\(\int \frac {\cos ^4(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{a+a \sec (c+d x)} \, dx\) [339]

Optimal result

Integrand size = 40, antiderivative size = 122 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {3 (B-C) x}{2 a}+\frac {(4 B-3 C) \sin (c+d x)}{a d}-\frac {3 (B-C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(4 B-3 C) \sin ^3(c+d x)}{3 a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4157, 4105, 3872, 2713, 2715, 8} \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {(4 B-3 C) \sin ^3(c+d x)}{3 a d}+\frac {(4 B-3 C) \sin (c+d x)}{a d}-\frac {3 (B-C) \sin (c+d x) \cos (c+d x)}{2 a d}-\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{d (a \sec (c+d x)+a)}-\frac {3 x (B-C)}{2 a} \]

[In]

Int[(Cos[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]

[Out]

(-3*(B - C)*x)/(2*a) + ((4*B - 3*C)*Sin[c + d*x])/(a*d) - (3*(B - C)*Cos[c + d*x]*Sin[c + d*x])/(2*a*d) - ((B
- C)*Cos[c + d*x]^2*Sin[c + d*x])/(d*(a + a*Sec[c + d*x])) - ((4*B - 3*C)*Sin[c + d*x]^3)/(3*a*d)

Rule 8

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

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 2715

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

Rule 3872

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(d*
Csc[e + f*x])^n, x], x] + Dist[b/d, Int[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]

Rule 4105

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(b*f*(
2*m + 1))), x] - Dist[1/(a^2*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*B*n - a*A*
(2*m + n + 1) + (A*b - a*B)*(m + n + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, n}, x] && NeQ[
A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)] &&  !GtQ[n, 0]

Rule 4157

Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(
x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.), x_Symbol] :> Dist[1/b^2, Int[(a + b*Csc[e + f*x])
^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {\cos ^3(c+d x) (B+C \sec (c+d x))}{a+a \sec (c+d x)} \, dx \\ & = -\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {\int \cos ^3(c+d x) (a (4 B-3 C)-3 a (B-C) \sec (c+d x)) \, dx}{a^2} \\ & = -\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac {(4 B-3 C) \int \cos ^3(c+d x) \, dx}{a}-\frac {(3 (B-C)) \int \cos ^2(c+d x) \, dx}{a} \\ & = -\frac {3 (B-C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(3 (B-C)) \int 1 \, dx}{2 a}-\frac {(4 B-3 C) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a d} \\ & = -\frac {3 (B-C) x}{2 a}+\frac {(4 B-3 C) \sin (c+d x)}{a d}-\frac {3 (B-C) \cos (c+d x) \sin (c+d x)}{2 a d}-\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac {(4 B-3 C) \sin ^3(c+d x)}{3 a d} \\ \end{align*}

Mathematica [B] (verified)

Leaf count is larger than twice the leaf count of optimal. \(249\) vs. \(2(122)=244\).

Time = 0.96 (sec) , antiderivative size = 249, normalized size of antiderivative = 2.04 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \left (-36 (B-C) d x \cos \left (\frac {d x}{2}\right )-36 (B-C) d x \cos \left (c+\frac {d x}{2}\right )+69 B \sin \left (\frac {d x}{2}\right )-60 C \sin \left (\frac {d x}{2}\right )+21 B \sin \left (c+\frac {d x}{2}\right )-12 C \sin \left (c+\frac {d x}{2}\right )+18 B \sin \left (c+\frac {3 d x}{2}\right )-9 C \sin \left (c+\frac {3 d x}{2}\right )+18 B \sin \left (2 c+\frac {3 d x}{2}\right )-9 C \sin \left (2 c+\frac {3 d x}{2}\right )-2 B \sin \left (2 c+\frac {5 d x}{2}\right )+3 C \sin \left (2 c+\frac {5 d x}{2}\right )-2 B \sin \left (3 c+\frac {5 d x}{2}\right )+3 C \sin \left (3 c+\frac {5 d x}{2}\right )+B \sin \left (3 c+\frac {7 d x}{2}\right )+B \sin \left (4 c+\frac {7 d x}{2}\right )\right )}{24 a d (1+\cos (c+d x))} \]

[In]

Integrate[(Cos[c + d*x]^4*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x]),x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-36*(B - C)*d*x*Cos[(d*x)/2] - 36*(B - C)*d*x*Cos[c + (d*x)/2] + 69*B*Sin[(d*x)/2]
- 60*C*Sin[(d*x)/2] + 21*B*Sin[c + (d*x)/2] - 12*C*Sin[c + (d*x)/2] + 18*B*Sin[c + (3*d*x)/2] - 9*C*Sin[c + (
3*d*x)/2] + 18*B*Sin[2*c + (3*d*x)/2] - 9*C*Sin[2*c + (3*d*x)/2] - 2*B*Sin[2*c + (5*d*x)/2] + 3*C*Sin[2*c + (5
*d*x)/2] - 2*B*Sin[3*c + (5*d*x)/2] + 3*C*Sin[3*c + (5*d*x)/2] + B*Sin[3*c + (7*d*x)/2] + B*Sin[4*c + (7*d*x)/
2]))/(24*a*d*(1 + Cos[c + d*x]))

Maple [A] (verified)

Time = 0.23 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.64

method	result	size
parallelrisch	\(\frac {\left (\left (-B +3 C \right ) \cos \left (2 d x +2 c \right )+B \cos \left (3 d x +3 c \right )+\left (17 B -6 C \right ) \cos \left (d x +c \right )+31 B -21 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-18 d x \left (B -C \right )}{12 d a}\)	\(78\)
derivativedivides	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {2 \left (\left (\frac {3 C}{2}-\frac {5 B}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {8 B}{3}+2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {C}{2}-\frac {3 B}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-3 \left (B -C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(122\)
default	\(\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) B -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) C -\frac {2 \left (\left (\frac {3 C}{2}-\frac {5 B}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}+\left (-\frac {8 B}{3}+2 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (\frac {C}{2}-\frac {3 B}{2}\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{\left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3}}-3 \left (B -C \right ) \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}\)	\(122\)
risch	\(-\frac {3 B x}{2 a}+\frac {3 x C}{2 a}-\frac {7 i B \,{\mathrm e}^{i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{i \left (d x +c \right )} C}{2 a d}+\frac {7 i B \,{\mathrm e}^{-i \left (d x +c \right )}}{8 a d}-\frac {i {\mathrm e}^{-i \left (d x +c \right )} C}{2 a d}+\frac {2 i B}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}-\frac {2 i C}{d a \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {B \sin \left (3 d x +3 c \right )}{12 a d}-\frac {B \sin \left (2 d x +2 c \right )}{4 a d}+\frac {\sin \left (2 d x +2 c \right ) C}{4 a d}\)	\(192\)
norman	\(\frac {\frac {\left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{a d}+\frac {3 \left (B -C \right ) x}{2 a}-\frac {4 \left (B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{a d}+\frac {9 \left (B -C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{2 a}+\frac {3 \left (B -C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{a}-\frac {3 \left (B -C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{a}-\frac {9 \left (B -C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}}{2 a}-\frac {3 \left (B -C \right ) x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{10}}{2 a}-\frac {2 \left (2 B -C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}+\frac {2 \left (4 B -3 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{a d}+\frac {2 \left (11 B -9 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{3 a d}-\frac {\left (25 B -21 C \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3 a d}}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \left (1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}\)	\(303\)

[In]

int(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/12*(((-B+3*C)*cos(2*d*x+2*c)+B*cos(3*d*x+3*c)+(17*B-6*C)*cos(d*x+c)+31*B-21*C)*tan(1/2*d*x+1/2*c)-18*d*x*(B-
C))/d/a

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 97, normalized size of antiderivative = 0.80 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {9 \, {\left (B - C\right )} d x \cos \left (d x + c\right ) + 9 \, {\left (B - C\right )} d x - {\left (2 \, B \cos \left (d x + c\right )^{3} - {\left (B - 3 \, C\right )} \cos \left (d x + c\right )^{2} + {\left (7 \, B - 3 \, C\right )} \cos \left (d x + c\right ) + 16 \, B - 12 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a d \cos \left (d x + c\right ) + a d\right )}} \]

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(9*(B - C)*d*x*cos(d*x + c) + 9*(B - C)*d*x - (2*B*cos(d*x + c)^3 - (B - 3*C)*cos(d*x + c)^2 + (7*B - 3*C
)*cos(d*x + c) + 16*B - 12*C)*sin(d*x + c))/(a*d*cos(d*x + c) + a*d)

Sympy [F]

\[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\int \frac {B \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx + \int \frac {C \cos ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}}{\sec {\left (c + d x \right )} + 1}\, dx}{a} \]

[In]

integrate(cos(d*x+c)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c)),x)

[Out]

(Integral(B*cos(c + d*x)**4*sec(c + d*x)/(sec(c + d*x) + 1), x) + Integral(C*cos(c + d*x)**4*sec(c + d*x)**2/(
sec(c + d*x) + 1), x))/a

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 310 vs. \(2 (116) = 232\).

Time = 0.33 (sec) , antiderivative size = 310, normalized size of antiderivative = 2.54 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {B {\left (\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {16 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a + \frac {3 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} - \frac {9 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {3 \, \sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, C {\left (\frac {\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a + \frac {2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} - \frac {3 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a} + \frac {\sin \left (d x + c\right )}{a {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

1/3*(B*((9*sin(d*x + c)/(cos(d*x + c) + 1) + 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d
*x + c) + 1)^5)/(a + 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin
(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 3*sin(d*x + c)/(a*(cos(d*x +
c) + 1))) - 3*C*((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*arctan(sin(d*x + c)/(cos(d*x + c) + 1)
)/a + sin(d*x + c)/(a*(cos(d*x + c) + 1))))/d

Giac [A] (verification not implemented)

none

Time = 0.28 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.24 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=-\frac {\frac {9 \, {\left (d x + c\right )} {\left (B - C\right )}}{a} - \frac {6 \, {\left (B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{a} - \frac {2 \, {\left (15 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 16 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a}}{6 \, d} \]

[In]

integrate(cos(d*x+c)^4*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

-1/6*(9*(d*x + c)*(B - C)/a - 6*(B*tan(1/2*d*x + 1/2*c) - C*tan(1/2*d*x + 1/2*c))/a - 2*(15*B*tan(1/2*d*x + 1/
2*c)^5 - 9*C*tan(1/2*d*x + 1/2*c)^5 + 16*B*tan(1/2*d*x + 1/2*c)^3 - 12*C*tan(1/2*d*x + 1/2*c)^3 + 9*B*tan(1/2*
d*x + 1/2*c) - 3*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^3*a))/d

Mupad [B] (verification not implemented)

Time = 17.00 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.13 \[ \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx=\frac {\left (5\,B-3\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {16\,B}{3}-4\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (3\,B-C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a\right )}-\frac {3\,x\,\left (B-C\right )}{2\,a}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (B-C\right )}{a\,d} \]

[In]

int((cos(c + d*x)^4*(B/cos(c + d*x) + C/cos(c + d*x)^2))/(a + a/cos(c + d*x)),x)

[Out]

(tan(c/2 + (d*x)/2)^5*(5*B - 3*C) + tan(c/2 + (d*x)/2)^3*((16*B)/3 - 4*C) + tan(c/2 + (d*x)/2)*(3*B - C))/(d*(
a + 3*a*tan(c/2 + (d*x)/2)^2 + 3*a*tan(c/2 + (d*x)/2)^4 + a*tan(c/2 + (d*x)/2)^6)) - (3*x*(B - C))/(2*a) + (ta
n(c/2 + (d*x)/2)*(B - C))/(a*d)