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]

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

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Rubi [A]  time = 0.07, antiderivative size = 52, normalized size of antiderivative = 1.00, 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 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]

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+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*}

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Mathematica [A]  time = 0.08, 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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fricas [A]  time = 0.43, size = 42, normalized size = 0.81 \[ \frac {{\left (2 \, B a + C b\right )} d x + {\left (C b \cos \left (d x + c\right ) + 2 \, C a + 2 \, B b\right )} \sin \left (d x + c\right )}{2 \, d} \]

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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giac [B]  time = 0.22, size = 121, normalized size = 2.33 \[ \frac {{\left (2 \, B a + C b\right )} {\left (d x + c\right )} + \frac {2 \, {\left (2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 2 \, C a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, B b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + C b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

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

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maple [A]  time = 0.13, size = 57, normalized size = 1.10 \[ \frac {C b \left (\frac {\cos \left (d x +c \right ) \sin \left (d x +c \right )}{2}+\frac {d x}{2}+\frac {c}{2}\right )+B b \sin \left (d x +c \right )+a C \sin \left (d x +c \right )+B \left (d x +c \right ) a}{d} \]

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*(d*x+c)*a)

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maxima [A]  time = 0.31, size = 55, normalized size = 1.06 \[ \frac {4 \, {\left (d x + c\right )} B a + {\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} C b + 4 \, C a \sin \left (d x + c\right ) + 4 \, B b \sin \left (d x + c\right )}{4 \, d} \]

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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mupad [B]  time = 1.66, size = 50, normalized size = 0.96 \[ B\,a\,x+\frac {C\,b\,x}{2}+\frac {B\,b\,\sin \left (c+d\,x\right )}{d}+\frac {C\,a\,\sin \left (c+d\,x\right )}{d}+\frac {C\,b\,\sin \left (2\,c+2\,d\,x\right )}{4\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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