\(\int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx\) [222]

Optimal result

Integrand size = 29, antiderivative size = 114 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {(3 a A+4 b B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(A b+a B) \tan (c+d x)}{d}+\frac {(3 a A+4 b B) \sec (c+d x) \tan (c+d x)}{8 d}+\frac {a A \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {(A b+a B) \tan ^3(c+d x)}{3 d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 114, 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} \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {(3 a A+4 b B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(a B+A b) \tan ^3(c+d x)}{3 d}+\frac {(a B+A b) \tan (c+d x)}{d}+\frac {(3 a A+4 b B) \tan (c+d x) \sec (c+d x)}{8 d}+\frac {a A \tan (c+d x) \sec ^3(c+d x)}{4 d} \]

[In]

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

[Out]

((3*a*A + 4*b*B)*ArcTanh[Sin[c + d*x]])/(8*d) + ((A*b + a*B)*Tan[c + d*x])/d + ((3*a*A + 4*b*B)*Sec[c + d*x]*T
an[c + d*x])/(8*d) + (a*A*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + ((A*b + 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 b+a B) \cos (c+d x)+b 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 b+a B)+(3 a A+4 b 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 b+a B) \int \sec ^4(c+d x) \, dx+\frac {1}{4} (3 a A+4 b B) \int \sec ^3(c+d x) \, dx \\ & = \frac {(3 a A+4 b 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} (3 a A+4 b B) \int \sec (c+d x) \, dx-\frac {(A b+a B) \text {Subst}\left (\int \left (1+x^2\right ) \, dx,x,-\tan (c+d x)\right )}{d} \\ & = \frac {(3 a A+4 b B) \text {arctanh}(\sin (c+d x))}{8 d}+\frac {(A b+a B) \tan (c+d x)}{d}+\frac {(3 a A+4 b 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 b+a B) \tan ^3(c+d x)}{3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.44 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.75 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 (3 a A+4 b B) \text {arctanh}(\sin (c+d x))+\sec (c+d x) \left (9 a A+12 b B+8 (A b+a B) (2+\cos (2 (c+d x))) \sec (c+d x)+6 a A \sec ^2(c+d x)\right ) \tan (c+d x)}{24 d} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 4.34 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.04

method	result	size
parts	\(\frac {a A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}-\frac {\left (A b +B a \right ) \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(119\)
derivativedivides	\(\frac {a A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-A b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(131\)
default	\(\frac {a A \left (-\left (-\frac {\left (\sec ^{3}\left (d x +c \right )\right )}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )-B a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )-A b \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+B b \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}\)	\(131\)
parallelrisch	\(\frac {-18 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a A +\frac {4 B b}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+18 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) \left (a A +\frac {4 B b}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+32 \left (A b +B a \right ) \sin \left (2 d x +2 c \right )+3 \left (3 a A +4 B b \right ) \sin \left (3 d x +3 c \right )+8 \left (A b +B a \right ) \sin \left (4 d x +4 c \right )+3 \sin \left (d x +c \right ) \left (11 a A +4 B b \right )}{12 d \left (\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )+3\right )}\)	\(192\)
risch	\(-\frac {i \left (9 A a \,{\mathrm e}^{7 i \left (d x +c \right )}+12 B b \,{\mathrm e}^{7 i \left (d x +c \right )}+33 A a \,{\mathrm e}^{5 i \left (d x +c \right )}+12 B b \,{\mathrm e}^{5 i \left (d x +c \right )}-48 A b \,{\mathrm e}^{4 i \left (d x +c \right )}-48 B a \,{\mathrm e}^{4 i \left (d x +c \right )}-33 A a \,{\mathrm e}^{3 i \left (d x +c \right )}-12 B b \,{\mathrm e}^{3 i \left (d x +c \right )}-64 A b \,{\mathrm e}^{2 i \left (d x +c \right )}-64 B a \,{\mathrm e}^{2 i \left (d x +c \right )}-9 a A \,{\mathrm e}^{i \left (d x +c \right )}-12 B b \,{\mathrm e}^{i \left (d x +c \right )}-16 A b -16 B a \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}+\frac {3 a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B b}{2 d}-\frac {3 a A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B b}{2 d}\)	\(266\)
norman	\(\frac {\frac {\left (5 a A -8 A b -8 B a +4 B b \right ) \left (\tan ^{11}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{4 d}+\frac {\left (5 a A +8 A b +8 B a +4 B b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}+\frac {\left (21 a A -8 A b -8 B a -12 B b \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (21 a A +8 A b +8 B a -12 B b \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d}+\frac {\left (39 a A -8 A b -8 B a +12 B b \right ) \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}+\frac {\left (39 a A +8 A b +8 B a +12 B b \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}-\frac {\left (3 a A +4 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {\left (3 a A +4 B b \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\)	\(282\)

[In]

int((a+cos(d*x+c)*b)*(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*b+B*a)/d*(-2/3-1/3*sec
(d*x+c)^2)*tan(d*x+c)+B*b/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 = 0.33 (sec) , antiderivative size = 136, normalized size of antiderivative = 1.19 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, A a + 4 \, B b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, A a + 4 \, B b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, A a + 4 \, B b\right )} \cos \left (d x + c\right )^{2} + 6 \, A a + 8 \, {\left (B a + A b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.20 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.43 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} B a + 16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} A b - 3 \, A a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} - 12 \, B b {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \]

[In]

integrate((a+b*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))*B*a + 16*(tan(d*x + c)^3 + 3*tan(d*x + c))*A*b - 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*b*(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 [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 304 vs. \(2 (106) = 212\).

Time = 0.30 (sec) , antiderivative size = 304, normalized size of antiderivative = 2.67 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {3 \, {\left (3 \, A a + 4 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 3 \, {\left (3 \, A a + 4 \, B b\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \]

[In]

integrate((a+b*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*b)*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 3*(3*A*a + 4*B*b)*log(abs(tan(1/2*d*x + 1/2*c) -
1)) + 2*(15*A*a*tan(1/2*d*x + 1/2*c)^7 - 24*B*a*tan(1/2*d*x + 1/2*c)^7 - 24*A*b*tan(1/2*d*x + 1/2*c)^7 + 12*B*
b*tan(1/2*d*x + 1/2*c)^7 + 9*A*a*tan(1/2*d*x + 1/2*c)^5 + 40*B*a*tan(1/2*d*x + 1/2*c)^5 + 40*A*b*tan(1/2*d*x +
1/2*c)^5 - 12*B*b*tan(1/2*d*x + 1/2*c)^5 + 9*A*a*tan(1/2*d*x + 1/2*c)^3 - 40*B*a*tan(1/2*d*x + 1/2*c)^3 - 40*
A*b*tan(1/2*d*x + 1/2*c)^3 - 12*B*b*tan(1/2*d*x + 1/2*c)^3 + 15*A*a*tan(1/2*d*x + 1/2*c) + 24*B*a*tan(1/2*d*x
+ 1/2*c) + 24*A*b*tan(1/2*d*x + 1/2*c) + 12*B*b*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 = 4.12 (sec) , antiderivative size = 194, normalized size of antiderivative = 1.70 \[ \int (a+b \cos (c+d x)) (A+B \cos (c+d x)) \sec ^5(c+d x) \, dx=\frac {\left (\frac {5\,A\,a}{4}-2\,A\,b-2\,B\,a+B\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {3\,A\,a}{4}+\frac {10\,A\,b}{3}+\frac {10\,B\,a}{3}-B\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,A\,a}{4}-\frac {10\,A\,b}{3}-\frac {10\,B\,a}{3}-B\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {5\,A\,a}{4}+2\,A\,b+2\,B\,a+B\,b\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,A\,a}{4}+B\,b\right )}{d} \]

[In]

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

[Out]

(tan(c/2 + (d*x)/2)*((5*A*a)/4 + 2*A*b + 2*B*a + B*b) + tan(c/2 + (d*x)/2)^7*((5*A*a)/4 - 2*A*b - 2*B*a + B*b)
- tan(c/2 + (d*x)/2)^3*((10*A*b)/3 - (3*A*a)/4 + (10*B*a)/3 + B*b) + tan(c/2 + (d*x)/2)^5*((3*A*a)/4 + (10*A*
b)/3 + (10*B*a)/3 - B*b))/(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)) + (atanh(tan(c/2 + (d*x)/2))*((3*A*a)/4 + B*b))/d