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

Optimal. Leaf size=163 \[ -\frac {(12 A+5 C) \sin ^3(c+d x)}{3 a^2 d}+\frac {(12 A+5 C) \sin (c+d x)}{a^2 d}-\frac {(5 A+2 C) \sin (c+d x) \cos (c+d x)}{a^2 d}-\frac {2 (5 A+2 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {x (5 A+2 C)}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-(5*A+2*C)*x/a^2+(12*A+5*C)*sin(d*x+c)/a^2/d-(5*A+2*C)*cos(d*x+c)*sin(d*x+c)/a^2/d-2/3*(5*A+2*C)*cos(d*x+c)^2*
sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(A+C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^2-1/3*(12*A+5*C)*sin(d*x+
c)^3/a^2/d

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Rubi [A]  time = 0.32, antiderivative size = 163, 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, 2633, 2635, 8} \[ -\frac {(12 A+5 C) \sin ^3(c+d x)}{3 a^2 d}+\frac {(12 A+5 C) \sin (c+d x)}{a^2 d}-\frac {(5 A+2 C) \sin (c+d x) \cos (c+d x)}{a^2 d}-\frac {2 (5 A+2 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {x (5 A+2 C)}{a^2}-\frac {(A+C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((5*A + 2*C)*x)/a^2) + ((12*A + 5*C)*Sin[c + d*x])/(a^2*d) - ((5*A + 2*C)*Cos[c + d*x]*Sin[c + d*x])/(a^2*d)
- (2*(5*A + 2*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((A + C)*Cos[c + d*x]^2*Sin[c +
d*x])/(3*d*(a + a*Sec[c + d*x])^2) - ((12*A + 5*C)*Sin[c + d*x]^3)/(3*a^2*d)

Rule 8

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

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

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 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 ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \frac {\cos ^3(c+d x) (-3 a (2 A+C)+a (4 A+C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {2 (5 A+2 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {\int \cos ^3(c+d x) \left (-3 a^2 (12 A+5 C)+6 a^2 (5 A+2 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {2 (5 A+2 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(2 (5 A+2 C)) \int \cos ^2(c+d x) \, dx}{a^2}+\frac {(12 A+5 C) \int \cos ^3(c+d x) \, dx}{a^2}\\ &=-\frac {(5 A+2 C) \cos (c+d x) \sin (c+d x)}{a^2 d}-\frac {2 (5 A+2 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(5 A+2 C) \int 1 \, dx}{a^2}-\frac {(12 A+5 C) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=-\frac {(5 A+2 C) x}{a^2}+\frac {(12 A+5 C) \sin (c+d x)}{a^2 d}-\frac {(5 A+2 C) \cos (c+d x) \sin (c+d x)}{a^2 d}-\frac {2 (5 A+2 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(A+C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(12 A+5 C) \sin ^3(c+d x)}{3 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 1.04, size = 349, normalized size = 2.14 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (-72 d x (5 A+2 C) \cos \left (c+\frac {d x}{2}\right )-156 A \sin \left (c+\frac {d x}{2}\right )+342 A \sin \left (c+\frac {3 d x}{2}\right )+118 A \sin \left (2 c+\frac {3 d x}{2}\right )+30 A \sin \left (2 c+\frac {5 d x}{2}\right )+30 A \sin \left (3 c+\frac {5 d x}{2}\right )-3 A \sin \left (3 c+\frac {7 d x}{2}\right )-3 A \sin \left (4 c+\frac {7 d x}{2}\right )+A \sin \left (4 c+\frac {9 d x}{2}\right )+A \sin \left (5 c+\frac {9 d x}{2}\right )-120 A d x \cos \left (c+\frac {3 d x}{2}\right )-120 A d x \cos \left (2 c+\frac {3 d x}{2}\right )-72 d x (5 A+2 C) \cos \left (\frac {d x}{2}\right )+516 A \sin \left (\frac {d x}{2}\right )-120 C \sin \left (c+\frac {d x}{2}\right )+164 C \sin \left (c+\frac {3 d x}{2}\right )+36 C \sin \left (2 c+\frac {3 d x}{2}\right )+12 C \sin \left (2 c+\frac {5 d x}{2}\right )+12 C \sin \left (3 c+\frac {5 d x}{2}\right )-48 C d x \cos \left (c+\frac {3 d x}{2}\right )-48 C d x \cos \left (2 c+\frac {3 d x}{2}\right )+264 C \sin \left (\frac {d x}{2}\right )\right )}{48 a^2 d (\sec (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*Sec[c + d*x]^2*(-72*(5*A + 2*C)*d*x*Cos[(d*x)/2] - 72*(5*A + 2*C)*d*x*Cos[c + (d*x)
/2] - 120*A*d*x*Cos[c + (3*d*x)/2] - 48*C*d*x*Cos[c + (3*d*x)/2] - 120*A*d*x*Cos[2*c + (3*d*x)/2] - 48*C*d*x*C
os[2*c + (3*d*x)/2] + 516*A*Sin[(d*x)/2] + 264*C*Sin[(d*x)/2] - 156*A*Sin[c + (d*x)/2] - 120*C*Sin[c + (d*x)/2
] + 342*A*Sin[c + (3*d*x)/2] + 164*C*Sin[c + (3*d*x)/2] + 118*A*Sin[2*c + (3*d*x)/2] + 36*C*Sin[2*c + (3*d*x)/
2] + 30*A*Sin[2*c + (5*d*x)/2] + 12*C*Sin[2*c + (5*d*x)/2] + 30*A*Sin[3*c + (5*d*x)/2] + 12*C*Sin[3*c + (5*d*x
)/2] - 3*A*Sin[3*c + (7*d*x)/2] - 3*A*Sin[4*c + (7*d*x)/2] + A*Sin[4*c + (9*d*x)/2] + A*Sin[5*c + (9*d*x)/2]))
/(48*a^2*d*(1 + Sec[c + d*x])^2)

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fricas [A]  time = 0.44, size = 148, normalized size = 0.91 \[ -\frac {3 \, {\left (5 \, A + 2 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (5 \, A + 2 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (5 \, A + 2 \, C\right )} d x - {\left (A \cos \left (d x + c\right )^{4} - A \cos \left (d x + c\right )^{3} + 3 \, {\left (2 \, A + C\right )} \cos \left (d x + c\right )^{2} + {\left (33 \, A + 14 \, C\right )} \cos \left (d x + c\right ) + 24 \, A + 10 \, C\right )} \sin \left (d x + c\right )}{3 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(3*(5*A + 2*C)*d*x*cos(d*x + c)^2 + 6*(5*A + 2*C)*d*x*cos(d*x + c) + 3*(5*A + 2*C)*d*x - (A*cos(d*x + c)^
4 - A*cos(d*x + c)^3 + 3*(2*A + C)*cos(d*x + c)^2 + (33*A + 14*C)*cos(d*x + c) + 24*A + 10*C)*sin(d*x + c))/(a
^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d)

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giac [A]  time = 0.28, size = 191, normalized size = 1.17 \[ -\frac {\frac {6 \, {\left (d x + c\right )} {\left (5 \, A + 2 \, C\right )}}{a^{2}} - \frac {4 \, {\left (15 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac {A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, A a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(6*(d*x + c)*(5*A + 2*C)/a^2 - 4*(15*A*tan(1/2*d*x + 1/2*c)^5 + 3*C*tan(1/2*d*x + 1/2*c)^5 + 20*A*tan(1/2
*d*x + 1/2*c)^3 + 6*C*tan(1/2*d*x + 1/2*c)^3 + 9*A*tan(1/2*d*x + 1/2*c) + 3*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*
d*x + 1/2*c)^2 + 1)^3*a^2) + (A*a^4*tan(1/2*d*x + 1/2*c)^3 + C*a^4*tan(1/2*d*x + 1/2*c)^3 - 27*A*a^4*tan(1/2*d
*x + 1/2*c) - 15*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [B]  time = 1.18, size = 322, normalized size = 1.98 \[ -\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{6 d \,a^{2}}-\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {9 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {5 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {40 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {4 C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 A \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {2 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {10 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) A}{d \,a^{2}}-\frac {4 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d/a^2*tan(1/2*d*x+1/2*c)^3*A-1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+9/2/d/a^2*A*tan(1/2*d*x+1/2*c)+5/2/d/a^2*C*
tan(1/2*d*x+1/2*c)+10/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^5*A+2/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)
^3*tan(1/2*d*x+1/2*c)^5*C+40/3/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*A+4/d/a^2/(1+tan(1/2*d*x+
1/2*c)^2)^3*C*tan(1/2*d*x+1/2*c)^3+6/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*A*tan(1/2*d*x+1/2*c)+2/d/a^2/(1+tan(1/2*
d*x+1/2*c)^2)^3*C*tan(1/2*d*x+1/2*c)-10/d/a^2*arctan(tan(1/2*d*x+1/2*c))*A-4/d/a^2*arctan(tan(1/2*d*x+1/2*c))*
C

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maxima [B]  time = 0.43, size = 325, normalized size = 1.99 \[ \frac {A {\left (\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {60 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} + C {\left (\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {24 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}} + \frac {12 \, \sin \left (d x + c\right )}{{\left (a^{2} + \frac {a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(A*(4*(9*sin(d*x + c)/(cos(d*x + c) + 1) + 20*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos
(d*x + c) + 1)^5)/(a^2 + 3*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a^2*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
+ a^2*sin(d*x + c)^6/(cos(d*x + c) + 1)^6) + (27*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x +
c) + 1)^3)/a^2 - 60*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2) + C*((15*sin(d*x + c)/(cos(d*x + c) + 1) - si
n(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 24*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a^2 + 12*sin(d*x + c)/((a^
2 + a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1))))/d

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mupad [B]  time = 2.68, size = 181, normalized size = 1.11 \[ \frac {\left (10\,A+2\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {40\,A}{3}+4\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,A+2\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {x\,\left (5\,A+2\,C\right )}{a^2}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (A+C\right )}{a^2}+\frac {5\,A+C}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (A+C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^5*(10*A + 2*C) + tan(c/2 + (d*x)/2)^3*((40*A)/3 + 4*C) + tan(c/2 + (d*x)/2)*(6*A + 2*C))/(
d*(3*a^2*tan(c/2 + (d*x)/2)^2 + 3*a^2*tan(c/2 + (d*x)/2)^4 + a^2*tan(c/2 + (d*x)/2)^6 + a^2)) - (x*(5*A + 2*C)
)/a^2 + (tan(c/2 + (d*x)/2)*((2*(A + C))/a^2 + (5*A + C)/(2*a^2)))/d - (tan(c/2 + (d*x)/2)^3*(A + C))/(6*a^2*d
)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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