3.355 \(\int \frac{\cos ^3(c+d x) (B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^3} \, dx\)

Optimal. Leaf size=187 \[ -\frac{8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac{(13 B-6 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{4 (19 B-9 C) \sin (c+d x) \cos (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{x (13 B-6 C)}{2 a^3}-\frac{(11 B-6 C) \sin (c+d x) \cos (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(B-C) \sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

((13*B - 6*C)*x)/(2*a^3) - (8*(19*B - 9*C)*Sin[c + d*x])/(15*a^3*d) + ((13*B - 6*C)*Cos[c + d*x]*Sin[c + d*x])
/(2*a^3*d) - ((B - C)*Cos[c + d*x]*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((11*B - 6*C)*Cos[c + d*x]*Sin
[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - (4*(19*B - 9*C)*Cos[c + d*x]*Sin[c + d*x])/(15*d*(a^3 + a^3*Sec[c
+ d*x]))

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Rubi [A]  time = 0.545247, antiderivative size = 187, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 6, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {4072, 4020, 3787, 2635, 8, 2637} \[ -\frac{8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac{(13 B-6 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac{4 (19 B-9 C) \sin (c+d x) \cos (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac{x (13 B-6 C)}{2 a^3}-\frac{(11 B-6 C) \sin (c+d x) \cos (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac{(B-C) \sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]

[Out]

((13*B - 6*C)*x)/(2*a^3) - (8*(19*B - 9*C)*Sin[c + d*x])/(15*a^3*d) + ((13*B - 6*C)*Cos[c + d*x]*Sin[c + d*x])
/(2*a^3*d) - ((B - C)*Cos[c + d*x]*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((11*B - 6*C)*Cos[c + d*x]*Sin
[c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - (4*(19*B - 9*C)*Cos[c + d*x]*Sin[c + d*x])/(15*d*(a^3 + a^3*Sec[c
+ d*x]))

Rule 4072

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]

Rule 4020

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 3787

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 2635

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] && Integer
Q[2*n]

Rule 8

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

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\cos ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^3} \, dx &=\int \frac{\cos ^2(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx\\ &=-\frac{(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac{\int \frac{\cos ^2(c+d x) (a (7 B-2 C)-4 a (B-C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac{(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac{\int \frac{\cos ^2(c+d x) \left (a^2 (43 B-18 C)-3 a^2 (11 B-6 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac{(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{\int \cos ^2(c+d x) \left (15 a^3 (13 B-6 C)-8 a^3 (19 B-9 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac{(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}-\frac{(8 (19 B-9 C)) \int \cos (c+d x) \, dx}{15 a^3}+\frac{(13 B-6 C) \int \cos ^2(c+d x) \, dx}{a^3}\\ &=-\frac{8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac{(13 B-6 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac{(13 B-6 C) \int 1 \, dx}{2 a^3}\\ &=\frac{(13 B-6 C) x}{2 a^3}-\frac{8 (19 B-9 C) \sin (c+d x)}{15 a^3 d}+\frac{(13 B-6 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac{(B-C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac{(11 B-6 C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac{4 (19 B-9 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end{align*}

Mathematica [B]  time = 0.718959, size = 435, normalized size = 2.33 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (600 d x (13 B-6 C) \cos \left (c+\frac{d x}{2}\right )+7560 B \sin \left (c+\frac{d x}{2}\right )-9230 B \sin \left (c+\frac{3 d x}{2}\right )+930 B \sin \left (2 c+\frac{3 d x}{2}\right )-2782 B \sin \left (2 c+\frac{5 d x}{2}\right )-750 B \sin \left (3 c+\frac{5 d x}{2}\right )-105 B \sin \left (3 c+\frac{7 d x}{2}\right )-105 B \sin \left (4 c+\frac{7 d x}{2}\right )+15 B \sin \left (4 c+\frac{9 d x}{2}\right )+15 B \sin \left (5 c+\frac{9 d x}{2}\right )+3900 B d x \cos \left (c+\frac{3 d x}{2}\right )+3900 B d x \cos \left (2 c+\frac{3 d x}{2}\right )+780 B d x \cos \left (2 c+\frac{5 d x}{2}\right )+780 B d x \cos \left (3 c+\frac{5 d x}{2}\right )+600 d x (13 B-6 C) \cos \left (\frac{d x}{2}\right )-12760 B \sin \left (\frac{d x}{2}\right )-4500 C \sin \left (c+\frac{d x}{2}\right )+4860 C \sin \left (c+\frac{3 d x}{2}\right )-900 C \sin \left (2 c+\frac{3 d x}{2}\right )+1452 C \sin \left (2 c+\frac{5 d x}{2}\right )+300 C \sin \left (3 c+\frac{5 d x}{2}\right )+60 C \sin \left (3 c+\frac{7 d x}{2}\right )+60 C \sin \left (4 c+\frac{7 d x}{2}\right )-1800 C d x \cos \left (c+\frac{3 d x}{2}\right )-1800 C d x \cos \left (2 c+\frac{3 d x}{2}\right )-360 C d x \cos \left (2 c+\frac{5 d x}{2}\right )-360 C d x \cos \left (3 c+\frac{5 d x}{2}\right )+7020 C \sin \left (\frac{d x}{2}\right )\right )}{480 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(600*(13*B - 6*C)*d*x*Cos[(d*x)/2] + 600*(13*B - 6*C)*d*x*Cos[c + (d*x)/2] + 3900*B
*d*x*Cos[c + (3*d*x)/2] - 1800*C*d*x*Cos[c + (3*d*x)/2] + 3900*B*d*x*Cos[2*c + (3*d*x)/2] - 1800*C*d*x*Cos[2*c
+ (3*d*x)/2] + 780*B*d*x*Cos[2*c + (5*d*x)/2] - 360*C*d*x*Cos[2*c + (5*d*x)/2] + 780*B*d*x*Cos[3*c + (5*d*x)/
2] - 360*C*d*x*Cos[3*c + (5*d*x)/2] - 12760*B*Sin[(d*x)/2] + 7020*C*Sin[(d*x)/2] + 7560*B*Sin[c + (d*x)/2] - 4
500*C*Sin[c + (d*x)/2] - 9230*B*Sin[c + (3*d*x)/2] + 4860*C*Sin[c + (3*d*x)/2] + 930*B*Sin[2*c + (3*d*x)/2] -
900*C*Sin[2*c + (3*d*x)/2] - 2782*B*Sin[2*c + (5*d*x)/2] + 1452*C*Sin[2*c + (5*d*x)/2] - 750*B*Sin[3*c + (5*d*
x)/2] + 300*C*Sin[3*c + (5*d*x)/2] - 105*B*Sin[3*c + (7*d*x)/2] + 60*C*Sin[3*c + (7*d*x)/2] - 105*B*Sin[4*c +
(7*d*x)/2] + 60*C*Sin[4*c + (7*d*x)/2] + 15*B*Sin[4*c + (9*d*x)/2] + 15*B*Sin[5*c + (9*d*x)/2]))/(480*a^3*d*(1
+ Cos[c + d*x])^3)

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Maple [A]  time = 0.102, size = 292, normalized size = 1.6 \begin{align*} -{\frac{B}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{C}{20\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}}+{\frac{2\,B}{3\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{C}{2\,d{a}^{3}} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{3}}-{\frac{31\,B}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }+{\frac{17\,C}{4\,d{a}^{3}}\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) }-7\,{\frac{ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}B}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{C \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}}{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}-5\,{\frac{B\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+2\,{\frac{C\tan \left ( 1/2\,dx+c/2 \right ) }{d{a}^{3} \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) ^{2}}}+13\,{\frac{B\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{d{a}^{3}}}-6\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) C}{d{a}^{3}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x)

[Out]

-1/20/d/a^3*tan(1/2*d*x+1/2*c)^5*B+1/20/d/a^3*C*tan(1/2*d*x+1/2*c)^5+2/3/d/a^3*tan(1/2*d*x+1/2*c)^3*B-1/2/d/a^
3*C*tan(1/2*d*x+1/2*c)^3-31/4/d/a^3*B*tan(1/2*d*x+1/2*c)+17/4/d/a^3*C*tan(1/2*d*x+1/2*c)-7/d/a^3/(1+tan(1/2*d*
x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*B+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*C*tan(1/2*d*x+1/2*c)^3-5/d/a^3/(1+tan(
1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)+2/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*C*tan(1/2*d*x+1/2*c)+13/d/a^3*B*ar
ctan(tan(1/2*d*x+1/2*c))-6/d/a^3*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [A]  time = 1.45823, size = 435, normalized size = 2.33 \begin{align*} -\frac{B{\left (\frac{60 \,{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac{2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac{\frac{465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{780 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} - 3 \, C{\left (\frac{40 \, \sin \left (d x + c\right )}{{\left (a^{3} + \frac{a^{3} \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{\frac{85 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac{120 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/60*(B*(60*(5*sin(d*x + c)/(cos(d*x + c) + 1) + 7*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^3 + 2*a^3*sin(d*x
+ c)^2/(cos(d*x + c) + 1)^2 + a^3*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (465*sin(d*x + c)/(cos(d*x + c) + 1)
- 40*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3 - 780*arctan(sin(d*x + c
)/(cos(d*x + c) + 1))/a^3) - 3*C*(40*sin(d*x + c)/((a^3 + a^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x +
c) + 1)) + (85*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + sin(d*x + c)^5/(cos(
d*x + c) + 1)^5)/a^3 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3))/d

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Fricas [A]  time = 0.520928, size = 495, normalized size = 2.65 \begin{align*} \frac{15 \,{\left (13 \, B - 6 \, C\right )} d x \cos \left (d x + c\right )^{3} + 45 \,{\left (13 \, B - 6 \, C\right )} d x \cos \left (d x + c\right )^{2} + 45 \,{\left (13 \, B - 6 \, C\right )} d x \cos \left (d x + c\right ) + 15 \,{\left (13 \, B - 6 \, C\right )} d x +{\left (15 \, B \cos \left (d x + c\right )^{4} - 15 \,{\left (3 \, B - 2 \, C\right )} \cos \left (d x + c\right )^{3} -{\left (479 \, B - 234 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \,{\left (239 \, B - 114 \, C\right )} \cos \left (d x + c\right ) - 304 \, B + 144 \, C\right )} \sin \left (d x + c\right )}{30 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

1/30*(15*(13*B - 6*C)*d*x*cos(d*x + c)^3 + 45*(13*B - 6*C)*d*x*cos(d*x + c)^2 + 45*(13*B - 6*C)*d*x*cos(d*x +
c) + 15*(13*B - 6*C)*d*x + (15*B*cos(d*x + c)^4 - 15*(3*B - 2*C)*cos(d*x + c)^3 - (479*B - 234*C)*cos(d*x + c)
^2 - 3*(239*B - 114*C)*cos(d*x + c) - 304*B + 144*C)*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c
)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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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(cos(d*x+c)**3*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.15859, size = 270, normalized size = 1.44 \begin{align*} \frac{\frac{30 \,{\left (d x + c\right )}{\left (13 \, B - 6 \, C\right )}}{a^{3}} - \frac{60 \,{\left (7 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 5 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, 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 )}^{2} a^{3}} - \frac{3 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 40 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 30 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 465 \, B a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 255 \, C a^{12} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{a^{15}}}{60 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

1/60*(30*(d*x + c)*(13*B - 6*C)/a^3 - 60*(7*B*tan(1/2*d*x + 1/2*c)^3 - 2*C*tan(1/2*d*x + 1/2*c)^3 + 5*B*tan(1/
2*d*x + 1/2*c) - 2*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^2*a^3) - (3*B*a^12*tan(1/2*d*x + 1/2*
c)^5 - 3*C*a^12*tan(1/2*d*x + 1/2*c)^5 - 40*B*a^12*tan(1/2*d*x + 1/2*c)^3 + 30*C*a^12*tan(1/2*d*x + 1/2*c)^3 +
465*B*a^12*tan(1/2*d*x + 1/2*c) - 255*C*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d