3.670 \(\int \frac{a^2-b^2 \cos ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx\)

Optimal. Leaf size=157 \[ \frac{2 \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \sqrt{a+b \cos (c+d x)}}-\frac{2 b \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 d}+\frac{4 a \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

[Out]

(4*a*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)*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*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)

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Rubi [A]  time = 0.233376, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.219, Rules used = {3016, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \sqrt{a+b \cos (c+d x)}}-\frac{2 b \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{3 d}+\frac{4 a \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*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)*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*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)

Rule 3016

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[
C/b^2, Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[-a + b*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x
] && EqQ[A*b^2 + 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^2-b^2 \cos ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx &=-\int (-a+b \cos (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 d}-\frac{2}{3} \int \frac{\frac{1}{2} \left (-3 a^2+b^2\right )-a b \cos (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 d}+\frac{1}{3} (2 a) \int \sqrt{a+b \cos (c+d x)} \, dx-\frac{1}{3} \left (-a^2+b^2\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx\\ &=-\frac{2 b \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac{\left (2 a \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 ) \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 \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 ) \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 \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.497941, size = 134, normalized size = 0.85 \[ \frac{2 \left (a^2-b^2\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-2 b \sin (c+d x) (a+b \cos (c+d x))+4 a (a+b) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{3 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a*(a + b)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a + b)] + 2*(a^2 - b^2)*Sqrt[(a
+ b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)] - 2*b*(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 = 0.718, size = 450, normalized size = 2.9 \begin{align*}{\frac{2}{3\,d}\sqrt{ \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a-b \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}} \left ( 4\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{5}{b}^{2}+2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}ab-6\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{3}{b}^{2}-\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{{\frac{1}{a-b} \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a-b \right ) }}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{-2\,{\frac{b}{a-b}}} \right ){a}^{2}+\sqrt{ \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}\sqrt{{\frac{1}{a-b} \left ( 2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a-b \right ) }}{\it EllipticF} \left ( \cos \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) ,\sqrt{-2\,{\frac{b}{a-b}}} \right ){b}^{2}-2\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{{\frac{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a-b}{a-b}}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{-2\,{\frac{b}{a-b}}} \right ){a}^{2}+2\,\sqrt{ \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}}\sqrt{{\frac{2\, \left ( \cos \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a-b}{a-b}}}{\it EllipticE} \left ( \cos \left ( 1/2\,dx+c/2 \right ) ,\sqrt{-2\,{\frac{b}{a-b}}} \right ) ab-2\,\cos \left ( 1/2\,dx+c/2 \right ) ab+2\,\cos \left ( 1/2\,dx+c/2 \right ){b}^{2} \right ){\frac{1}{\sqrt{-2\,b \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{2}}}} \left ( \sin \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) \right ) ^{-1}{\frac{1}{\sqrt{-2\, \left ( \sin \left ( 1/2\,dx+c/2 \right ) \right ) ^{2}b+a+b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a^2-b^2*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*cos(1/2*d*x+1/2*c)^5*b^2+2*cos(1/2*d*x+1/2*
c)^3*a*b-6*cos(1/2*d*x+1/2*c)^3*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))*a^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))*b^2-2*(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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2+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)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b-2*
cos(1/2*d*x+1/2*c)*a*b+2*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{b^{2} \cos \left (d x + c\right )^{2} - a^{2}}{\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^2-b^2*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-integrate((b^2*cos(d*x + c)^2 - a^2)/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 (-\sqrt{b \cos \left (d x + c\right ) + a}{\left (b \cos \left (d x + c\right ) - a\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(b*cos(d*x + c) + a)*(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**2-b**2*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{b^{2} \cos \left (d x + c\right )^{2} - a^{2}}{\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^2-b^2*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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