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]

-2/3*b*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/d+4/3*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin
(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2/3*(a^2-b^2)
*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*c
os(d*x+c))/(a+b))^(1/2)/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.23, antiderivative size = 157, normalized size of antiderivative = 1.00, 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 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]

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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.51, 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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fricas [F]  time = 0.68, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\sqrt {b \cos \left (d x + c\right ) + a} {\left (b \cos \left (d x + c\right ) - a\right )}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int -\frac {b^{2} \cos \left (d x + c\right )^{2} - a^{2}}{\sqrt {b \cos \left (d x + c\right ) + a}}\,{d x} \]

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)

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maple [B]  time = 2.74, size = 450, normalized size = 2.87 \[ \frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}+2 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b -6 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{2}+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) b^{2}-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{2}+2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a b -2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a b +2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((-cos(d*x+c)^2*b^2+a^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*sin(1/2*d*x+1/2*c)^4*b+(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 = 0.00, size = 0, normalized size = 0.00 \[ -\int \frac {b^{2} \cos \left (d x + c\right )^{2} - a^{2}}{\sqrt {b \cos \left (d x + c\right ) + a}}\,{d x} \]

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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mupad [B]  time = 1.81, size = 166, normalized size = 1.06 \[ \frac {2\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\,\sqrt {\frac {a+b\,\cos \left (c+d\,x\right )}{a+b}}}{d\,\sqrt {a+b\,\cos \left (c+d\,x\right )}}-\frac {2\,\sqrt {\frac {a+b\,\cos \left (c+d\,x\right )}{a+b}}\,\left (\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\,\left (2\,a^2+b^2\right )-2\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\,\left (a+b\right )\right )}{3\,d\,\sqrt {a+b\,\cos \left (c+d\,x\right )}}-\frac {2\,b\,\sin \left (c+d\,x\right )\,\sqrt {a+b\,\cos \left (c+d\,x\right )}}{3\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*a^2*ellipticF(c/2 + (d*x)/2, (2*b)/(a + b))*((a + b*cos(c + d*x))/(a + b))^(1/2))/(d*(a + b*cos(c + d*x))^(
1/2)) - (2*((a + b*cos(c + d*x))/(a + b))^(1/2)*(ellipticF(c/2 + (d*x)/2, (2*b)/(a + b))*(2*a^2 + b^2) - 2*a*e
llipticE(c/2 + (d*x)/2, (2*b)/(a + b))*(a + b)))/(3*d*(a + b*cos(c + d*x))^(1/2)) - (2*b*sin(c + d*x)*(a + b*c
os(c + d*x))^(1/2))/(3*d)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a - b \cos {\left (c + d x \right )}\right ) \sqrt {a + b \cos {\left (c + d x \right )}}\, dx \]

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]

Integral((a - b*cos(c + d*x))*sqrt(a + b*cos(c + d*x)), x)

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