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]

(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)

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Rubi [A]  time = 0.124678, antiderivative size = 94, normalized size of antiderivative = 1., 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 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]

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 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]

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*}

Mathematica [A]  time = 0.189075, 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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Maple [A]  time = 0.049, size = 116, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{a}^{2}B \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +2\,{a}^{2}C \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +2\,{a}^{2}B\sin \left ( dx+c \right ) +{a}^{2}C\sin \left ( dx+c \right ) +{a}^{2}B \left ( dx+c \right ) \right ) } \end{align*}

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)+a^2*B*(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*a^2*B*sin(d*x+c)+a^2*C*sin(d*x+c)+a^2*B*(d*x+c))

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Maxima [A]  time = 1.05685, size = 149, normalized size = 1.59 \begin{align*} \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} \end{align*}

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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Fricas [A]  time = 1.64236, size = 165, normalized size = 1.76 \begin{align*} \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} \end{align*}

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

[Out]

Timed out

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Giac [A]  time = 1.76221, size = 192, normalized size = 2.04 \begin{align*} \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} \end{align*}

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