\(\int \frac {\cos (c+d x) (A+C \cos ^2(c+d x))}{\sqrt {a+b \cos (c+d x)}} \, dx\) [650]

Optimal result

Integrand size = 33, antiderivative size = 233 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 \left (8 a^2 C+3 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (15 A b^2+8 a^2 C+7 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {a+b \cos (c+d x)}}-\frac {8 a C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d} \]

[Out]

-8/15*a*C*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^2/d+2/5*C*cos(d*x+c)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d+2/15*
(8*a^2*C+3*b^2*(5*A+3*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)
*(b/(a+b))^(1/2))*(a+b*cos(d*x+c))^(1/2)/b^3/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/15*a*(15*A*b^2+8*C*a^2+7*C*b^2
)*(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)*(b/(a+b))^(1/2))*((a+b*
cos(d*x+c))/(a+b))^(1/2)/b^3/d/(a+b*cos(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.41 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3129, 3102, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {2 a \left (8 a^2 C+15 A b^2+7 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (8 a^2 C+3 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{15 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {8 a C \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{15 b^2 d}+\frac {2 C \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b d} \]

[In]

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

[Out]

(2*(8*a^2*C + 3*b^2*(5*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)])/(15*b^3*d*Sqr
t[(a + b*Cos[c + d*x])/(a + b)]) - (2*a*(15*A*b^2 + 8*a^2*C + 7*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*Elli
pticF[(c + d*x)/2, (2*b)/(a + b)])/(15*b^3*d*Sqrt[a + b*Cos[c + d*x]]) - (8*a*C*Sqrt[a + b*Cos[c + d*x]]*Sin[c
+ d*x])/(15*b^2*d) + (2*C*Cos[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*b*d)

Rule 2732

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 2734

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[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 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/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 2742

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[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 2831

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[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 3102

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 2))), x] + Dist[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]

Rule 3129

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

Rubi steps \begin{align*} \text {integral}& = \frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac {2 \int \frac {a C+\frac {1}{2} b (5 A+3 C) \cos (c+d x)-2 a C \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{5 b} \\ & = -\frac {8 a C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac {4 \int \frac {\frac {a b C}{2}+\frac {1}{4} \left (8 a^2 C+3 b^2 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^2} \\ & = -\frac {8 a C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}-\frac {\left (a \left (15 A b^2+8 a^2 C+7 b^2 C\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^3}+\frac {\left (8 a^2 C+3 b^2 (5 A+3 C)\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{15 b^3} \\ & = -\frac {8 a C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d}+\frac {\left (\left (8 a^2 C+3 b^2 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{15 b^3 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (a \left (15 A b^2+8 a^2 C+7 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{15 b^3 \sqrt {a+b \cos (c+d x)}} \\ & = \frac {2 \left (8 a^2 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 )}{15 b^3 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 a \left (15 A b^2+8 a^2 C+7 b^2 C\right ) \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^3 d \sqrt {a+b \cos (c+d x)}}-\frac {8 a C \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{15 b^2 d}+\frac {2 C \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b d} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.59 (sec) , antiderivative size = 190, normalized size of antiderivative = 0.82 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=\frac {2 (a+b) \left (15 A b^2+8 a^2 C+9 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-2 a \left (15 A b^2+8 a^2 C+7 b^2 C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )+b C \left (-8 a^2+3 b^2-2 a b \cos (c+d x)+3 b^2 \cos (2 (c+d x))\right ) \sin (c+d x)}{15 b^3 d \sqrt {a+b \cos (c+d x)}} \]

[In]

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

[Out]

(2*(a + b)*(15*A*b^2 + 8*a^2*C + 9*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticE[(c + d*x)/2, (2*b)/(a +
b)] - 2*a*(15*A*b^2 + 8*a^2*C + 7*b^2*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a +
b)] + b*C*(-8*a^2 + 3*b^2 - 2*a*b*Cos[c + d*x] + 3*b^2*Cos[2*(c + d*x)])*Sin[c + d*x])/(15*b^3*d*Sqrt[a + b*C
os[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(886\) vs. \(2(271)=542\).

Time = 15.41 (sec) , antiderivative size = 887, normalized size of antiderivative = 3.81

method	result	size
parts	\(\text {Expression too large to display}\)	\(887\)
default	\(\text {Expression too large to display}\)	\(892\)

[In]

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

[Out]

2*A*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(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-EllipticE(cos(1/2*d*x+1/2*c),(
-2*b/(a-b))^(1/2))*a+EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(
1/2*d*x+1/2*c)^2)^(1/2)/b/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d-2/15*C*((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*cos(1/2*d*x+1/2*c)^7*b^3-4*cos(1/2*d*x+1/2*c)^5*a*b^2-48*cos(1/2
*d*x+1/2*c)^5*b^3-8*cos(1/2*d*x+1/2*c)^3*a^2*b+6*cos(1/2*d*x+1/2*c)^3*a*b^2+30*cos(1/2*d*x+1/2*c)^3*b^3-8*a^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))-7*a*b^2*(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))+8*(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^3-8*(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^2*b+9*(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*b^2-9*(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+8*cos(1/2*d*x+1/2*c)*a^2*b-2*cos(1/2*d*x+1/2*c)*a*b^2-6*cos(1/2*d*x+1/2*c)*b^3)/b^3/(-2*sin(1/2*d*x+1/2*c)
^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(-2*b*sin(1/2*d*x+1/2*c)^2+a+b)^(1/2)/d

Fricas [C] (verification not implemented)

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

Time = 0.12 (sec) , antiderivative size = 473, normalized size of antiderivative = 2.03 \[ \int \frac {\cos (c+d x) \left (A+C \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx=-\frac {2 \, \sqrt {2} {\left (-8 i \, C a^{3} - 3 i \, {\left (5 \, A + 2 \, C\right )} a b^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) + 2 \, \sqrt {2} {\left (8 i \, C a^{3} + 3 i \, {\left (5 \, A + 2 \, C\right )} a b^{2}\right )} \sqrt {b} {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right ) + 3 \, \sqrt {2} {\left (-8 i \, C a^{2} b - 3 i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) + 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) + 3 \, \sqrt {2} {\left (8 i \, C a^{2} b + 3 i \, {\left (5 \, A + 3 \, C\right )} b^{3}\right )} \sqrt {b} {\rm weierstrassZeta}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, {\rm weierstrassPInverse}\left (\frac {4 \, {\left (4 \, a^{2} - 3 \, b^{2}\right )}}{3 \, b^{2}}, -\frac {8 \, {\left (8 \, a^{3} - 9 \, a b^{2}\right )}}{27 \, b^{3}}, \frac {3 \, b \cos \left (d x + c\right ) - 3 i \, b \sin \left (d x + c\right ) + 2 \, a}{3 \, b}\right )\right ) - 6 \, {\left (3 \, C b^{3} \cos \left (d x + c\right ) - 4 \, C a b^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a} \sin \left (d x + c\right )}{45 \, b^{4} d} \]

[In]

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

[Out]

-1/45*(2*sqrt(2)*(-8*I*C*a^3 - 3*I*(5*A + 2*C)*a*b^2)*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) + 2*sqrt(2)*(8*I*C*a^3 + 3*I*(5
*A + 2*C)*a*b^2)*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*co
s(d*x + c) - 3*I*b*sin(d*x + c) + 2*a)/b) + 3*sqrt(2)*(-8*I*C*a^2*b - 3*I*(5*A + 3*C)*b^3)*sqrt(b)*weierstrass
Zeta(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(2)*(8*I*C*a^2*b + 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, weierstrassPInver
se(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))
- 6*(3*C*b^3*cos(d*x + c) - 4*C*a*b^2)*sqrt(b*cos(d*x + c) + a)*sin(d*x + c))/(b^4*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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