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

Optimal. Leaf size=88 \[ \frac{(3 A+4 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{A \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{1}{8} x (3 A+4 C)-\frac{B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.0825976, antiderivative size = 88, normalized size of antiderivative = 1., 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 = {4047, 2633, 4045, 2635, 8} \[ \frac{(3 A+4 C) \sin (c+d x) \cos (c+d x)}{8 d}+\frac{A \sin (c+d x) \cos ^3(c+d x)}{4 d}+\frac{1}{8} x (3 A+4 C)-\frac{B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 4045

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*(csc[(e_.) + (f_.)*(x_)]^2*(C_.) + (A_)), x_Symbol] :> Simp[(A*Cot[e
+ f*x]*(b*Csc[e + f*x])^m)/(f*m), x] + Dist[(C*m + A*(m + 1))/(b^2*m), Int[(b*Csc[e + f*x])^(m + 2), x], x] /
; FreeQ[{b, e, f, A, C}, x] && NeQ[C*m + A*(m + 1), 0] && LeQ[m, -1]

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]

Rubi steps

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

Mathematica [A]  time = 0.167745, size = 70, normalized size = 0.8 \[ \frac{24 (A+C) \sin (2 (c+d x))+3 A \sin (4 (c+d x))+36 A c+36 A d x-32 B \sin ^3(c+d x)+96 B \sin (c+d x)+48 c C+48 C d x}{96 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.053, size = 84, normalized size = 1. \begin{align*}{\frac{1}{d} \left ( 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 \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+C \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \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)+C*sec(d*x+c)^2),x)

[Out]

1/d*(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*(2+cos(d*x+c)^2)*sin(d*x+c)+C*(1/2*c
os(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 0.930469, size = 104, normalized size = 1.18 \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 - 32 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B + 24 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C}{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)+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 - 32*(sin(d*x + c)^3 - 3*sin(d*x + c))*B + 2
4*(2*d*x + 2*c + sin(2*d*x + 2*c))*C)/d

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Fricas [A]  time = 0.488391, size = 163, normalized size = 1.85 \begin{align*} \frac{3 \,{\left (3 \, A + 4 \, C\right )} d x +{\left (6 \, A \cos \left (d x + c\right )^{3} + 8 \, B \cos \left (d x + c\right )^{2} + 3 \,{\left (3 \, A + 4 \, C\right )} \cos \left (d x + c\right ) + 16 \, B\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)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.19107, size = 270, normalized size = 3.07 \begin{align*} \frac{3 \,{\left (d x + c\right )}{\left (3 \, A + 4 \, C\right )} - \frac{2 \,{\left (15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 24 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} + 12 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{7} - 9 \, A \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 )^{5} + 12 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 9 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 40 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 12 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 15 \, A \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 24 \, 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}}}{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),x, algorithm="giac")

[Out]

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