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

Optimal. Leaf size=92 \[ \frac{\sin (c+d x) (2 a A+3 a C+3 b B)}{3 d}+\frac{(a B+A b) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} x (a B+A b+2 b C)+\frac{a A \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.179562, antiderivative size = 92, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 39, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.128, Rules used = {4074, 4047, 2637, 4045, 8} \[ \frac{\sin (c+d x) (2 a A+3 a C+3 b B)}{3 d}+\frac{(a B+A b) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} x (a B+A b+2 b C)+\frac{a A \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((A*b + a*B + 2*b*C)*x)/2 + ((2*a*A + 3*b*B + 3*a*C)*Sin[c + d*x])/(3*d) + ((A*b + a*B)*Cos[c + d*x]*Sin[c + d
*x])/(2*d) + (a*A*Cos[c + d*x]^2*Sin[c + d*x])/(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 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

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 8

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

Rubi steps

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

Mathematica [A]  time = 0.183066, size = 85, normalized size = 0.92 \[ \frac{3 \sin (c+d x) (3 a A+4 a C+4 b B)+3 (a B+A b) \sin (2 (c+d x))+a A \sin (3 (c+d x))+6 a B c+6 a B d x+6 A b c+6 A b d x+12 b C d x}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.061, size = 102, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{Aa \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+Ab \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +Ba \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +Bb\sin \left ( dx+c \right ) +aC\sin \left ( dx+c \right ) +Cb \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.0218, size = 132, normalized size = 1.43 \begin{align*} -\frac{4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} A a - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} A b - 12 \,{\left (d x + c\right )} C b - 12 \, C a \sin \left (d x + c\right ) - 12 \, B b \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 0.509012, size = 173, normalized size = 1.88 \begin{align*} \frac{3 \,{\left (B a +{\left (A + 2 \, C\right )} b\right )} d x +{\left (2 \, A a \cos \left (d x + c\right )^{2} + 2 \,{\left (2 \, A + 3 \, C\right )} a + 6 \, B b + 3 \,{\left (B a + A b\right )} \cos \left (d x + c\right )\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)^3*(a+b*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/6*(3*(B*a + (A + 2*C)*b)*d*x + (2*A*a*cos(d*x + c)^2 + 2*(2*A + 3*C)*a + 6*B*b + 3*(B*a + A*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)**3*(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 = 1.17735, size = 306, normalized size = 3.33 \begin{align*} \frac{3 \,{\left (B a + A b + 2 \, C b\right )}{\left (d x + c\right )} + \frac{2 \,{\left (6 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 4 \, A 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} + 12 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, B 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 )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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