3.97 \(\int \cos ^2(c+d x) \sqrt{a+a \cos (c+d x)} \, dx\)

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

[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)

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Rubi [A]  time = 0.114655, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {2759, 2751, 2646} \[ \frac{2 \sin (c+d x) (a \cos (c+d x)+a)^{3/2}}{5 a d}-\frac{4 \sin (c+d x) \sqrt{a \cos (c+d x)+a}}{15 d}+\frac{14 a \sin (c+d x)}{15 d \sqrt{a \cos (c+d x)+a}} \]

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 2759

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)]

Rule 2751

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*Sin
[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 2646

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]

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*}

Mathematica [A]  time = 0.0988188, size = 68, normalized size = 0.79 \[ \frac{\left (30 \sin \left (\frac{1}{2} (c+d x)\right )+5 \sin \left (\frac{3}{2} (c+d x)\right )+3 \sin \left (\frac{5}{2} (c+d x)\right )\right ) \sec \left (\frac{1}{2} (c+d x)\right ) \sqrt{a (\cos (c+d x)+1)}}{30 d} \]

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 = 0.818, size = 71, normalized size = 0.8 \begin{align*}{\frac{2\,a\sqrt{2}}{15\,d}\cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \left ( 12\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-4\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+7 \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(cos(d*x+c)^2*(a+cos(d*x+c)*a)^(1/2),x)

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

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Maxima [A]  time = 1.92158, size = 69, normalized size = 0.8 \begin{align*} \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{align*}

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 = 1.52935, size = 142, normalized size = 1.65 \begin{align*} \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{align*}

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

[Out]

Timed out