3.335 \(\int \frac{\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=331 \[ -\frac{2 a^2 (A b-a B) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-5 a^3 B+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (2 a^2 A b-8 a^3 B+9 a b^2 B-3 A b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (2 a^3 A b+15 a^2 b^2 B-8 a^4 B-6 a A b^3-3 b^4 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

[Out]

(-2*(2*a^3*A*b - 6*a*A*b^3 - 8*a^4*B + 15*a^2*b^2*B - 3*b^4*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
(2*b)/(a + b)])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(2*a^2*A*b - 3*A*b^3 - 8*a^3*
B + 9*a*b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*b^3*(a^2 - b^2)*d*
Sqrt[a + b*Cos[c + d*x]]) - (2*a^2*(A*b - a*B)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2))
+ (2*a*(2*a^2*A*b - 6*A*b^3 - 5*a^3*B + 9*a*b^2*B)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x
]])

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Rubi [A]  time = 0.552717, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.212, Rules used = {2988, 3021, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 a^2 (A b-a B) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-5 a^3 B+9 a b^2 B-6 A b^3\right ) \sin (c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (2 a^2 A b-8 a^3 B+9 a b^2 B-3 A b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (2 a^3 A b+15 a^2 b^2 B-8 a^4 B-6 a A b^3-3 b^4 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(2*a^3*A*b - 6*a*A*b^3 - 8*a^4*B + 15*a^2*b^2*B - 3*b^4*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2,
(2*b)/(a + b)])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(2*a^2*A*b - 3*A*b^3 - 8*a^3*
B + 9*a*b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*b^3*(a^2 - b^2)*d*
Sqrt[a + b*Cos[c + d*x]]) - (2*a^2*(A*b - a*B)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2))
+ (2*a*(2*a^2*A*b - 6*A*b^3 - 5*a^3*B + 9*a*b^2*B)*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x
]])

Rule 2988

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

Rule 3021

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

Rule 2752

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

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2653

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

Rubi steps

\begin{align*} \int \frac{\cos ^2(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx &=-\frac{2 a^2 (A b-a B) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 \int \frac{\frac{3}{2} a b (A b-a B)+\frac{1}{2} \left (2 a^2-3 b^2\right ) (A b-a B) \cos (c+d x)+\frac{3}{2} b \left (a^2-b^2\right ) B \cos ^2(c+d x)}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 b^2 \left (a^2-b^2\right )}\\ &=-\frac{2 a^2 (A b-a B) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}-\frac{4 \int \frac{-\frac{1}{4} b^2 \left (a^2 A b+3 A b^3+2 a^3 B-6 a b^2 B\right )+\frac{1}{4} b \left (2 a^3 A b-6 a A b^3-8 a^4 B+15 a^2 b^2 B-3 b^4 B\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{2 a^2 (A b-a B) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{\left (2 a^2 A b-3 A b^3-8 a^3 B+9 a b^2 B\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )}-\frac{\left (2 a^3 A b-6 a A b^3-8 a^4 B+15 a^2 b^2 B-3 b^4 B\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{2 a^2 (A b-a B) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}-\frac{\left (\left (2 a^3 A b-6 a A b^3-8 a^4 B+15 a^2 b^2 B-3 b^4 B\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{3 b^3 \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{\left (\left (2 a^2 A b-3 A b^3-8 a^3 B+9 a b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}}\right ) \int \frac{1}{\sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}}} \, dx}{3 b^3 \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}\\ &=-\frac{2 \left (2 a^3 A b-6 a A b^3-8 a^4 B+15 a^2 b^2 B-3 b^4 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 \left (2 a^2 A b-3 A b^3-8 a^3 B+9 a b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^3 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{2 a^2 (A b-a B) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}+\frac{2 a \left (2 a^2 A b-6 A b^3-5 a^3 B+9 a b^2 B\right ) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 2.16107, size = 274, normalized size = 0.83 \[ \frac{2 \left (\frac{\left (\frac{a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (b^2 \left (a^2 A b+2 a^3 B-6 a b^2 B+3 A b^3\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (-2 a^3 A b-15 a^2 b^2 B+8 a^4 B+6 a A b^3+3 b^4 B\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )\right )}{(a-b)^2 (a+b)}-\frac{a b \sin (c+d x) \left (b \left (-2 a^2 A b+5 a^3 B-9 a b^2 B+6 A b^3\right ) \cos (c+d x)+a \left (-a^2 A b+4 a^3 B-8 a b^2 B+5 A b^3\right )\right )}{\left (a^2-b^2\right )^2}\right )}{3 b^3 d (a+b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*((((a + b*Cos[c + d*x])/(a + b))^(3/2)*(b^2*(a^2*A*b + 3*A*b^3 + 2*a^3*B - 6*a*b^2*B)*EllipticF[(c + d*x)/2
, (2*b)/(a + b)] + (-2*a^3*A*b + 6*a*A*b^3 + 8*a^4*B - 15*a^2*b^2*B + 3*b^4*B)*((a + b)*EllipticE[(c + d*x)/2,
(2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])))/((a - b)^2*(a + b)) - (a*b*(a*(-(a^2*A*b) + 5*A*b
^3 + 4*a^3*B - 8*a*b^2*B) + b*(-2*a^2*A*b + 6*A*b^3 + 5*a^3*B - 9*a*b^2*B)*Cos[c + d*x])*Sin[c + d*x])/(a^2 -
b^2)^2))/(3*b^3*d*(a + b*Cos[c + d*x])^(3/2))

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Maple [B]  time = 15.549, size = 950, normalized size = 2.9 \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)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x)

[Out]

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/b^3/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2
*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(A*b*Ell
ipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-3*B*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a+B*Ellipti
cE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-B*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b)-2*a/b^3*(2*A
*b-3*B*a)/sin(1/2*d*x+1/2*c)^2/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)/(a^2-b^2)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(
1/2*d*x+1/2*c)^2)^(1/2)*((sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*Elli
pticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(
a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b+2*b*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^2)+2*a^2*(A*b-B*a)/b^3*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2
*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2*(a-b)/b)^2+8/3*b*sin(1/2*d*x+1/2*c)^2/(a-b)^2/(a+b)^2*cos(1/2*d*x+1/2*c
)*a/(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*b^3)*(sin(1
/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d
*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-4/3*a/(a-b)/(a+b)^2*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/
2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))))/sin(1
/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (B \cos \left (d x + c\right )^{3} + A \cos \left (d x + c\right )^{2}\right )} \sqrt{b \cos \left (d x + c\right ) + a}}{b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \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)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((B*cos(d*x + c)^3 + A*cos(d*x + c)^2)*sqrt(b*cos(d*x + c) + a)/(b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x
+ c)^2 + 3*a^2*b*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)**2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

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