3.298 \(\int \cos (c+d x) \sqrt{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=231 \[ -\frac{2 \left (a^2-b^2\right ) (5 A b-2 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-2 a^2 B+5 a A b+9 b^2 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (5 A b-2 a B) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b d}+\frac{2 B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d} \]

[Out]

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

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Rubi [A]  time = 0.407589, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.258, Rules used = {2968, 3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \left (a^2-b^2\right ) (5 A b-2 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-2 a^2 B+5 a A b+9 b^2 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (5 A b-2 a B) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b d}+\frac{2 B \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]

[Out]

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

Rule 2968

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

Rule 3023

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

Rule 2753

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

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 \cos (c+d x) \sqrt{a+b \cos (c+d x)} (A+B \cos (c+d x)) \, dx &=\int \sqrt{a+b \cos (c+d x)} \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx\\ &=\frac{2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac{2 \int \sqrt{a+b \cos (c+d x)} \left (\frac{3 b B}{2}+\frac{1}{2} (5 A b-2 a B) \cos (c+d x)\right ) \, dx}{5 b}\\ &=\frac{2 (5 A b-2 a B) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac{4 \int \frac{\frac{1}{4} b (5 A b+7 a B)+\frac{1}{4} \left (5 a A b-2 a^2 B+9 b^2 B\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b}\\ &=\frac{2 (5 A b-2 a B) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}-\frac{\left (\left (a^2-b^2\right ) (5 A b-2 a B)\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b^2}+\frac{\left (5 a A b-2 a^2 B+9 b^2 B\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{15 b^2}\\ &=\frac{2 (5 A b-2 a B) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}+\frac{\left (\left (5 a A b-2 a^2 B+9 b^2 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}{15 b^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (\left (a^2-b^2\right ) (5 A b-2 a B) \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}{15 b^2 \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (5 a A b-2 a^2 B+9 b^2 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 \left (a^2-b^2\right ) (5 A b-2 a B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 (5 A b-2 a B) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac{2 B (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d}\\ \end{align*}

Mathematica [A]  time = 0.823999, size = 179, normalized size = 0.77 \[ \frac{2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (\left (-2 a^2 B+5 a A b+9 b^2 B\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )+b^2 (7 a B+5 A b) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )+2 b \sin (c+d x) (a+b \cos (c+d x)) (a B+5 A b+3 b B \cos (c+d x))}{15 b^2 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*(A + B*Cos[c + d*x]),x]

[Out]

(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(5*A*b + 7*a*B)*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + (5*a*A*b -
2*a^2*B + 9*b^2*B)*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))
+ 2*b*(a + b*Cos[c + d*x])*(5*A*b + a*B + 3*b*B*Cos[c + d*x])*Sin[c + d*x])/(15*b^2*d*Sqrt[a + b*Cos[c + d*x]]
)

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Maple [B]  time = 4.45, size = 993, normalized size = 4.3 \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)*(a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x)

[Out]

-2/15*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^6+(20*A*b^3+16*B*a*b^2+24*B*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-10*A*a*b^2-10*A*b^3-2*B*a^2*b-
8*B*a*b^2-6*B*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-5*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+5*A*b^3*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^
(1/2))+5*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))*a^2*b-5*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))*a*b^2+2*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-2*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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-
b))^(1/2))*b^2-2*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)*EllipticE(
cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3+2*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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b+9*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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^2-9*B*(si
n(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^3)/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{\left (B \cos \left (d x + c\right ) + A\right )} \sqrt{b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c) + a)*cos(d*x + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\right )\right )} \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(cos(d*x+c)*(a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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

[Out]

Timed out