\(\int \frac {\cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx\) [524]

Optimal result

Integrand size = 23, antiderivative size = 165 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {4 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (2 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b d} \]

[Out]

2/3*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/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)/b^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)+2/3*(2*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*cos(d*x+c))/(a+b))^(1/2)/b^2/d/(a+b*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2870, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 \left (2 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 b^2 d \sqrt {a+b \cos (c+d x)}}-\frac {4 a \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b d} \]

[In]

Int[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*b^2*d*Sqrt[(a + b*Cos[c + d*x])/(a +
b)]) + (2*(2*a^2 + b^2)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*b^2*d*Sqr
t[a + b*Cos[c + d*x]]) + (2*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*b*d)

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]

Rule 2870

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[(
-d^2)*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[b*(d^2*(m + 1) + c^2*(m + 2)) - d*(a*d - 2*b*c*(m + 2))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c
, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&  !LtQ[m, -1]

Rubi steps \begin{align*} \text {integral}& = \frac {2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b d}+\frac {2 \int \frac {\frac {b}{2}-a \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b} \\ & = \frac {2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b d}+\frac {1}{3} \left (1+\frac {2 a^2}{b^2}\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx-\frac {(2 a) \int \sqrt {a+b \cos (c+d x)} \, dx}{3 b^2} \\ & = \frac {2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b 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 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (\left (1+\frac {2 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 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (1+\frac {2 a^2}{b^2}\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.54 (sec) , antiderivative size = 137, normalized size of antiderivative = 0.83 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {-4 a (a+b) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+2 \left (2 a^2+b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+2 b (a+b \cos (c+d x)) \sin (c+d x)}{3 b^2 d \sqrt {a+b \cos (c+d x)}} \]

[In]

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(452\) vs. \(2(207)=414\).

Time = 4.11 (sec) , antiderivative size = 453, normalized size of antiderivative = 2.75

method	result	size
default	\(-\frac {2 \sqrt {\left (2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}+2 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a b -6 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b^{2}+2 a^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )+b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right )-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a^{2}+2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {\frac {2 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a -b}{a -b}}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {-\frac {2 b}{a -b}}\right ) a b -2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) a b +2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) b^{2}\right )}{3 b^{2} \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) b +\left (a +b \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-2 b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+a +b}\, d}\)	\(453\)

[In]

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

[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*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+2*a^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))+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*(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*(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*cos(1/2*d*x+1/2*c)*a*b+2*cos(1/2*d*x+1/2*c)*b^2)/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*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 398, normalized size of antiderivative = 2.41 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {-6 i \, \sqrt {2} a b^{\frac {3}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) + 6 i \, \sqrt {2} a b^{\frac {3}{2}} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) + 6 \, \sqrt {b \cos \left (d x + c\right ) + a} b^{2} \sin \left (d x + c\right ) + \sqrt {2} {\left (-4 i \, a^{2} - 3 i \, b^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) + \sqrt {2} {\left (4 i \, a^{2} + 3 i \, b^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )}{9 \, b^{3} d} \]

[In]

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

[Out]

1/9*(-6*I*sqrt(2)*a*b^(3/2)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassP
Inverse(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)) + 6*I*sqrt(2)*a*b^(3/2)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstras
sPInverse(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)) + 6*sqrt(b*cos(d*x + c) + a)*b^2*sin(d*x + c) + sqrt(2)*(-4*I*a^2 - 3*I*b^2)*sqrt(b)*weierstrassPInver
se(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)
+ sqrt(2)*(4*I*a^2 + 3*I*b^2)*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))/(b^3*d)

Sympy [F]

\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int \frac {\cos ^{2}{\left (c + d x \right )}}{\sqrt {a + b \cos {\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

Integral(cos(c + d*x)**2/sqrt(a + b*cos(c + d*x)), x)

Maxima [F]

\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )^{2}}{\sqrt {b \cos \left (d x + c\right ) + a}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 14.34 (sec) , antiderivative size = 116, normalized size of antiderivative = 0.70 \[ \int \frac {\cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2\,\sin \left (c+d\,x\right )\,\sqrt {a+b\,\cos \left (c+d\,x\right )}}{3\,b\,d}+\frac {2\,\sqrt {\frac {a+b\,\cos \left (c+d\,x\right )}{a+b}}\,\left (\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\,\left (2\,a^2+b^2\right )-2\,a\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |\frac {2\,b}{a+b}\right )\,\left (a+b\right )\right )}{3\,b^2\,d\,\sqrt {a+b\,\cos \left (c+d\,x\right )}} \]

[In]

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

[Out]

(2*sin(c + d*x)*(a + b*cos(c + d*x))^(1/2))/(3*b*d) + (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*ellipticE(c/2 + (d*x)/2, (2*b)/(a + b))*(a + b)))/(3*b^2*d*(a + b
*cos(c + d*x))^(1/2))