3.183 \(\int \frac{\cos ^{\frac{5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=112 \[ -\frac{5 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{5 \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d (\cos (c+d x)+1)}-\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

[Out]

(4*EllipticE[(c + d*x)/2, 2])/(a^2*d) - (5*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) - (5*Sqrt[Cos[c + d*x]]*Sin[c
+ d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - (Cos[c + d*x]^(3/2)*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

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Rubi [A]  time = 0.183698, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2765, 2977, 2748, 2641, 2639} \[ -\frac{5 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac{4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{5 \sin (c+d x) \sqrt{\cos (c+d x)}}{3 a^2 d (\cos (c+d x)+1)}-\frac{\sin (c+d x) \cos ^{\frac{3}{2}}(c+d x)}{3 d (a \cos (c+d x)+a)^2} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^(5/2)/(a + a*Cos[c + d*x])^2,x]

[Out]

(4*EllipticE[(c + d*x)/2, 2])/(a^2*d) - (5*EllipticF[(c + d*x)/2, 2])/(3*a^2*d) - (5*Sqrt[Cos[c + d*x]]*Sin[c
+ d*x])/(3*a^2*d*(1 + Cos[c + d*x])) - (Cos[c + d*x]^(3/2)*Sin[c + d*x])/(3*d*(a + a*Cos[c + d*x])^2)

Rule 2765

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

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*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*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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{\cos ^{\frac{5}{2}}(c+d x)}{(a+a \cos (c+d x))^2} \, dx &=-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{\int \frac{\sqrt{\cos (c+d x)} \left (\frac{3 a}{2}-\frac{7}{2} a \cos (c+d x)\right )}{a+a \cos (c+d x)} \, dx}{3 a^2}\\ &=-\frac{5 \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{\int \frac{\frac{5 a^2}{2}-6 a^2 \cos (c+d x)}{\sqrt{\cos (c+d x)}} \, dx}{3 a^4}\\ &=-\frac{5 \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}-\frac{5 \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^2}+\frac{2 \int \sqrt{\cos (c+d x)} \, dx}{a^2}\\ &=\frac{4 E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{a^2 d}-\frac{5 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^2 d}-\frac{5 \sqrt{\cos (c+d x)} \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac{\cos ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{3 d (a+a \cos (c+d x))^2}\\ \end{align*}

Mathematica [C]  time = 2.36648, size = 319, normalized size = 2.85 \[ \frac{\cos ^4\left (\frac{1}{2} (c+d x)\right ) \left (-\frac{\csc \left (\frac{c}{2}\right ) \sec \left (\frac{c}{2}\right ) \sqrt{\cos (c+d x)} \left (20 \cos \left (\frac{1}{2} (c-d x)\right )+16 \cos \left (\frac{1}{2} (3 c+d x)\right )+9 \cos \left (\frac{1}{2} (c+3 d x)\right )+3 \cos \left (\frac{1}{2} (5 c+3 d x)\right )\right ) \sec ^3\left (\frac{1}{2} (c+d x)\right )}{2 d}+\frac{4 i \sqrt{2} e^{-i (c+d x)} \left (12 \left (-1+e^{2 i c}\right ) \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (-\frac{1}{4},\frac{1}{2},\frac{3}{4},-e^{2 i (c+d x)}\right )+5 \left (-1+e^{2 i c}\right ) e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \text{Hypergeometric2F1}\left (\frac{1}{4},\frac{1}{2},\frac{5}{4},-e^{2 i (c+d x)}\right )+12 \left (1+e^{2 i (c+d x)}\right )\right )}{\left (-1+e^{2 i c}\right ) d \sqrt{e^{-i (c+d x)} \left (1+e^{2 i (c+d x)}\right )}}\right )}{3 a^2 (\cos (c+d x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^(5/2)/(a + a*Cos[c + d*x])^2,x]

[Out]

(Cos[(c + d*x)/2]^4*(((4*I)*Sqrt[2]*(12*(1 + E^((2*I)*(c + d*x))) + 12*(-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]]*(20*Cos[(c - d*x)/2] + 16*Cos
[(3*c + d*x)/2] + 9*Cos[(c + 3*d*x)/2] + 3*Cos[(5*c + 3*d*x)/2])*Csc[c/2]*Sec[c/2]*Sec[(c + d*x)/2]^3)/(2*d)))
/(3*a^2*(1 + Cos[c + d*x])^2)

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Maple [A]  time = 2.42, size = 257, normalized size = 2.3 \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 ( 24\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{6}+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 ) ^{3}+24\,\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 ) ^{3}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{2} \right ) -38\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+15\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}-1 \right ){\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 ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-3} \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)^(5/2)/(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)*(24*cos(1/2*d*x+1/2*c)^6+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)^3+24*(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)^3*EllipticE(cos(1/2*d*x+1/2*c),2^(1
/2))-38*cos(1/2*d*x+1/2*c)^4+15*cos(1/2*d*x+1/2*c)^2-1)/a^2/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/
2)/cos(1/2*d*x+1/2*c)^3/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{\cos \left (d x + c\right )^{\frac{5}{2}}}{{\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)^(5/2)/(a+a*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^(5/2)/(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{\cos \left (d x + c\right )^{\frac{5}{2}}}{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)^(5/2)/(a+a*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

integral(cos(d*x + c)^(5/2)/(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)**(5/2)/(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{\cos \left (d x + c\right )^{\frac{5}{2}}}{{\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)^(5/2)/(a+a*cos(d*x+c))^2,x, algorithm="giac")

[Out]

integrate(cos(d*x + c)^(5/2)/(a*cos(d*x + c) + a)^2, x)