3.3.84 \(\int \sqrt {\cos (c+d x)} (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [284]

3.3.84.1 Optimal result
3.3.84.2 Mathematica [A] (verified)
3.3.84.3 Rubi [A] (verified)
3.3.84.4 Maple [B] (verified)
3.3.84.5 Fricas [C] (verification not implemented)
3.3.84.6 Sympy [F(-1)]
3.3.84.7 Maxima [F]
3.3.84.8 Giac [F]
3.3.84.9 Mupad [B] (verification not implemented)

3.3.84.1 Optimal result

Integrand size = 30, antiderivative size = 87 \[ \int \sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {6 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 B \sqrt {\cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 C \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d} \]

output
6/5*C*(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))/d+2/3*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))/d+2/5*C*cos(d*x+c)^(3/2)*sin(d*x+c)/ 
d+2/3*B*sin(d*x+c)*cos(d*x+c)^(1/2)/d
 
3.3.84.2 Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.76 \[ \int \sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \left (9 C E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 B \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (5 B+3 C \cos (c+d x)) \sin (c+d x)\right )}{15 d} \]

input
Integrate[Sqrt[Cos[c + d*x]]*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
 
output
(2*(9*C*EllipticE[(c + d*x)/2, 2] + 5*B*EllipticF[(c + d*x)/2, 2] + Sqrt[C 
os[c + d*x]]*(5*B + 3*C*Cos[c + d*x])*Sin[c + d*x]))/(15*d)
 
3.3.84.3 Rubi [A] (verified)

Time = 0.47 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.02, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {3042, 3489, 3042, 3227, 3042, 3115, 3042, 3119, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3489

\(\displaystyle \int \cos ^{\frac {3}{2}}(c+d x) (B+C \cos (c+d x))dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

\(\displaystyle B \int \cos ^{\frac {3}{2}}(c+d x)dx+C \int \cos ^{\frac {5}{2}}(c+d x)dx\)

\(\Big \downarrow \) 3042

\(\displaystyle B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+C \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\)

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

\(\displaystyle B \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+C \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle B \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+C \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\)

input
Int[Sqrt[Cos[c + d*x]]*(B*Cos[c + d*x] + C*Cos[c + d*x]^2),x]
 
output
B*((2*EllipticF[(c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x 
])/(3*d)) + C*((6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2) 
*Sin[c + d*x])/(5*d))
 

3.3.84.3.1 Defintions of rubi rules used

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 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 3227
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3489
Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((B_.)*sin[(e_.) + (f_.)*(x_)] + 
(C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[1/b   Int[(b*Sin[e + f* 
x])^(m + 1)*(B + C*Sin[e + f*x]), x], x] /; FreeQ[{b, e, f, B, C, m}, x]
 
3.3.84.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(261\) vs. \(2(127)=254\).

Time = 11.14 (sec) , antiderivative size = 262, normalized size of antiderivative = 3.01

method result size
default \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-24 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) C \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (20 B +24 C \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-10 B -6 C \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(262\)
parts \(-\frac {2 B \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (4 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}-\frac {2 C \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, \left (-8 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-3 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{5 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\) \(383\)

input
int(cos(d*x+c)^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x,method=_RETURNVERBOSE 
)
 
output
-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-24*cos(1/2 
*d*x+1/2*c)*C*sin(1/2*d*x+1/2*c)^6+(20*B+24*C)*sin(1/2*d*x+1/2*c)^4*cos(1/ 
2*d*x+1/2*c)+(-10*B-6*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*B*(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))-9*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))/(-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
 
3.3.84.5 Fricas [C] (verification not implemented)

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

Time = 0.10 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.57 \[ \int \sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (3 \, C \cos \left (d x + c\right ) + 5 \, B\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} B {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 i \, \sqrt {2} C {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 i \, \sqrt {2} C {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{15 \, d} \]

input
integrate(cos(d*x+c)^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fri 
cas")
 
output
1/15*(2*(3*C*cos(d*x + c) + 5*B)*sqrt(cos(d*x + c))*sin(d*x + c) - 5*I*sqr 
t(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) + 5*I*sqr 
t(2)*B*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 9*I*sqr 
t(2)*C*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I* 
sin(d*x + c))) - 9*I*sqrt(2)*C*weierstrassZeta(-4, 0, weierstrassPInverse( 
-4, 0, cos(d*x + c) - I*sin(d*x + c))))/d
 
3.3.84.6 Sympy [F(-1)]

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

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

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

input
integrate(cos(d*x+c)^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="max 
ima")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*sqrt(cos(d*x + c)), x)
 
3.3.84.8 Giac [F]

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

input
integrate(cos(d*x+c)^(1/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="gia 
c")
 
output
integrate((C*cos(d*x + c)^2 + B*cos(d*x + c))*sqrt(cos(d*x + c)), x)
 
3.3.84.9 Mupad [B] (verification not implemented)

Time = 1.39 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.92 \[ \int \sqrt {\cos (c+d x)} \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2\,B\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,B\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d}-\frac {2\,C\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

input
int(cos(c + d*x)^(1/2)*(B*cos(c + d*x) + C*cos(c + d*x)^2),x)
 
output
(2*B*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (2*B*cos(c + d*x)^(1/2)*sin(c + 
d*x))/(3*d) - (2*C*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 1 
1/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2))