3.260 \(\int \frac{(B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^5(c+d x)}{a+a \cos (c+d x)} \, dx\)

Optimal. Leaf size=131 \[ \frac{(4 B-3 C) \tan ^3(c+d x)}{3 a d}+\frac{(4 B-3 C) \tan (c+d x)}{a d}-\frac{3 (B-C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{3 (B-C) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac{(B-C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)} \]

[Out]

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

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Rubi [A]  time = 0.264745, antiderivative size = 131, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {3029, 2978, 2748, 3767, 3768, 3770} \[ \frac{(4 B-3 C) \tan ^3(c+d x)}{3 a d}+\frac{(4 B-3 C) \tan (c+d x)}{a d}-\frac{3 (B-C) \tanh ^{-1}(\sin (c+d x))}{2 a d}-\frac{3 (B-C) \tan (c+d x) \sec (c+d x)}{2 a d}-\frac{(B-C) \tan (c+d x) \sec ^2(c+d x)}{d (a \cos (c+d x)+a)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3029

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

Rule 2978

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 2748

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 3767

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 3768

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 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [B]  time = 4.39558, size = 490, normalized size = 3.74 \[ \frac{\cos \left (\frac{1}{2} (c+d x)\right ) \left (\sec \left (\frac{c}{2}\right ) \sec (c) \sec ^3(c+d x) \left (-24 B \sin \left (c-\frac{d x}{2}\right )-6 B \sin \left (c+\frac{d x}{2}\right )-24 B \sin \left (2 c+\frac{d x}{2}\right )+21 B \sin \left (c+\frac{3 d x}{2}\right )+9 B \sin \left (2 c+\frac{3 d x}{2}\right )-9 B \sin \left (3 c+\frac{3 d x}{2}\right )+7 B \sin \left (c+\frac{5 d x}{2}\right )+B \sin \left (2 c+\frac{5 d x}{2}\right )-3 B \sin \left (3 c+\frac{5 d x}{2}\right )-9 B \sin \left (4 c+\frac{5 d x}{2}\right )+16 B \sin \left (2 c+\frac{7 d x}{2}\right )+10 B \sin \left (3 c+\frac{7 d x}{2}\right )+6 B \sin \left (4 c+\frac{7 d x}{2}\right )+6 (B+C) \sin \left (\frac{d x}{2}\right )+3 (13 B-9 C) \sin \left (\frac{3 d x}{2}\right )+12 C \sin \left (c-\frac{d x}{2}\right )+6 C \sin \left (c+\frac{d x}{2}\right )+24 C \sin \left (2 c+\frac{d x}{2}\right )-9 C \sin \left (c+\frac{3 d x}{2}\right )-9 C \sin \left (2 c+\frac{3 d x}{2}\right )+9 C \sin \left (3 c+\frac{3 d x}{2}\right )-3 C \sin \left (c+\frac{5 d x}{2}\right )+3 C \sin \left (2 c+\frac{5 d x}{2}\right )+3 C \sin \left (3 c+\frac{5 d x}{2}\right )+9 C \sin \left (4 c+\frac{5 d x}{2}\right )-12 C \sin \left (2 c+\frac{7 d x}{2}\right )-6 C \sin \left (3 c+\frac{7 d x}{2}\right )-6 C \sin \left (4 c+\frac{7 d x}{2}\right )\right )+144 (B-C) \cos \left (\frac{1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )-\log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )\right )}{48 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*(144*(B - C)*Cos[(c + d*x)/2]*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sec[c]*Sec[c + d*x]^3*(6*(B + C)*Sin[(d*x)/2] + 3*(13*B - 9*C)*Sin[(3*d*x)/
2] - 24*B*Sin[c - (d*x)/2] + 12*C*Sin[c - (d*x)/2] - 6*B*Sin[c + (d*x)/2] + 6*C*Sin[c + (d*x)/2] - 24*B*Sin[2*
c + (d*x)/2] + 24*C*Sin[2*c + (d*x)/2] + 21*B*Sin[c + (3*d*x)/2] - 9*C*Sin[c + (3*d*x)/2] + 9*B*Sin[2*c + (3*d
*x)/2] - 9*C*Sin[2*c + (3*d*x)/2] - 9*B*Sin[3*c + (3*d*x)/2] + 9*C*Sin[3*c + (3*d*x)/2] + 7*B*Sin[c + (5*d*x)/
2] - 3*C*Sin[c + (5*d*x)/2] + B*Sin[2*c + (5*d*x)/2] + 3*C*Sin[2*c + (5*d*x)/2] - 3*B*Sin[3*c + (5*d*x)/2] + 3
*C*Sin[3*c + (5*d*x)/2] - 9*B*Sin[4*c + (5*d*x)/2] + 9*C*Sin[4*c + (5*d*x)/2] + 16*B*Sin[2*c + (7*d*x)/2] - 12
*C*Sin[2*c + (7*d*x)/2] + 10*B*Sin[3*c + (7*d*x)/2] - 6*C*Sin[3*c + (7*d*x)/2] + 6*B*Sin[4*c + (7*d*x)/2] - 6*
C*Sin[4*c + (7*d*x)/2])))/(48*a*d*(1 + Cos[c + d*x]))

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Maple [B]  time = 0.066, size = 340, normalized size = 2.6 \begin{align*}{\frac{B}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{C}{da}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-{\frac{B}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}+{\frac{C}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}-{\frac{B}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{3\,B}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{3\,C}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }-{\frac{5\,B}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{3\,C}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-{\frac{B}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}+{\frac{B}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{C}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{3\,B}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{3\,C}{2\,da}\ln \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{5\,B}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}+{\frac{3\,C}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a/d*B*tan(1/2*d*x+1/2*c)-1/a/d*C*tan(1/2*d*x+1/2*c)-1/3/a/d*B/(tan(1/2*d*x+1/2*c)-1)^3+1/2/a/d/(tan(1/2*d*x+
1/2*c)-1)^2*C-1/a/d*B/(tan(1/2*d*x+1/2*c)-1)^2+3/2/a/d*B*ln(tan(1/2*d*x+1/2*c)-1)-3/2/a/d*ln(tan(1/2*d*x+1/2*c
)-1)*C-5/2/a/d*B/(tan(1/2*d*x+1/2*c)-1)+3/2/a/d/(tan(1/2*d*x+1/2*c)-1)*C-1/3/a/d*B/(tan(1/2*d*x+1/2*c)+1)^3+1/
a/d*B/(tan(1/2*d*x+1/2*c)+1)^2-1/2/a/d/(tan(1/2*d*x+1/2*c)+1)^2*C-3/2/a/d*B*ln(tan(1/2*d*x+1/2*c)+1)+3/2/a/d*l
n(tan(1/2*d*x+1/2*c)+1)*C-5/2/a/d*B/(tan(1/2*d*x+1/2*c)+1)+3/2/a/d/(tan(1/2*d*x+1/2*c)+1)*C

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Maxima [B]  time = 1.20924, size = 497, normalized size = 3.79 \begin{align*} \frac{B{\left (\frac{2 \,{\left (\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}}\right )}}{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 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a} + \frac{9 \, \log \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a} + \frac{6 \, \sin \left (d x + c\right )}{a{\left (\cos \left (d x + c\right ) + 1\right )}}\right )} - 3 \, C{\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 )}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(B*(2*(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*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6) - 9*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a + 9*log(sin(d*x + c)/(cos(d
*x + c) + 1) - 1)/a + 6*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 3*C*(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*si
n(d*x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 1.76172, size = 419, normalized size = 3.2 \begin{align*} -\frac{9 \,{\left ({\left (B - C\right )} \cos \left (d x + c\right )^{4} +{\left (B - C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \,{\left ({\left (B - C\right )} \cos \left (d x + c\right )^{4} +{\left (B - C\right )} \cos \left (d x + c\right )^{3}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (4 \,{\left (4 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{3} +{\left (7 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{2} -{\left (B - 3 \, C\right )} \cos \left (d x + c\right ) + 2 \, B\right )} \sin \left (d x + c\right )}{12 \,{\left (a d \cos \left (d x + c\right )^{4} + a d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(9*((B - C)*cos(d*x + c)^4 + (B - C)*cos(d*x + c)^3)*log(sin(d*x + c) + 1) - 9*((B - C)*cos(d*x + c)^4 +
(B - C)*cos(d*x + c)^3)*log(-sin(d*x + c) + 1) - 2*(4*(4*B - 3*C)*cos(d*x + c)^3 + (7*B - 3*C)*cos(d*x + c)^2
- (B - 3*C)*cos(d*x + c) + 2*B)*sin(d*x + c))/(a*d*cos(d*x + c)^4 + a*d*cos(d*x + c)^3)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.34002, size = 246, normalized size = 1.88 \begin{align*} -\frac{\frac{9 \,{\left (B - C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1 \right |}\right )}{a} - \frac{9 \,{\left (B - C\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(9*(B - C)*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a - 9*(B - C)*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a - 6*(B*t
an(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