3.786 \(\int \frac{-\frac{5 B}{3}+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=28 \[ -\frac{2 B \sin (c+d x)}{3 d (a \cos (c+d x)+a)^{5/2}} \]

[Out]

(-2*B*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^(5/2))

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Rubi [A]  time = 0.0423808, antiderivative size = 28, normalized size of antiderivative = 1., number of steps used = 1, number of rules used = 1, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.034, Rules used = {2749} \[ -\frac{2 B \sin (c+d x)}{3 d (a \cos (c+d x)+a)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[((-5*B)/3 + B*Cos[c + d*x])/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(-2*B*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^(5/2))

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{5 B}{3}+B \cos (c+d x)}{(a+a \cos (c+d x))^{5/2}} \, dx &=-\frac{2 B \sin (c+d x)}{3 d (a+a \cos (c+d x))^{5/2}}\\ \end{align*}

Mathematica [A]  time = 0.0671509, size = 28, normalized size = 1. \[ -\frac{2 B \sin (c+d x)}{3 d (a (\cos (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[((-5*B)/3 + B*Cos[c + d*x])/(a + a*Cos[c + d*x])^(5/2),x]

[Out]

(-2*B*Sin[c + d*x])/(3*d*(a*(1 + Cos[c + d*x]))^(5/2))

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Maple [A]  time = 0.828, size = 48, normalized size = 1.7 \begin{align*} -{\frac{B\sqrt{2}}{6\,{a}^{2}d}\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{\frac{1}{\sqrt{ \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-5/3*B+B*cos(d*x+c))/(a+cos(d*x+c)*a)^(5/2),x)

[Out]

-1/6/cos(1/2*d*x+1/2*c)^3/a^2*sin(1/2*d*x+1/2*c)*B*2^(1/2)/(cos(1/2*d*x+1/2*c)^2*a)^(1/2)/d

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Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5/3*B+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 1.3452, size = 169, normalized size = 6.04 \begin{align*} -\frac{2 \, \sqrt{a \cos \left (d x + c\right ) + a} B \sin \left (d x + c\right )}{3 \,{\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((-5/3*B+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-2/3*sqrt(a*cos(d*x + c) + a)*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 [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5/3*B+B*cos(d*x+c))/(a+a*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [B]  time = 3.49017, size = 80, normalized size = 2.86 \begin{align*} -\frac{\sqrt{a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + a}{\left (\frac{\sqrt{2} B \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2}}{a^{3}} + \frac{\sqrt{2} B}{a^{3}}\right )} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((-5/3*B+B*cos(d*x+c))/(a+a*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/6*sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a)*(sqrt(2)*B*tan(1/2*d*x + 1/2*c)^2/a^3 + sqrt(2)*B/a^3)*tan(1/2*d*x + 1
/2*c)/d