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

Optimal. Leaf size=107 \[ \frac{2 \left (a^2 C+3 a b B+b^2 C\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (2 a^2 B+2 a b C+b^2 B\right )+\frac{b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.159257, antiderivative size = 107, 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, 2753, 2734} \[ \frac{2 \left (a^2 C+3 a b B+b^2 C\right ) \sin (c+d x)}{3 d}+\frac{1}{2} x \left (2 a^2 B+2 a b C+b^2 B\right )+\frac{b (2 a C+3 b B) \sin (c+d x) \cos (c+d x)}{6 d}+\frac{C \sin (c+d x) (a+b \cos (c+d x))^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a^2*B + b^2*B + 2*a*b*C)*x)/2 + (2*(3*a*b*B + a^2*C + b^2*C)*Sin[c + d*x])/(3*d) + (b*(3*b*B + 2*a*C)*Cos[
c + d*x]*Sin[c + d*x])/(6*d) + (C*(a + b*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 2753

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[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rule 2734

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

Rubi steps

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

Mathematica [A]  time = 0.216213, size = 90, normalized size = 0.84 \[ \frac{6 (c+d x) \left (2 a^2 B+2 a b C+b^2 B\right )+3 \left (4 a^2 C+8 a b B+3 b^2 C\right ) \sin (c+d x)+3 b (2 a C+b B) \sin (2 (c+d x))+b^2 C \sin (3 (c+d x))}{12 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(6*(2*a^2*B + b^2*B + 2*a*b*C)*(c + d*x) + 3*(8*a*b*B + 4*a^2*C + 3*b^2*C)*Sin[c + d*x] + 3*b*(b*B + 2*a*C)*Si
n[2*(c + d*x)] + b^2*C*Sin[3*(c + d*x)])/(12*d)

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Maple [A]  time = 0.039, size = 114, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({\frac{{b}^{2}C \left ( 2+ \left ( \cos \left ( dx+c \right ) \right ) ^{2} \right ) \sin \left ( dx+c \right ) }{3}}+{b}^{2}B \left ({\frac{\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) }{2}}+{\frac{dx}{2}}+{\frac{c}{2}} \right ) +2\,abC \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +2\,abB\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+b*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*b^2*C*(2+cos(d*x+c)^2)*sin(d*x+c)+b^2*B*(1/2*cos(d*x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+2*a*b*C*(1/2*cos(d*
x+c)*sin(d*x+c)+1/2*d*x+1/2*c)+2*a*b*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 = 1.035, size = 146, normalized size = 1.36 \begin{align*} \frac{12 \,{\left (d x + c\right )} B a^{2} + 6 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C a b + 3 \,{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B b^{2} - 4 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} C b^{2} + 12 \, C a^{2} \sin \left (d x + c\right ) + 24 \, B a b \sin \left (d x + c\right )}{12 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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Fricas [A]  time = 1.4496, size = 201, normalized size = 1.88 \begin{align*} \frac{3 \,{\left (2 \, B a^{2} + 2 \, C a b + B b^{2}\right )} d x +{\left (2 \, C b^{2} \cos \left (d x + c\right )^{2} + 6 \, C a^{2} + 12 \, B a b + 4 \, C b^{2} + 3 \,{\left (2 \, C a b + B b^{2}\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((a+b*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*(2*B*a^2 + 2*C*a*b + B*b^2)*d*x + (2*C*b^2*cos(d*x + c)^2 + 6*C*a^2 + 12*B*a*b + 4*C*b^2 + 3*(2*C*a*b +
B*b^2)*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((a+b*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 [B]  time = 1.56496, size = 343, normalized size = 3.21 \begin{align*} \frac{3 \,{\left (2 \, B a^{2} + 2 \, C a b + B b^{2}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 12 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 24 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, C b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, C a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 12 \, B a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, B b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 6 \, C b^{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+b*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*(2*B*a^2 + 2*C*a*b + B*b^2)*(d*x + c) + 2*(6*C*a^2*tan(1/2*d*x + 1/2*c)^5 + 12*B*a*b*tan(1/2*d*x + 1/2*
c)^5 - 6*C*a*b*tan(1/2*d*x + 1/2*c)^5 - 3*B*b^2*tan(1/2*d*x + 1/2*c)^5 + 6*C*b^2*tan(1/2*d*x + 1/2*c)^5 + 12*C
*a^2*tan(1/2*d*x + 1/2*c)^3 + 24*B*a*b*tan(1/2*d*x + 1/2*c)^3 + 4*C*b^2*tan(1/2*d*x + 1/2*c)^3 + 6*C*a^2*tan(1
/2*d*x + 1/2*c) + 12*B*a*b*tan(1/2*d*x + 1/2*c) + 6*C*a*b*tan(1/2*d*x + 1/2*c) + 3*B*b^2*tan(1/2*d*x + 1/2*c)
+ 6*C*b^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d