3.783 \(\int \frac{-\frac{3 B}{2}+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=26 \[ -\frac{B \sin (c+d x)}{2 d (a \cos (c+d x)+a)^3} \]

[Out]

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

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Rubi [A]  time = 0.0328949, antiderivative size = 26, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.037, Rules used = {2749} \[ -\frac{B \sin (c+d x)}{2 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Int[((-3*B)/2 + B*Cos[c + d*x])/(a + a*Cos[c + d*x])^3,x]

[Out]

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

Rule 2749

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

Rubi steps

\begin{align*} \int \frac{-\frac{3 B}{2}+B \cos (c+d x)}{(a+a \cos (c+d x))^3} \, dx &=-\frac{B \sin (c+d x)}{2 d (a+a \cos (c+d x))^3}\\ \end{align*}

Mathematica [A]  time = 0.103346, size = 27, normalized size = 1.04 \[ -\frac{B \sin (c+d x)}{2 a^3 d (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[((-3*B)/2 + B*Cos[c + d*x])/(a + a*Cos[c + d*x])^3,x]

[Out]

-(B*Sin[c + d*x])/(2*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 0.06, size = 48, normalized size = 1.9 \begin{align*}{\frac{B}{8\,d{a}^{3}} \left ( - \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}-2\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}-\tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3/2*B+B*cos(d*x+c))/(a+cos(d*x+c)*a)^3,x)

[Out]

1/8/d*B/a^3*(-tan(1/2*d*x+1/2*c)^5-2*tan(1/2*d*x+1/2*c)^3-tan(1/2*d*x+1/2*c))

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Maxima [B]  time = 1.06946, size = 155, normalized size = 5.96 \begin{align*} -\frac{\frac{B{\left (\frac{15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac{2 \, B{\left (\frac{5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{40 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3/2*B+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/40*(B*(15*sin(d*x + c)/(cos(d*x + c) + 1) + 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(
d*x + c) + 1)^5)/a^3 - 2*B*(5*sin(d*x + c)/(cos(d*x + c) + 1) - sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/a^3)/d

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Fricas [B]  time = 1.33963, size = 135, normalized size = 5.19 \begin{align*} -\frac{B \sin \left (d x + c\right )}{2 \,{\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3/2*B+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*B*sin(d*x + c)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)

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Sympy [A]  time = 3.66162, size = 80, normalized size = 3.08 \begin{align*} \begin{cases} - \frac{B \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{3} d} - \frac{B \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{4 a^{3} d} - \frac{B \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{8 a^{3} d} & \text{for}\: d \neq 0 \\\frac{x \left (B \cos{\left (c \right )} - \frac{3 B}{2}\right )}{\left (a \cos{\left (c \right )} + a\right )^{3}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3/2*B+B*cos(d*x+c))/(a+a*cos(d*x+c))**3,x)

[Out]

Piecewise((-B*tan(c/2 + d*x/2)**5/(8*a**3*d) - B*tan(c/2 + d*x/2)**3/(4*a**3*d) - B*tan(c/2 + d*x/2)/(8*a**3*d
), Ne(d, 0)), (x*(B*cos(c) - 3*B/2)/(a*cos(c) + a)**3, True))

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Giac [A]  time = 1.32672, size = 63, normalized size = 2.42 \begin{align*} -\frac{B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 2 \, B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{8 \, a^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-3/2*B+B*cos(d*x+c))/(a+a*cos(d*x+c))^3,x, algorithm="giac")

[Out]

-1/8*(B*tan(1/2*d*x + 1/2*c)^5 + 2*B*tan(1/2*d*x + 1/2*c)^3 + B*tan(1/2*d*x + 1/2*c))/(a^3*d)