\(\int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx\) [368]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [F(-1)]
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 33, antiderivative size = 61 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}+\frac {2 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b (a+b) d} \]

[Out]

2*B*(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/d+2*(A*b-B*a)*(cos
(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/b/(a+b)/d

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3081, 2720, 2884} \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\frac {2 (A b-a B) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}+\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d} \]

[In]

Int[(A + B*Cos[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])),x]

[Out]

(2*B*EllipticF[(c + d*x)/2, 2])/(b*d) + (2*(A*b - a*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(b*(a + b)*d
)

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2884

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

Rule 3081

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

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

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.95 \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\frac {2 \left ((a+b) B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(A b-a B) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )\right )}{b (a+b) d} \]

[In]

Integrate[(A + B*Cos[c + d*x])/(Sqrt[Cos[c + d*x]]*(a + b*Cos[c + d*x])),x]

[Out]

(2*((a + b)*B*EllipticF[(c + d*x)/2, 2] + (A*b - a*B)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(b*(a + b)*d
)

Maple [A] (verified)

Time = 3.30 (sec) , antiderivative size = 217, normalized size of antiderivative = 3.56

method result size
default \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (A \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) b +B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a -B F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b -B \Pi \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) a \right )}{\left (a -b \right ) b \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(217\)

[In]

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

[Out]

-2*((2*cos(1/2*d*x+1/2*c)^2-1)*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+1)^(1/2)*(A*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*b+B*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a-B
*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b-B*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a)/(a-b)/b/(-2*si
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+B \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\int \frac {A+B\,\cos \left (c+d\,x\right )}{\sqrt {\cos \left (c+d\,x\right )}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]

[In]

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

[Out]

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