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

Optimal. Leaf size=54 \[ -\frac{B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d}+\frac{C \sin (c+d x) \cos (c+d x)}{2 d}+\frac{C x}{2} \]

[Out]

(C*x)/2 + (B*Sin[c + d*x])/d + (C*Cos[c + d*x]*Sin[c + d*x])/(2*d) - (B*Sin[c + d*x]^3)/(3*d)

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Rubi [A]  time = 0.051129, antiderivative size = 54, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.179, Rules used = {4047, 2633, 12, 2635, 8} \[ -\frac{B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d}+\frac{C \sin (c+d x) \cos (c+d x)}{2 d}+\frac{C x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*x)/2 + (B*Sin[c + d*x])/d + (C*Cos[c + d*x]*Sin[c + d*x])/(2*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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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]

Rubi steps

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

Mathematica [A]  time = 0.0624231, size = 57, normalized size = 1.06 \[ -\frac{B \sin ^3(c+d x)}{3 d}+\frac{B \sin (c+d x)}{d}+\frac{C (c+d x)}{2 d}+\frac{C \sin (2 (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(C*(c + d*x))/(2*d) + (B*Sin[c + d*x])/d - (B*Sin[c + d*x]^3)/(3*d) + (C*Sin[2*(c + d*x)])/(4*d)

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Maple [A]  time = 0.046, size = 49, normalized size = 0.9 \begin{align*}{\frac{1}{d} \left ({\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*(B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

1/d*(1/3*B*(2+cos(d*x+c)^2)*sin(d*x+c)+C*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c))

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Maxima [A]  time = 0.936716, size = 62, normalized size = 1.15 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} B - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(4*(sin(d*x + c)^3 - 3*sin(d*x + c))*B - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*C)/d

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Fricas [A]  time = 0.481587, size = 105, normalized size = 1.94 \begin{align*} \frac{3 \, C d x +{\left (2 \, B \cos \left (d x + c\right )^{2} + 3 \, C \cos \left (d x + c\right ) + 4 \, B\right )} \sin \left (d x + c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [B]  time = 1.22505, size = 132, normalized size = 2.44 \begin{align*} \frac{3 \,{\left (d x + c\right )} C + \frac{2 \,{\left (6 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, C \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, 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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(d*x + c)*C + 2*(6*B*tan(1/2*d*x + 1/2*c)^5 - 3*C*tan(1/2*d*x + 1/2*c)^5 + 4*B*tan(1/2*d*x + 1/2*c)^3 +
6*B*tan(1/2*d*x + 1/2*c) + 3*C*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d