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

Optimal. Leaf size=185 \[ -\frac{a \left (3 a^2-4 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac{\left (3 a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}+\frac{a^2 \left (3 a^2-5 b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2}-\frac{a^2 \sin (c+d x) \sqrt{\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.458616, antiderivative size = 185, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {2792, 3059, 2639, 3002, 2641, 2805} \[ -\frac{a \left (3 a^2-4 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d \left (a^2-b^2\right )}+\frac{\left (3 a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 d \left (a^2-b^2\right )}+\frac{a^2 \left (3 a^2-5 b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 d (a-b) (a+b)^2}-\frac{a^2 \sin (c+d x) \sqrt{\cos (c+d x)}}{b d \left (a^2-b^2\right ) (a+b \cos (c+d x))} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1))/(
d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^(m - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2
- b^2, 0] && NeQ[c^2 - d^2, 0]

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]

Rule 3002

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

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 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rubi steps

\begin{align*} \int \frac{\cos ^{\frac{5}{2}}(c+d x)}{(a+b \cos (c+d x))^2} \, dx &=-\frac{a^2 \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}-\frac{\int \frac{\frac{a^2}{2}-a b \cos (c+d x)-\frac{1}{2} \left (3 a^2-2 b^2\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b \left (a^2-b^2\right )}\\ &=-\frac{a^2 \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\int \frac{-\frac{a^2 b}{2}-\frac{1}{2} a \left (3 a^2-4 b^2\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^2 \left (a^2-b^2\right )}+\frac{\left (3 a^2-2 b^2\right ) \int \sqrt{\cos (c+d x)} \, dx}{2 b^2 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b^2\right ) d}-\frac{a^2 \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}+\frac{\left (a^2 \left (3 a^2-5 b^2\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 b^3 \left (a^2-b^2\right )}-\frac{\left (a \left (3 a^2-4 b^2\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{2 b^3 \left (a^2-b^2\right )}\\ &=\frac{\left (3 a^2-2 b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^2 \left (a^2-b^2\right ) d}-\frac{a \left (3 a^2-4 b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{b^3 \left (a^2-b^2\right ) d}+\frac{a^2 \left (3 a^2-5 b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{(a-b) b^3 (a+b)^2 d}-\frac{a^2 \sqrt{\cos (c+d x)} \sin (c+d x)}{b \left (a^2-b^2\right ) d (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.74727, size = 255, normalized size = 1.38 \[ \frac{\frac{4 a^2 \sin (c+d x) \sqrt{\cos (c+d x)}}{\left (b^2-a^2\right ) (a+b \cos (c+d x))}+\frac{\frac{2 \left (a^2-2 b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{2 \left (3 a^2-2 b^2\right ) \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt{\sin ^2(c+d x)}}+4 a \left (2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{2 a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{(a-b) (a+b)}}{4 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((4*a^2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/((-a^2 + b^2)*(a + b*Cos[c + d*x])) + ((2*(a^2 - 2*b^2)*EllipticPi[(2
*b)/(a + b), (c + d*x)/2, 2])/(a + b) + 4*a*(2*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c +
d*x)/2, 2])/(a + b)) + (2*(3*a^2 - 2*b^2)*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*Ell
ipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (2*a^2 - b^2)*EllipticPi[-(b/a), -ArcSin[Sqrt[Cos[c + d*x]]], -1])*Si
n[c + d*x])/(a*b^2*Sqrt[Sin[c + d*x]^2]))/((a - b)*(a + b)))/(4*b*d)

________________________________________________________________________________________

Maple [B]  time = 8.118, size = 815, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2/b^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)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(2*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2
))*a+EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b)-12/b^2*a^2/(-2*a*b+2*b^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)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2
*b/(a-b),2^(1/2))-2/b^3*a^3*(-1/a*b^2/(a^2-b^2)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)
^2)^(1/2)/(2*b*cos(1/2*d*x+1/2*c)^2+a-b)-1/2/a/(a+b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^
(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-1/2*b/(a^2-b^
2)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c
)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+1/2*b/(a^2-b^2)/a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*
x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))
-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x
+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+
2*b^2)*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x
+1/2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (b \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+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [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+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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+b*cos(d*x+c))**2,x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{\frac{5}{2}}}{{\left (b \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+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

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