\(\int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx\) [65]

Optimal result

Integrand size = 25, antiderivative size = 75 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 (3 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 C \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 b d} \] Output:

2/3*(3*A+C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d/(b*c 
os(d*x+c))^(1/2)+2/3*C*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/b/d
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.77 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 (3 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+C \sin (2 (c+d x))}{3 d \sqrt {b \cos (c+d x)}} \] Input:

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

Output:

(2*(3*A + C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + C*Sin[2*(c + d 
*x)])/(3*d*Sqrt[b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.35 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3493, 3042, 3121, 3042, 3120}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3493

\(\displaystyle \frac {1}{3} (3 A+C) \int \frac {1}{\sqrt {b \cos (c+d x)}}dx+\frac {2 C \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{3} (3 A+C) \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 C \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 b d}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {(3 A+C) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 C \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(3 A+C) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 C \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 b d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {2 (3 A+C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 C \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 b d}\)

Input:

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

Output:

(2*(3*A + C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos 
[c + d*x]]) + (2*C*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(3*b*d)
 

Defintions of rubi rules used

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, 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 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3493

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_) + (C_.)*sin[(e_.) + (f_.)*( 
x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f 
*(m + 2))), x] + Simp[(A*(m + 2) + C*(m + 1))/(m + 2)   Int[(b*Sin[e + f*x] 
)^m, x], x] /; FreeQ[{b, e, f, A, C, m}, x] &&  !LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(224\) vs. \(2(66)=132\).

Time = 0.00 (sec) , antiderivative size = 225, normalized size of antiderivative = 3.00

method	result	size
parts	\(\frac {2 A \sqrt {\cos \left (d x +c \right )}\, \operatorname {InverseJacobiAM}\left (\frac {d x}{2}+\frac {c}{2}, \sqrt {2}\right )}{d \sqrt {b \cos \left (d x +c \right )}}-\frac {2 C \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\)	\(225\)
default	\(-\frac {2 \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (4 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+3 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-2 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\)	\(236\)

Input:

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

Output:

2*A/d/(b*cos(d*x+c))^(1/2)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c, 
2^(1/2))-2/3*C*(b*(-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)* 
(4*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)-2*sin(1/2*d*x+1/2*c)^2*cos(1/2* 
d*x+1/2*c)+(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*E 
llipticF(cos(1/2*d*x+1/2*c),2^(1/2)))/(-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)/(b*(-1+2*cos(1/2*d*x+1/2*c)^2))^(1 
/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.08 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.23 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=-\frac {2 \, {\left (\sqrt {\frac {1}{2}} {\left (3 i \, A + i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {\frac {1}{2}} {\left (-3 i \, A - i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - \sqrt {b \cos \left (d x + c\right )} C \sin \left (d x + c\right )\right )}}{3 \, b d} \] Input:

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

Output:

-2/3*(sqrt(1/2)*(3*I*A + I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + 
 c) + I*sin(d*x + c)) + sqrt(1/2)*(-3*I*A - I*C)*sqrt(b)*weierstrassPInver 
se(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - sqrt(b*cos(d*x + c))*C*sin(d*x 
+ c))/(b*d)
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [B] (verification not implemented)

Time = 43.43 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.25 \[ \int \frac {A+C \cos ^2(c+d x)}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2\,C\,\sin \left (c+d\,x\right )\,\sqrt {b\,\cos \left (c+d\,x\right )}}{3\,b\,d}+\frac {2\,A\,\sqrt {\cos \left (c+d\,x\right )}\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d\,\sqrt {b\,\cos \left (c+d\,x\right )}}+\frac {2\,C\,\sqrt {\cos \left (c+d\,x\right )}\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d\,\sqrt {b\,\cos \left (c+d\,x\right )}} \] Input:

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

Output:

(2*C*sin(c + d*x)*(b*cos(c + d*x))^(1/2))/(3*b*d) + (2*A*cos(c + d*x)^(1/2 
)*ellipticF(c/2 + (d*x)/2, 2))/(d*(b*cos(c + d*x))^(1/2)) + (2*C*cos(c + d 
*x)^(1/2)*ellipticF(c/2 + (d*x)/2, 2))/(3*d*(b*cos(c + d*x))^(1/2))
 

Reduce [F]

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

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

Output:

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