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

Optimal. Leaf size=215 \[ -\frac {32 (54 A+5 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A+2 C) \sin (c+d x) \cos (c+d x)}{2 a^4 d}-\frac {16 (54 A+5 C) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(129 A+10 C) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {x (21 A+2 C)}{2 a^4}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {2 A \sin (c+d x) \cos (c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

[Out]

1/2*(21*A+2*C)*x/a^4-32/105*(54*A+5*C)*sin(d*x+c)/a^4/d+1/2*(21*A+2*C)*cos(d*x+c)*sin(d*x+c)/a^4/d-1/105*(129*
A+10*C)*cos(d*x+c)*sin(d*x+c)/a^4/d/(1+sec(d*x+c))^2-16/105*(54*A+5*C)*cos(d*x+c)*sin(d*x+c)/a^4/d/(1+sec(d*x+
c))-1/7*(A+C)*cos(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^4-2/5*A*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^3

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Rubi [A]  time = 0.63, antiderivative size = 215, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {4085, 4020, 3787, 2635, 8, 2637} \[ -\frac {32 (54 A+5 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A+2 C) \sin (c+d x) \cos (c+d x)}{2 a^4 d}-\frac {16 (54 A+5 C) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)}-\frac {(129 A+10 C) \sin (c+d x) \cos (c+d x)}{105 a^4 d (\sec (c+d x)+1)^2}+\frac {x (21 A+2 C)}{2 a^4}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{7 d (a \sec (c+d x)+a)^4}-\frac {2 A \sin (c+d x) \cos (c+d x)}{5 a d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((21*A + 2*C)*x)/(2*a^4) - (32*(54*A + 5*C)*Sin[c + d*x])/(105*a^4*d) + ((21*A + 2*C)*Cos[c + d*x]*Sin[c + d*x
])/(2*a^4*d) - ((129*A + 10*C)*Cos[c + d*x]*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])^2) - (16*(54*A + 5*C)*
Cos[c + d*x]*Sin[c + d*x])/(105*a^4*d*(1 + Sec[c + d*x])) - ((A + C)*Cos[c + d*x]*Sin[c + d*x])/(7*d*(a + a*Se
c[c + d*x])^4) - (2*A*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]

Rule 4085

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> -Simp[(a*(A + C)*Cot[e + f*x]*(a + b*Csc[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[b*C*n + A*b*(
2*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, 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+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^4} \, dx &=-\frac {(A+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 C)+a (5 A-2 C) \sec (c+d x))}{(a+a \sec (c+d x))^3} \, dx}{7 a^2}\\ &=-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 A \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+10 C)+56 a^2 A \sec (c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx}{35 a^4}\\ &=-\frac {(129 A+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 A \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+50 C)+3 a^3 (129 A+10 C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{105 a^6}\\ &=-\frac {(129 A+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 A \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {16 (54 A+5 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+2 C)+32 a^4 (54 A+5 C) \sec (c+d x)\right ) \, dx}{105 a^8}\\ &=-\frac {(129 A+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 A \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {16 (54 A+5 C) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(21 A+2 C) \int \cos ^2(c+d x) \, dx}{a^4}-\frac {(32 (54 A+5 C)) \int \cos (c+d x) \, dx}{105 a^4}\\ &=-\frac {32 (54 A+5 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A+2 C) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac {(129 A+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 A \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {16 (54 A+5 C) \cos (c+d x) \sin (c+d x)}{105 d \left (a^4+a^4 \sec (c+d x)\right )}+\frac {(21 A+2 C) \int 1 \, dx}{2 a^4}\\ &=\frac {(21 A+2 C) x}{2 a^4}-\frac {32 (54 A+5 C) \sin (c+d x)}{105 a^4 d}+\frac {(21 A+2 C) \cos (c+d x) \sin (c+d x)}{2 a^4 d}-\frac {(129 A+10 C) \cos (c+d x) \sin (c+d x)}{105 a^4 d (1+\sec (c+d x))^2}-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{7 d (a+a \sec (c+d x))^4}-\frac {2 A \cos (c+d x) \sin (c+d x)}{5 a d (a+a \sec (c+d x))^3}-\frac {16 (54 A+5 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 [B]  time = 2.45, size = 505, normalized size = 2.35 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^7\left (\frac {1}{2} (c+d x)\right ) \left (14700 d x (21 A+2 C) \cos \left (c+\frac {d x}{2}\right )+386190 A \sin \left (c+\frac {d x}{2}\right )-422478 A \sin \left (c+\frac {3 d x}{2}\right )+132930 A \sin \left (2 c+\frac {3 d x}{2}\right )-181461 A \sin \left (2 c+\frac {5 d x}{2}\right )+3675 A \sin \left (3 c+\frac {5 d x}{2}\right )-36003 A \sin \left (3 c+\frac {7 d x}{2}\right )-9555 A \sin \left (4 c+\frac {7 d x}{2}\right )-945 A \sin \left (4 c+\frac {9 d x}{2}\right )-945 A \sin \left (5 c+\frac {9 d x}{2}\right )+105 A \sin \left (5 c+\frac {11 d x}{2}\right )+105 A \sin \left (6 c+\frac {11 d x}{2}\right )+185220 A d x \cos \left (c+\frac {3 d x}{2}\right )+185220 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+61740 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+61740 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+8820 A d x \cos \left (3 c+\frac {7 d x}{2}\right )+8820 A d x \cos \left (4 c+\frac {7 d x}{2}\right )+14700 d x (21 A+2 C) \cos \left (\frac {d x}{2}\right )-539490 A \sin \left (\frac {d x}{2}\right )+66080 C \sin \left (c+\frac {d x}{2}\right )-57120 C \sin \left (c+\frac {3 d x}{2}\right )+30240 C \sin \left (2 c+\frac {3 d x}{2}\right )-22400 C \sin \left (2 c+\frac {5 d x}{2}\right )+6720 C \sin \left (3 c+\frac {5 d x}{2}\right )-4160 C \sin \left (3 c+\frac {7 d x}{2}\right )+17640 C d x \cos \left (c+\frac {3 d x}{2}\right )+17640 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+5880 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+5880 C d x \cos \left (3 c+\frac {5 d x}{2}\right )+840 C d x \cos \left (3 c+\frac {7 d x}{2}\right )+840 C d x \cos \left (4 c+\frac {7 d x}{2}\right )-79520 C \sin \left (\frac {d x}{2}\right )\right )}{107520 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^7*(14700*(21*A + 2*C)*d*x*Cos[(d*x)/2] + 14700*(21*A + 2*C)*d*x*Cos[c + (d*x)/2] +
185220*A*d*x*Cos[c + (3*d*x)/2] + 17640*C*d*x*Cos[c + (3*d*x)/2] + 185220*A*d*x*Cos[2*c + (3*d*x)/2] + 17640*C
*d*x*Cos[2*c + (3*d*x)/2] + 61740*A*d*x*Cos[2*c + (5*d*x)/2] + 5880*C*d*x*Cos[2*c + (5*d*x)/2] + 61740*A*d*x*C
os[3*c + (5*d*x)/2] + 5880*C*d*x*Cos[3*c + (5*d*x)/2] + 8820*A*d*x*Cos[3*c + (7*d*x)/2] + 840*C*d*x*Cos[3*c +
(7*d*x)/2] + 8820*A*d*x*Cos[4*c + (7*d*x)/2] + 840*C*d*x*Cos[4*c + (7*d*x)/2] - 539490*A*Sin[(d*x)/2] - 79520*
C*Sin[(d*x)/2] + 386190*A*Sin[c + (d*x)/2] + 66080*C*Sin[c + (d*x)/2] - 422478*A*Sin[c + (3*d*x)/2] - 57120*C*
Sin[c + (3*d*x)/2] + 132930*A*Sin[2*c + (3*d*x)/2] + 30240*C*Sin[2*c + (3*d*x)/2] - 181461*A*Sin[2*c + (5*d*x)
/2] - 22400*C*Sin[2*c + (5*d*x)/2] + 3675*A*Sin[3*c + (5*d*x)/2] + 6720*C*Sin[3*c + (5*d*x)/2] - 36003*A*Sin[3
*c + (7*d*x)/2] - 4160*C*Sin[3*c + (7*d*x)/2] - 9555*A*Sin[4*c + (7*d*x)/2] - 945*A*Sin[4*c + (9*d*x)/2] - 945
*A*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]))/(107520*a^4*d)

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fricas [A]  time = 0.44, size = 234, normalized size = 1.09 \[ \frac {105 \, {\left (21 \, A + 2 \, C\right )} d x \cos \left (d x + c\right )^{4} + 420 \, {\left (21 \, A + 2 \, C\right )} d x \cos \left (d x + c\right )^{3} + 630 \, {\left (21 \, A + 2 \, C\right )} d x \cos \left (d x + c\right )^{2} + 420 \, {\left (21 \, A + 2 \, C\right )} d x \cos \left (d x + c\right ) + 105 \, {\left (21 \, A + 2 \, C\right )} d x + {\left (105 \, A \cos \left (d x + c\right )^{5} - 420 \, A \cos \left (d x + c\right )^{4} - 4 \, {\left (1509 \, A + 130 \, C\right )} \cos \left (d x + c\right )^{3} - 4 \, {\left (3411 \, A + 310 \, C\right )} \cos \left (d x + c\right )^{2} - {\left (11619 \, A + 1070 \, C\right )} \cos \left (d x + c\right ) - 3456 \, A - 320 \, C\right )} \sin \left (d x + c\right )}{210 \, {\left (a^{4} d \cos \left (d x + c\right )^{4} + 4 \, a^{4} d \cos \left (d x + c\right )^{3} + 6 \, a^{4} d \cos \left (d x + c\right )^{2} + 4 \, a^{4} d \cos \left (d x + c\right ) + a^{4} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/210*(105*(21*A + 2*C)*d*x*cos(d*x + c)^4 + 420*(21*A + 2*C)*d*x*cos(d*x + c)^3 + 630*(21*A + 2*C)*d*x*cos(d*
x + c)^2 + 420*(21*A + 2*C)*d*x*cos(d*x + c) + 105*(21*A + 2*C)*d*x + (105*A*cos(d*x + c)^5 - 420*A*cos(d*x +
c)^4 - 4*(1509*A + 130*C)*cos(d*x + c)^3 - 4*(3411*A + 310*C)*cos(d*x + c)^2 - (11619*A + 1070*C)*cos(d*x + c)
- 3456*A - 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 = 0.77, size = 207, normalized size = 0.96 \[ \frac {\frac {420 \, {\left (d x + c\right )} {\left (21 \, A + 2 \, C\right )}}{a^{4}} - \frac {840 \, {\left (9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{4}} + \frac {15 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 15 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 189 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 105 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 1365 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 385 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 11655 \, A a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1575 \, C a^{24} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{28}}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+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 + 2*C)/a^4 - 840*(9*A*tan(1/2*d*x + 1/2*c)^3 + 7*A*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*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 - 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 + 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) - 1575*C*a^24*tan(1/2*d*x + 1/2*c))/a^28)/d

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maple [A]  time = 1.16, size = 264, normalized size = 1.23 \[ \frac {\left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{56 d \,a^{4}}+\frac {C \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{56 d \,a^{4}}-\frac {9 A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{40 d \,a^{4}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{4}}+\frac {13 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{8 d \,a^{4}}+\frac {11 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{24 d \,a^{4}}-\frac {111 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {15 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{8 d \,a^{4}}-\frac {9 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {7 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{4} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {21 A \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{4}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(A+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*C*tan(1/2*d*x+1/2*c)^7-9/40/d/a^4*A*tan(1/2*d*x+1/2*c)^5-1/8/d/a^
4*C*tan(1/2*d*x+1/2*c)^5+13/8/d/a^4*tan(1/2*d*x+1/2*c)^3*A+11/24/d/a^4*C*tan(1/2*d*x+1/2*c)^3-111/8/d/a^4*A*ta
n(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*tan(1/2*d*x+1/2*c)^3*A-7/d
/a^4/(1+tan(1/2*d*x+1/2*c)^2)^2*A*tan(1/2*d*x+1/2*c)+21/d/a^4*A*arctan(tan(1/2*d*x+1/2*c))+2/d/a^4*arctan(tan(
1/2*d*x+1/2*c))*C

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maxima [A]  time = 0.55, size = 318, normalized size = 1.48 \[ -\frac {3 \, A {\left (\frac {280 \, {\left (\frac {7 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {9 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{4} + \frac {2 \, a^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {3885 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {455 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {63 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {5 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {5880 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )} + 5 \, C {\left (\frac {\frac {315 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {77 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {21 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {3 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}}}{a^{4}} - \frac {336 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{4}}\right )}}{840 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+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) + 5*C*((315*sin(d*x + c)/(cos(d*x +
c) + 1) - 77*sin(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 = 2.70, size = 249, normalized size = 1.16 \[ \frac {x\,\left (21\,A+2\,C\right )}{2\,a^4}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^4}+\frac {6\,A+2\,C}{8\,a^4}+\frac {15\,A-C}{24\,a^4}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {3\,\left (A+C\right )}{40\,a^4}+\frac {6\,A+2\,C}{40\,a^4}\right )}{d}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {5\,\left (A+C\right )}{4\,a^4}+\frac {3\,\left (6\,A+2\,C\right )}{4\,a^4}+\frac {3\,\left (15\,A-C\right )}{8\,a^4}+\frac {20\,A-4\,C}{8\,a^4}\right )}{d}-\frac {9\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+7\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^4\right )}+\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (A+C\right )}{56\,a^4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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