\(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^{3/2}} \, dx\) [657]

Optimal result

Integrand size = 35, antiderivative size = 375 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {2 \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{5 b^4 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (5 A b^2+2 \left (4 a^2+b^2\right ) C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{5 b^4 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \]

[Out]

-2*(A*b^2+C*a^2)*cos(d*x+c)^2*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)-2/5*a*(5*A*b^2+8*C*a^2-3*C*b^2)*
sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b^3/(a^2-b^2)/d+2/5*(5*A*b^2+6*C*a^2-C*b^2)*cos(d*x+c)*sin(d*x+c)*(a+b*cos(d
*x+c))^(1/2)/b^2/(a^2-b^2)/d+2/5*(2*a^2*b^2*(5*A-4*C)+16*a^4*C-b^4*(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^4/(a^2-b^2)/d/(
(a+b*cos(d*x+c))/(a+b))^(1/2)-4/5*a*(5*A*b^2+2*(4*a^2+b^2)*C)*(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^4/d/(a+b*cos(d*x+c))^(1
/2)

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 375, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3127, 3128, 3102, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=-\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (6 a^2 C+5 A b^2-b^2 C\right ) \sin (c+d x) \cos (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b^2 d \left (a^2-b^2\right )}-\frac {4 a \left (2 C \left (4 a^2+b^2\right )+5 A b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{5 b^4 d \sqrt {a+b \cos (c+d x)}}-\frac {2 a \left (8 a^2 C+5 A b^2-3 b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{5 b^3 d \left (a^2-b^2\right )}+\frac {2 \left (16 a^4 C+2 a^2 b^2 (5 A-4 C)-b^4 (5 A+3 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{5 b^4 d \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

[In]

Int[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^(3/2),x]

[Out]

(2*(2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C))*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/
(a + b)])/(5*b^4*(a^2 - b^2)*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (4*a*(5*A*b^2 + 2*(4*a^2 + b^2)*C)*Sqrt[(
a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(5*b^4*d*Sqrt[a + b*Cos[c + d*x]]) - (2*(A
*b^2 + a^2*C)*Cos[c + d*x]^2*Sin[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) - (2*a*(5*A*b^2 + 8*a^2*
C - 3*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*b^3*(a^2 - b^2)*d) + (2*(5*A*b^2 + 6*a^2*C - b^2*C)*Cos
[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(5*b^2*(a^2 - b^2)*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 3127

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

Rule 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*((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)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x
], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, 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 \left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 \int \frac {\cos (c+d x) \left (2 \left (A b^2+a^2 C\right )-\frac {1}{2} a b (A+C) \cos (c+d x)-\frac {1}{2} \left (5 A b^2+6 a^2 C-b^2 C\right ) \cos ^2(c+d x)\right )}{\sqrt {a+b \cos (c+d x)}} \, dx}{b \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {4 \int \frac {-\frac {1}{2} a \left (b^2 (5 A-C)+6 a^2 C\right )+\frac {1}{4} b \left (5 A b^2+2 a^2 C+3 b^2 C\right ) \cos (c+d x)+\frac {3}{4} a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{5 b^2 \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {-\frac {3}{8} a b \left (5 A b^2+\left (4 a^2+b^2\right ) C\right )-\frac {3}{8} \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}-\frac {\left (2 a \left (5 A b^2+2 \left (4 a^2+b^2\right ) C\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{5 b^4}+\frac {\left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{5 b^4 \left (a^2-b^2\right )} \\ & = -\frac {2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d}+\frac {\left (\left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (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}{5 b^4 \left (a^2-b^2\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (2 a \left (5 A b^2+2 \left (4 a^2+b^2\right ) 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}{5 b^4 \sqrt {a+b \cos (c+d x)}} \\ & = \frac {2 \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (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 )}{5 b^4 \left (a^2-b^2\right ) d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (5 A b^2+2 \left (4 a^2+b^2\right ) 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 )}{5 b^4 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}-\frac {2 a \left (5 A b^2+8 a^2 C-3 b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^3 \left (a^2-b^2\right ) d}+\frac {2 \left (5 A b^2+6 a^2 C-b^2 C\right ) \cos (c+d x) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \\ \end{align*}

Mathematica [A] (verified)

Time = 2.76 (sec) , antiderivative size = 289, normalized size of antiderivative = 0.77 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\frac {\frac {2 a b^2 \left (5 A b^2+\left (4 a^2+b^2\right ) C\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )}{(a-b) (a+b)}+\frac {2 \left (2 a^2 b^2 (5 A-4 C)+16 a^4 C-b^4 (5 A+3 C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),\frac {2 b}{a+b}\right )\right )}{(a-b) (a+b)}+\frac {10 a^2 b \left (A b^2+a^2 C\right ) \sin (c+d x)}{-a^2+b^2}-6 a b C (a+b \cos (c+d x)) \sin (c+d x)+b^2 C (a+b \cos (c+d x)) \sin (2 (c+d x))}{5 b^4 d \sqrt {a+b \cos (c+d x)}} \]

[In]

Integrate[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^(3/2),x]

[Out]

((2*a*b^2*(5*A*b^2 + (4*a^2 + b^2)*C)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)]
)/((a - b)*(a + b)) + (2*(2*a^2*b^2*(5*A - 4*C) + 16*a^4*C - b^4*(5*A + 3*C))*Sqrt[(a + b*Cos[c + d*x])/(a + b
)]*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)]))/((a - b)*(a + b)
) + (10*a^2*b*(A*b^2 + a^2*C)*Sin[c + d*x])/(-a^2 + b^2) - 6*a*b*C*(a + b*Cos[c + d*x])*Sin[c + d*x] + b^2*C*(
a + b*Cos[c + d*x])*Sin[2*(c + d*x)])/(5*b^4*d*Sqrt[a + b*Cos[c + d*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1292\) vs. \(2(411)=822\).

Time = 20.86 (sec) , antiderivative size = 1293, normalized size of antiderivative = 3.45

method	result	size
default	\(\text {Expression too large to display}\)	\(1293\)
parts	\(\text {Expression too large to display}\)	\(1817\)

[In]

int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+cos(d*x+c)*b)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16*C/b*(-1/10/b*cos(1/2*d*x+1/2*c)^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)-1/60/b^2*(-4*a+12*b)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/
2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/60/b^2*(-4*a+12*b)*(a-b)*(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)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1
/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/60*(4*a^2-15*a*b+27*b^2)/b^3*(a-b)*(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)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos
(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))-8/b^2*C*(a+3*b)*(-1/6/b
*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/6*(a-b)/b*(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)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2
)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/12/b^2*(-2*a+6*b)*(a-b)*(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)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(E
llipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))))-2*a^2*(A*b^
2+C*a^2)/b^4/sin(1/2*d*x+1/2*c)^2/(2*b*sin(1/2*d*x+1/2*c)^2-a-b)/(a^2-b^2)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*si
n(1/2*d*x+1/2*c)^2)^(1/2)*(2*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2*b+EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b
))^(1/2))*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*a-EllipticE(cos(1/2
*d*x+1/2*c),(-2*b/(a-b))^(1/2))*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/
2)*b)-2/b^4*(A*b^2+C*a^2+2*C*a*b+3*C*b^2)*(a-b)*(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)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(
a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2)))-2*(A*a*b^2+A*b^3+C*a^3+C*a^2*b+C*a*b^2+C*b^3)/b
^4*(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)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*
sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(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 = 0.19 (sec) , antiderivative size = 818, normalized size of antiderivative = 2.18 \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/15*(6*(8*C*a^4*b^2 + (5*A - 3*C)*a^2*b^4 - (C*a^2*b^4 - C*b^6)*cos(d*x + c)^2 + 2*(C*a^3*b^3 - C*a*b^5)*cos
(d*x + c))*sqrt(b*cos(d*x + c) + a)*sin(d*x + c) + (sqrt(2)*(-32*I*C*a^5*b - 4*I*(5*A - 7*C)*a^3*b^3 + I*(25*A
+ 9*C)*a*b^5)*cos(d*x + c) + sqrt(2)*(-32*I*C*a^6 - 4*I*(5*A - 7*C)*a^4*b^2 + I*(25*A + 9*C)*a^2*b^4))*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(2)*(32*I*C*a^5*b + 4*I*(5*A - 7*C)*a^3*b^3 - I*(25*A + 9*C)*a*b^5)*cos(d*x + c) + s
qrt(2)*(32*I*C*a^6 + 4*I*(5*A - 7*C)*a^4*b^2 - I*(25*A + 9*C)*a^2*b^4))*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(2)*
(16*I*C*a^4*b^2 + 2*I*(5*A - 4*C)*a^2*b^4 - I*(5*A + 3*C)*b^6)*cos(d*x + c) + sqrt(2)*(16*I*C*a^5*b + 2*I*(5*A
- 4*C)*a^3*b^3 - I*(5*A + 3*C)*a*b^5))*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(2)*(-16*I*C*a^4*b^2 - 2*I*(5*A - 4*C)*a^2*b^4 + I*(5*A + 3*C)*b^6)*cos(d*x
+ c) + sqrt(2)*(-16*I*C*a^5*b - 2*I*(5*A - 4*C)*a^3*b^3 + I*(5*A + 3*C)*a*b^5))*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)))/((a^2*b^6 - b^8)*d*cos(d*x + c) + (a^3*b^5
- a*b^7)*d)

Sympy [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**2*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(3/2),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx=\int \frac {{\cos \left (c+d\,x\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

[In]

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

[Out]

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