\(\int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx\) [708]

Optimal result

Integrand size = 35, antiderivative size = 85 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}-\frac {2 a C \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b^2 d}+\frac {2 \left (A b^2+a^2 C\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^2 (a+b) d} \] Output:

2*C*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/b/d-2*a*C*InverseJacobiAM(1/2*d* 
x+1/2*c,2^(1/2))/b^2/d+2*(A*b^2+C*a^2)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/( 
a+b),2^(1/2))/b^2/(a+b)/d
 

Mathematica [A] (warning: unable to verify)

Time = 1.24 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.49 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\frac {\frac {(2 A+C) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {C \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b^2 \sqrt {\sin ^2(c+d x)}}}{d} \] Input:

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

Output:

(((2*A + C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (C*(-2*a* 
b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin 
[Sqrt[Cos[c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[ 
Cos[c + d*x]]], -1])*Sin[c + d*x])/(a*b^2*Sqrt[Sin[c + d*x]^2]))/d
 

Rubi [A] (verified)

Time = 0.71 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.257, Rules used = {3042, 3539, 25, 3042, 3119, 3481, 3042, 3120, 3284}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx\)

\(\Big \downarrow \) 3539

\(\displaystyle \frac {C \int \sqrt {\cos (c+d x)}dx}{b}-\frac {\int -\frac {A b-a C \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {\int \frac {A b-a C \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}+\frac {C \int \sqrt {\cos (c+d x)}dx}{b}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {A b-a C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {C \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {\int \frac {A b-a C \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\)

\(\Big \downarrow \) 3481

\(\displaystyle \frac {\frac {\left (a^2 C+A b^2\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx}{b}-\frac {a C \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{b}}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (a^2 C+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {a C \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{b}}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {\left (a^2 C+A b^2\right ) \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{b}-\frac {2 a C \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {\frac {2 \left (a^2 C+A b^2\right ) \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b d (a+b)}-\frac {2 a C \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{b d}}{b}+\frac {2 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{b d}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3119

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

rule 3120

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 3284

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 3481

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) 
+ (f_.)*(x_)]))/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[ 
B/d   Int[(a + b*Sin[e + f*x])^m, x], x] - Simp[(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]
 

rule 3539

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(258\) vs. \(2(90)=180\).

Time = 6.33 (sec) , antiderivative size = 259, normalized size of antiderivative = 3.05

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {1-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (A \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) b^{2}-C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a^{2}+C \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b -C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a b +C \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b^{2}+C \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) a^{2}\right )}{b^{2} \left (a -b \right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(259\)

Input:

int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(1/2)/(a+b*cos(d*x+c)),x,method=_RETURNV 
ERBOSE)
 

Output:

-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)*(1-2*cos(1/2*d*x+1/2*c)^2)^(1/2)*(A*EllipticPi(cos(1/2*d*x+1 
/2*c),-2*b/(a-b),2^(1/2))*b^2-C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2+ 
C*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b-C*EllipticE(cos(1/2*d*x+1/2*c) 
,2^(1/2))*a*b+C*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*b^2+C*EllipticPi(cos 
(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^2)/b^2/(a-b)/(-2*sin(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. \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{\sqrt {\cos \left (c+d\,x\right )}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \] Input:

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(cos(c + d*x))/(cos(c + d*x)**2*b + cos(c + d*x)*a),x)*a + int((sq 
rt(cos(c + d*x))*cos(c + d*x))/(cos(c + d*x)*b + a),x)*c