3.780 \(\int (a+a \cos (c+d x)) (-\frac{B}{2}+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=37 \[ \frac{a B \sin (c+d x)}{2 d}+\frac{a B \sin (c+d x) \cos (c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.0201143, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.04, Rules used = {2734} \[ \frac{a B \sin (c+d x)}{2 d}+\frac{a B \sin (c+d x) \cos (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Cos[c + d*x])*(-B/2 + B*Cos[c + d*x]),x]

[Out]

(a*B*Sin[c + d*x])/(2*d) + (a*B*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]

Rubi steps

\begin{align*} \int (a+a \cos (c+d x)) \left (-\frac{B}{2}+B \cos (c+d x)\right ) \, dx &=\frac{a B \sin (c+d x)}{2 d}+\frac{a B \cos (c+d x) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.0505303, size = 29, normalized size = 0.78 \[ \frac{a B (2 \sin (c+d x)+\sin (2 (c+d x))+2 c)}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Cos[c + d*x])*(-B/2 + B*Cos[c + d*x]),x]

[Out]

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

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Maple [A]  time = 0.044, size = 51, normalized size = 1.4 \begin{align*}{\frac{1}{2\,d} \left ( 2\,aB \left ( 1/2\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) +1/2\,dx+c/2 \right ) +aB\sin \left ( dx+c \right ) -aB \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+cos(d*x+c)*a)*(-1/2*B+B*cos(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.05095, size = 61, normalized size = 1.65 \begin{align*} \frac{{\left (2 \, d x + 2 \, c + \sin \left (2 \, d x + 2 \, c\right )\right )} B a - 2 \,{\left (d x + c\right )} B a + 2 \, B a \sin \left (d x + c\right )}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(-1/2*B+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.38892, size = 61, normalized size = 1.65 \begin{align*} \frac{{\left (B a \cos \left (d x + c\right ) + B a\right )} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(-1/2*B+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 0.396081, size = 87, normalized size = 2.35 \begin{align*} \begin{cases} \frac{B a x \sin ^{2}{\left (c + d x \right )}}{2} + \frac{B a x \cos ^{2}{\left (c + d x \right )}}{2} - \frac{B a x}{2} + \frac{B a \sin{\left (c + d x \right )} \cos{\left (c + d x \right )}}{2 d} + \frac{B a \sin{\left (c + d x \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (B \cos{\left (c \right )} - \frac{B}{2}\right ) \left (a \cos{\left (c \right )} + a\right ) & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(-1/2*B+B*cos(d*x+c)),x)

[Out]

Piecewise((B*a*x*sin(c + d*x)**2/2 + B*a*x*cos(c + d*x)**2/2 - B*a*x/2 + B*a*sin(c + d*x)*cos(c + d*x)/(2*d) +
B*a*sin(c + d*x)/(2*d), Ne(d, 0)), (x*(B*cos(c) - B/2)*(a*cos(c) + a), True))

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Giac [A]  time = 1.74801, size = 41, normalized size = 1.11 \begin{align*} \frac{B a \sin \left (2 \, d x + 2 \, c\right )}{4 \, d} + \frac{B a \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*cos(d*x+c))*(-1/2*B+B*cos(d*x+c)),x, algorithm="giac")

[Out]

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