3.1.97 \(\int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx\) [97]

Optimal. Leaf size=86 \[ \frac {14 a \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {4 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d} \]

[Out]

2/5*(a+a*cos(d*x+c))^(3/2)*sin(d*x+c)/a/d+14/15*a*sin(d*x+c)/d/(a+a*cos(d*x+c))^(1/2)-4/15*sin(d*x+c)*(a+a*cos
(d*x+c))^(1/2)/d

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Rubi [A]
time = 0.08, antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {2838, 2830, 2725} \begin {gather*} \frac {2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}-\frac {4 \sin (c+d x) \sqrt {a \cos (c+d x)+a}}{15 d}+\frac {14 a \sin (c+d x)}{15 d \sqrt {a \cos (c+d x)+a}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2*Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(14*a*Sin[c + d*x])/(15*d*Sqrt[a + a*Cos[c + d*x]]) - (4*Sqrt[a + a*Cos[c + d*x]]*Sin[c + d*x])/(15*d) + (2*(a
+ a*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*a*d)

Rule 2725

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos[c + d*x]/(d*Sqrt[a + b*Sin[c + d*x
]])), x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2830

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

Rule 2838

Int[sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-Cos[e + f*x])*(
(a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[1/(b*(m + 2)), Int[(a + b*Sin[e + f*x])^m*(b*(m + 1) -
a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && EqQ[a^2 - b^2, 0] &&  !LtQ[m, -2^(-1)]

Rubi steps

\begin {align*} \int \cos ^2(c+d x) \sqrt {a+a \cos (c+d x)} \, dx &=\frac {2 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d}+\frac {2 \int \left (\frac {3 a}{2}-a \cos (c+d x)\right ) \sqrt {a+a \cos (c+d x)} \, dx}{5 a}\\ &=-\frac {4 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d}+\frac {7}{15} \int \sqrt {a+a \cos (c+d x)} \, dx\\ &=\frac {14 a \sin (c+d x)}{15 d \sqrt {a+a \cos (c+d x)}}-\frac {4 \sqrt {a+a \cos (c+d x)} \sin (c+d x)}{15 d}+\frac {2 (a+a \cos (c+d x))^{3/2} \sin (c+d x)}{5 a d}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 68, normalized size = 0.79 \begin {gather*} \frac {\sqrt {a (1+\cos (c+d x))} \sec \left (\frac {1}{2} (c+d x)\right ) \left (30 \sin \left (\frac {1}{2} (c+d x)\right )+5 \sin \left (\frac {3}{2} (c+d x)\right )+3 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{30 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2*Sqrt[a + a*Cos[c + d*x]],x]

[Out]

(Sqrt[a*(1 + Cos[c + d*x])]*Sec[(c + d*x)/2]*(30*Sin[(c + d*x)/2] + 5*Sin[(3*(c + d*x))/2] + 3*Sin[(5*(c + d*x
))/2]))/(30*d)

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Maple [A]
time = 0.10, size = 71, normalized size = 0.83

method	result	size
default	\(\frac {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (12 \left (\cos ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-4 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+7\right ) \sqrt {2}}{15 \sqrt {a \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, d}\)	\(71\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(a+a*cos(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2/15*cos(1/2*d*x+1/2*c)*a*sin(1/2*d*x+1/2*c)*(12*cos(1/2*d*x+1/2*c)^4-4*cos(1/2*d*x+1/2*c)^2+7)*2^(1/2)/(a*cos
(1/2*d*x+1/2*c)^2)^(1/2)/d

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Maxima [A]
time = 0.56, size = 51, normalized size = 0.59 \begin {gather*} \frac {{\left (3 \, \sqrt {2} \sin \left (\frac {5}{2} \, d x + \frac {5}{2} \, c\right ) + 5 \, \sqrt {2} \sin \left (\frac {3}{2} \, d x + \frac {3}{2} \, c\right ) + 30 \, \sqrt {2} \sin \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )} \sqrt {a}}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/30*(3*sqrt(2)*sin(5/2*d*x + 5/2*c) + 5*sqrt(2)*sin(3/2*d*x + 3/2*c) + 30*sqrt(2)*sin(1/2*d*x + 1/2*c))*sqrt(
a)/d

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Fricas [A]
time = 0.40, size = 52, normalized size = 0.60 \begin {gather*} \frac {2 \, \sqrt {a \cos \left (d x + c\right ) + a} {\left (3 \, \cos \left (d x + c\right )^{2} + 4 \, \cos \left (d x + c\right ) + 8\right )} \sin \left (d x + c\right )}{15 \, {\left (d \cos \left (d x + c\right ) + d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

2/15*sqrt(a*cos(d*x + c) + a)*(3*cos(d*x + c)^2 + 4*cos(d*x + c) + 8)*sin(d*x + c)/(d*cos(d*x + c) + d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \sqrt {a \left (\cos {\left (c + d x \right )} + 1\right )} \cos ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(a+a*cos(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(a*(cos(c + d*x) + 1))*cos(c + d*x)**2, x)

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Giac [A]
time = 0.48, size = 75, normalized size = 0.87 \begin {gather*} \frac {\sqrt {2} {\left (3 \, \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 ) + 5 \, \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 ) + 30 \, \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 )\right )} \sqrt {a}}{30 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(a+a*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

1/30*sqrt(2)*(3*sgn(cos(1/2*d*x + 1/2*c))*sin(5/2*d*x + 5/2*c) + 5*sgn(cos(1/2*d*x + 1/2*c))*sin(3/2*d*x + 3/2
*c) + 30*sgn(cos(1/2*d*x + 1/2*c))*sin(1/2*d*x + 1/2*c))*sqrt(a)/d

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\cos \left (c+d\,x\right )}^2\,\sqrt {a+a\,\cos \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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