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

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

[Out]

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

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Rubi [A]  time = 0.0424421, 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} \[ \frac{2 B \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.103002, size = 45, normalized size = 1.61 \[ \frac{8 a B \sin \left (\frac{1}{2} (c+d x)\right ) \cos ^3\left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)}}{5 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.658, size = 48, normalized size = 1.7 \begin{align*}{\frac{8\,{a}^{2}B\sqrt{2}}{5\,d} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{5}\sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\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((a+cos(d*x+c)*a)^(3/2)*(-3/5*B+B*cos(d*x+c)),x)

[Out]

8/5*cos(1/2*d*x+1/2*c)^5*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 [B]  time = 1.90831, size = 124, normalized size = 4.43 \begin{align*} \frac{{\left (\sqrt{2} a \sin \left (\frac{5}{2} \, d x + \frac{5}{2} \, c\right ) + 5 \, \sqrt{2} a \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 20 \, \sqrt{2} a \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} B \sqrt{a} - 2 \,{\left (\sqrt{2} a \sin \left (\frac{3}{2} \, d x + \frac{3}{2} \, c\right ) + 9 \, \sqrt{2} a \sin \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )\right )} B \sqrt{a}}{10 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/10*((sqrt(2)*a*sin(5/2*d*x + 5/2*c) + 5*sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 20*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*
B*sqrt(a) - 2*(sqrt(2)*a*sin(3/2*d*x + 3/2*c) + 9*sqrt(2)*a*sin(1/2*d*x + 1/2*c))*B*sqrt(a))/d

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Fricas [A]  time = 1.37911, size = 95, normalized size = 3.39 \begin{align*} \frac{2 \,{\left (B a \cos \left (d x + c\right ) + B a\right )} \sqrt{a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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

[Out]

Timed out