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

Optimal. Leaf size=52 \[ \frac{(a C+b B) \sin (c+d x)}{d}+\frac{1}{2} x (2 a B+b C)+\frac{b C \sin (c+d x) \cos (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.065021, antiderivative size = 52, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 36, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.056, Rules used = {3029, 2734} \[ \frac{(a C+b B) \sin (c+d x)}{d}+\frac{1}{2} x (2 a B+b C)+\frac{b C \sin (c+d x) \cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*a*B + b*C)*x)/2 + ((b*B + a*C)*Sin[c + d*x])/d + (b*C*Cos[c + d*x]*Sin[c + d*x])/(2*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 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)) \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)) (B+C \cos (c+d x)) \, dx\\ &=\frac{1}{2} (2 a B+b C) x+\frac{(b B+a C) \sin (c+d x)}{d}+\frac{b C \cos (c+d x) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0811117, size = 51, normalized size = 0.98 \[ \frac{4 (a C+b B) \sin (c+d x)+4 a B d x+b C \sin (2 (c+d x))+2 b c C+2 b C d x}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.035, size = 57, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ( Cb \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 ) +Ba \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))*(B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c),x)

[Out]

1/d*(C*b*(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)+B*a*(d*x+c))

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Maxima [A]  time = 1.02688, size = 74, normalized size = 1.42 \begin{align*} \frac{4 \,{\left (d x + c\right )} B a +{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b + 4 \, C a \sin \left (d x + c\right ) + 4 \, B b \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(4*(d*x + c)*B*a + (2*d*x + 2*c + sin(2*d*x + 2*c))*C*b + 4*C*a*sin(d*x + c) + 4*B*b*sin(d*x + c))/d

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Fricas [A]  time = 1.67619, size = 104, normalized size = 2. \begin{align*} \frac{{\left (2 \, B a + C b\right )} d x +{\left (C b \cos \left (d x + c\right ) + 2 \, C a + 2 \, B b\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (B + C \cos{\left (c + d x \right )}\right ) \left (a + b \cos{\left (c + d x \right )}\right ) \cos{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B]  time = 1.32393, size = 163, normalized size = 3.13 \begin{align*} \frac{{\left (2 \, B a + C b\right )}{\left (d x + c\right )} + \frac{2 \,{\left (2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - C b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2 \, C a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, B b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + C 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 )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((2*B*a + C*b)*(d*x + c) + 2*(2*C*a*tan(1/2*d*x + 1/2*c)^3 + 2*B*b*tan(1/2*d*x + 1/2*c)^3 - C*b*tan(1/2*d*
x + 1/2*c)^3 + 2*C*a*tan(1/2*d*x + 1/2*c) + 2*B*b*tan(1/2*d*x + 1/2*c) + C*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*
x + 1/2*c)^2 + 1)^2)/d