\(\int \sqrt {a+b \cos (c+d x)} (A+B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\) [1016]

Optimal result

Integrand size = 35, antiderivative size = 237 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \left (3 b^2 (5 A+3 C)+a (5 b B-2 a C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) (5 b B-2 a C) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 (5 b B-2 a C) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b d}+\frac {2 C (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{5 b d} \] Output:

2/15*(3*b^2*(5*A+3*C)+a*(5*B*b-2*C*a))*(a+b*cos(d*x+c))^(1/2)*EllipticE(si 
n(1/2*d*x+1/2*c),2^(1/2)*(b/(a+b))^(1/2))/b^2/d/((a+b*cos(d*x+c))/(a+b))^( 
1/2)-2/15*(a^2-b^2)*(5*B*b-2*C*a)*((a+b*cos(d*x+c))/(a+b))^(1/2)*InverseJa 
cobiAM(1/2*d*x+1/2*c,2^(1/2)*(b/(a+b))^(1/2))/b^2/d/(a+b*cos(d*x+c))^(1/2) 
+2/15*(5*B*b-2*C*a)*(a+b*cos(d*x+c))^(1/2)*sin(d*x+c)/b/d+2/5*C*(a+b*cos(d 
*x+c))^(3/2)*sin(d*x+c)/b/d
 

Mathematica [A] (verified)

Time = 1.95 (sec) , antiderivative size = 189, normalized size of antiderivative = 0.80 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx=\frac {2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 (15 a A+5 b B+7 a C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+\left (15 A b^2+5 a b B-2 a^2 C+9 b^2 C\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )\right )+2 b (a+b \cos (c+d x)) (5 b B+a C+3 b C \cos (c+d x)) \sin (c+d x)}{15 b^2 d \sqrt {a+b \cos (c+d x)}} \] Input:

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

Output:

(2*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(15*a*A + 5*b*B + 7*a*C)*Ellipt 
icF[(c + d*x)/2, (2*b)/(a + b)] + (15*A*b^2 + 5*a*b*B - 2*a^2*C + 9*b^2*C) 
*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, 
 (2*b)/(a + b)])) + 2*b*(a + b*Cos[c + d*x])*(5*b*B + a*C + 3*b*C*Cos[c + 
d*x])*Sin[c + d*x])/(15*b^2*d*Sqrt[a + b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 1.22 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.03, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3042, 3502, 27, 3042, 3232, 27, 3042, 3231, 3042, 3134, 3042, 3132, 3142, 3042, 3140}

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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3232

\(\displaystyle \frac {\frac {2}{3} \int \frac {b (15 a A+5 b B+7 a C)+\left (3 (5 A+3 C) b^2+a (5 b B-2 a C)\right ) \cos (c+d x)}{2 \sqrt {a+b \cos (c+d x)}}dx+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {1}{3} \int \frac {b (15 a A+5 b B+7 a C)+\left (3 (5 A+3 C) b^2+a (5 b B-2 a C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}}dx+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {1}{3} \int \frac {b (15 a A+5 b B+7 a C)+\left (3 (5 A+3 C) b^2+a (5 b B-2 a C)\right ) \sin \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (5 b B-2 a C) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 d}}{5 b}+\frac {2 C \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{5 b d}\)

\(\Big \downarrow \) 3231

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3134

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3132

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

\(\Big \downarrow \) 3142

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3140

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

Input:

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

Output:

(2*C*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(5*b*d) + (((2*(3*b^2*(5*A + 
 3*C) + a*(5*b*B - 2*a*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, 
 (2*b)/(a + b)])/(b*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2) 
*(5*b*B - 2*a*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, 
 (2*b)/(a + b)])/(b*d*Sqrt[a + b*Cos[c + d*x]]))/3 + (2*(5*b*B - 2*a*C)*Sq 
rt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*d))/(5*b)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3132

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a 
 + b]/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, 
b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3134

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[a + 
b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c + d*x])/(a + b)]   Int[Sqrt[a/(a + b) + ( 
b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2 
, 0] &&  !GtQ[a + b, 0]
 

rule 3140

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*S 
qrt[a + b]))*EllipticF[(1/2)*(c - Pi/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[ 
{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]
 

rule 3142

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[Sqrt[(a 
 + b*Sin[c + d*x])/(a + b)]/Sqrt[a + b*Sin[c + d*x]]   Int[1/Sqrt[a/(a + b) 
 + (b/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - 
 b^2, 0] &&  !GtQ[a + b, 0]
 

rule 3231

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + ( 
f_.)*(x_)]], x_Symbol] :> Simp[(b*c - a*d)/b   Int[1/Sqrt[a + b*Sin[e + f*x 
]], x], x] + Simp[d/b   Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a, b 
, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]
 

rule 3232

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)]), x_Symbol] :> Simp[(-d)*Cos[e + f*x]*((a + b*Sin[e + f*x])^m/( 
f*(m + 1))), x] + Simp[1/(m + 1)   Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[b* 
d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ 
[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 
 0] && IntegerQ[2*m]
 

rule 3502

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1186\) vs. \(2(226)=452\).

Time = 23.72 (sec) , antiderivative size = 1187, normalized size of antiderivative = 5.01

method	result	size
default	\(\text {Expression too large to display}\)	\(1187\)
parts	\(\text {Expression too large to display}\)	\(1289\)

Input:

int((a+b*cos(d*x+c))^(1/2)*(A+B*cos(d*x+c)+C*cos(d*x+c)^2),x,method=_RETUR 
NVERBOSE)
 

Output:

-2/15*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(24*C*co 
s(1/2*d*x+1/2*c)^7*b^3+20*B*cos(1/2*d*x+1/2*c)^5*b^3-48*C*cos(1/2*d*x+1/2* 
c)^5*b^3-30*B*cos(1/2*d*x+1/2*c)^3*b^3+30*C*cos(1/2*d*x+1/2*c)^3*b^3+10*B* 
cos(1/2*d*x+1/2*c)*b^3-6*C*cos(1/2*d*x+1/2*c)*b^3-15*A*(sin(1/2*d*x+1/2*c) 
^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d 
*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+5*B*b^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2 
*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2 
*b/(a-b))^(1/2))+2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c) 
^2+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3- 
2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1 
/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3-9*C*(sin(1/2*d*x+ 
1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)*EllipticE(cos 
(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+16*C*cos(1/2*d*x+1/2*c)^5*a*b^2+10 
*B*cos(1/2*d*x+1/2*c)^3*a*b^2+2*C*cos(1/2*d*x+1/2*c)^3*a^2*b-24*C*cos(1/2* 
d*x+1/2*c)^3*a*b^2-10*B*cos(1/2*d*x+1/2*c)*a*b^2-2*C*cos(1/2*d*x+1/2*c)*a^ 
2*b+8*C*cos(1/2*d*x+1/2*c)*a*b^2-5*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*co 
s(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a 
-b))^(1/2))*a*b^2-2*C*a*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2 
*c)^2+a-b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b 
^2+2*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-...
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 512, normalized size of antiderivative = 2.16 \[ \int \sqrt {a+b \cos (c+d x)} \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx =\text {Too large to display} \] Input:

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

Output:

-2/45*(sqrt(1/2)*(4*I*C*a^3 - 10*I*B*a^2*b + 3*I*(5*A + C)*a*b^2 + 15*I*B* 
b^3)*sqrt(b)*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9 
*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3*I*b*sin(d*x + c) + 2*a)/b) + sqrt(1 
/2)*(-4*I*C*a^3 + 10*I*B*a^2*b - 3*I*(5*A + C)*a*b^2 - 15*I*B*b^3)*sqrt(b) 
*weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 
 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b) + 3*sqrt(1/2)*(2*I*C 
*a^2*b - 5*I*B*a*b^2 - 3*I*(5*A + 3*C)*b^3)*sqrt(b)*weierstrassZeta(4/3*(4 
*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4 
*a^2 - 3*b^2)/b^2, -8/27*(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) + 3* 
I*b*sin(d*x + c) + 2*a)/b)) + 3*sqrt(1/2)*(-2*I*C*a^2*b + 5*I*B*a*b^2 + 3* 
I*(5*A + 3*C)*b^3)*sqrt(b)*weierstrassZeta(4/3*(4*a^2 - 3*b^2)/b^2, -8/27* 
(8*a^3 - 9*a*b^2)/b^3, weierstrassPInverse(4/3*(4*a^2 - 3*b^2)/b^2, -8/27* 
(8*a^3 - 9*a*b^2)/b^3, 1/3*(3*b*cos(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b 
)) - 3*(3*C*b^3*cos(d*x + c) + C*a*b^2 + 5*B*b^3)*sqrt(b*cos(d*x + c) + a) 
*sin(d*x + c))/(b^3*d)
 

Sympy [F]

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

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

Output:

Integral(sqrt(a + b*cos(c + d*x))*(A + B*cos(c + d*x) + C*cos(c + d*x)**2) 
, x)
                                                                                    
                                                                                    
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(cos(c + d*x)*b + a),x)*a + int(sqrt(cos(c + d*x)*b + a)*cos(c + d 
*x),x)*b + int(sqrt(cos(c + d*x)*b + a)*cos(c + d*x)**2,x)*c