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

Optimal. Leaf size=174 \[ \frac{(4 A-4 B+3 C) \sin ^3(c+d x)}{3 a d}-\frac{(4 A-4 B+3 C) \sin (c+d x)}{a d}+\frac{(5 A-4 B+4 C) \sin (c+d x) \cos ^3(c+d x)}{4 a d}+\frac{3 (5 A-4 B+4 C) \sin (c+d x) \cos (c+d x)}{8 a d}-\frac{(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{d (a \sec (c+d x)+a)}+\frac{3 x (5 A-4 B+4 C)}{8 a} \]

[Out]

(3*(5*A - 4*B + 4*C)*x)/(8*a) - ((4*A - 4*B + 3*C)*Sin[c + d*x])/(a*d) + (3*(5*A - 4*B + 4*C)*Cos[c + d*x]*Sin
[c + d*x])/(8*a*d) + ((5*A - 4*B + 4*C)*Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d) - ((A - B + C)*Cos[c + d*x]^3*Sin
[c + d*x])/(d*(a + a*Sec[c + d*x])) + ((4*A - 4*B + 3*C)*Sin[c + d*x]^3)/(3*a*d)

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Rubi [A]  time = 0.212429, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 41, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.122, Rules used = {4084, 3787, 2635, 8, 2633} \[ \frac{(4 A-4 B+3 C) \sin ^3(c+d x)}{3 a d}-\frac{(4 A-4 B+3 C) \sin (c+d x)}{a d}+\frac{(5 A-4 B+4 C) \sin (c+d x) \cos ^3(c+d x)}{4 a d}+\frac{3 (5 A-4 B+4 C) \sin (c+d x) \cos (c+d x)}{8 a d}-\frac{(A-B+C) \sin (c+d x) \cos ^3(c+d x)}{d (a \sec (c+d x)+a)}+\frac{3 x (5 A-4 B+4 C)}{8 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(5*A - 4*B + 4*C)*x)/(8*a) - ((4*A - 4*B + 3*C)*Sin[c + d*x])/(a*d) + (3*(5*A - 4*B + 4*C)*Cos[c + d*x]*Sin
[c + d*x])/(8*a*d) + ((5*A - 4*B + 4*C)*Cos[c + d*x]^3*Sin[c + d*x])/(4*a*d) - ((A - B + C)*Cos[c + d*x]^3*Sin
[c + d*x])/(d*(a + a*Sec[c + d*x])) + ((4*A - 4*B + 3*C)*Sin[c + d*x]^3)/(3*a*d)

Rule 4084

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

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 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 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]

Rubi steps

\begin{align*} \int \frac{\cos ^4(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{a+a \sec (c+d x)} \, dx &=-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{\int \cos ^4(c+d x) (a (5 A-4 B+4 C)-a (4 A-4 B+3 C) \sec (c+d x)) \, dx}{a^2}\\ &=-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}-\frac{(4 A-4 B+3 C) \int \cos ^3(c+d x) \, dx}{a}+\frac{(5 A-4 B+4 C) \int \cos ^4(c+d x) \, dx}{a}\\ &=\frac{(5 A-4 B+4 C) \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{(3 (5 A-4 B+4 C)) \int \cos ^2(c+d x) \, dx}{4 a}+\frac{(4 A-4 B+3 C) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{a d}\\ &=-\frac{(4 A-4 B+3 C) \sin (c+d x)}{a d}+\frac{3 (5 A-4 B+4 C) \cos (c+d x) \sin (c+d x)}{8 a d}+\frac{(5 A-4 B+4 C) \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{(4 A-4 B+3 C) \sin ^3(c+d x)}{3 a d}+\frac{(3 (5 A-4 B+4 C)) \int 1 \, dx}{8 a}\\ &=\frac{3 (5 A-4 B+4 C) x}{8 a}-\frac{(4 A-4 B+3 C) \sin (c+d x)}{a d}+\frac{3 (5 A-4 B+4 C) \cos (c+d x) \sin (c+d x)}{8 a d}+\frac{(5 A-4 B+4 C) \cos ^3(c+d x) \sin (c+d x)}{4 a d}-\frac{(A-B+C) \cos ^3(c+d x) \sin (c+d x)}{d (a+a \sec (c+d x))}+\frac{(4 A-4 B+3 C) \sin ^3(c+d x)}{3 a d}\\ \end{align*}

Mathematica [B]  time = 1.00996, size = 393, normalized size = 2.26 \[ \frac{\sec \left (\frac{c}{2}\right ) \cos \left (\frac{1}{2} (c+d x)\right ) \left (72 d x (5 A-4 B+4 C) \cos \left (c+\frac{d x}{2}\right )+72 d x (5 A-4 B+4 C) \cos \left (\frac{d x}{2}\right )-168 A \sin \left (c+\frac{d x}{2}\right )-120 A \sin \left (c+\frac{3 d x}{2}\right )-120 A \sin \left (2 c+\frac{3 d x}{2}\right )+40 A \sin \left (2 c+\frac{5 d x}{2}\right )+40 A \sin \left (3 c+\frac{5 d x}{2}\right )-5 A \sin \left (3 c+\frac{7 d x}{2}\right )-5 A \sin \left (4 c+\frac{7 d x}{2}\right )+3 A \sin \left (4 c+\frac{9 d x}{2}\right )+3 A \sin \left (5 c+\frac{9 d x}{2}\right )-552 A \sin \left (\frac{d x}{2}\right )+168 B \sin \left (c+\frac{d x}{2}\right )+144 B \sin \left (c+\frac{3 d x}{2}\right )+144 B \sin \left (2 c+\frac{3 d x}{2}\right )-16 B \sin \left (2 c+\frac{5 d x}{2}\right )-16 B \sin \left (3 c+\frac{5 d x}{2}\right )+8 B \sin \left (3 c+\frac{7 d x}{2}\right )+8 B \sin \left (4 c+\frac{7 d x}{2}\right )+552 B \sin \left (\frac{d x}{2}\right )-96 C \sin \left (c+\frac{d x}{2}\right )-72 C \sin \left (c+\frac{3 d x}{2}\right )-72 C \sin \left (2 c+\frac{3 d x}{2}\right )+24 C \sin \left (2 c+\frac{5 d x}{2}\right )+24 C \sin \left (3 c+\frac{5 d x}{2}\right )-480 C \sin \left (\frac{d x}{2}\right )\right )}{192 a d (\cos (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]*Sec[c/2]*(72*(5*A - 4*B + 4*C)*d*x*Cos[(d*x)/2] + 72*(5*A - 4*B + 4*C)*d*x*Cos[c + (d*x)/2]
- 552*A*Sin[(d*x)/2] + 552*B*Sin[(d*x)/2] - 480*C*Sin[(d*x)/2] - 168*A*Sin[c + (d*x)/2] + 168*B*Sin[c + (d*x)/
2] - 96*C*Sin[c + (d*x)/2] - 120*A*Sin[c + (3*d*x)/2] + 144*B*Sin[c + (3*d*x)/2] - 72*C*Sin[c + (3*d*x)/2] - 1
20*A*Sin[2*c + (3*d*x)/2] + 144*B*Sin[2*c + (3*d*x)/2] - 72*C*Sin[2*c + (3*d*x)/2] + 40*A*Sin[2*c + (5*d*x)/2]
- 16*B*Sin[2*c + (5*d*x)/2] + 24*C*Sin[2*c + (5*d*x)/2] + 40*A*Sin[3*c + (5*d*x)/2] - 16*B*Sin[3*c + (5*d*x)/
2] + 24*C*Sin[3*c + (5*d*x)/2] - 5*A*Sin[3*c + (7*d*x)/2] + 8*B*Sin[3*c + (7*d*x)/2] - 5*A*Sin[4*c + (7*d*x)/2
] + 8*B*Sin[4*c + (7*d*x)/2] + 3*A*Sin[4*c + (9*d*x)/2] + 3*A*Sin[5*c + (9*d*x)/2]))/(192*a*d*(1 + Cos[c + d*x
]))

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Maple [B]  time = 0.111, size = 526, normalized size = 3. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/a/d*A*tan(1/2*d*x+1/2*c)+1/a/d*B*tan(1/2*d*x+1/2*c)-1/a/d*C*tan(1/2*d*x+1/2*c)-25/4/a/d/(1+tan(1/2*d*x+1/2*
c)^2)^4*tan(1/2*d*x+1/2*c)^7*A-3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*C+5/a/d/(1+tan(1/2*d*x+1/
2*c)^2)^4*tan(1/2*d*x+1/2*c)^7*B-115/12/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*A-7/a/d/(1+tan(1/2
*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*C+31/3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^5*B-109/12/a/d/
(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*A-5/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*C+25/3
/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*tan(1/2*d*x+1/2*c)^3*B-7/4/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*A*tan(1/2*d*x+1/2*c)
-1/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*C*tan(1/2*d*x+1/2*c)+3/a/d/(1+tan(1/2*d*x+1/2*c)^2)^4*B*tan(1/2*d*x+1/2*c)+1
5/4/a/d*A*arctan(tan(1/2*d*x+1/2*c))-3/a/d*arctan(tan(1/2*d*x+1/2*c))*B+3/a/d*arctan(tan(1/2*d*x+1/2*c))*C

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Maxima [B]  time = 1.45223, size = 709, normalized size = 4.07 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(A*((21*sin(d*x + c)/(cos(d*x + c) + 1) + 109*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 115*sin(d*x + c)^5/(
cos(d*x + c) + 1)^5 + 75*sin(d*x + c)^7/(cos(d*x + c) + 1)^7)/(a + 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6
*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 4*a*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x +
c) + 1)^8) - 45*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 12*sin(d*x + c)/(a*(cos(d*x + c) + 1))) - 4*B*((9
*sin(d*x + c)/(cos(d*x + c) + 1) + 16*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 15*sin(d*x + c)^5/(cos(d*x + c) +
1)^5)/(a + 3*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + a*sin(d*x + c)^
6/(cos(d*x + c) + 1)^6) - 9*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + 3*sin(d*x + c)/(a*(cos(d*x + c) + 1)))
+ 12*C*((sin(d*x + c)/(cos(d*x + c) + 1) + 3*sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/(a + 2*a*sin(d*x + c)^2/(co
s(d*x + c) + 1)^2 + a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4) - 3*arctan(sin(d*x + c)/(cos(d*x + c) + 1))/a + sin
(d*x + c)/(a*(cos(d*x + c) + 1))))/d

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Fricas [A]  time = 0.50898, size = 344, normalized size = 1.98 \begin{align*} \frac{9 \,{\left (5 \, A - 4 \, B + 4 \, C\right )} d x \cos \left (d x + c\right ) + 9 \,{\left (5 \, A - 4 \, B + 4 \, C\right )} d x +{\left (6 \, A \cos \left (d x + c\right )^{4} - 2 \,{\left (A - 4 \, B\right )} \cos \left (d x + c\right )^{3} +{\left (13 \, A - 4 \, B + 12 \, C\right )} \cos \left (d x + c\right )^{2} -{\left (19 \, A - 28 \, B + 12 \, C\right )} \cos \left (d x + c\right ) - 64 \, A + 64 \, B - 48 \, C\right )} \sin \left (d x + c\right )}{24 \,{\left (a d \cos \left (d x + c\right ) + a d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(9*(5*A - 4*B + 4*C)*d*x*cos(d*x + c) + 9*(5*A - 4*B + 4*C)*d*x + (6*A*cos(d*x + c)^4 - 2*(A - 4*B)*cos(d
*x + c)^3 + (13*A - 4*B + 12*C)*cos(d*x + c)^2 - (19*A - 28*B + 12*C)*cos(d*x + c) - 64*A + 64*B - 48*C)*sin(d
*x + c))/(a*d*cos(d*x + c) + a*d)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.27924, size = 336, normalized size = 1.93 \begin{align*} \frac{\frac{9 \,{\left (d x + c\right )}{\left (5 \, A - 4 \, B + 4 \, C\right )}}{a} - \frac{24 \,{\left (A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )}}{a} - \frac{2 \,{\left (75 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 60 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 36 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 115 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 124 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 84 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 109 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 100 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 60 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 21 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 36 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, 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 )}^{4} a}}{24 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(9*(d*x + c)*(5*A - 4*B + 4*C)/a - 24*(A*tan(1/2*d*x + 1/2*c) - B*tan(1/2*d*x + 1/2*c) + C*tan(1/2*d*x +
1/2*c))/a - 2*(75*A*tan(1/2*d*x + 1/2*c)^7 - 60*B*tan(1/2*d*x + 1/2*c)^7 + 36*C*tan(1/2*d*x + 1/2*c)^7 + 115*A
*tan(1/2*d*x + 1/2*c)^5 - 124*B*tan(1/2*d*x + 1/2*c)^5 + 84*C*tan(1/2*d*x + 1/2*c)^5 + 109*A*tan(1/2*d*x + 1/2
*c)^3 - 100*B*tan(1/2*d*x + 1/2*c)^3 + 60*C*tan(1/2*d*x + 1/2*c)^3 + 21*A*tan(1/2*d*x + 1/2*c) - 36*B*tan(1/2*
d*x + 1/2*c) + 12*C*tan(1/2*d*x + 1/2*c))/((tan(1/2*d*x + 1/2*c)^2 + 1)^4*a))/d