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

Optimal. Leaf size=94 \[ \frac {2 a^2 (3 B+2 C) \sin (c+d x)}{3 d}+\frac {a^2 (3 B+2 C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {1}{2} a^2 x (3 B+2 C)+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.079, Rules used = {3029, 2751, 2644} \[ \frac {2 a^2 (3 B+2 C) \sin (c+d x)}{3 d}+\frac {a^2 (3 B+2 C) \sin (c+d x) \cos (c+d x)}{6 d}+\frac {1}{2} a^2 x (3 B+2 C)+\frac {C \sin (c+d x) (a \cos (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2644

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[((2*a^2 + b^2)*x)/2, x] + (-Simp[(2*a*b*Cos[c
+ d*x])/d, x] - Simp[(b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d), x]) /; FreeQ[{a, b, c, d}, x]

Rule 2751

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(b*(m + 1)), Int[(a + b*Sin
[e + f*x])^m, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] &&  !LtQ[m,
-2^(-1)]

Rule 3029

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])
^(m + 1)*(c + d*Sin[e + f*x])^n*(b*B - a*C + b*C*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

Rubi steps

\begin {align*} \int (a+a \cos (c+d x))^2 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec (c+d x) \, dx &=\int (a+a \cos (c+d x))^2 (B+C \cos (c+d x)) \, dx\\ &=\frac {C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}+\frac {1}{3} (3 B+2 C) \int (a+a \cos (c+d x))^2 \, dx\\ &=\frac {1}{2} a^2 (3 B+2 C) x+\frac {2 a^2 (3 B+2 C) \sin (c+d x)}{3 d}+\frac {a^2 (3 B+2 C) \cos (c+d x) \sin (c+d x)}{6 d}+\frac {C (a+a \cos (c+d x))^2 \sin (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]  time = 0.19, size = 61, normalized size = 0.65 \[ \frac {a^2 (3 (8 B+7 C) \sin (c+d x)+3 (B+2 C) \sin (2 (c+d x))+18 B d x+C \sin (3 (c+d x))+12 C d x)}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(18*B*d*x + 12*C*d*x + 3*(8*B + 7*C)*Sin[c + d*x] + 3*(B + 2*C)*Sin[2*(c + d*x)] + C*Sin[3*(c + d*x)]))/(
12*d)

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fricas [A]  time = 0.44, size = 70, normalized size = 0.74 \[ \frac {3 \, {\left (3 \, B + 2 \, C\right )} a^{2} d x + {\left (2 \, C a^{2} \cos \left (d x + c\right )^{2} + 3 \, {\left (B + 2 \, C\right )} a^{2} \cos \left (d x + c\right ) + 2 \, {\left (6 \, B + 5 \, C\right )} a^{2}\right )} \sin \left (d x + c\right )}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(3*B + 2*C)*a^2*d*x + (2*C*a^2*cos(d*x + c)^2 + 3*(B + 2*C)*a^2*cos(d*x + c) + 2*(6*B + 5*C)*a^2)*sin(d
*x + c))/d

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giac [A]  time = 0.38, size = 142, normalized size = 1.51 \[ \frac {3 \, {\left (3 \, B a^{2} + 2 \, C a^{2}\right )} {\left (d x + c\right )} + \frac {2 \, {\left (9 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 24 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 16 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 18 \, C a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*(3*(3*B*a^2 + 2*C*a^2)*(d*x + c) + 2*(9*B*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 24
*B*a^2*tan(1/2*d*x + 1/2*c)^3 + 16*C*a^2*tan(1/2*d*x + 1/2*c)^3 + 15*B*a^2*tan(1/2*d*x + 1/2*c) + 18*C*a^2*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 = 0.21, size = 116, normalized size = 1.23 \[ \frac {\frac {a^{2} C \left (2+\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )}{3}+B \,a^{2} \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 a^{2} C \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+2 B \,a^{2} \sin \left (d x +c \right )+a^{2} C \sin \left (d x +c \right )+B \,a^{2} \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.51, size = 110, normalized size = 1.17 \[ \frac {3 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a^{2} + 12 \, {\left (d x + c\right )} B a^{2} - 4 \, {\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C a^{2} + 6 \, {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a^{2} + 24 \, B a^{2} \sin \left (d x + c\right ) + 12 \, C a^{2} \sin \left (d x + c\right )}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(3*(2*d*x + 2*c + sin(2*d*x + 2*c))*B*a^2 + 12*(d*x + c)*B*a^2 - 4*(sin(d*x + c)^3 - 3*sin(d*x + c))*C*a^
2 + 6*(2*d*x + 2*c + sin(2*d*x + 2*c))*C*a^2 + 24*B*a^2*sin(d*x + c) + 12*C*a^2*sin(d*x + c))/d

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mupad [B]  time = 1.09, size = 98, normalized size = 1.04 \[ \frac {3\,B\,a^2\,x}{2}+C\,a^2\,x+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )}{d}+\frac {7\,C\,a^2\,\sin \left (c+d\,x\right )}{4\,d}+\frac {B\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{4\,d}+\frac {C\,a^2\,\sin \left (2\,c+2\,d\,x\right )}{2\,d}+\frac {C\,a^2\,\sin \left (3\,c+3\,d\,x\right )}{12\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(3*B*a^2*x)/2 + C*a^2*x + (2*B*a^2*sin(c + d*x))/d + (7*C*a^2*sin(c + d*x))/(4*d) + (B*a^2*sin(2*c + 2*d*x))/(
4*d) + (C*a^2*sin(2*c + 2*d*x))/(2*d) + (C*a^2*sin(3*c + 3*d*x))/(12*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int B \cos {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 2 B \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 2 C \cos ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int C \cos ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(B*cos(c + d*x)*sec(c + d*x), x) + Integral(2*B*cos(c + d*x)**2*sec(c + d*x), x) + Integral(B*co
s(c + d*x)**3*sec(c + d*x), x) + Integral(C*cos(c + d*x)**2*sec(c + d*x), x) + Integral(2*C*cos(c + d*x)**3*se
c(c + d*x), x) + Integral(C*cos(c + d*x)**4*sec(c + d*x), x))

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