\(\int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx\) [18]

Optimal result

Integrand size = 31, antiderivative size = 144 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {a^2 (7 A+8 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (4 A+5 B) \tan (c+d x)}{3 d}+\frac {a^2 (7 A+8 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (5 A+4 B) \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{4 d} \]

[Out]

1/8*a^2*(7*A+8*B)*arctanh(sin(d*x+c))/d+1/3*a^2*(4*A+5*B)*tan(d*x+c)/d+1/8*a^2*(7*A+8*B)*sec(d*x+c)*tan(d*x+c)
/d+1/12*a^2*(5*A+4*B)*sec(d*x+c)^2*tan(d*x+c)/d+1/4*A*(a^2+a^2*cos(d*x+c))*sec(d*x+c)^3*tan(d*x+c)/d

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3054, 3047, 3100, 2827, 3853, 3855, 3852, 8} \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {a^2 (7 A+8 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (4 A+5 B) \tan (c+d x)}{3 d}+\frac {a^2 (5 A+4 B) \tan (c+d x) \sec ^2(c+d x)}{12 d}+\frac {a^2 (7 A+8 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {A \tan (c+d x) \sec ^3(c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{4 d} \]

[In]

Int[(a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]

[Out]

(a^2*(7*A + 8*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^2*(4*A + 5*B)*Tan[c + d*x])/(3*d) + (a^2*(7*A + 8*B)*Sec[c
+ d*x]*Tan[c + d*x])/(8*d) + (a^2*(5*A + 4*B)*Sec[c + d*x]^2*Tan[c + d*x])/(12*d) + (A*(a^2 + a^2*Cos[c + d*x]
)*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 3853

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Csc[c + d*x])^(n - 1)/(d*(n
- 1))), x] + Dist[b^2*((n - 2)/(n - 1)), Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n,
1] && IntegerQ[2*n]

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 \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int (a+a \cos (c+d x)) (a (5 A+4 B)+2 a (A+2 B) \cos (c+d x)) \sec ^4(c+d x) \, dx \\ & = \frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{4} \int \left (a^2 (5 A+4 B)+\left (2 a^2 (A+2 B)+a^2 (5 A+4 B)\right ) \cos (c+d x)+2 a^2 (A+2 B) \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx \\ & = \frac {a^2 (5 A+4 B) \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{12} \int \left (3 a^2 (7 A+8 B)+4 a^2 (4 A+5 B) \cos (c+d x)\right ) \sec ^3(c+d x) \, dx \\ & = \frac {a^2 (5 A+4 B) \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{3} \left (a^2 (4 A+5 B)\right ) \int \sec ^2(c+d x) \, dx+\frac {1}{4} \left (a^2 (7 A+8 B)\right ) \int \sec ^3(c+d x) \, dx \\ & = \frac {a^2 (7 A+8 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (5 A+4 B) \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {1}{8} \left (a^2 (7 A+8 B)\right ) \int \sec (c+d x) \, dx-\frac {\left (a^2 (4 A+5 B)\right ) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{3 d} \\ & = \frac {a^2 (7 A+8 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a^2 (4 A+5 B) \tan (c+d x)}{3 d}+\frac {a^2 (7 A+8 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a^2 (5 A+4 B) \sec ^2(c+d x) \tan (c+d x)}{12 d}+\frac {A \left (a^2+a^2 \cos (c+d x)\right ) \sec ^3(c+d x) \tan (c+d x)}{4 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.82 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.56 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {a^2 \left (3 (7 A+8 B) \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (48 (A+B)+3 (7 A+8 B) \sec (c+d x)+6 A \sec ^3(c+d x)+8 (2 A+B) \tan ^2(c+d x)\right )\right )}{24 d} \]

[In]

Integrate[(a + a*Cos[c + d*x])^2*(A + B*Cos[c + d*x])*Sec[c + d*x]^5,x]

[Out]

(a^2*(3*(7*A + 8*B)*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(48*(A + B) + 3*(7*A + 8*B)*Sec[c + d*x] + 6*A*Sec[c
+ d*x]^3 + 8*(2*A + B)*Tan[c + d*x]^2)))/(24*d)

Maple [A] (verified)

Time = 4.48 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.04

method	result	size
parts	\(\frac {A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}+\frac {\left (A \,a^{2}+2 B \,a^{2}\right ) \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}-\frac {\left (2 A \,a^{2}+B \,a^{2}\right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B \,a^{2} \tan \left (d x +c \right )}{d}\)	\(150\)
parallelrisch	\(\frac {16 \left (-\frac {21 \left (A +\frac {8 B}{7}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{32}+\frac {21 \left (A +\frac {8 B}{7}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{32}+\left (A +\frac {7 B}{8}\right ) \sin \left (2 d x +2 c \right )+\frac {3 \left (\frac {7 A}{8}+B \right ) \sin \left (3 d x +3 c \right )}{8}+\frac {\left (A +\frac {5 B}{4}\right ) \sin \left (4 d x +4 c \right )}{4}+\frac {45 \left (A +\frac {8 B}{15}\right ) \sin \left (d x +c \right )}{64}\right ) a^{2}}{3 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(176\)
derivativedivides	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )-2 A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(187\)
default	\(\frac {A \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+B \,a^{2} \tan \left (d x +c \right )-2 A \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+2 B \,a^{2} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+A \,a^{2} \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B \,a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(187\)
norman	\(\frac {\frac {a^{2} \left (3 A -8 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}+\frac {a^{2} \left (7 A +8 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}-\frac {a^{2} \left (7 A +8 B \right ) \left (\tan ^{13}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a^{2} \left (25 A +24 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a^{2} \left (53 A -104 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a^{2} \left (71 A +40 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {a^{2} \left (85 A +56 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {a^{2} \left (7 A +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a^{2} \left (7 A +8 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(267\)
risch	\(-\frac {i a^{2} \left (21 A \,{\mathrm e}^{7 i \left (d x +c \right )}+24 B \,{\mathrm e}^{7 i \left (d x +c \right )}-24 B \,{\mathrm e}^{6 i \left (d x +c \right )}+45 A \,{\mathrm e}^{5 i \left (d x +c \right )}+24 B \,{\mathrm e}^{5 i \left (d x +c \right )}-96 A \,{\mathrm e}^{4 i \left (d x +c \right )}-120 B \,{\mathrm e}^{4 i \left (d x +c \right )}-45 A \,{\mathrm e}^{3 i \left (d x +c \right )}-24 B \,{\mathrm e}^{3 i \left (d x +c \right )}-128 A \,{\mathrm e}^{2 i \left (d x +c \right )}-136 B \,{\mathrm e}^{2 i \left (d x +c \right )}-21 A \,{\mathrm e}^{i \left (d x +c \right )}-24 B \,{\mathrm e}^{i \left (d x +c \right )}-32 A -40 B \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {7 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{d}-\frac {7 A \,a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{d}\)	\(274\)

[In]

int((a+cos(d*x+c)*a)^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x,method=_RETURNVERBOSE)

[Out]

A*a^2/d*(-(-1/4*sec(d*x+c)^3-3/8*sec(d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c)))+(A*a^2+2*B*a^2)/d*(1/2*
sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))-(2*A*a^2+B*a^2)/d*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+B*a^
2/d*tan(d*x+c)

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 145, normalized size of antiderivative = 1.01 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (7 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (7 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, {\left (4 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} + 3 \, {\left (7 \, A + 8 \, B\right )} a^{2} \cos \left (d x + c\right )^{2} + 8 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 6 \, A a^{2}\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="fricas")

[Out]

1/48*(3*(7*A + 8*B)*a^2*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(7*A + 8*B)*a^2*cos(d*x + c)^4*log(-sin(d*x +
c) + 1) + 2*(8*(4*A + 5*B)*a^2*cos(d*x + c)^3 + 3*(7*A + 8*B)*a^2*cos(d*x + c)^2 + 8*(2*A + B)*a^2*cos(d*x +
c) + 6*A*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**2*(A+B*cos(d*x+c))*sec(d*x+c)**5,x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.60 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {32 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a^{2} + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a^{2} - 3 \, A a^{2} {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, A a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 24 \, B a^{2} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + 48 \, B a^{2} \tan \left (d x + c\right )}{48 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="maxima")

[Out]

1/48*(32*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*a^2 + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*B*a^2 - 3*A*a^2*(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)) - 12*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
)) - 24*B*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)) + 48*B*a^2
*tan(d*x + c))/d

Giac [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.47 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (7 \, A a^{2} + 8 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (7 \, A a^{2} + 8 \, B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (21 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 77 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 88 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 83 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 136 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 75 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 72 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[In]

integrate((a+a*cos(d*x+c))^2*(A+B*cos(d*x+c))*sec(d*x+c)^5,x, algorithm="giac")

[Out]

1/24*(3*(7*A*a^2 + 8*B*a^2)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(7*A*a^2 + 8*B*a^2)*log(abs(tan(1/2*d*x + 1
/2*c) - 1)) - 2*(21*A*a^2*tan(1/2*d*x + 1/2*c)^7 + 24*B*a^2*tan(1/2*d*x + 1/2*c)^7 - 77*A*a^2*tan(1/2*d*x + 1/
2*c)^5 - 88*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 83*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 136*B*a^2*tan(1/2*d*x + 1/2*c)^3
- 75*A*a^2*tan(1/2*d*x + 1/2*c) - 72*B*a^2*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 = 1.94 (sec) , antiderivative size = 183, normalized size of antiderivative = 1.27 \[ \int (a+a \cos (c+d x))^2 (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {\left (-\frac {7\,A\,a^2}{4}-2\,B\,a^2\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {77\,A\,a^2}{12}+\frac {22\,B\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {83\,A\,a^2}{12}-\frac {34\,B\,a^2}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {25\,A\,a^2}{4}+6\,B\,a^2\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {2\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {7\,A}{8}+B\right )}{d} \]

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^2)/cos(c + d*x)^5,x)

[Out]

(tan(c/2 + (d*x)/2)*((25*A*a^2)/4 + 6*B*a^2) - tan(c/2 + (d*x)/2)^7*((7*A*a^2)/4 + 2*B*a^2) + tan(c/2 + (d*x)/
2)^5*((77*A*a^2)/12 + (22*B*a^2)/3) - tan(c/2 + (d*x)/2)^3*((83*A*a^2)/12 + (34*B*a^2)/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)) + (2*a^2*atanh(tan(c/2
+ (d*x)/2))*((7*A)/8 + B))/d