3.194 \(\int \frac{\sqrt{\cos (c+d x)}}{(a+a \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=155 \[ \frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{15 a d (a \cos (c+d x)+a)^2}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3} \]

[Out]

EllipticE[(c + d*x)/2, 2]/(10*a^3*d) + EllipticF[(c + d*x)/2, 2]/(6*a^3*d) + (Sqrt[Cos[c + d*x]]*Sin[c + d*x])
/(5*d*(a + a*Cos[c + d*x])^3) + (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - (Sqrt[Cos[
c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*Cos[c + d*x]))

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Rubi [A]  time = 0.305266, antiderivative size = 155, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2764, 2978, 2748, 2641, 2639} \[ \frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}-\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{10 d \left (a^3 \cos (c+d x)+a^3\right )}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{15 a d (a \cos (c+d x)+a)^2}+\frac{\sin (c+d x) \sqrt{\cos (c+d x)}}{5 d (a \cos (c+d x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

EllipticE[(c + d*x)/2, 2]/(10*a^3*d) + EllipticF[(c + d*x)/2, 2]/(6*a^3*d) + (Sqrt[Cos[c + d*x]]*Sin[c + d*x])
/(5*d*(a + a*Cos[c + d*x])^3) + (Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(15*a*d*(a + a*Cos[c + d*x])^2) - (Sqrt[Cos[
c + d*x]]*Sin[c + d*x])/(10*d*(a^3 + a^3*Cos[c + d*x]))

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 2978

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

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, 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]

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(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))^3} \, dx &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\int \frac{\frac{a}{2}+\frac{3}{2} a \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))^2} \, dx}{5 a^2}\\ &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}+\frac{\int \frac{2 a^2+\frac{1}{2} a^2 \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+a \cos (c+d x))} \, dx}{15 a^4}\\ &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \frac{\frac{5 a^3}{4}+\frac{3}{4} a^3 \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{15 a^6}\\ &=\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}+\frac{\int \sqrt{\cos (c+d x)} \, dx}{20 a^3}+\frac{\int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{12 a^3}\\ &=\frac{E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{10 a^3 d}+\frac{F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{6 a^3 d}+\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{5 d (a+a \cos (c+d x))^3}+\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{15 a d (a+a \cos (c+d x))^2}-\frac{\sqrt{\cos (c+d x)} \sin (c+d x)}{10 d \left (a^3+a^3 \cos (c+d x)\right )}\\ \end{align*}

Mathematica [C]  time = 3.06659, size = 334, normalized size = 2.15 \[ \frac{\cos ^6\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \left (4 \cos \left (\frac{1}{2} (c-d x)\right )+26 \cos \left (\frac{1}{2} (3 c+d x)\right )+10 \cos \left (\frac{1}{2} (c+3 d x)\right )+5 \cos \left (\frac{1}{2} (5 c+3 d x)\right )+3 \cos \left (\frac{1}{2} (3 c+5 d x)\right )\right ) \sec ^5\left (\frac{1}{2} (c+d x)\right )}{8 d}+\frac{4 i \sqrt{2} e^{-i (c+d x)} \left (3 \left (-1+e^{2 i c}\right ) \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )-5 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+3 \left (1+e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) d \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{15 a^3 (\cos (c+d x)+1)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cos[(c + d*x)/2]^6*(((4*I)*Sqrt[2]*(3*(1 + E^((2*I)*(c + d*x))) + 3*(-1 + E^((2*I)*c))*Sqrt[1 + E^((2*I)*(c +
d*x))]*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c + d*x))] - 5*E^(I*(c + d*x))*(-1 + E^((2*I)*c))*Sqrt[1
+ E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/4, 1/2, 5/4, -E^((2*I)*(c + d*x))]))/(d*E^(I*(c + d*x))*(-1 + E^((2
*I)*c))*Sqrt[(1 + E^((2*I)*(c + d*x)))/E^(I*(c + d*x))]) - (Sqrt[Cos[c + d*x]]*(4*Cos[(c - d*x)/2] + 26*Cos[(3
*c + d*x)/2] + 10*Cos[(c + 3*d*x)/2] + 5*Cos[(5*c + 3*d*x)/2] + 3*Cos[(3*c + 5*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[
(c + d*x)/2]^5)/(8*d)))/(15*a^3*(1 + Cos[c + d*x])^3)

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Maple [A]  time = 2.42, size = 270, normalized size = 1.7 \begin{align*}{\frac{1}{60\,{a}^{3}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 ( 12\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{8}-10\,\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 ) ^{5}+6\,\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} \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -22\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+6\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+7\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-3 \right ) \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-5}{\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)^3,x)

[Out]

1/60*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(12*cos(1/2*d*x+1/2*c)^8-10*(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)^5+6*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)*cos(1/2*d*x+1/2*c)^5*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2))-22*cos(1/2*d*x+1/2*c)^6+6*cos(1/2*d*x+1/2*c)^4+7*cos(1/2*d*x+1/2*c)^2-3)/a^3/cos(1/2*d*x+1/2*c)^5/(-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 )}^{3}}\,{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))^3,x, algorithm="maxima")

[Out]

integrate(sqrt(cos(d*x + c))/(a*cos(d*x + c) + a)^3, 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^{3} \cos \left (d x + c\right )^{3} + 3 \, a^{3} \cos \left (d x + c\right )^{2} + 3 \, a^{3} \cos \left (d x + c\right ) + a^{3}}, 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))^3,x, algorithm="fricas")

[Out]

integral(sqrt(cos(d*x + c))/(a^3*cos(d*x + c)^3 + 3*a^3*cos(d*x + c)^2 + 3*a^3*cos(d*x + c) + a^3), 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))**3,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 )}^{3}}\,{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))^3,x, algorithm="giac")

[Out]

integrate(sqrt(cos(d*x + c))/(a*cos(d*x + c) + a)^3, x)