3.61 \(\int \cos ^3(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\)

Optimal. Leaf size=56 \[ \frac {(A+C) \sin (c+d x)}{d}-\frac {A \sin ^3(c+d x)}{3 d}+\frac {B \sin (c+d x) \cos (c+d x)}{2 d}+\frac {B x}{2} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 56, 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 = {4047, 2635, 8, 4044, 3013} \[ \frac {(A+C) \sin (c+d x)}{d}-\frac {A \sin ^3(c+d x)}{3 d}+\frac {B \sin (c+d x) \cos (c+d x)}{2 d}+\frac {B x}{2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

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

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 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]

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 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]

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 53, normalized size = 0.95 \[ \frac {3 (3 A+4 C) \sin (c+d x)+A \sin (3 (c+d x))+3 B \sin (2 (c+d x))+6 B c+6 B d x}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^3*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]

[Out]

(6*B*c + 6*B*d*x + 3*(3*A + 4*C)*Sin[c + d*x] + 3*B*Sin[2*(c + d*x)] + A*Sin[3*(c + d*x)])/(12*d)

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fricas [A]  time = 0.42, size = 45, normalized size = 0.80 \[ \frac {3 \, B d x + {\left (2 \, A \cos \left (d x + c\right )^{2} + 3 \, B \cos \left (d x + c\right ) + 4 \, A + 6 \, C\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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giac [B]  time = 0.24, size = 138, normalized size = 2.46 \[ \frac {3 \, {\left (d x + c\right )} B + \frac {2 \, {\left (6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 4 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 12 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 6 \, C \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(d*x + c)*B + 2*(6*A*tan(1/2*d*x + 1/2*c)^5 - 3*B*tan(1/2*d*x + 1/2*c)^5 + 6*C*tan(1/2*d*x + 1/2*c)^5 +
4*A*tan(1/2*d*x + 1/2*c)^3 + 12*C*tan(1/2*d*x + 1/2*c)^3 + 6*A*tan(1/2*d*x + 1/2*c) + 3*B*tan(1/2*d*x + 1/2*c
) + 6*C*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 1.36, size = 57, normalized size = 1.02 \[ \frac {\frac {A \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+C \sin \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)

[Out]

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

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maxima [A]  time = 0.35, size = 55, normalized size = 0.98 \[ -\frac {4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A - 3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B - 12 \, C \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+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))*A - 3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B - 12*C*sin(d*x + c))/d

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mupad [B]  time = 2.53, size = 66, normalized size = 1.18 \[ \frac {B\,x}{2}+\frac {2\,A\,\sin \left (c+d\,x\right )}{3\,d}+\frac {C\,\sin \left (c+d\,x\right )}{d}+\frac {B\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )}{2\,d}+\frac {A\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^3*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (A + B \sec {\left (c + d x \right )} + C \sec ^{2}{\left (c + d x \right )}\right ) \cos ^{3}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**3*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)

[Out]

Integral((A + B*sec(c + d*x) + C*sec(c + d*x)**2)*cos(c + d*x)**3, x)

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