\(\int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx\) [9]

Optimal result

Integrand size = 29, antiderivative size = 106 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {a (3 A+4 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (A+B) \tan (c+d x)}{d}+\frac {a (3 A+4 B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a (A+B) \tan ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.19 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3047, 3100, 2827, 3852, 3853, 3855} \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {a (3 A+4 B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a (A+B) \tan ^3(c+d x)}{3 d}+\frac {a (A+B) \tan (c+d x)}{d}+\frac {a (3 A+4 B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[In]

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

[Out]

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

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.73 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {a \left (3 (3 A+4 B) \text {arctanh}(\sin (c+d x))+\sec (c+d x) \left (9 A+12 B+8 (A+B) (2+\cos (2 (c+d x))) \sec (c+d x)+6 A \sec ^2(c+d x)\right ) \tan (c+d x)\right )}{24 d} \]

[In]

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

[Out]

(a*(3*(3*A + 4*B)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*(9*A + 12*B + 8*(A + B)*(2 + Cos[2*(c + d*x)])*Sec[c +
d*x] + 6*A*Sec[c + d*x]^2)*Tan[c + d*x]))/(24*d)

Maple [A] (verified)

Time = 3.06 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.12

method	result	size
parts	\(\frac {a A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (a A +B a \right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(119\)
derivativedivides	\(\frac {-a A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(131\)
default	\(\frac {-a A \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B a \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+a A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}\)	\(131\)
parallelrisch	\(\frac {8 \left (-\frac {9 \left (A +\frac {4 B}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{16}+\frac {9 \left (A +\frac {4 B}{3}\right ) \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{16}+\left (A +B \right ) \sin \left (2 d x +2 c \right )+\frac {3 \left (\frac {3 A}{4}+B \right ) \sin \left (3 d x +3 c \right )}{8}+\frac {\left (A +B \right ) \sin \left (4 d x +4 c \right )}{4}+\frac {33 \sin \left (d x +c \right ) \left (A +\frac {4 B}{11}\right )}{32}\right ) a}{3 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(170\)
norman	\(\frac {-\frac {a \left (3 A +4 B \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {a \left (13 A -20 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {a \left (13 A +12 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {a \left (29 A -4 B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {a \left (31 A +4 B \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {a \left (47 A +20 B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {a \left (3 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {a \left (3 A +4 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(226\)
risch	\(-\frac {i a \left (9 A \,{\mathrm e}^{7 i \left (d x +c \right )}+12 B \,{\mathrm e}^{7 i \left (d x +c \right )}+33 A \,{\mathrm e}^{5 i \left (d x +c \right )}+12 B \,{\mathrm e}^{5 i \left (d x +c \right )}-48 A \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B \,{\mathrm e}^{4 i \left (d x +c \right )}-33 A \,{\mathrm e}^{3 i \left (d x +c \right )}-12 B \,{\mathrm e}^{3 i \left (d x +c \right )}-64 A \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B \,{\mathrm e}^{2 i \left (d x +c \right )}-9 A \,{\mathrm e}^{i \left (d x +c \right )}-12 B \,{\mathrm e}^{i \left (d x +c \right )}-16 A -16 B \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{2 d}-\frac {3 a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{2 d}\)	\(253\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.30 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.20 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (A + B\right )} a \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A + 4 \, B\right )} a \cos \left (d x + c\right )^{2} + 8 \, {\left (A + B\right )} a \cos \left (d x + c\right ) + 6 \, A a\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

1/48*(3*(3*A + 4*B)*a*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 3*(3*A + 4*B)*a*cos(d*x + c)^4*log(-sin(d*x + c)
+ 1) + 2*(16*(A + B)*a*cos(d*x + c)^3 + 3*(3*A + 4*B)*a*cos(d*x + c)^2 + 8*(A + B)*a*cos(d*x + c) + 6*A*a)*sin
(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.54 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a - 3 \, A a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B a {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 188, normalized size of antiderivative = 1.77 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, A a + 4 \, B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, A a + 4 \, B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 49 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 28 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 31 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 52 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 39 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 36 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[In]

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

[Out]

1/24*(3*(3*A*a + 4*B*a)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*A*a + 4*B*a)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) - 2*(9*A*a*tan(1/2*d*x + 1/2*c)^7 + 12*B*a*tan(1/2*d*x + 1/2*c)^7 - 49*A*a*tan(1/2*d*x + 1/2*c)^5 - 28*B*a
*tan(1/2*d*x + 1/2*c)^5 + 31*A*a*tan(1/2*d*x + 1/2*c)^3 + 52*B*a*tan(1/2*d*x + 1/2*c)^3 - 39*A*a*tan(1/2*d*x +
1/2*c) - 36*B*a*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 = 1.67 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.57 \[ \int (a+a \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {\left (-\frac {3\,A\,a}{4}-B\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {49\,A\,a}{12}+\frac {7\,B\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (-\frac {31\,A\,a}{12}-\frac {13\,B\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {13\,A\,a}{4}+3\,B\,a\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (3\,A+4\,B\right )}{4\,d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((13*A*a)/4 + 3*B*a) - tan(c/2 + (d*x)/2)^7*((3*A*a)/4 + B*a) - tan(c/2 + (d*x)/2)^3*((31*
A*a)/12 + (13*B*a)/3) + tan(c/2 + (d*x)/2)^5*((49*A*a)/12 + (7*B*a)/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)) + (a*atanh(tan(c/2 + (d*x)/2))*(3*A + 4*B)
)/(4*d)