\(\int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx\) [45]

Optimal result

Integrand size = 31, antiderivative size = 107 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {(3 A-2 B) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {2 (A-B) \tan (c+d x)}{a d}+\frac {(3 A-2 B) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))} \]

[Out]

1/2*(3*A-2*B)*arctanh(sin(d*x+c))/a/d-2*(A-B)*tan(d*x+c)/a/d+1/2*(3*A-2*B)*sec(d*x+c)*tan(d*x+c)/a/d-(A-B)*sec
(d*x+c)*tan(d*x+c)/d/(a+a*cos(d*x+c))

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 107, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3057, 2827, 3853, 3855, 3852, 8} \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {(3 A-2 B) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {2 (A-B) \tan (c+d x)}{a d}+\frac {(3 A-2 B) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac {(A-B) \tan (c+d x) \sec (c+d x)}{d (a \cos (c+d x)+a)} \]

[In]

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

[Out]

((3*A - 2*B)*ArcTanh[Sin[c + d*x]])/(2*a*d) - (2*(A - B)*Tan[c + d*x])/(a*d) + ((3*A - 2*B)*Sec[c + d*x]*Tan[c
+ d*x])/(2*a*d) - ((A - B)*Sec[c + d*x]*Tan[c + d*x])/(d*(a + a*Cos[c + d*x]))

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 3057

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*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*
x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Dist[1/(a*(2*m + 1)*(b*c - a*d)), Int[(a + b*Sin[e + f*x])^(m +
1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b*d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*
(m + n + 2)*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] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c,
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-B) \sec (c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac {\int (a (3 A-2 B)-2 a (A-B) \cos (c+d x)) \sec ^3(c+d x) \, dx}{a^2} \\ & = -\frac {(A-B) \sec (c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac {(3 A-2 B) \int \sec ^3(c+d x) \, dx}{a}-\frac {(2 (A-B)) \int \sec ^2(c+d x) \, dx}{a} \\ & = \frac {(3 A-2 B) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))}+\frac {(3 A-2 B) \int \sec (c+d x) \, dx}{2 a}+\frac {(2 (A-B)) \text {Subst}(\int 1 \, dx,x,-\tan (c+d x))}{a d} \\ & = \frac {(3 A-2 B) \text {arctanh}(\sin (c+d x))}{2 a d}-\frac {2 (A-B) \tan (c+d x)}{a d}+\frac {(3 A-2 B) \sec (c+d x) \tan (c+d x)}{2 a d}-\frac {(A-B) \sec (c+d x) \tan (c+d x)}{d (a+a \cos (c+d x))} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(289\) vs. \(2(107)=214\).

Time = 2.94 (sec) , antiderivative size = 289, normalized size of antiderivative = 2.70 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \left (4 (-A+B) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \left ((-6 A+4 B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+6 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )-4 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )+\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {A}{\left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}-\frac {4 (A-B) \sin (d x)}{\left (\cos \left (\frac {c}{2}\right )-\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+\sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}\right )\right )}{2 a d (1+\cos (c+d x))} \]

[In]

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

[Out]

(Cos[(c + d*x)/2]*(4*(-A + B)*Sec[c/2]*Sin[(d*x)/2] + Cos[(c + d*x)/2]*((-6*A + 4*B)*Log[Cos[(c + d*x)/2] - Si
n[(c + d*x)/2]] + 6*A*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]] - 4*B*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]
+ A/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 - A/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - (4*(A - B)*Sin[d*x])
/((Cos[c/2] - Sin[c/2])*(Cos[c/2] + Sin[c/2])*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])))))/(2*a*d*(1 + Cos[c + d*x]))

Maple [A] (verified)

Time = 1.22 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.18

method	result	size
parallelrisch	\(\frac {-3 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {2 B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \left (1+\cos \left (2 d x +2 c \right )\right ) \left (A -\frac {2 B}{3}\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\left (-2 B +2 A \right ) \cos \left (2 d x +2 c \right )+\left (1+\cos \left (d x +c \right )\right ) \left (A -2 B \right )\right )}{2 a d \left (1+\cos \left (2 d x +2 c \right )\right )}\)	\(126\)
derivativedivides	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-\frac {3 A}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (\frac {3 A}{2}-B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-\frac {3 A}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-\frac {3 A}{2}+B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\)	\(142\)
default	\(\frac {-A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}-\frac {-\frac {3 A}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1}+\left (\frac {3 A}{2}-B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\frac {A}{2 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {-\frac {3 A}{2}+B}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1}+\left (-\frac {3 A}{2}+B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d a}\)	\(142\)
norman	\(\frac {\frac {\left (3 A -B \right ) \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {\left (4 A -3 B \right ) \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (A -B \right ) \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {\left (2 A -3 B \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {\left (3 A -2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 a d}+\frac {\left (3 A -2 B \right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 a d}\)	\(186\)
risch	\(-\frac {i \left (3 A \,{\mathrm e}^{4 i \left (d x +c \right )}-2 B \,{\mathrm e}^{4 i \left (d x +c \right )}+3 A \,{\mathrm e}^{3 i \left (d x +c \right )}-2 B \,{\mathrm e}^{3 i \left (d x +c \right )}+5 A \,{\mathrm e}^{2 i \left (d x +c \right )}-6 B \,{\mathrm e}^{2 i \left (d x +c \right )}+A \,{\mathrm e}^{i \left (d x +c \right )}-2 B \,{\mathrm e}^{i \left (d x +c \right )}+4 A -4 B \right )}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 a d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right ) B}{a d}+\frac {3 A \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 a d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right ) B}{a d}\)	\(226\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.46 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {{\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{3} + {\left (3 \, A - 2 \, B\right )} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, {\left (A - B\right )} \cos \left (d x + c\right )^{2} + {\left (A - 2 \, B\right )} \cos \left (d x + c\right ) - A\right )} \sin \left (d x + c\right )}{4 \, {\left (a d \cos \left (d x + c\right )^{3} + a d \cos \left (d x + c\right )^{2}\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\int \frac {A \sec ^{3}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}}{\cos {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

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

[Out]

(Integral(A*sec(c + d*x)**3/(cos(c + d*x) + 1), x) + Integral(B*cos(c + d*x)*sec(c + d*x)**3/(cos(c + d*x) + 1
), x))/a

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 282 vs. \(2 (103) = 206\).

Time = 0.21 (sec) , antiderivative size = 282, normalized size of antiderivative = 2.64 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=-\frac {A {\left (\frac {2 \, {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a - \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac {3 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac {2 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} + 2 \, B {\left (\frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} - \frac {\log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} - \frac {2 \, \sin \left (d x + c\right )}{{\left (a - \frac {a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} - \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{2 \, d} \]

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 0.32 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.47 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {\frac {{\left (3 \, A - 2 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac {{\left (3 \, A - 2 \, B\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a} - \frac {2 \, {\left (A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} + \frac {2 \, {\left (3 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2} a}}{2 \, d} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.43 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.11 \[ \int \frac {(A+B \cos (c+d x)) \sec ^3(c+d x)}{a+a \cos (c+d x)} \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (3\,A-2\,B\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-2\,B\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}+\frac {2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (\frac {3\,A}{2}-B\right )}{a\,d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (A-B\right )}{a\,d} \]

[In]

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

[Out]

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