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

Optimal. Leaf size=82 \[ \frac{a (2 A+3 (B+C)) \sin (c+d x)}{3 d}+\frac{a (A+B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} a x (A+B+2 C)+\frac{a A \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

[Out]

(a*(A + B + 2*C)*x)/2 + (a*(2*A + 3*(B + C))*Sin[c + d*x])/(3*d) + (a*(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 = 0.17611, antiderivative size = 82, 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} \[ \frac{a (2 A+3 (B+C)) \sin (c+d x)}{3 d}+\frac{a (A+B) \sin (c+d x) \cos (c+d x)}{2 d}+\frac{1}{2} a x (A+B+2 C)+\frac{a A \sin (c+d x) \cos ^2(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(A + B + 2*C)*x)/2 + (a*(2*A + 3*(B + C))*Sin[c + d*x])/(3*d) + (a*(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+a \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 (A+B)-a (2 A+3 (B+C)) \sec (c+d x)-3 a 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 (A+B)-3 a C \sec ^2(c+d x)\right ) \, dx+\frac{1}{3} (a (2 A+3 (B+C))) \int \cos (c+d x) \, dx\\ &=\frac{a (2 A+3 (B+C)) \sin (c+d x)}{3 d}+\frac{a (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 (A+B+2 C)) \int 1 \, dx\\ &=\frac{1}{2} a (A+B+2 C) x+\frac{a (2 A+3 (B+C)) \sin (c+d x)}{3 d}+\frac{a (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 = 0.228515, size = 64, normalized size = 0.78 \[ \frac{a (3 (3 A+4 (B+C)) \sin (c+d x)+3 (A+B) \sin (2 (c+d x))+A \sin (3 (c+d x))+6 A d x+6 B d x+12 C d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.087, size = 102, normalized size = 1.2 \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}}+Aa \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 ) +Ba\sin \left ( dx+c \right ) +aC\sin \left ( dx+c \right ) +aC \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+a*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*a*(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*a*sin(d*x+c)+a*C*sin(d*x+c)+a*C*(d*x+c))

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Maxima [A]  time = 0.93669, size = 132, normalized size = 1.61 \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 )} A a - 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 12 \,{\left (d x + c\right )} C a - 12 \, B a \sin \left (d x + c\right ) - 12 \, C a \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+a*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))*A*a - 3*(2*d*x + 2*c + sin
(2*d*x + 2*c))*B*a - 12*(d*x + c)*C*a - 12*B*a*sin(d*x + c) - 12*C*a*sin(d*x + c))/d

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Fricas [A]  time = 0.497724, size = 162, normalized size = 1.98 \begin{align*} \frac{3 \,{\left (A + B + 2 \, C\right )} a d x +{\left (2 \, A a \cos \left (d x + c\right )^{2} + 3 \,{\left (A + B\right )} a \cos \left (d x + c\right ) + 2 \,{\left (2 \, A + 3 \, B + 3 \, C\right )} a\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+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="fricas")

[Out]

1/6*(3*(A + B + 2*C)*a*d*x + (2*A*a*cos(d*x + c)^2 + 3*(A + B)*a*cos(d*x + c) + 2*(2*A + 3*B + 3*C)*a)*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+a*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 = 1.2487, size = 231, normalized size = 2.82 \begin{align*} \frac{3 \,{\left (A a + B a + 2 \, C a\right )}{\left (d x + c\right )} + \frac{2 \,{\left (3 \, 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} + 4 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 12 \, B 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} + 9 \, A a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 9 \, 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 )\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+a*sec(d*x+c))*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="giac")

[Out]

1/6*(3*(A*a + B*a + 2*C*a)*(d*x + c) + 2*(3*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 + 4*A*a*tan(1/2*d*x + 1/2*c)^3 + 12*B*a*tan(1/2*d*x + 1/2*c)^3 + 12*C*a*tan(1/2*d*x + 1
/2*c)^3 + 9*A*a*tan(1/2*d*x + 1/2*c) + 9*B*a*tan(1/2*d*x + 1/2*c) + 6*C*a*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x +
1/2*c)^2 + 1)^3)/d