3.868 \(\int \cos ^4(c+d x) (a+b \sec (c+d x)) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=116 \[ \frac{\sin (c+d x) (a B+A b+b C)}{d}+\frac{\sin (c+d x) \cos (c+d x) (3 a A+4 a C+4 b B)}{8 d}-\frac{(a B+A b) \sin ^3(c+d x)}{3 d}+\frac{1}{8} x (3 a A+4 a C+4 b B)+\frac{a A \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

[Out]

((3*a*A + 4*b*B + 4*a*C)*x)/8 + ((A*b + a*B + b*C)*Sin[c + d*x])/d + ((3*a*A + 4*b*B + 4*a*C)*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)

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Rubi [A]  time = 0.214684, antiderivative size = 116, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {4074, 4047, 2635, 8, 4044, 3013} \[ \frac{\sin (c+d x) (a B+A b+b C)}{d}+\frac{\sin (c+d x) \cos (c+d x) (3 a A+4 a C+4 b B)}{8 d}-\frac{(a B+A b) \sin ^3(c+d x)}{3 d}+\frac{1}{8} x (3 a A+4 a C+4 b B)+\frac{a A \sin (c+d x) \cos ^3(c+d x)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a*A + 4*b*B + 4*a*C)*x)/8 + ((A*b + a*B + b*C)*Sin[c + d*x])/d + ((3*a*A + 4*b*B + 4*a*C)*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 4074

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_)), 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) + (n*(a*C + B*b) + A*a*(n + 1))*Csc[e + f*x]
+ b*C*n*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B, C}, x] && LtQ[n, -1]

Rule 4047

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(
C_.)), x_Symbol] :> Dist[B/b, Int[(b*Csc[e + f*x])^(m + 1), x], x] + Int[(b*Csc[e + f*x])^m*(A + C*Csc[e + f*x
]^2), x] /; FreeQ[{b, e, f, A, B, C, m}, 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 4044

Int[csc[(e_.) + (f_.)*(x_)]^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Int[(C + A*Sin[e + f*
x]^2)/Sin[e + f*x]^(m + 2), x] /; FreeQ[{e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && ILtQ[(m + 1)/2, 0]

Rule 3013

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Dist[f^(-1), Subst[I
nt[(1 - x^2)^((m - 1)/2)*(A + C - C*x^2), x], x, Cos[e + f*x]], x] /; FreeQ[{e, f, A, C}, x] && IGtQ[(m + 1)/2
, 0]

Rubi steps

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

Mathematica [A]  time = 0.315989, size = 117, normalized size = 1.01 \[ \frac{24 \sin (c+d x) (3 a B+3 A b+4 b C)+24 \sin (2 (c+d x)) (a (A+C)+b B)+3 a A \sin (4 (c+d x))+36 a A c+36 a A d x+8 a B \sin (3 (c+d x))+48 a c C+48 a C d x+8 A b \sin (3 (c+d x))+48 b B c+48 b B d x}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(36*a*A*c + 48*b*B*c + 48*a*c*C + 36*a*A*d*x + 48*b*B*d*x + 48*a*C*d*x + 24*(3*A*b + 3*a*B + 4*b*C)*Sin[c + d*
x] + 24*(b*B + a*(A + C))*Sin[2*(c + d*x)] + 8*A*b*Sin[3*(c + d*x)] + 8*a*B*Sin[3*(c + d*x)] + 3*a*A*Sin[4*(c
+ d*x)])/(96*d)

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Maple [A]  time = 0.073, size = 141, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ( Aa \left ({\frac{\sin \left ( dx+c \right ) }{4} \left ( \left ( \cos \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\cos \left ( dx+c \right ) }{2}} \right ) }+{\frac{3\,dx}{8}}+{\frac{3\,c}{8}} \right ) +{\frac{Ab \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{Ba \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+Bb \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +aC \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +C\sin \left ( dx+c \right ) b \right ) } \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))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),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*
B*a*(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)+a*C*(1/2*cos(d*x+c)*sin(d*x+c)+1
/2*d*x+1/2*c)+C*sin(d*x+c)*b)

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Maxima [A]  time = 1.03443, size = 178, normalized size = 1.53 \begin{align*} \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 + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C 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 \, C b \sin \left (d x + c\right )}{96 \, 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))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),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
+ 24*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*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 + 96*C*b*sin(d*x + c))/d

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Fricas [A]  time = 0.518022, size = 239, normalized size = 2.06 \begin{align*} \frac{3 \,{\left ({\left (3 \, A + 4 \, C\right )} 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 + 8 \,{\left (2 \, A + 3 \, C\right )} b + 3 \,{\left ({\left (3 \, A + 4 \, C\right )} a + 4 \, B b\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{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))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/24*(3*((3*A + 4*C)*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 + 8*(2*A +
3*C)*b + 3*((3*A + 4*C)*a + 4*B*b)*cos(d*x + c))*sin(d*x + c))/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))*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.16558, size = 529, normalized size = 4.56 \begin{align*} \frac{3 \,{\left (3 \, A a + 4 \, C 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} + 12 \, C 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} - 24 \, C 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} + 12 \, C 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} - 72 \, C 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} - 12 \, C 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} - 72 \, C 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 ) - 12 \, C 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 ) - 24 \, C 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} \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))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/24*(3*(3*A*a + 4*C*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 +
12*C*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 - 24*C*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 + 12*C*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 - 72*C*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 - 12*C*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 - 72*C*b*tan(1/2*d*x + 1/2*c)^3 - 15*A*a*tan(1/2*d*x + 1/2*c) - 24*B*a*ta
n(1/2*d*x + 1/2*c) - 12*C*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) -
24*C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^4)/d