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

Optimal. Leaf size=183 \[ -\frac {2 (76 A+11 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 A+2 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {(76 A+11 C) \sin (c+d x) \cos (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {x (13 A+2 C)}{2 a^3}-\frac {(11 A+C) \sin (c+d x) \cos (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

[Out]

1/2*(13*A+2*C)*x/a^3-2/15*(76*A+11*C)*sin(d*x+c)/a^3/d+1/2*(13*A+2*C)*cos(d*x+c)*sin(d*x+c)/a^3/d-1/5*(A+C)*co
s(d*x+c)*sin(d*x+c)/d/(a+a*sec(d*x+c))^3-1/15*(11*A+C)*cos(d*x+c)*sin(d*x+c)/a/d/(a+a*sec(d*x+c))^2-1/15*(76*A
+11*C)*cos(d*x+c)*sin(d*x+c)/d/(a^3+a^3*sec(d*x+c))

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Rubi [A]  time = 0.47, antiderivative size = 183, normalized size of antiderivative = 1.00, number of steps used = 7, 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} \[ -\frac {2 (76 A+11 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 A+2 C) \sin (c+d x) \cos (c+d x)}{2 a^3 d}-\frac {(76 A+11 C) \sin (c+d x) \cos (c+d x)}{15 d \left (a^3 \sec (c+d x)+a^3\right )}+\frac {x (13 A+2 C)}{2 a^3}-\frac {(11 A+C) \sin (c+d x) \cos (c+d x)}{15 a d (a \sec (c+d x)+a)^2}-\frac {(A+C) \sin (c+d x) \cos (c+d x)}{5 d (a \sec (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((13*A + 2*C)*x)/(2*a^3) - (2*(76*A + 11*C)*Sin[c + d*x])/(15*a^3*d) + ((13*A + 2*C)*Cos[c + d*x]*Sin[c + d*x]
)/(2*a^3*d) - ((A + C)*Cos[c + d*x]*Sin[c + d*x])/(5*d*(a + a*Sec[c + d*x])^3) - ((11*A + C)*Cos[c + d*x]*Sin[
c + d*x])/(15*a*d*(a + a*Sec[c + d*x])^2) - ((76*A + 11*C)*Cos[c + d*x]*Sin[c + d*x])/(15*d*(a^3 + a^3*Sec[c +
d*x]))

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))^3} \, dx &=-\frac {(A+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 A+2 C)+a (4 A-C) \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx}{5 a^2}\\ &=-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 A+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 A+8 C)+3 a^2 (11 A+C) \sec (c+d x)\right )}{a+a \sec (c+d x)} \, dx}{15 a^4}\\ &=-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 A+C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(76 A+11 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 A+2 C)+2 a^3 (76 A+11 C) \sec (c+d x)\right ) \, dx}{15 a^6}\\ &=-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 A+C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(76 A+11 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(13 A+2 C) \int \cos ^2(c+d x) \, dx}{a^3}-\frac {(2 (76 A+11 C)) \int \cos (c+d x) \, dx}{15 a^3}\\ &=-\frac {2 (76 A+11 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 A+2 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 A+C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(76 A+11 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}+\frac {(13 A+2 C) \int 1 \, dx}{2 a^3}\\ &=\frac {(13 A+2 C) x}{2 a^3}-\frac {2 (76 A+11 C) \sin (c+d x)}{15 a^3 d}+\frac {(13 A+2 C) \cos (c+d x) \sin (c+d x)}{2 a^3 d}-\frac {(A+C) \cos (c+d x) \sin (c+d x)}{5 d (a+a \sec (c+d x))^3}-\frac {(11 A+C) \cos (c+d x) \sin (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {(76 A+11 C) \cos (c+d x) \sin (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )}\\ \end {align*}

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Mathematica [B]  time = 1.55, size = 385, normalized size = 2.10 \[ \frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (600 d x (13 A+2 C) \cos \left (c+\frac {d x}{2}\right )+7560 A \sin \left (c+\frac {d x}{2}\right )-9230 A \sin \left (c+\frac {3 d x}{2}\right )+930 A \sin \left (2 c+\frac {3 d x}{2}\right )-2782 A \sin \left (2 c+\frac {5 d x}{2}\right )-750 A \sin \left (3 c+\frac {5 d x}{2}\right )-105 A \sin \left (3 c+\frac {7 d x}{2}\right )-105 A \sin \left (4 c+\frac {7 d x}{2}\right )+15 A \sin \left (4 c+\frac {9 d x}{2}\right )+15 A \sin \left (5 c+\frac {9 d x}{2}\right )+3900 A d x \cos \left (c+\frac {3 d x}{2}\right )+3900 A d x \cos \left (2 c+\frac {3 d x}{2}\right )+780 A d x \cos \left (2 c+\frac {5 d x}{2}\right )+780 A d x \cos \left (3 c+\frac {5 d x}{2}\right )+600 d x (13 A+2 C) \cos \left (\frac {d x}{2}\right )-12760 A \sin \left (\frac {d x}{2}\right )+2160 C \sin \left (c+\frac {d x}{2}\right )-1840 C \sin \left (c+\frac {3 d x}{2}\right )+720 C \sin \left (2 c+\frac {3 d x}{2}\right )-512 C \sin \left (2 c+\frac {5 d x}{2}\right )+600 C d x \cos \left (c+\frac {3 d x}{2}\right )+600 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+120 C d x \cos \left (2 c+\frac {5 d x}{2}\right )+120 C d x \cos \left (3 c+\frac {5 d x}{2}\right )-2960 C \sin \left (\frac {d x}{2}\right )\right )}{3840 a^3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c/2]*Sec[(c + d*x)/2]^5*(600*(13*A + 2*C)*d*x*Cos[(d*x)/2] + 600*(13*A + 2*C)*d*x*Cos[c + (d*x)/2] + 3900
*A*d*x*Cos[c + (3*d*x)/2] + 600*C*d*x*Cos[c + (3*d*x)/2] + 3900*A*d*x*Cos[2*c + (3*d*x)/2] + 600*C*d*x*Cos[2*c
+ (3*d*x)/2] + 780*A*d*x*Cos[2*c + (5*d*x)/2] + 120*C*d*x*Cos[2*c + (5*d*x)/2] + 780*A*d*x*Cos[3*c + (5*d*x)/
2] + 120*C*d*x*Cos[3*c + (5*d*x)/2] - 12760*A*Sin[(d*x)/2] - 2960*C*Sin[(d*x)/2] + 7560*A*Sin[c + (d*x)/2] + 2
160*C*Sin[c + (d*x)/2] - 9230*A*Sin[c + (3*d*x)/2] - 1840*C*Sin[c + (3*d*x)/2] + 930*A*Sin[2*c + (3*d*x)/2] +
720*C*Sin[2*c + (3*d*x)/2] - 2782*A*Sin[2*c + (5*d*x)/2] - 512*C*Sin[2*c + (5*d*x)/2] - 750*A*Sin[3*c + (5*d*x
)/2] - 105*A*Sin[3*c + (7*d*x)/2] - 105*A*Sin[4*c + (7*d*x)/2] + 15*A*Sin[4*c + (9*d*x)/2] + 15*A*Sin[5*c + (9
*d*x)/2]))/(3840*a^3*d)

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fricas [A]  time = 0.42, size = 184, normalized size = 1.01 \[ \frac {15 \, {\left (13 \, A + 2 \, C\right )} d x \cos \left (d x + c\right )^{3} + 45 \, {\left (13 \, A + 2 \, C\right )} d x \cos \left (d x + c\right )^{2} + 45 \, {\left (13 \, A + 2 \, C\right )} d x \cos \left (d x + c\right ) + 15 \, {\left (13 \, A + 2 \, C\right )} d x + {\left (15 \, A \cos \left (d x + c\right )^{4} - 45 \, A \cos \left (d x + c\right )^{3} - {\left (479 \, A + 64 \, C\right )} \cos \left (d x + c\right )^{2} - 3 \, {\left (239 \, A + 34 \, C\right )} \cos \left (d x + c\right ) - 304 \, A - 44 \, C\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

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))^3,x, algorithm="fricas")

[Out]

1/30*(15*(13*A + 2*C)*d*x*cos(d*x + c)^3 + 45*(13*A + 2*C)*d*x*cos(d*x + c)^2 + 45*(13*A + 2*C)*d*x*cos(d*x +
c) + 15*(13*A + 2*C)*d*x + (15*A*cos(d*x + c)^4 - 45*A*cos(d*x + c)^3 - (479*A + 64*C)*cos(d*x + c)^2 - 3*(239
*A + 34*C)*cos(d*x + c) - 304*A - 44*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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giac [A]  time = 1.82, size = 174, normalized size = 0.95 \[ \frac {\frac {30 \, {\left (d x + c\right )} {\left (13 \, A + 2 \, C\right )}}{a^{3}} - \frac {60 \, {\left (7 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 5 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2} a^{3}} - \frac {3 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 20 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 465 \, A a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 105 \, C a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]

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))^3,x, algorithm="giac")

[Out]

1/60*(30*(d*x + c)*(13*A + 2*C)/a^3 - 60*(7*A*tan(1/2*d*x + 1/2*c)^3 + 5*A*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x
+ 1/2*c)^2 + 1)^2*a^3) - (3*A*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*A*a^12*tan(1
/2*d*x + 1/2*c)^3 - 20*C*a^12*tan(1/2*d*x + 1/2*c)^3 + 465*A*a^12*tan(1/2*d*x + 1/2*c) + 105*C*a^12*tan(1/2*d*
x + 1/2*c))/a^15)/d

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maple [A]  time = 1.36, size = 224, normalized size = 1.22 \[ -\frac {A \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}-\frac {C \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{20 d \,a^{3}}+\frac {2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d \,a^{3}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3 d \,a^{3}}-\frac {31 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}-\frac {7 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}-\frac {5 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{3} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{2}}+\frac {13 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{3}}+\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{3}} \]

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))^3,x)

[Out]

-1/20/d/a^3*A*tan(1/2*d*x+1/2*c)^5-1/20/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*A+1/3/d/a^
3*C*tan(1/2*d*x+1/2*c)^3-31/4/d/a^3*A*tan(1/2*d*x+1/2*c)-7/4/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*A-5/d/a^3/(1+tan(1/2*d*x+1/2*c)^2)^2*A*tan(1/2*d*x+1/2*c)+13/d/a^3*arctan(ta
n(1/2*d*x+1/2*c))*A+2/d/a^3*arctan(tan(1/2*d*x+1/2*c))*C

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maxima [A]  time = 0.54, size = 276, normalized size = 1.51 \[ -\frac {A {\left (\frac {60 \, {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {7 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{3} + \frac {2 \, a^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {465 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {40 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {780 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )} + C {\left (\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {120 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{3}}\right )}}{60 \, d} \]

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))^3,x, algorithm="maxima")

[Out]

-1/60*(A*(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) + C*((105*sin(d*x + c)/(cos(d*x + c) + 1) - 20*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 - 120*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^3))/d

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mupad [B]  time = 2.69, size = 184, normalized size = 1.01 \[ \frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {A+C}{4\,a^3}+\frac {5\,A+C}{12\,a^3}\right )}{d}+\frac {x\,\left (13\,A+2\,C\right )}{2\,a^3}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,\left (A+C\right )}{2\,a^3}+\frac {3\,\left (5\,A+C\right )}{4\,a^3}+\frac {10\,A-2\,C}{4\,a^3}\right )}{d}-\frac {7\,A\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+5\,A\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^3\right )}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (A+C\right )}{20\,a^3\,d} \]

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))^3,x)

[Out]

(tan(c/2 + (d*x)/2)^3*((A + C)/(4*a^3) + (5*A + C)/(12*a^3)))/d + (x*(13*A + 2*C))/(2*a^3) - (tan(c/2 + (d*x)/
2)*((3*(A + C))/(2*a^3) + (3*(5*A + C))/(4*a^3) + (10*A - 2*C)/(4*a^3)))/d - (5*A*tan(c/2 + (d*x)/2) + 7*A*tan
(c/2 + (d*x)/2)^3)/(d*(2*a^3*tan(c/2 + (d*x)/2)^2 + a^3*tan(c/2 + (d*x)/2)^4 + a^3)) - (tan(c/2 + (d*x)/2)^5*(
A + C))/(20*a^3*d)

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

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))**3,x)

[Out]

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

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