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

Optimal. Leaf size=136 \[ \frac{\left (a^2 B+2 a A b+b^2 B\right ) \sin (c+d x)}{d}+\frac{\left (3 a^2 A+8 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (3 a^2 A+8 a b B+4 A b^2\right )+\frac{a^2 A \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac{a (a B+2 A b) \sin ^3(c+d x)}{3 d} \]

[Out]

((3*a^2*A + 4*A*b^2 + 8*a*b*B)*x)/8 + ((2*a*A*b + a^2*B + b^2*B)*Sin[c + d*x])/d + ((3*a^2*A + 4*A*b^2 + 8*a*b
*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*A*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (a*(2*A*b + a*B)*Sin[c + d*
x]^3)/(3*d)

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Rubi [A]  time = 0.259949, antiderivative size = 136, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.194, Rules used = {4024, 4047, 2635, 8, 4044, 3013} \[ \frac{\left (a^2 B+2 a A b+b^2 B\right ) \sin (c+d x)}{d}+\frac{\left (3 a^2 A+8 a b B+4 A b^2\right ) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{1}{8} x \left (3 a^2 A+8 a b B+4 A b^2\right )+\frac{a^2 A \sin (c+d x) \cos ^3(c+d x)}{4 d}-\frac{a (a B+2 A b) \sin ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2*A + 4*A*b^2 + 8*a*b*B)*x)/8 + ((2*a*A*b + a^2*B + b^2*B)*Sin[c + d*x])/d + ((3*a^2*A + 4*A*b^2 + 8*a*b
*B)*Cos[c + d*x]*Sin[c + d*x])/(8*d) + (a^2*A*Cos[c + d*x]^3*Sin[c + d*x])/(4*d) - (a*(2*A*b + a*B)*Sin[c + d*
x]^3)/(3*d)

Rule 4024

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^2*(csc[(e_.) + (f_.)*(x_)]*(B_
.) + (A_)), x_Symbol] :> Simp[(a^2*A*Cos[e + f*x]*(d*Csc[e + f*x])^(n + 1))/(d*f*n), x] + Dist[1/(d*n), Int[(d
*Csc[e + f*x])^(n + 1)*(a*(2*A*b + a*B)*n + (2*a*b*B*n + A*(b^2*n + a^2*(n + 1)))*Csc[e + f*x] + b^2*B*n*Csc[e
+ f*x]^2), x], x] /; FreeQ[{a, b, d, e, f, A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LeQ[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))^2 (A+B \sec (c+d x)) \, dx &=\frac{a^2 A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{1}{4} \int \cos ^3(c+d x) \left (-4 a (2 A b+a B)+\left (A \left (-3 a^2-4 b^2\right )-8 a b B\right ) \sec (c+d x)-4 b^2 B \sec ^2(c+d x)\right ) \, dx\\ &=\frac{a^2 A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{1}{4} \int \cos ^3(c+d x) \left (-4 a (2 A b+a B)-4 b^2 B \sec ^2(c+d x)\right ) \, dx-\frac{1}{4} \left (-3 a^2 A-4 A b^2-8 a b B\right ) \int \cos ^2(c+d x) \, dx\\ &=\frac{\left (3 a^2 A+4 A b^2+8 a b B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{1}{4} \int \cos (c+d x) \left (-4 b^2 B-4 a (2 A b+a B) \cos ^2(c+d x)\right ) \, dx-\frac{1}{8} \left (-3 a^2 A-4 A b^2-8 a b B\right ) \int 1 \, dx\\ &=\frac{1}{8} \left (3 a^2 A+4 A b^2+8 a b B\right ) x+\frac{\left (3 a^2 A+4 A b^2+8 a b B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 A \cos ^3(c+d x) \sin (c+d x)}{4 d}+\frac{\operatorname{Subst}\left (\int \left (-4 b^2 B-4 a (2 A b+a B)+4 a (2 A b+a B) x^2\right ) \, dx,x,-\sin (c+d x)\right )}{4 d}\\ &=\frac{1}{8} \left (3 a^2 A+4 A b^2+8 a b B\right ) x+\frac{\left (2 a A b+a^2 B+b^2 B\right ) \sin (c+d x)}{d}+\frac{\left (3 a^2 A+4 A b^2+8 a b B\right ) \cos (c+d x) \sin (c+d x)}{8 d}+\frac{a^2 A \cos ^3(c+d x) \sin (c+d x)}{4 d}-\frac{a (2 A b+a B) \sin ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.447607, size = 118, normalized size = 0.87 \[ \frac{12 (c+d x) \left (3 a^2 A+8 a b B+4 A b^2\right )+24 \left (3 a^2 B+6 a A b+4 b^2 B\right ) \sin (c+d x)+24 \left (a^2 A+2 a b B+A b^2\right ) \sin (2 (c+d x))+3 a^2 A \sin (4 (c+d x))+8 a (a B+2 A b) \sin (3 (c+d x))}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(12*(3*a^2*A + 4*A*b^2 + 8*a*b*B)*(c + d*x) + 24*(6*a*A*b + 3*a^2*B + 4*b^2*B)*Sin[c + d*x] + 24*(a^2*A + A*b^
2 + 2*a*b*B)*Sin[2*(c + d*x)] + 8*a*(2*A*b + a*B)*Sin[3*(c + d*x)] + 3*a^2*A*Sin[4*(c + d*x)])/(96*d)

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Maple [A]  time = 0.068, size = 152, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{2}A \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{B{a}^{2} \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{\frac{2\,Aab \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+2\,Bab \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +A{b}^{2} \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +B{b}^{2}\sin \left ( dx+c \right ) \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))^2*(A+B*sec(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 0.992416, size = 192, normalized size = 1.41 \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^{2} - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B a^{2} - 64 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a b + 48 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a b + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b^{2} + 96 \, B b^{2} \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))^2*(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^2 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B
*a^2 - 64*(sin(d*x + c)^3 - 3*sin(d*x + c))*A*a*b + 48*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a*b + 24*(2*d*x + 2*
c + sin(2*d*x + 2*c))*A*b^2 + 96*B*b^2*sin(d*x + c))/d

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Fricas [A]  time = 0.492599, size = 274, normalized size = 2.01 \begin{align*} \frac{3 \,{\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )} d x +{\left (6 \, A a^{2} \cos \left (d x + c\right )^{3} + 16 \, B a^{2} + 32 \, A a b + 24 \, B b^{2} + 8 \,{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\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))^2*(A+B*sec(d*x+c)),x, algorithm="fricas")

[Out]

1/24*(3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*d*x + (6*A*a^2*cos(d*x + c)^3 + 16*B*a^2 + 32*A*a*b + 24*B*b^2 + 8*(B*a^
2 + 2*A*a*b)*cos(d*x + c)^2 + 3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*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))**2*(A+B*sec(d*x+c)),x)

[Out]

Timed out

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Giac [B]  time = 1.1952, size = 590, normalized size = 4.34 \begin{align*} \frac{3 \,{\left (3 \, A a^{2} + 8 \, B a b + 4 \, A b^{2}\right )}{\left (d x + c\right )} - \frac{2 \,{\left (15 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 24 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 48 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 24 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 24 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 9 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 40 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 80 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 24 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 72 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 80 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 24 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 72 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 48 \, A a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 12 \, A b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, B b^{2} \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))^2*(A+B*sec(d*x+c)),x, algorithm="giac")

[Out]

1/24*(3*(3*A*a^2 + 8*B*a*b + 4*A*b^2)*(d*x + c) - 2*(15*A*a^2*tan(1/2*d*x + 1/2*c)^7 - 24*B*a^2*tan(1/2*d*x +
1/2*c)^7 - 48*A*a*b*tan(1/2*d*x + 1/2*c)^7 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^7 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^7
- 24*B*b^2*tan(1/2*d*x + 1/2*c)^7 - 9*A*a^2*tan(1/2*d*x + 1/2*c)^5 - 40*B*a^2*tan(1/2*d*x + 1/2*c)^5 - 80*A*a
*b*tan(1/2*d*x + 1/2*c)^5 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^5 + 12*A*b^2*tan(1/2*d*x + 1/2*c)^5 - 72*B*b^2*tan(1
/2*d*x + 1/2*c)^5 + 9*A*a^2*tan(1/2*d*x + 1/2*c)^3 - 40*B*a^2*tan(1/2*d*x + 1/2*c)^3 - 80*A*a*b*tan(1/2*d*x +
1/2*c)^3 - 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 - 12*A*b^2*tan(1/2*d*x + 1/2*c)^3 - 72*B*b^2*tan(1/2*d*x + 1/2*c)^3
- 15*A*a^2*tan(1/2*d*x + 1/2*c) - 24*B*a^2*tan(1/2*d*x + 1/2*c) - 48*A*a*b*tan(1/2*d*x + 1/2*c) - 24*B*a*b*ta
n(1/2*d*x + 1/2*c) - 12*A*b^2*tan(1/2*d*x + 1/2*c) - 24*B*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 +
1)^4)/d