3.1.63 \(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{\sqrt {b \cos (c+d x)}} \, dx\) [63]

3.1.63.1 Optimal result
3.1.63.2 Mathematica [A] (verified)
3.1.63.3 Rubi [A] (verified)
3.1.63.4 Maple [B] (verified)
3.1.63.5 Fricas [C] (verification not implemented)
3.1.63.6 Sympy [F(-1)]
3.1.63.7 Maxima [F]
3.1.63.8 Giac [F]
3.1.63.9 Mupad [F(-1)]

3.1.63.1 Optimal result

Integrand size = 33, antiderivative size = 112 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {2 (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 (7 A+5 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 b d}+\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^3 d} \]

output
2/7*C*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/b^3/d+2/21*(7*A+5*C)*(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))*cos 
(d*x+c)^(1/2)/d/(b*cos(d*x+c))^(1/2)+2/21*(7*A+5*C)*sin(d*x+c)*(b*cos(d*x+ 
c))^(1/2)/b/d
 
3.1.63.2 Mathematica [A] (verified)

Time = 0.49 (sec) , antiderivative size = 77, normalized size of antiderivative = 0.69 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {4 (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+(14 A+13 C+3 C \cos (2 (c+d x))) \sin (2 (c+d x))}{42 d \sqrt {b \cos (c+d x)}} \]

input
Integrate[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/Sqrt[b*Cos[c + d*x]],x]
 
output
(4*(7*A + 5*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + (14*A + 13*C 
 + 3*C*Cos[2*(c + d*x)])*Sin[2*(c + d*x)])/(42*d*Sqrt[b*Cos[c + d*x]])
 
3.1.63.3 Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.03, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.273, Rules used = {2030, 3042, 3493, 3042, 3115, 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.

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

\(\Big \downarrow \) 2030

\(\displaystyle \frac {\int (b \cos (c+d x))^{3/2} \left (C \cos ^2(c+d x)+A\right )dx}{b^2}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3493

\(\displaystyle \frac {\frac {1}{7} (7 A+5 C) \int (b \cos (c+d x))^{3/2}dx+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} (7 A+5 C) \int \left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}dx+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^2}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {\frac {1}{7} (7 A+5 C) \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \cos (c+d x)}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} (7 A+5 C) \left (\frac {1}{3} b^2 \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^2}\)

\(\Big \downarrow \) 3121

\(\displaystyle \frac {\frac {1}{7} (7 A+5 C) \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^2}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{7} (7 A+5 C) \left (\frac {b^2 \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^2}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {\frac {1}{7} (7 A+5 C) \left (\frac {2 b^2 \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 b \sin (c+d x) \sqrt {b \cos (c+d x)}}{3 d}\right )+\frac {2 C \sin (c+d x) (b \cos (c+d x))^{5/2}}{7 b d}}{b^2}\)

input
Int[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/Sqrt[b*Cos[c + d*x]],x]
 
output
((2*C*(b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b*d) + ((7*A + 5*C)*((2*b^2* 
Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*d*Sqrt[b*Cos[c + d*x]]) + 
 (2*b*Sqrt[b*Cos[c + d*x]]*Sin[c + d*x])/(3*d)))/7)/b^2
 

3.1.63.3.1 Defintions of rubi rules used

rule 2030
Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042
Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3115
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* 
x]*((b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[b^2*((n - 1)/n)   Int[(b*Sin 
[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[ 
2*n]
 

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]
 
3.1.63.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(292\) vs. \(2(124)=248\).

Time = 10.06 (sec) , antiderivative size = 293, normalized size of antiderivative = 2.62

method result size
default \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (48 C \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (28 A +56 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-14 A -16 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+7 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{21 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(293\)
parts \(-\frac {2 A \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}-\frac {2 C \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (48 \left (\cos ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-120 \left (\cos ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+128 \left (\cos ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-72 \left (\cos ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+5 \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}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(396\)

input
int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(cos(d*x+c)*b)^(1/2),x,method=_RETURNV 
ERBOSE)
 
output
-2/21*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*(48*C*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8-72*cos(1/2*d*x+1/2*c)*C*sin(1/2*d*x+1/ 
2*c)^6+(28*A+56*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-14*A-16*C)*si 
n(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+7*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2* 
sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+5*C*(s 
in(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(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)/((2*cos(1/2*d*x+1/2*c)^2-1)*b)^(1/2)/d
 
3.1.63.5 Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.97 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {b \cos (c+d x)}} \, dx=\frac {\sqrt {2} {\left (-7 i \, A - 5 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {2} {\left (7 i \, A + 5 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 2 \, {\left (3 \, C \cos \left (d x + c\right )^{2} + 7 \, A + 5 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{21 \, b d} \]

input
integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/2),x, algorith 
m="fricas")
 
output
1/21*(sqrt(2)*(-7*I*A - 5*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x 
+ c) + I*sin(d*x + c)) + sqrt(2)*(7*I*A + 5*I*C)*sqrt(b)*weierstrassPInver 
se(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 2*(3*C*cos(d*x + c)^2 + 7*A + 5 
*C)*sqrt(b*cos(d*x + c))*sin(d*x + c))/(b*d)
 
3.1.63.6 Sympy [F(-1)]

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

input
integrate(cos(d*x+c)**2*(A+C*cos(d*x+c)**2)/(b*cos(d*x+c))**(1/2),x)
 
output
Timed out
 
3.1.63.7 Maxima [F]

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

input
integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/2),x, algorith 
m="maxima")
 
output
integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^2/sqrt(b*cos(d*x + c)), x)
 
3.1.63.8 Giac [F]

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

input
integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(b*cos(d*x+c))^(1/2),x, algorith 
m="giac")
 
output
integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^2/sqrt(b*cos(d*x + c)), x)
 
3.1.63.9 Mupad [F(-1)]

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

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