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

Optimal. Leaf size=239 \[ -\frac {8 (216 A-83 B+20 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B+2 C) \sin (c+d x) \cos (c+d x)}{2 a^4 d}-\frac {4 (216 A-83 B+20 C) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(129 A-52 B+10 C) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {x (21 A-8 B+2 C)}{2 a^4}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {(2 A-B) \sin (c+d x) \cos (c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

[Out]

1/2*(21*A-8*B+2*C)*x/a^4-8/105*(216*A-83*B+20*C)*sin(d*x+c)/a^4/d+1/2*(21*A-8*B+2*C)*cos(d*x+c)*sin(d*x+c)/a^4
/d-1/105*(129*A-52*B+10*C)*cos(d*x+c)*sin(d*x+c)/a^4/d/(1+sec(d*x+c))^2-4/105*(216*A-83*B+20*C)*cos(d*x+c)*sin
(d*x+c)/a^4/d/(1+sec(d*x+c))-1/7*(A-B+C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^4-1/5*(2*A-B)*cos(d*x+c)*sin
(d*x+c)/a/d/(a+a*sec(d*x+c))^3

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Rubi [A]  time = 0.70, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 41, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.146, Rules used = {4084, 4020, 3787, 2635, 8, 2637} \[ -\frac {8 (216 A-83 B+20 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B+2 C) \sin (c+d x) \cos (c+d x)}{2 a^4 d}-\frac {4 (216 A-83 B+20 C) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(129 A-52 B+10 C) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {x (21 A-8 B+2 C)}{2 a^4}-\frac {(A-B+C) \sin (c+d x) \cos (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {(2 A-B) \sin (c+d x) \cos (c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((21*A - 8*B + 2*C)*x)/(2*a^4) - (8*(216*A - 83*B + 20*C)*Sin[c + d*x])/(105*a^4*d) + ((21*A - 8*B + 2*C)*Cos[
c + d*x]*Sin[c + d*x])/(2*a^4*d) - ((129*A - 52*B + 10*C)*Cos[c + d*x]*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d
*x])^2) - (4*(216*A - 83*B + 20*C)*Cos[c + d*x]*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A - B + C)*Co
s[c + d*x]*Sin[c + d*x])/(7*d*(a + a*Sec[c + d*x])^4) - ((2*A - B)*Cos[c + d*x]*Sin[c + d*x])/(5*a*d*(a + a*Se
c[c + d*x])^3)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 2637

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

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 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 4084

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

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}+\frac {\int \frac {\cos ^2(c+d x) (a (9 A-2 B+2 C)-a (5 A-5 B-2 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x) \left (a^2 (73 A-24 B+10 C)-28 a^2 (2 A-B) \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(129 A-52 B+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}+\frac {\int \frac {\cos ^2(c+d x) \left (a^3 (477 A-176 B+50 C)-3 a^3 (129 A-52 B+10 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(129 A-52 B+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {\int \cos ^2(c+d x) \left (105 a^4 (21 A-8 B+2 C)-8 a^4 (216 A-83 B+20 C) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac {(129 A-52 B+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(21 A-8 B+2 C) \int \cos ^2(c+d x) \, dx}{a^4}-\frac {(8 (216 A-83 B+20 C)) \int \cos (c+d x) \, dx}{105 a^4}\\ &=-\frac {8 (216 A-83 B+20 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac {(129 A-52 B+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(21 A-8 B+2 C) \int 1 \, dx}{2 a^4}\\ &=\frac {(21 A-8 B+2 C) x}{2 a^4}-\frac {8 (216 A-83 B+20 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B+2 C) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac {(129 A-52 B+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {(2 A-B) \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {4 (216 A-83 B+20 C) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [A]  time = 5.38, size = 345, normalized size = 1.44 \[ \frac {4 \cos \left (\frac {1}{2} (c+d x)\right ) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right ) \left (210 \cos ^7\left (\frac {1}{2} (c+d x)\right ) (4 (B-4 A) \sin (c+d x)+2 d x (21 A-8 B+2 C)+A \sin (2 (c+d x)))+4 \tan \left (\frac {c}{2}\right ) (447 A-286 B+160 C) \cos ^5\left (\frac {1}{2} (c+d x)\right )-6 \tan \left (\frac {c}{2}\right ) (39 A-32 B+25 C) \cos ^3\left (\frac {1}{2} (c+d x)\right )+15 \tan \left (\frac {c}{2}\right ) (A-B+C) \cos \left (\frac {1}{2} (c+d x)\right )+15 \sec \left (\frac {c}{2}\right ) (A-B+C) \sin \left (\frac {d x}{2}\right )-8 \sec \left (\frac {c}{2}\right ) (1653 A-764 B+260 C) \sin \left (\frac {d x}{2}\right ) \cos ^6\left (\frac {1}{2} (c+d x)\right )+4 \sec \left (\frac {c}{2}\right ) (447 A-286 B+160 C) \sin \left (\frac {d x}{2}\right ) \cos ^4\left (\frac {1}{2} (c+d x)\right )-6 \sec \left (\frac {c}{2}\right ) (39 A-32 B+25 C) \sin \left (\frac {d x}{2}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right )\right )}{105 a^4 d (\cos (c+d x)+1)^4 (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]*(C + B*Cos[c + d*x] + A*Cos[c + d*x]^2)*(15*(A - B + C)*Sec[c/2]*Sin[(d*x)/2] - 6*(39*A -
32*B + 25*C)*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + 4*(447*A - 286*B + 160*C)*Cos[(c + d*x)/2]^4*Sec[c/2]*
Sin[(d*x)/2] - 8*(1653*A - 764*B + 260*C)*Cos[(c + d*x)/2]^6*Sec[c/2]*Sin[(d*x)/2] + 210*Cos[(c + d*x)/2]^7*(2
*(21*A - 8*B + 2*C)*d*x + 4*(-4*A + B)*Sin[c + d*x] + A*Sin[2*(c + d*x)]) + 15*(A - B + C)*Cos[(c + d*x)/2]*Ta
n[c/2] - 6*(39*A - 32*B + 25*C)*Cos[(c + d*x)/2]^3*Tan[c/2] + 4*(447*A - 286*B + 160*C)*Cos[(c + d*x)/2]^5*Tan
[c/2]))/(105*a^4*d*(1 + Cos[c + d*x])^4*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)]))

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fricas [A]  time = 0.43, size = 267, normalized size = 1.12 \[ \frac {105 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} d x \cos \left (d x + c\right )^{4} + 420 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} d x \cos \left (d x + c\right )^{3} + 630 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} d x \cos \left (d x + c\right )^{2} + 420 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} d x \cos \left (d x + c\right ) + 105 \, {\left (21 \, A - 8 \, B + 2 \, C\right )} d x + {\left (105 \, A \cos \left (d x + c\right )^{5} - 210 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (1509 \, A - 592 \, B + 130 \, C\right )} \cos \left (d x + c\right )^{3} - 4 \, {\left (3411 \, A - 1318 \, B + 310 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (11619 \, A - 4472 \, B + 1070 \, C\right )} \cos \left (d x + c\right ) - 3456 \, A + 1328 \, B - 320 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(105*(21*A - 8*B + 2*C)*d*x*cos(d*x + c)^4 + 420*(21*A - 8*B + 2*C)*d*x*cos(d*x + c)^3 + 630*(21*A - 8*B
+ 2*C)*d*x*cos(d*x + c)^2 + 420*(21*A - 8*B + 2*C)*d*x*cos(d*x + c) + 105*(21*A - 8*B + 2*C)*d*x + (105*A*cos
(d*x + c)^5 - 210*(2*A - B)*cos(d*x + c)^4 - 4*(1509*A - 592*B + 130*C)*cos(d*x + c)^3 - 4*(3411*A - 1318*B +
310*C)*cos(d*x + c)^2 - (11619*A - 4472*B + 1070*C)*cos(d*x + c) - 3456*A + 1328*B - 320*C)*sin(d*x + c))/(a^4
*d*cos(d*x + c)^4 + 4*a^4*d*cos(d*x + c)^3 + 6*a^4*d*cos(d*x + c)^2 + 4*a^4*d*cos(d*x + c) + a^4*d)

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giac [A]  time = 0.30, size = 302, normalized size = 1.26 \[ \frac {\frac {420 \, {\left (d x + c\right )} {\left (21 \, A - 8 \, B + 2 \, C\right )}}{a^{4}} - \frac {840 \, {\left (9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 189 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 147 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1365 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 805 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 385 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11655 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5145 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(420*(d*x + c)*(21*A - 8*B + 2*C)/a^4 - 840*(9*A*tan(1/2*d*x + 1/2*c)^3 - 2*B*tan(1/2*d*x + 1/2*c)^3 + 7
*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^4) + (15*A*a^24*tan(1/2*
d*x + 1/2*c)^7 - 15*B*a^24*tan(1/2*d*x + 1/2*c)^7 + 15*C*a^24*tan(1/2*d*x + 1/2*c)^7 - 189*A*a^24*tan(1/2*d*x
+ 1/2*c)^5 + 147*B*a^24*tan(1/2*d*x + 1/2*c)^5 - 105*C*a^24*tan(1/2*d*x + 1/2*c)^5 + 1365*A*a^24*tan(1/2*d*x +
1/2*c)^3 - 805*B*a^24*tan(1/2*d*x + 1/2*c)^3 + 385*C*a^24*tan(1/2*d*x + 1/2*c)^3 - 11655*A*a^24*tan(1/2*d*x +
1/2*c) + 5145*B*a^24*tan(1/2*d*x + 1/2*c) - 1575*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 1.22, size = 429, normalized size = 1.79 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}-\frac {B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}+\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {9 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {7 B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{8 d \,a^{4}}+\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{8 d \,a^{4}}-\frac {23 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}+\frac {11 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{24 d \,a^{4}}-\frac {111 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {49 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {15 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {9 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {21 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}}-\frac {8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{4}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/56/d/a^4*tan(1/2*d*x+1/2*c)^7*A-1/56/d/a^4*B*tan(1/2*d*x+1/2*c)^7+1/56/d/a^4*C*tan(1/2*d*x+1/2*c)^7-9/40/d/a
^4*A*tan(1/2*d*x+1/2*c)^5+7/40/d/a^4*B*tan(1/2*d*x+1/2*c)^5-1/8/d/a^4*tan(1/2*d*x+1/2*c)^5*C+13/8/d/a^4*tan(1/
2*d*x+1/2*c)^3*A-23/24/d/a^4*B*tan(1/2*d*x+1/2*c)^3+11/24/d/a^4*tan(1/2*d*x+1/2*c)^3*C-111/8/d/a^4*A*tan(1/2*d
*x+1/2*c)+49/8/d/a^4*B*tan(1/2*d*x+1/2*c)-15/8/d/a^4*C*tan(1/2*d*x+1/2*c)-9/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*t
an(1/2*d*x+1/2*c)^3*A+2/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*tan(1/2*d*x+1/2*c)^3*B-7/d/a^4/(1+tan(1/2*d*x+1/2*c)^
2)^2*A*tan(1/2*d*x+1/2*c)+2/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)+21/d/a^4*A*arctan(tan(1/2*d*
x+1/2*c))-8/d/a^4*arctan(tan(1/2*d*x+1/2*c))*B+2/d/a^4*arctan(tan(1/2*d*x+1/2*c))*C

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maxima [B]  time = 0.49, size = 474, normalized size = 1.98 \[ -\frac {3 \, A {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} + \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {5880 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - B {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} + \frac {a^{4} \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 {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {6720 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} + 5 \, C {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/840*(3*A*(280*(7*sin(d*x + c)/(cos(d*x + c) + 1) + 9*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^4 + 2*a^4*sin(
d*x + c)^2/(cos(d*x + c) + 1)^2 + a^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) + (3885*sin(d*x + c)/(cos(d*x + c)
+ 1) - 455*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 63*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 5*sin(d*x + c)^7/(co
s(d*x + c) + 1)^7)/a^4 - 5880*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4) - B*(1680*sin(d*x + c)/((a^4 + a^4*
sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)) + (5145*sin(d*x + c)/(cos(d*x + c) + 1) - 805*sin(d*x
+ c)^3/(cos(d*x + c) + 1)^3 + 147*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 15*sin(d*x + c)^7/(cos(d*x + c) + 1)^
7)/a^4 - 6720*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4) + 5*C*((315*sin(d*x + c)/(cos(d*x + c) + 1) - 77*si
n(d*x + c)^3/(cos(d*x + c) + 1)^3 + 21*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 3*sin(d*x + c)^7/(cos(d*x + c) +
1)^7)/a^4 - 336*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^4))/d

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mupad [B]  time = 3.27, size = 285, normalized size = 1.19 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {6\,A-4\,B+2\,C}{8\,a^4}-\frac {5\,B-15\,A+C}{24\,a^4}+\frac {A-B+C}{4\,a^4}\right )}{d}-\frac {\left (9\,A-2\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (7\,A-2\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {6\,A-4\,B+2\,C}{40\,a^4}+\frac {3\,\left (A-B+C\right )}{40\,a^4}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (6\,A-4\,B+2\,C\right )}{4\,a^4}-\frac {3\,\left (5\,B-15\,A+C\right )}{8\,a^4}+\frac {5\,\left (A-B+C\right )}{4\,a^4}+\frac {20\,A-4\,C}{8\,a^4}\right )}{d}+\frac {x\,\left (21\,A-8\,B+2\,C\right )}{2\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A-B+C\right )}{56\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^3*((6*A - 4*B + 2*C)/(8*a^4) - (5*B - 15*A + C)/(24*a^4) + (A - B + C)/(4*a^4)))/d - (tan(
c/2 + (d*x)/2)^3*(9*A - 2*B) + tan(c/2 + (d*x)/2)*(7*A - 2*B))/(d*(2*a^4*tan(c/2 + (d*x)/2)^2 + a^4*tan(c/2 +
(d*x)/2)^4 + a^4)) - (tan(c/2 + (d*x)/2)^5*((6*A - 4*B + 2*C)/(40*a^4) + (3*(A - B + C))/(40*a^4)))/d - (tan(c
/2 + (d*x)/2)*((3*(6*A - 4*B + 2*C))/(4*a^4) - (3*(5*B - 15*A + C))/(8*a^4) + (5*(A - B + C))/(4*a^4) + (20*A
- 4*C)/(8*a^4)))/d + (x*(21*A - 8*B + 2*C))/(2*a^4) + (tan(c/2 + (d*x)/2)^7*(A - B + C))/(56*a^4*d)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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