3.284 \(\int \cos ^4(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)) \, dx\)

Optimal. Leaf size=105 \[ -\frac {(a B+A b) \sin ^3(c+d x)}{3 d}+\frac {(a B+A b) \sin (c+d x)}{d}+\frac {(3 a A+4 b B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x (3 a A+4 b B)+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[Out]

1/8*(3*A*a+4*B*b)*x+(A*b+B*a)*sin(d*x+c)/d+1/8*(3*A*a+4*B*b)*cos(d*x+c)*sin(d*x+c)/d+1/4*a*A*cos(d*x+c)^3*sin(
d*x+c)/d-1/3*(A*b+B*a)*sin(d*x+c)^3/d

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Rubi [A]  time = 0.14, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {3996, 3787, 2633, 2635, 8} \[ -\frac {(a B+A b) \sin ^3(c+d x)}{3 d}+\frac {(a B+A b) \sin (c+d x)}{d}+\frac {(3 a A+4 b B) \sin (c+d x) \cos (c+d x)}{8 d}+\frac {1}{8} x (3 a A+4 b B)+\frac {a A \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a*A + 4*b*B)*x)/8 + ((A*b + a*B)*Sin[c + d*x])/d + ((3*a*A + 4*b*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a*
A*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - ((A*b + a*B)*Sin[c + d*x]^3)/(3*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 3996

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(B_.)
+ (A_)), x_Symbol] :> Simp[(A*a*Cot[e + f*x]*(d*Csc[e + f*x])^n)/(f*n), x] + Dist[1/(d*n), Int[(d*Csc[e + f*x
])^(n + 1)*Simp[n*(B*a + A*b) + (B*b*n + A*a*(n + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B},
x] && NeQ[A*b - a*B, 0] && LeQ[n, -1]

Rubi steps

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

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Mathematica [A]  time = 0.24, size = 91, normalized size = 0.87 \[ \frac {-32 (a B+A b) \sin ^3(c+d x)+96 (a B+A b) \sin (c+d x)+24 (a A+b B) \sin (2 (c+d x))+3 a A \sin (4 (c+d x))+36 a A c+36 a A d x+48 b B c+48 b B d x}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*a*A*c + 48*b*B*c + 36*a*A*d*x + 48*b*B*d*x + 96*(A*b + a*B)*Sin[c + d*x] - 32*(A*b + a*B)*Sin[c + d*x]^3 +
24*(a*A + b*B)*Sin[2*(c + d*x)] + 3*a*A*Sin[4*(c + d*x)])/(96*d)

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fricas [A]  time = 0.44, size = 81, normalized size = 0.77 \[ \frac {3 \, {\left (3 \, A a + 4 \, B b\right )} d x + {\left (6 \, A a \cos \left (d x + c\right )^{3} + 8 \, {\left (B a + A b\right )} \cos \left (d x + c\right )^{2} + 16 \, B a + 16 \, A b + 3 \, {\left (3 \, A a + 4 \, B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*A*a + 4*B*b)*d*x + (6*A*a*cos(d*x + c)^3 + 8*(B*a + A*b)*cos(d*x + c)^2 + 16*B*a + 16*A*b + 3*(3*A*
a + 4*B*b)*cos(d*x + c))*sin(d*x + c))/d

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giac [B]  time = 0.24, size = 272, normalized size = 2.59 \[ \frac {3 \, {\left (3 \, A a + 4 \, B b\right )} {\left (d x + c\right )} - \frac {2 \, {\left (15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 9 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 40 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 15 \, A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 24 \, A b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 12 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{4}}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(3*(3*A*a + 4*B*b)*(d*x + c) - 2*(15*A*a*tan(1/2*d*x + 1/2*c)^7 - 24*B*a*tan(1/2*d*x + 1/2*c)^7 - 24*A*b*
tan(1/2*d*x + 1/2*c)^7 + 12*B*b*tan(1/2*d*x + 1/2*c)^7 - 9*A*a*tan(1/2*d*x + 1/2*c)^5 - 40*B*a*tan(1/2*d*x + 1
/2*c)^5 - 40*A*b*tan(1/2*d*x + 1/2*c)^5 + 12*B*b*tan(1/2*d*x + 1/2*c)^5 + 9*A*a*tan(1/2*d*x + 1/2*c)^3 - 40*B*
a*tan(1/2*d*x + 1/2*c)^3 - 40*A*b*tan(1/2*d*x + 1/2*c)^3 - 12*B*b*tan(1/2*d*x + 1/2*c)^3 - 15*A*a*tan(1/2*d*x
+ 1/2*c) - 24*B*a*tan(1/2*d*x + 1/2*c) - 24*A*b*tan(1/2*d*x + 1/2*c) - 12*B*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d
*x + 1/2*c)^2 + 1)^4)/d

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maple [A]  time = 1.52, size = 107, normalized size = 1.02 \[ \frac {a A \left (\frac {\left (\cos ^{3}\left (d x +c \right )+\frac {3 \cos \left (d x +c \right )}{2}\right ) \sin \left (d x +c \right )}{4}+\frac {3 d x}{8}+\frac {3 c}{8}\right )+\frac {A b \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+\frac {a B \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*A*(1/4*(cos(d*x+c)^3+3/2*cos(d*x+c))*sin(d*x+c)+3/8*d*x+3/8*c)+1/3*A*b*(2+cos(d*x+c)^2)*sin(d*x+c)+1/3*
a*B*(2+cos(d*x+c)^2)*sin(d*x+c)+B*b*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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maxima [A]  time = 2.05, size = 101, normalized size = 0.96 \[ \frac {3 \, {\left (12 \, d x + 12 \, c + \sin \left (4 \, d x + 4 \, c\right ) + 8 \, \sin \left (2 \, d x + 2 \, c\right )\right )} A a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a - 32 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A b + 24 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(12*d*x + 12*c + sin(4*d*x + 4*c) + 8*sin(2*d*x + 2*c))*A*a - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B*a
- 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*b + 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*b)/d

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mupad [B]  time = 2.13, size = 117, normalized size = 1.11 \[ \frac {3\,A\,a\,x}{8}+\frac {B\,b\,x}{2}+\frac {3\,A\,b\,\sin \left (c+d\,x\right )}{4\,d}+\frac {3\,B\,a\,\sin \left (c+d\,x\right )}{4\,d}+\frac {A\,a\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {A\,a\,\sin \left (4\,c+4\,d\,x\right )}{32\,d}+\frac {A\,b\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,a\,\sin \left (3\,c+3\,d\,x\right )}{12\,d}+\frac {B\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*A*a*x)/8 + (B*b*x)/2 + (3*A*b*sin(c + d*x))/(4*d) + (3*B*a*sin(c + d*x))/(4*d) + (A*a*sin(2*c + 2*d*x))/(4*
d) + (A*a*sin(4*c + 4*d*x))/(32*d) + (A*b*sin(3*c + 3*d*x))/(12*d) + (B*a*sin(3*c + 3*d*x))/(12*d) + (B*b*sin(
2*c + 2*d*x))/(4*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \sec {\left (c + d x \right )}\right ) \left (a + b \sec {\left (c + d x \right )}\right ) \cos ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**4*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)),x)

[Out]

Integral((A + B*sec(c + d*x))*(a + b*sec(c + d*x))*cos(c + d*x)**4, x)

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