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

Optimal. Leaf size=57 \[ \frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

EllipticF[(c + d*x)/2, 2]/(3*a^2*d) + (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

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Rubi [A]  time = 0.0544056, antiderivative size = 57, 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 = {2764, 21, 2641} \[ \frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

EllipticF[(c + d*x)/2, 2]/(3*a^2*d) + (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

Rule 2764

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)),
Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a*d*n - b*c*(m + 1) - b*d*(m + n + 1)*Sin[
e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] && LtQ[m, -1] && LtQ[0, n, 1] && (IntegersQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \frac{\sqrt{\cos (c+d x)}}{(a+a \cos (c+d x))^2} \, dx &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{\frac{a}{2}+\frac{1}{2} a \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{3 a^2}\\ &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}+\frac{\int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}\\ &=\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end{align*}

Mathematica [A]  time = 0.20425, size = 63, normalized size = 1.11 \[ \frac{4 \cos ^4\left (\frac{1}{2} (c+d x)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )+\sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Cos[(c + d*x)/2]^4*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*a^2*d*(1 + Cos[c + d*x])
^2)

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Maple [B]  time = 2.062, size = 188, normalized size = 3.3 \begin{align*} -{\frac{1}{6\,{a}^{2}d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 2\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{-2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1}{\it EllipticF} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}+2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}-3\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}+1 \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1
/2*c)^2+1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*cos(1/2*d*x+1/2*c)^3+2*cos(1/2*d*x+1/2*c)^4-3*cos(1/2*d
*x+1/2*c)^2+1)/a^2/cos(1/2*d*x+1/2*c)^3/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c
)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cos \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{\cos \left (d x + c\right )}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(cos(d*x + c))/(a^2*cos(d*x + c)^2 + 2*a^2*cos(d*x + c) + a^2), x)

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{\cos \left (d x + c\right )}}{{\left (a \cos \left (d x + c\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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