3.651 \(\int \frac{A+C \cos ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=174 \[ \frac{2 \left (2 a^2 C+b^2 (3 A+C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^2 d \sqrt{a+b \cos (c+d x)}}-\frac{4 a C \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 C \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 b d} \]

[Out]

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

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Rubi [A]  time = 0.205793, antiderivative size = 174, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3024, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (2 a^2 C+b^2 (3 A+C)\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^2 d \sqrt{a+b \cos (c+d x)}}-\frac{4 a C \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 C \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3024

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

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

Mathematica [A]  time = 0.683545, size = 148, normalized size = 0.85 \[ \frac{2 \left (C \left (2 a^2+b^2\right )+3 A b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+2 b C \sin (c+d x) (a+b \cos (c+d x))-4 a C (a+b) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^2 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.378, size = 532, normalized size = 3.1 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(4*C*cos(1/2*d*x+1/2*c)^5*b^2+3*A*b^2*(sin(1/
2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(
1/2))+2*C*cos(1/2*d*x+1/2*c)^3*a*b-6*C*cos(1/2*d*x+1/2*c)^3*b^2+2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d
*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2+C*b^2*(sin(1/2*d*x+1/2*c)
^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-2*C*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-
b))^(1/2))*a^2+2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2
*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b-2*C*cos(1/2*d*x+1/2*c)*a*b+2*C*cos(1/2*d*x+1/2*c)*b^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*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{C \cos \left (d x + c\right )^{2} + A}{\sqrt{b \cos \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((C*cos(d*x + c)^2 + A)/sqrt(b*cos(d*x + c) + a), 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((A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(1/2),x)

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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