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

Optimal. Leaf size=170 \[ -\frac {4 (3 B-2 C) \sin ^3(c+d x)}{3 a^2 d}+\frac {4 (3 B-2 C) \sin (c+d x)}{a^2 d}-\frac {(10 B-7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {(10 B-7 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {x (10 B-7 C)}{2 a^2}-\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

[Out]

-1/2*(10*B-7*C)*x/a^2+4*(3*B-2*C)*sin(d*x+c)/a^2/d-1/2*(10*B-7*C)*cos(d*x+c)*sin(d*x+c)/a^2/d-1/3*(10*B-7*C)*c
os(d*x+c)^2*sin(d*x+c)/a^2/d/(1+sec(d*x+c))-1/3*(B-C)*cos(d*x+c)^2*sin(d*x+c)/d/(a+a*sec(d*x+c))^2-4/3*(3*B-2*
C)*sin(d*x+c)^3/a^2/d

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Rubi [A]  time = 0.40, antiderivative size = 170, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, Rules used = {4072, 4020, 3787, 2633, 2635, 8} \[ -\frac {4 (3 B-2 C) \sin ^3(c+d x)}{3 a^2 d}+\frac {4 (3 B-2 C) \sin (c+d x)}{a^2 d}-\frac {(10 B-7 C) \sin (c+d x) \cos (c+d x)}{2 a^2 d}-\frac {(10 B-7 C) \sin (c+d x) \cos ^2(c+d x)}{3 a^2 d (\sec (c+d x)+1)}-\frac {x (10 B-7 C)}{2 a^2}-\frac {(B-C) \sin (c+d x) \cos ^2(c+d x)}{3 d (a \sec (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((10*B - 7*C)*x)/(2*a^2) + (4*(3*B - 2*C)*Sin[c + d*x])/(a^2*d) - ((10*B - 7*C)*Cos[c + d*x]*Sin[c + d*x])/(2
*a^2*d) - ((10*B - 7*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*a^2*d*(1 + Sec[c + d*x])) - ((B - C)*Cos[c + d*x]^2*Si
n[c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) - (4*(3*B - 2*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 4072

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

Rubi steps

\begin {align*} \int \frac {\cos ^4(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^2} \, dx &=\int \frac {\cos ^3(c+d x) (B+C \sec (c+d x))}{(a+a \sec (c+d x))^2} \, dx\\ &=-\frac {(B-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 B-C)-4 a (B-C) \sec (c+d x))}{a+a \sec (c+d x)} \, dx}{3 a^2}\\ &=-\frac {(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-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 (12 a^2 (3 B-2 C)-3 a^2 (10 B-7 C) \sec (c+d x)\right ) \, dx}{3 a^4}\\ &=-\frac {(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(10 B-7 C) \int \cos ^2(c+d x) \, dx}{a^2}+\frac {(4 (3 B-2 C)) \int \cos ^3(c+d x) \, dx}{a^2}\\ &=-\frac {(10 B-7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {(10 B-7 C) \int 1 \, dx}{2 a^2}-\frac {(4 (3 B-2 C)) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a^2 d}\\ &=-\frac {(10 B-7 C) x}{2 a^2}+\frac {4 (3 B-2 C) \sin (c+d x)}{a^2 d}-\frac {(10 B-7 C) \cos (c+d x) \sin (c+d x)}{2 a^2 d}-\frac {(10 B-7 C) \cos ^2(c+d x) \sin (c+d x)}{3 a^2 d (1+\sec (c+d x))}-\frac {(B-C) \cos ^2(c+d x) \sin (c+d x)}{3 d (a+a \sec (c+d x))^2}-\frac {4 (3 B-2 C) \sin ^3(c+d x)}{3 a^2 d}\\ \end {align*}

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Mathematica [B]  time = 0.78, size = 369, normalized size = 2.17 \[ \frac {\sec \left (\frac {c}{2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \left (-36 d x (10 B-7 C) \cos \left (c+\frac {d x}{2}\right )-156 B \sin \left (c+\frac {d x}{2}\right )+342 B \sin \left (c+\frac {3 d x}{2}\right )+118 B \sin \left (2 c+\frac {3 d x}{2}\right )+30 B \sin \left (2 c+\frac {5 d x}{2}\right )+30 B \sin \left (3 c+\frac {5 d x}{2}\right )-3 B \sin \left (3 c+\frac {7 d x}{2}\right )-3 B \sin \left (4 c+\frac {7 d x}{2}\right )+B \sin \left (4 c+\frac {9 d x}{2}\right )+B \sin \left (5 c+\frac {9 d x}{2}\right )-120 B d x \cos \left (c+\frac {3 d x}{2}\right )-120 B d x \cos \left (2 c+\frac {3 d x}{2}\right )-36 d x (10 B-7 C) \cos \left (\frac {d x}{2}\right )+516 B \sin \left (\frac {d x}{2}\right )+147 C \sin \left (c+\frac {d x}{2}\right )-239 C \sin \left (c+\frac {3 d x}{2}\right )-63 C \sin \left (2 c+\frac {3 d x}{2}\right )-15 C \sin \left (2 c+\frac {5 d x}{2}\right )-15 C \sin \left (3 c+\frac {5 d x}{2}\right )+3 C \sin \left (3 c+\frac {7 d x}{2}\right )+3 C \sin \left (4 c+\frac {7 d x}{2}\right )+84 C d x \cos \left (c+\frac {3 d x}{2}\right )+84 C d x \cos \left (2 c+\frac {3 d x}{2}\right )-381 C \sin \left (\frac {d x}{2}\right )\right )}{48 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(-36*(10*B - 7*C)*d*x*Cos[(d*x)/2] - 36*(10*B - 7*C)*d*x*Cos[c + (d*x)/2] - 120*B*d
*x*Cos[c + (3*d*x)/2] + 84*C*d*x*Cos[c + (3*d*x)/2] - 120*B*d*x*Cos[2*c + (3*d*x)/2] + 84*C*d*x*Cos[2*c + (3*d
*x)/2] + 516*B*Sin[(d*x)/2] - 381*C*Sin[(d*x)/2] - 156*B*Sin[c + (d*x)/2] + 147*C*Sin[c + (d*x)/2] + 342*B*Sin
[c + (3*d*x)/2] - 239*C*Sin[c + (3*d*x)/2] + 118*B*Sin[2*c + (3*d*x)/2] - 63*C*Sin[2*c + (3*d*x)/2] + 30*B*Sin
[2*c + (5*d*x)/2] - 15*C*Sin[2*c + (5*d*x)/2] + 30*B*Sin[3*c + (5*d*x)/2] - 15*C*Sin[3*c + (5*d*x)/2] - 3*B*Si
n[3*c + (7*d*x)/2] + 3*C*Sin[3*c + (7*d*x)/2] - 3*B*Sin[4*c + (7*d*x)/2] + 3*C*Sin[4*c + (7*d*x)/2] + B*Sin[4*
c + (9*d*x)/2] + B*Sin[5*c + (9*d*x)/2]))/(48*a^2*d*(1 + Cos[c + d*x])^2)

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fricas [A]  time = 0.49, size = 157, normalized size = 0.92 \[ -\frac {3 \, {\left (10 \, B - 7 \, C\right )} d x \cos \left (d x + c\right )^{2} + 6 \, {\left (10 \, B - 7 \, C\right )} d x \cos \left (d x + c\right ) + 3 \, {\left (10 \, B - 7 \, C\right )} d x - {\left (2 \, B \cos \left (d x + c\right )^{4} - {\left (2 \, B - 3 \, C\right )} \cos \left (d x + c\right )^{3} + 6 \, {\left (2 \, B - C\right )} \cos \left (d x + c\right )^{2} + {\left (66 \, B - 43 \, C\right )} \cos \left (d x + c\right ) + 48 \, B - 32 \, C\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(10*B - 7*C)*d*x*cos(d*x + c)^2 + 6*(10*B - 7*C)*d*x*cos(d*x + c) + 3*(10*B - 7*C)*d*x - (2*B*cos(d*x
+ c)^4 - (2*B - 3*C)*cos(d*x + c)^3 + 6*(2*B - C)*cos(d*x + c)^2 + (66*B - 43*C)*cos(d*x + c) + 48*B - 32*C)*s
in(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 = 0.34, size = 192, normalized size = 1.13 \[ -\frac {\frac {3 \, {\left (d x + c\right )} {\left (10 \, B - 7 \, C\right )}}{a^{2}} - \frac {2 \, {\left (30 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 15 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 18 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 9 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3} a^{2}} + \frac {B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 27 \, B a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 21 \, C a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(d*x + c)*(10*B - 7*C)/a^2 - 2*(30*B*tan(1/2*d*x + 1/2*c)^5 - 15*C*tan(1/2*d*x + 1/2*c)^5 + 40*B*tan(1
/2*d*x + 1/2*c)^3 - 24*C*tan(1/2*d*x + 1/2*c)^3 + 18*B*tan(1/2*d*x + 1/2*c) - 9*C*tan(1/2*d*x + 1/2*c))/((tan(
1/2*d*x + 1/2*c)^2 + 1)^3*a^2) + (B*a^4*tan(1/2*d*x + 1/2*c)^3 - C*a^4*tan(1/2*d*x + 1/2*c)^3 - 27*B*a^4*tan(1
/2*d*x + 1/2*c) + 21*C*a^4*tan(1/2*d*x + 1/2*c))/a^6)/d

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maple [B]  time = 1.12, size = 322, normalized size = 1.89 \[ -\frac {B \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {C \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{6 d \,a^{2}}+\frac {9 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}-\frac {7 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d \,a^{2}}+\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {5 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {40 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{3 d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}+\frac {6 B \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {3 C \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d \,a^{2} \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{3}}-\frac {10 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) B}{d \,a^{2}}+\frac {7 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) C}{d \,a^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6/d/a^2*B*tan(1/2*d*x+1/2*c)^3+1/6/d/a^2*C*tan(1/2*d*x+1/2*c)^3+9/2/d/a^2*B*tan(1/2*d*x+1/2*c)-7/2/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*B-5/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*B-8/d/a^2/(1+tan(1/2*d*x+
1/2*c)^2)^3*tan(1/2*d*x+1/2*c)^3*C+6/d/a^2/(1+tan(1/2*d*x+1/2*c)^2)^3*B*tan(1/2*d*x+1/2*c)-3/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))*B+7/d/a^2*arctan(tan(1/2*d*x+1/2*c))*
C

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maxima [B]  time = 0.55, size = 372, normalized size = 2.19 \[ \frac {B {\left (\frac {4 \, {\left (\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {15 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{2} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {a^{2} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}} + \frac {\frac {27 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {60 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )} - C {\left (\frac {6 \, {\left (\frac {3 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}\right )}}{a^{2} + \frac {2 \, a^{2} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {a^{2} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}}} + \frac {\frac {21 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {42 \, \arctan \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a^{2}}\right )}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(B*(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*(6*(3*sin(d*x + c)/(cos(d*x + c) + 1) + 5
*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a^2 + 2*a^2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + a^2*sin(d*x + c)^4/(c
os(d*x + c) + 1)^4) + (21*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 42*arct
an(sin(d*x + c)/(cos(d*x + c) + 1))/a^2))/d

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mupad [B]  time = 2.90, size = 187, normalized size = 1.10 \[ \frac {\left (10\,B-5\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {40\,B}{3}-8\,C\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (6\,B-3\,C\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left (a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+3\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+a^2\right )}-\frac {x\,\left (10\,B-7\,C\right )}{2\,a^2}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {2\,\left (B-C\right )}{a^2}+\frac {5\,B-3\,C}{2\,a^2}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (B-C\right )}{6\,a^2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(tan(c/2 + (d*x)/2)^5*(10*B - 5*C) + tan(c/2 + (d*x)/2)^3*((40*B)/3 - 8*C) + tan(c/2 + (d*x)/2)*(6*B - 3*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*(10*B - 7*C
))/(2*a^2) + (tan(c/2 + (d*x)/2)*((2*(B - C))/a^2 + (5*B - 3*C)/(2*a^2)))/d - (tan(c/2 + (d*x)/2)^3*(B - C))/(
6*a^2*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)**4*(B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**2,x)

[Out]

Timed out

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