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]

2/5*B*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/d

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Rubi [A]  time = 0.04, antiderivative size = 28, normalized size of antiderivative = 1.00, 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*}

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Mathematica [A]  time = 0.12, 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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fricas [A]  time = 1.03, size = 36, normalized size = 1.29 \[ \frac {2 \, {\left (B a \cos \left (d x + c\right ) + B a\right )} \sqrt {a \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{5 \, d} \]

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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giac [B]  time = 0.45, size = 86, normalized size = 3.07 \[ \frac {1}{10} \, \sqrt {2} {\left (\frac {B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right )}{d} + \frac {3 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right )}{d} + \frac {2 \, B a \mathrm {sgn}\left (\cos \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{d}\right )} \sqrt {a} \]

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]

1/10*sqrt(2)*(B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c)/d + 3*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(3/2*d
*x + 3/2*c)/d + 2*B*a*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c)/d)*sqrt(a)

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maple [A]  time = 0.30, size = 48, normalized size = 1.71 \[ \frac {8 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2} \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) B \sqrt {2}}{5 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*cos(d*x+c))^(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)/(a*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

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maxima [B]  time = 0.54, size = 92, normalized size = 3.29 \[ \frac {{\left (\sqrt {2} a \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 20 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a} - 2 \, {\left (\sqrt {2} a \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 9 \, \sqrt {2} a \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} B \sqrt {a}}{10 \, d} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.04 \[ \int -\left (\frac {3\,B}{5}-B\,\cos \left (c+d\,x\right )\right )\,{\left (a+a\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {B \left (\int \left (- 3 a \sqrt {a \cos {\left (c + d x \right )} + a}\right )\, dx + \int 2 a \sqrt {a \cos {\left (c + d x \right )} + a} \cos {\left (c + d x \right )}\, dx + \int 5 a \sqrt {a \cos {\left (c + d x \right )} + a} \cos ^{2}{\left (c + d x \right )}\, dx\right )}{5} \]

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]

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

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