d x +c \right )-3 A \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+3 B \,a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,a^{3} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{3} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(219\)
risch	\(-\frac {i a^{3} \left (45 A \,{\mathrm e}^{7 i \left (d x +c \right )}+36 B \,{\mathrm e}^{7 i \left (d x +c \right )}-24 A \,{\mathrm e}^{6 i \left (d x +c \right )}-72 B \,{\mathrm e}^{6 i \left (d x +c \right )}+69 A \,{\mathrm e}^{5 i \left (d x +c \right )}+36 B \,{\mathrm e}^{5 i \left (d x +c \right )}-216 A \,{\mathrm e}^{4 i \left (d x +c \right )}-264 B \,{\mathrm e}^{4 i \left (d x +c \right )}-69 A \,{\mathrm e}^{3 i \left (d x +c \right )}-36 B \,{\mathrm e}^{3 i \left (d x +c \right )}-264 A \,{\mathrm e}^{2 i \left (d x +c \right )}-280 B \,{\mathrm e}^{2 i \left (d x +c \right )}-45 A \,{\mathrm e}^{i \left (d x +c \right )}-36 B \,{\mathrm e}^{i \left (d x +c \right )}-72 A -88 B \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {15 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}-\frac {15 A \,a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {5 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}\)	\(287\)
norman	\(\frac {\frac {19 a^{3} \left (3 A +4 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}-\frac {5 a^{3} \left (3 A +4 B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {5 a^{3} \left (3 A +4 B \right ) \left (\tan ^{15}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a^{3} \left (49 A +44 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{3} \left (81 A +44 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}-\frac {a^{3} \left (111 A +404 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a^{3} \left (369 A +236 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a^{3} \left (-52 B +57 A \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {5 a^{3} \left (3 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {5 a^{3} \left (3 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(294\)

[In]

int((a+cos(d*x+c)*a)^3*(A+B*cos(d*x+c))*sec(d*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

10*(-3/4*(A+4/3*B)*(3/4+1/4*cos(4*d*x+4*c)+cos(2*d*x+2*c))*ln(tan(1/2*d*x+1/2*c)-1)+3/4*(3/4+1/4*cos(4*d*x+4*c
)+cos(2*d*x+2*c))*(A+4/3*B)*ln(tan(1/2*d*x+1/2*c)+1)+(A+13/15*B)*sin(2*d*x+2*c)+(3/8*A+3/10*B)*sin(3*d*x+3*c)+
(3/10*A+11/30*B)*sin(4*d*x+4*c)+23/40*(A+12/23*B)*sin(d*x+c))*a^3/d/(cos(4*d*x+4*c)+4*cos(2*d*x+2*c)+3)

Fricas [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.94 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {15 \, {\left (3 \, A + 4 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (3 \, A + 4 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (9 \, A + 11 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 9 \, {\left (5 \, A + 4 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 6 \, A a^{3}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(15*(3*A + 4*B)*a^3*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 15*(3*A + 4*B)*a^3*cos(d*x + c)^4*log(-sin(d*x
+ c) + 1) + 2*(8*(9*A + 11*B)*a^3*cos(d*x + c)^3 + 9*(5*A + 4*B)*a^3*cos(d*x + c)^2 + 8*(3*A + B)*a^3*cos(d*x
+ c) + 6*A*a^3)*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**3*(A+B*cos(d*x+c))*sec(d*x+c)**5,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.75 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {48 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{3} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{3} - 3 \, A a^{3} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, A a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 36 \, B a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 24 \, B a^{3} {\left (\log \left (\sin \left (d x + c\right ) + 1\right ) - \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, A a^{3} \tan \left (d x + c\right ) + 144 \, B a^{3} \tan \left (d x + c\right )}{48 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(48*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^3 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^3 - 3*A*a^3*(2*(3*
sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin
(d*x + c) - 1)) - 36*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
)) - 36*B*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)) + 24*B*a^3
*(log(sin(d*x + c) + 1) - log(sin(d*x + c) - 1)) + 48*A*a^3*tan(d*x + c) + 144*B*a^3*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.35 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.38 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {15 \, {\left (3 \, A a^{3} + 4 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 15 \, {\left (3 \, A a^{3} + 4 \, B a^{3}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (45 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 60 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 165 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 220 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 219 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 292 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 147 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 132 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(15*(3*A*a^3 + 4*B*a^3)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 15*(3*A*a^3 + 4*B*a^3)*log(abs(tan(1/2*d*x +
1/2*c) - 1)) - 2*(45*A*a^3*tan(1/2*d*x + 1/2*c)^7 + 60*B*a^3*tan(1/2*d*x + 1/2*c)^7 - 165*A*a^3*tan(1/2*d*x +
1/2*c)^5 - 220*B*a^3*tan(1/2*d*x + 1/2*c)^5 + 219*A*a^3*tan(1/2*d*x + 1/2*c)^3 + 292*B*a^3*tan(1/2*d*x + 1/2*
c)^3 - 147*A*a^3*tan(1/2*d*x + 1/2*c) - 132*B*a^3*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d

Mupad [B] (verification not implemented)

Time = 2.90 (sec) , antiderivative size = 185, normalized size of antiderivative = 1.20 \[ \int (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {\left (-\frac {15\,A\,a^3}{4}-5\,B\,a^3\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {55\,A\,a^3}{4}+\frac {55\,B\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {73\,A\,a^3}{4}-\frac {73\,B\,a^3}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {49\,A\,a^3}{4}+11\,B\,a^3\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {5\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A+4\,B\right )}{4\,d} \]

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^3)/cos(c + d*x)^5,x)

[Out]

(tan(c/2 + (d*x)/2)*((49*A*a^3)/4 + 11*B*a^3) - tan(c/2 + (d*x)/2)^7*((15*A*a^3)/4 + 5*B*a^3) + tan(c/2 + (d*x
)/2)^5*((55*A*a^3)/4 + (55*B*a^3)/3) - tan(c/2 + (d*x)/2)^3*((73*A*a^3)/4 + (73*B*a^3)/3))/(d*(6*tan(c/2 + (d*
x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (5*a^3*atanh(tan(c/2
+ (d*x)/2))*(3*A + 4*B))/(4*d)