\(\int \cos (c+d x) \sqrt {b \cos (c+d x)} (A+C \cos ^2(c+d x)) \, dx\) [38]

Optimal result

Integrand size = 31, antiderivative size = 110 \[ \int \cos (c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {2 b (7 A+5 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d \sqrt {b \cos (c+d x)}}+\frac {2 (7 A+5 C) \sqrt {b \cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 C (b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b^2 d} \] Output:

2/21*b*(7*A+5*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d 
/(b*cos(d*x+c))^(1/2)+2/21*(7*A+5*C)*(b*cos(d*x+c))^(1/2)*sin(d*x+c)/d+2/7 
*C*(b*cos(d*x+c))^(5/2)*sin(d*x+c)/b^2/d
 

Mathematica [A] (verified)

Time = 0.34 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.81 \[ \int \cos (c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\frac {(b \cos (c+d x))^{3/2} \left (4 (7 A+5 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+2 \sqrt {\cos (c+d x)} (14 A+13 C+3 C \cos (2 (c+d x))) \sin (c+d x)\right )}{42 b d \cos ^{\frac {3}{2}}(c+d x)} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.51 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.290, 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.

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

\(\Big \downarrow \) 2030

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3493

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3115

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3121

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3120

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

Input:

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

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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(293\) vs. \(2(97)=194\).

Time = 1.58 (sec) , antiderivative size = 294, normalized size of antiderivative = 2.67

method	result	size
default	\(-\frac {2 \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b \left (48 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-72 C \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (28 A +56 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-14 A -16 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+7 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+5 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{21 \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\)	\(294\)
parts	\(-\frac {2 A \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b \left (4 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3 \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}-\frac {2 C \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b \left (48 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}-120 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}+128 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-72 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+16 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{21 \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\)	\(398\)

Input:

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

Output:

-2/21*(b*(-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*b*(48*C*s 
in(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)-72*C*sin(1/2*d*x+1/2*c)^6*cos(1/2*d 
*x+1/2*c)+(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) 
*sin(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 
*(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(c 
os(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 = 0.09 (sec) , antiderivative size = 106, normalized size of antiderivative = 0.96 \[ \int \cos (c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=-\frac {2 \, {\left (\sqrt {\frac {1}{2}} {\left (7 i \, A + 5 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) + \sqrt {\frac {1}{2}} {\left (-7 i \, A - 5 i \, C\right )} \sqrt {b} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - {\left (3 \, C \cos \left (d x + c\right )^{2} + 7 \, A + 5 \, C\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{21 \, d} \] Input:

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

Output:

-2/21*(sqrt(1/2)*(7*I*A + 5*I*C)*sqrt(b)*weierstrassPInverse(-4, 0, cos(d* 
x + c) + I*sin(d*x + c)) + sqrt(1/2)*(-7*I*A - 5*I*C)*sqrt(b)*weierstrassP 
Inverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - (3*C*cos(d*x + c)^2 + 7*A 
+ 5*C)*sqrt(b*cos(d*x + c))*sin(d*x + c))/d
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int \cos (c+d x) \sqrt {b \cos (c+d x)} \left (A+C \cos ^2(c+d x)\right ) \, dx=\sqrt {b}\, \left (\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}d x \right ) c \right ) \] Input:

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