3.6.25 \(\int \frac {\cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\) [525]

Optimal. Leaf size=122 \[ \frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}} \]

[Out]

2*(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*c
os(d*x+c))^(1/2)/b/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2*a*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*Ellipt
icF(sin(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b))^(1/2)/b/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]
time = 0.09, antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2831, 2742, 2740, 2734, 2732} \begin {gather*} \frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

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

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2831

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]

Rubi steps

\begin {align*} \int \frac {\cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx &=\frac {\int \sqrt {a+b \cos (c+d x)} \, dx}{b}-\frac {a \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{b}\\ &=\frac {\sqrt {a+b \cos (c+d x)} \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{b \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a \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}{b \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{b d \sqrt {a+b \cos (c+d x)}}\\ \end {align*}

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Mathematica [A]
time = 2.55, size = 86, normalized size = 0.70 \begin {gather*} \frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \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 d \sqrt {a+b \cos (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]/Sqrt[a + b*Cos[c + d*x]],x]

[Out]

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

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Maple [A]
time = 0.22, size = 220, normalized size = 1.80

method	result	size
default	\(\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a -b}{a -b}}\, \left (\EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a -\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a +\EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) b \right )}{\sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +a +b}\, d}\)	\(220\)
risch	\(-\frac {i \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right ) \sqrt {2}\, {\mathrm e}^{-i \left (d x +c \right )}}{b d \sqrt {\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right ) {\mathrm e}^{-i \left (d x +c \right )}}}-\frac {i \left (-\frac {2 \left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right )}{b \sqrt {\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right ) {\mathrm e}^{i \left (d x +c \right )}}}+\frac {2 \left (a +\sqrt {a^{2}-b^{2}}\right ) \sqrt {\frac {\left ({\mathrm e}^{i \left (d x +c \right )}+\frac {a +\sqrt {a^{2}-b^{2}}}{b}\right ) b}{a +\sqrt {a^{2}-b^{2}}}}\, \sqrt {\frac {{\mathrm e}^{i \left (d x +c \right )}-\frac {-a +\sqrt {a^{2}-b^{2}}}{b}}{-\frac {a +\sqrt {a^{2}-b^{2}}}{b}-\frac {-a +\sqrt {a^{2}-b^{2}}}{b}}}\, \sqrt {-\frac {b \,{\mathrm e}^{i \left (d x +c \right )}}{a +\sqrt {a^{2}-b^{2}}}}\, \left (\left (-\frac {a +\sqrt {a^{2}-b^{2}}}{b}-\frac {-a +\sqrt {a^{2}-b^{2}}}{b}\right ) \EllipticE \left (\sqrt {\frac {\left ({\mathrm e}^{i \left (d x +c \right )}+\frac {a +\sqrt {a^{2}-b^{2}}}{b}\right ) b}{a +\sqrt {a^{2}-b^{2}}}}, \sqrt {-\frac {a +\sqrt {a^{2}-b^{2}}}{b \left (-\frac {a +\sqrt {a^{2}-b^{2}}}{b}-\frac {-a +\sqrt {a^{2}-b^{2}}}{b}\right )}}\right )+\frac {\left (-a +\sqrt {a^{2}-b^{2}}\right ) \EllipticF \left (\sqrt {\frac {\left ({\mathrm e}^{i \left (d x +c \right )}+\frac {a +\sqrt {a^{2}-b^{2}}}{b}\right ) b}{a +\sqrt {a^{2}-b^{2}}}}, \sqrt {-\frac {a +\sqrt {a^{2}-b^{2}}}{b \left (-\frac {a +\sqrt {a^{2}-b^{2}}}{b}-\frac {-a +\sqrt {a^{2}-b^{2}}}{b}\right )}}\right )}{b}\right )}{b \sqrt {b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+{\mathrm e}^{i \left (d x +c \right )} b}}\right ) \sqrt {2}\, \sqrt {\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right ) {\mathrm e}^{i \left (d x +c \right )}}\, {\mathrm e}^{-i \left (d x +c \right )}}{d \sqrt {\left (b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{i \left (d x +c \right )}+b \right ) {\mathrm e}^{-i \left (d x +c \right )}}}\)	\(749\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(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-EllipticE(cos(1/2*d*x+1/2*c),(-2
*b/(a-b))^(1/2))*a+EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/
2*d*x+1/2*c)^2)^(1/2)/b/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.12, size = 355, normalized size = 2.91 \begin {gather*} \frac {2 i \, \sqrt {2} a \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) - 2 i \, \sqrt {2} a \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) + 3 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) - 3 i \, \sqrt {2} b^{\frac {3}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right )}{3 \, b^{2} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*(2*I*sqrt(2)*a*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*
cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) - 2*I*sqrt(2)*a*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^
2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b) + 3*I*sqrt(2)*b^(3/2)*wei
erstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2
, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b)) - 3*I*sqrt(2)*b^(3/2)*wei
erstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2
, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b)))/(b^2*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\cos {\left (c + d x \right )}}{\sqrt {a + b \cos {\left (c + d x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)/(a+b*cos(d*x+c))**(1/2),x)

[Out]

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.66, size = 80, normalized size = 0.66 \begin {gather*} \frac {2\,\left (\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\,\left (a+b\right )-a\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\right )\,\sqrt {\frac {a+b\,\cos \left (c+d\,x\right )}{a+b}}}{b\,d\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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