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]

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

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Rubi [A]  time = 0.03, antiderivative size = 26, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.13, 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]

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

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fricas [B]  time = 0.75, size = 56, normalized size = 2.15 \[ -\frac {B \sin \left (d x + c\right )}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]

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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giac [A]  time = 0.36, size = 47, normalized size = 1.81 \[ -\frac {B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 2 \, B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + B \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{8 \, a^{3} d} \]

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)

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maple [A]  time = 0.07, size = 48, normalized size = 1.85 \[ \frac {B \left (-\left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-2 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 d \,a^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-3/2*B+B*cos(d*x+c))/(a+a*cos(d*x+c))^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 = 0.34, size = 115, normalized size = 4.42 \[ -\frac {\frac {B {\left (\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {2 \, B {\left (\frac {5 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {\sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{40 \, d} \]

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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mupad [B]  time = 0.64, size = 33, normalized size = 1.27 \[ -\frac {B\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2}{8\,a^3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.37, size = 80, normalized size = 3.08 \[ \begin {cases} - \frac {B \tan ^{5}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{3} d} - \frac {B \tan ^{3}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{4 a^{3} d} - \frac {B \tan {\left (\frac {c}{2} + \frac {d x}{2} \right )}}{8 a^{3} d} & \text {for}\: d \neq 0 \\\frac {x \left (B \cos {\relax (c )} - \frac {3 B}{2}\right )}{\left (a \cos {\relax (c )} + a\right )^{3}} & \text {otherwise} \end {cases} \]

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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