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

Optimal. Leaf size=223 \[ -\frac {8 (216 A-83 B) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B) \sin (c+d x) \cos (c+d x)}{2 a^4 d}-\frac {4 (216 A-83 B) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(129 A-52 B) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {x (21 A-8 B)}{2 a^4}-\frac {(2 A-B) \sin (c+d x) \cos (c+d x)}{5 a d (a \sec (c+d x)+a)^3}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.65, antiderivative size = 223, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 5, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.161, Rules used = {4020, 3787, 2635, 8, 2637} \[ -\frac {8 (216 A-83 B) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B) \sin (c+d x) \cos (c+d x)}{2 a^4 d}-\frac {4 (216 A-83 B) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(129 A-52 B) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {x (21 A-8 B)}{2 a^4}-\frac {(2 A-B) \sin (c+d x) \cos (c+d x)}{5 a d (a \sec (c+d x)+a)^3}-\frac {(A-B) \sin (c+d x) \cos (c+d x)}{7 d (a \sec (c+d x)+a)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((21*A - 8*B)*x)/(2*a^4) - (8*(216*A - 83*B)*Sin[c + d*x])/(105*a^4*d) + ((21*A - 8*B)*Cos[c + d*x]*Sin[c + d*
x])/(2*a^4*d) - ((129*A - 52*B)*Cos[c + d*x]*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - (4*(216*A - 83*B
)*Cos[c + d*x]*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A - B)*Cos[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*Sec[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]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) (A+B \sec (c+d x))}{(a+a \sec (c+d x))^4} \, dx &=-\frac {(A-B) \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)-5 a (A-B) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A-B) \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)-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) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B) \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)-3 a^3 (129 A-52 B) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(129 A-52 B) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B) \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) \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)-8 a^4 (216 A-83 B) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac {(129 A-52 B) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B) \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) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}-\frac {(8 (216 A-83 B)) \int \cos (c+d x) \, dx}{105 a^4}+\frac {(21 A-8 B) \int \cos ^2(c+d x) \, dx}{a^4}\\ &=-\frac {8 (216 A-83 B) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac {(129 A-52 B) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B) \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) \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) \int 1 \, dx}{2 a^4}\\ &=\frac {(21 A-8 B) x}{2 a^4}-\frac {8 (216 A-83 B) \sin (c+d x)}{105 a^4 d}+\frac {(21 A-8 B) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac {(129 A-52 B) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A-B) \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) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 1.20, size = 555, normalized size = 2.49 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (14700 d x (21 A-8 B) \cos \left (c+\frac {d x}{2}\right )+14700 d x (21 A-8 B) \cos \left (\frac {d x}{2}\right )+386190 A \sin \left (c+\frac {d x}{2}\right )-422478 A \sin \left (c+\frac {3 d x}{2}\right )+132930 A \sin \left (2 c+\frac {3 d x}{2}\right )-181461 A \sin \left (2 c+\frac {5 d x}{2}\right )+3675 A \sin \left (3 c+\frac {5 d x}{2}\right )-36003 A \sin \left (3 c+\frac {7 d x}{2}\right )-9555 A \sin \left (4 c+\frac {7 d x}{2}\right )-945 A \sin \left (4 c+\frac {9 d x}{2}\right )-945 A \sin \left (5 c+\frac {9 d x}{2}\right )+105 A \sin \left (5 c+\frac {11 d x}{2}\right )+105 A \sin \left (6 c+\frac {11 d x}{2}\right )+185220 A d x \cos \left (c+\frac {3 d x}{2}\right )+185220 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+61740 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+61740 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+8820 A d x \cos \left (3 c+\frac {7 d x}{2}\right )+8820 A d x \cos \left (4 c+\frac {7 d x}{2}\right )-539490 A \sin \left (\frac {d x}{2}\right )-184520 B \sin \left (c+\frac {d x}{2}\right )+184464 B \sin \left (c+\frac {3 d x}{2}\right )-72240 B \sin \left (2 c+\frac {3 d x}{2}\right )+77168 B \sin \left (2 c+\frac {5 d x}{2}\right )-8400 B \sin \left (3 c+\frac {5 d x}{2}\right )+15164 B \sin \left (3 c+\frac {7 d x}{2}\right )+2940 B \sin \left (4 c+\frac {7 d x}{2}\right )+420 B \sin \left (4 c+\frac {9 d x}{2}\right )+420 B \sin \left (5 c+\frac {9 d x}{2}\right )-70560 B d x \cos \left (c+\frac {3 d x}{2}\right )-70560 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-23520 B d x \cos \left (2 c+\frac {5 d x}{2}\right )-23520 B d x \cos \left (3 c+\frac {5 d x}{2}\right )-3360 B d x \cos \left (3 c+\frac {7 d x}{2}\right )-3360 B d x \cos \left (4 c+\frac {7 d x}{2}\right )+243320 B \sin \left (\frac {d x}{2}\right )\right )}{6720 a^4 d (\cos (c+d x)+1)^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(14700*(21*A - 8*B)*d*x*Cos[(d*x)/2] + 14700*(21*A - 8*B)*d*x*Cos[c + (d*x)/2] + 18
5220*A*d*x*Cos[c + (3*d*x)/2] - 70560*B*d*x*Cos[c + (3*d*x)/2] + 185220*A*d*x*Cos[2*c + (3*d*x)/2] - 70560*B*d
*x*Cos[2*c + (3*d*x)/2] + 61740*A*d*x*Cos[2*c + (5*d*x)/2] - 23520*B*d*x*Cos[2*c + (5*d*x)/2] + 61740*A*d*x*Co
s[3*c + (5*d*x)/2] - 23520*B*d*x*Cos[3*c + (5*d*x)/2] + 8820*A*d*x*Cos[3*c + (7*d*x)/2] - 3360*B*d*x*Cos[3*c +
(7*d*x)/2] + 8820*A*d*x*Cos[4*c + (7*d*x)/2] - 3360*B*d*x*Cos[4*c + (7*d*x)/2] - 539490*A*Sin[(d*x)/2] + 2433
20*B*Sin[(d*x)/2] + 386190*A*Sin[c + (d*x)/2] - 184520*B*Sin[c + (d*x)/2] - 422478*A*Sin[c + (3*d*x)/2] + 1844
64*B*Sin[c + (3*d*x)/2] + 132930*A*Sin[2*c + (3*d*x)/2] - 72240*B*Sin[2*c + (3*d*x)/2] - 181461*A*Sin[2*c + (5
*d*x)/2] + 77168*B*Sin[2*c + (5*d*x)/2] + 3675*A*Sin[3*c + (5*d*x)/2] - 8400*B*Sin[3*c + (5*d*x)/2] - 36003*A*
Sin[3*c + (7*d*x)/2] + 15164*B*Sin[3*c + (7*d*x)/2] - 9555*A*Sin[4*c + (7*d*x)/2] + 2940*B*Sin[4*c + (7*d*x)/2
] - 945*A*Sin[4*c + (9*d*x)/2] + 420*B*Sin[4*c + (9*d*x)/2] - 945*A*Sin[5*c + (9*d*x)/2] + 420*B*Sin[5*c + (9*
d*x)/2] + 105*A*Sin[5*c + (11*d*x)/2] + 105*A*Sin[6*c + (11*d*x)/2]))/(6720*a^4*d*(1 + Cos[c + d*x])^4)

________________________________________________________________________________________

fricas [A]  time = 0.45, size = 240, normalized size = 1.08 \[ \frac {105 \, {\left (21 \, A - 8 \, B\right )} d x \cos \left (d x + c\right )^{4} + 420 \, {\left (21 \, A - 8 \, B\right )} d x \cos \left (d x + c\right )^{3} + 630 \, {\left (21 \, A - 8 \, B\right )} d x \cos \left (d x + c\right )^{2} + 420 \, {\left (21 \, A - 8 \, B\right )} d x \cos \left (d x + c\right ) + 105 \, {\left (21 \, A - 8 \, B\right )} d x + {\left (105 \, A \cos \left (d x + c\right )^{5} - 210 \, {\left (2 \, A - B\right )} \cos \left (d x + c\right )^{4} - 4 \, {\left (1509 \, A - 592 \, B\right )} \cos \left (d x + c\right )^{3} - 4 \, {\left (3411 \, A - 1318 \, B\right )} \cos \left (d x + c\right )^{2} - {\left (11619 \, A - 4472 \, B\right )} \cos \left (d x + c\right ) - 3456 \, A + 1328 \, B\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(105*(21*A - 8*B)*d*x*cos(d*x + c)^4 + 420*(21*A - 8*B)*d*x*cos(d*x + c)^3 + 630*(21*A - 8*B)*d*x*cos(d*
x + c)^2 + 420*(21*A - 8*B)*d*x*cos(d*x + c) + 105*(21*A - 8*B)*d*x + (105*A*cos(d*x + c)^5 - 210*(2*A - B)*co
s(d*x + c)^4 - 4*(1509*A - 592*B)*cos(d*x + c)^3 - 4*(3411*A - 1318*B)*cos(d*x + c)^2 - (11619*A - 4472*B)*cos
(d*x + c) - 3456*A + 1328*B)*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)

________________________________________________________________________________________

giac [A]  time = 0.28, size = 233, normalized size = 1.04 \[ \frac {\frac {420 \, {\left (d x + c\right )} {\left (21 \, A - 8 \, B\right )}}{a^{4}} - \frac {840 \, {\left (9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 189 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 147 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1365 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 805 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11655 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5145 \, B a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/840*(420*(d*x + c)*(21*A - 8*B)/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 - 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 + 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 - 11655*A*a^24*tan(1/2*d*x + 1/
2*c) + 5145*B*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

________________________________________________________________________________________

maple [A]  time = 1.16, size = 332, normalized size = 1.49 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}-\frac {B \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {9 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {7 B \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}+\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{8 d \,a^{4}}-\frac {23 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}-\frac {111 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}+\frac {49 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {9 A \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {2 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {21 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}}-\frac {8 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(A+B*sec(d*x+c))/(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)^7-9/40/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+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)^3-111/8/d/a^4*A*t
an(1/2*d*x+1/2*c)+49/8/d/a^4*B*tan(1/2*d*x+1/2*c)-9/d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*A*tan(1/2*d*x+1/2*c)^3+2/
d/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*B*tan(1/2*d*x+1/2*c)^3-7/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

________________________________________________________________________________________

maxima [A]  time = 0.71, size = 364, normalized size = 1.63 \[ -\frac {3 \, A {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} + \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {5880 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} - B {\left (\frac {1680 \, \sin \left (d x + c\right )}{{\left (a^{4} + \frac {a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}} + \frac {\frac {5145 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {805 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {147 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {15 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {6720 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+B*sec(d*x+c))/(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))/d

________________________________________________________________________________________

mupad [B]  time = 2.00, size = 179, normalized size = 0.80 \[ \frac {\frac {21\,A\,d\,x}{2}-4\,B\,d\,x}{a^4\,d}-\frac {\left (9\,A-2\,B\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (7\,A-2\,B\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{a^4\,d\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {111\,A}{8}-\frac {49\,B}{8}\right )}{a^4\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {13\,A}{8}-\frac {23\,B}{24}\right )}{a^4\,d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {9\,A}{40}-\frac {7\,B}{40}\right )}{a^4\,d}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {A}{56}-\frac {B}{56}\right )}{a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {A \cos ^{2}{\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx + \int \frac {B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}}{\sec ^{4}{\left (c + d x \right )} + 4 \sec ^{3}{\left (c + d x \right )} + 6 \sec ^{2}{\left (c + d x \right )} + 4 \sec {\left (c + d x \right )} + 1}\, dx}{a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(A*cos(c + d*x)**2/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c + d*x) + 1), x)
+ Integral(B*cos(c + d*x)**2*sec(c + d*x)/(sec(c + d*x)**4 + 4*sec(c + d*x)**3 + 6*sec(c + d*x)**2 + 4*sec(c
+ d*x) + 1), x))/a**4

________________________________________________________________________________________