3.8.100 \(\int \cos (c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\) [800]

3.8.100.1 Optimal result
3.8.100.2 Mathematica [A] (verified)
3.8.100.3 Rubi [A] (verified)
3.8.100.4 Maple [A] (verified)
3.8.100.5 Fricas [C] (verification not implemented)
3.8.100.6 Sympy [F(-1)]
3.8.100.7 Maxima [F]
3.8.100.8 Giac [F]
3.8.100.9 Mupad [F(-1)]

3.8.100.1 Optimal result

Integrand size = 29, antiderivative size = 139 \[ \int \cos (c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {6 B \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d \sqrt {\cos (c+d x)}}+\frac {2 A b \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 A \sqrt {b \cos (c+d x)} \sin (c+d x)}{3 d}+\frac {2 B (b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d} \]

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

Time = 0.28 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.65 \[ \int \cos (c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {2 (b \cos (c+d x))^{3/2} \left (9 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+5 A \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (5 A+3 B \cos (c+d x)) \sin (c+d x)\right )}{15 b d \cos ^{\frac {3}{2}}(c+d x)} \]

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

Time = 0.61 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.09, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.345, Rules used = {2030, 3042, 3227, 3042, 3115, 3042, 3121, 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.

\(\displaystyle \int \cos (c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx\)

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3227

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3120

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

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

3.8.100.3.1 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 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 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 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]
 
3.8.100.4 Maple [A] (verified)

Time = 6.96 (sec) , antiderivative size = 271, normalized size of antiderivative = 1.95

method result size
default \(-\frac {2 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b \left (-24 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (20 A +24 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-10 A -6 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(271\)
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 )}\, b \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 B \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 )}\, b \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 {-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}\) \(401\)

input
int(cos(d*x+c)*(cos(d*x+c)*b)^(1/2)*(A+B*cos(d*x+c)),x,method=_RETURNVERBO 
SE)
 
output
-2/15*((2*cos(1/2*d*x+1/2*c)^2-1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(-24*B*c 
os(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+(20*A+24*B)*sin(1/2*d*x+1/2*c)^4*co 
s(1/2*d*x+1/2*c)+(-10*A-6*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+5*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))-9*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)*EllipticE(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
 
3.8.100.5 Fricas [C] (verification not implemented)

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

Time = 0.11 (sec) , antiderivative size = 151, normalized size of antiderivative = 1.09 \[ \int \cos (c+d x) \sqrt {b \cos (c+d x)} (A+B \cos (c+d x)) \, dx=\frac {-5 i \, \sqrt {2} A \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + 5 i \, \sqrt {2} A \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 9 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 9 i \, \sqrt {2} B \sqrt {b} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (3 \, B \cos \left (d x + c\right ) + 5 \, A\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{15 \, d} \]

input
integrate(cos(d*x+c)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="f 
ricas")
 
output
1/15*(-5*I*sqrt(2)*A*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*s 
in(d*x + c)) + 5*I*sqrt(2)*A*sqrt(b)*weierstrassPInverse(-4, 0, cos(d*x + 
c) - I*sin(d*x + c)) + 9*I*sqrt(2)*B*sqrt(b)*weierstrassZeta(-4, 0, weiers 
trassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 9*I*sqrt(2)*B*sqrt( 
b)*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin( 
d*x + c))) + 2*(3*B*cos(d*x + c) + 5*A)*sqrt(b*cos(d*x + c))*sin(d*x + c)) 
/d
 
3.8.100.6 Sympy [F(-1)]

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

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

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

input
integrate(cos(d*x+c)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="m 
axima")
 
output
integrate((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*cos(d*x + c), x)
 
3.8.100.8 Giac [F]

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

input
integrate(cos(d*x+c)*(b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)),x, algorithm="g 
iac")
 
output
integrate((B*cos(d*x + c) + A)*sqrt(b*cos(d*x + c))*cos(d*x + c), x)
 
3.8.100.9 Mupad [F(-1)]

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

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