3.321 \(\int \frac{\cos (c+d x) (A+B \cos (c+d x))}{\sqrt{a+b \cos (c+d x)}} \, dx\)
Optimal. Leaf size=183 \[ -\frac{2 \left (-2 a^2 B+3 a A b-b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 (3 A b-2 a B) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 B \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 b d} \]
[Out]
(2*(3*A*b - 2*a*B)*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*(3*a*A*b - 2*a^2*B - b^2*B)*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*B*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*b*d)
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Rubi [A] time = 0.291797, antiderivative size = 183, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.226, Rules used =
{2968, 3023, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \left (-2 a^2 B+3 a A b-b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 (3 A b-2 a B) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 B \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 b d} \]
Antiderivative was successfully verified.
[In]
Int[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/Sqrt[a + b*Cos[c + d*x]],x]
[Out]
(2*(3*A*b - 2*a*B)*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*(3*a*A*b - 2*a^2*B - b^2*B)*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*B*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*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 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 (c+d x) (A+B \cos (c+d x))}{\sqrt{a+b \cos (c+d x)}} \, dx &=\int \frac{A \cos (c+d x)+B \cos ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=\frac{2 B \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 b d}+\frac{2 \int \frac{\frac{b B}{2}+\frac{1}{2} (3 A b-2 a B) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 b}\\ &=\frac{2 B \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 b d}+\frac{(3 A b-2 a B) \int \sqrt{a+b \cos (c+d x)} \, dx}{3 b^2}-\frac{\left (3 a A b-2 a^2 B-b^2 B\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 b^2}\\ &=\frac{2 B \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 b d}+\frac{\left ((3 A b-2 a B) \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 A b-2 a^2 B-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^2 \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 (3 A b-2 a B) \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 A b-2 a^2 B-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^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 B \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 b d}\\ \end{align*}
Mathematica [A] time = 0.656256, size = 154, normalized size = 0.84 \[ \frac{2 \left (2 a^2 B-3 a A b+b^2 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-2 (a+b) (2 a B-3 A b) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+2 b B \sin (c+d x) (a+b \cos (c+d x))}{3 b^2 d \sqrt{a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cos[c + d*x]*(A + B*Cos[c + d*x]))/Sqrt[a + b*Cos[c + d*x]],x]
[Out]
(-2*(a + b)*(-3*A*b + 2*a*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a + b)] + 2*(-3*
a*A*b + 2*a^2*B + b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)] + 2*b*B*(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 = 4.369, size = 671, normalized size = 3.7 \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))/(a+b*cos(d*x+c))^(1/2),x)
[Out]
2/3*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*B*cos(1/2*d*x+1/2*c)^5*b^2+3*A*a*b*(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(
1/2))-3*A*(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)*EllipticE(cos(1/2*d*x+1/2*
c),(-2*b/(a-b))^(1/2))*a*b+3*A*(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)*Ellip
ticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^2-2*B*cos(1/2*d*x+1/2*c)^3*a*b+6*B*cos(1/2*d*x+1/2*c)^3*b^2-2*B*
(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)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(
a-b))^(1/2))*a^2-B*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)*EllipticF(cos
(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+2*B*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2-2*B*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b+2*B*cos(1/2*d*x+1/2*c)*a*b-2*B*c
os(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{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\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(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((B*cos(d*x + c) + A)*cos(d*x + c)/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{B \cos \left (d x + c\right )^{2} + A \cos \left (d x + c\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))/(a+b*cos(d*x+c))^(1/2),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))/(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{{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )}{\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(cos(d*x+c)*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((B*cos(d*x + c) + A)*cos(d*x + c)/sqrt(b*cos(d*x + c) + a), x)