3.1061 \(\int \frac{a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\)
Optimal. Leaf size=167 \[ -\frac{2 C \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 (3 b B-2 a C) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 b C \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 d} \]
[Out]
(2*(3*b*B - 2*a*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(3*d*Sqrt[(a + b*Cos[c + d*
x])/(a + b)]) - (2*(a^2 - b^2)*C*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*
d*Sqrt[a + b*Cos[c + d*x]]) + (2*b*C*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)
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Rubi [A] time = 0.342166, antiderivative size = 167, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 50, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.14, Rules used =
{3015, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 C \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \sqrt{a+b \cos (c+d x)}}+\frac{2 (3 b B-2 a C) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 b C \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*Cos[c + d*x]^2)/Sqrt[a + b*Cos[c + d*x]],x]
[Out]
(2*(3*b*B - 2*a*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(3*d*Sqrt[(a + b*Cos[c + d*
x])/(a + b)]) - (2*(a^2 - b^2)*C*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*
d*Sqrt[a + b*Cos[c + d*x]]) + (2*b*C*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)
Rule 3015
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f
_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*B - a*C + b*C*Sin[e + f*x], x],
x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
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 \frac{a b B-a^2 C+b^2 B \cos (c+d x)+b^2 C \cos ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx &=\frac{\int \sqrt{a+b \cos (c+d x)} \left (b^2 (b B-a C)+b^3 C \cos (c+d x)\right ) \, dx}{b^2}\\ &=\frac{2 b C \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac{2 \int \frac{\frac{1}{2} b^2 \left (3 a b B-3 a^2 C+b^2 C\right )+\frac{1}{2} b^3 (3 b B-2 a C) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 b^2}\\ &=\frac{2 b C \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 d}-\frac{1}{3} \left (\left (a^2-b^2\right ) C\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx+\frac{1}{3} (3 b B-2 a C) \int \sqrt{a+b \cos (c+d x)} \, dx\\ &=\frac{2 b C \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac{\left ((3 b B-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 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (\left (a^2-b^2\right ) C \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{2 (3 b B-2 a C) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 \left (a^2-b^2\right ) C \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 b C \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 d}\\ \end{align*}
Mathematica [A] time = 0.633121, size = 144, normalized size = 0.86 \[ \frac{-2 C \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-2 (a+b) (2 a C-3 b B) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+2 b C \sin (c+d x) (a+b \cos (c+d x))}{3 d \sqrt{a+b \cos (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(a*b*B - a^2*C + b^2*B*Cos[c + d*x] + b^2*C*Cos[c + d*x]^2)/Sqrt[a + b*Cos[c + d*x]],x]
[Out]
(-2*(a + b)*(-3*b*B + 2*a*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - 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*d*Sqrt[a + b*Cos[c + d*x]])
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Maple [B] time = 1.104, size = 598, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a*b*B-a^2*C+b^2*B*cos(d*x+c)+b^2*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*B*(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-3*B*(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))*b^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-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*(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^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)/
(-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 b^{2} \cos \left (d x + c\right )^{2} + B b^{2} \cos \left (d x + c\right ) - C a^{2} + B a b}{\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*b*B-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate((C*b^2*cos(d*x + c)^2 + B*b^2*cos(d*x + c) - C*a^2 + B*a*b)/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 ({\left (C b \cos \left (d x + c\right ) - C a + B b\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((a*b*B-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
integral((C*b*cos(d*x + c) - C*a + B*b)*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*b*B-a**2*C+b**2*B*cos(d*x+c)+b**2*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 b^{2} \cos \left (d x + c\right )^{2} + B b^{2} \cos \left (d x + c\right ) - C a^{2} + B a b}{\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*b*B-a^2*C+b^2*B*cos(d*x+c)+b^2*C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate((C*b^2*cos(d*x + c)^2 + B*b^2*cos(d*x + c) - C*a^2 + B*a*b)/sqrt(b*cos(d*x + c) + a), x)