3.664 \(\int \frac {\cos ^2(c+d x) (A+C \cos ^2(c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=392 \[ -\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}+\frac {2 \left (16 a^4 C+2 a^2 b^2 (A-8 C)-b^4 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {4 a \left (8 a^4 C+a^2 b^2 (A-14 C)-b^4 (3 A-4 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {4 a \left (-3 a^4 C+5 a^2 b^2 C+2 A b^4\right ) \sin (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}} \]

[Out]

-2/3*(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))^(3/2)-4/3*a*(2*A*b^4-3*C*a^4+5*C*a^2
*b^2)*sin(d*x+c)/b^3/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)+2/3*(A*b^2+2*C*a^2-C*b^2)*sin(d*x+c)*(a+b*cos(d*x+c)
)^(1/2)/b^3/(a^2-b^2)/d-4/3*a*(a^2*b^2*(A-14*C)-b^4*(3*A-4*C)+8*a^4*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)^2/d/((a+b*
cos(d*x+c))/(a+b))^(1/2)+2/3*(2*a^2*b^2*(A-8*C)+16*a^4*C-b^4*(3*A+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/(a^2-b^2)/d/(
a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.83, antiderivative size = 392, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {3048, 3031, 3023, 2752, 2663, 2661, 2655, 2653} \[ -\frac {2 \left (a^2 C+A b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}+\frac {2 \left (2 a^2 C+A b^2-b^2 C\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}-\frac {4 a \left (5 a^2 b^2 C-3 a^4 C+2 A b^4\right ) \sin (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (2 a^2 b^2 (A-8 C)+16 a^4 C-b^4 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}-\frac {4 a \left (a^2 b^2 (A-14 C)+8 a^4 C-b^4 (3 A-4 C)\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(-4*a*(a^2*b^2*(A - 14*C) - b^4*(3*A - 4*C) + 8*a^4*C)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(
a + b)])/(3*b^4*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) + (2*(2*a^2*b^2*(A - 8*C) + 16*a^4*C - b^4
*(3*A + C))*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(3*b^4*(a^2 - b^2)*d*Sqr
t[a + b*Cos[c + d*x]]) - (2*(A*b^2 + a^2*C)*Cos[c + d*x]^2*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x
])^(3/2)) - (4*a*(2*A*b^4 - 3*a^4*C + 5*a^2*b^2*C)*Sin[c + d*x])/(3*b^3*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x
]]) + (2*(A*b^2 + 2*a^2*C - b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(3*b^3*(a^2 - b^2)*d)

Rule 2653

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

Rule 2655

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

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

Rule 2663

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*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

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 3023

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 3031

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

Rule 3048

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*Sin[
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]

Rubi steps

\begin {align*} \int \frac {\cos ^2(c+d x) \left (A+C \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{5/2}} \, dx &=-\frac {2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \int \frac {\cos (c+d x) \left (2 \left (A b^2+a^2 C\right )-\frac {3}{2} a b (A+C) \cos (c+d x)-\frac {3}{2} \left (A b^2+2 a^2 C-b^2 C\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 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)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {4 \int \frac {-\frac {1}{2} b \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right )-\frac {1}{2} a \left (2 A b^4-\left (6 a^4-11 a^2 b^2+3 b^4\right ) C\right ) \cos (c+d x)-\frac {3}{4} b \left (a^2-b^2\right ) \left (A b^2+2 a^2 C-b^2 C\right ) \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {\frac {3}{8} b^2 \left (4 a^4 C-b^4 (3 A+C)-a^2 b^2 (A+7 C)\right )+\frac {3}{4} a b \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 C\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{9 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac {2 \left (A b^2+a^2 C\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {\left (2 a \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 C\right )\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{3 b^4 \left (a^2-b^2\right )^2}+\frac {\left (2 a^2 b^2 (A-8 C)+16 a^4 C-b^4 (3 A+C)\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 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)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {\left (2 a \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 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}{3 b^4 \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {\left (\left (2 a^2 b^2 (A-8 C)+16 a^4 C-b^4 (3 A+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}{3 b^4 \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\\ &=-\frac {4 a \left (a^2 b^2 (A-14 C)-b^4 (3 A-4 C)+8 a^4 C\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 \left (2 a^2 b^2 (A-8 C)+16 a^4 C-b^4 (3 A+C)\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 \left (a^2-b^2\right ) 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)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {4 a \left (2 A b^4-3 a^4 C+5 a^2 b^2 C\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (A b^2+2 a^2 C-b^2 C\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [A]  time = 2.79, size = 306, normalized size = 0.78 \[ \frac {2 \left (\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (b \left (-4 a^4 b C+a^2 b^3 (A+7 C)+b^5 (3 A+C)\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-2 a \left (8 a^4 C+a^2 b^2 (A-14 C)+b^4 (4 C-3 A)\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )\right )}{(a-b)^2 (a+b)}+\frac {b \sin (c+d x) \left (16 a^6 C+2 a^4 A b^2-25 a^4 b^2 C-10 a^2 A b^4+C \left (b^3-a^2 b\right )^2 \cos (2 (c+d x))+4 a b \left (5 a^4 C+a^2 b^2 (A-8 C)+b^4 (C-3 A)\right ) \cos (c+d x)+b^6 C\right )}{2 \left (a^2-b^2\right )^2}\right )}{3 b^4 d (a+b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^2*(A + C*Cos[c + d*x]^2))/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(2*((((a + b*Cos[c + d*x])/(a + b))^(3/2)*(b*(-4*a^4*b*C + b^5*(3*A + C) + a^2*b^3*(A + 7*C))*EllipticF[(c + d
*x)/2, (2*b)/(a + b)] - 2*a*(a^2*b^2*(A - 14*C) + 8*a^4*C + b^4*(-3*A + 4*C))*((a + b)*EllipticE[(c + d*x)/2,
(2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])))/((a - b)^2*(a + b)) + (b*(2*a^4*A*b^2 - 10*a^2*A*b
^4 + 16*a^6*C - 25*a^4*b^2*C + b^6*C + 4*a*b*(a^2*b^2*(A - 8*C) + 5*a^4*C + b^4*(-3*A + C))*Cos[c + d*x] + (-(
a^2*b) + b^3)^2*C*Cos[2*(c + d*x)])*Sin[c + d*x])/(2*(a^2 - b^2)^2)))/(3*b^4*d*(a + b*Cos[c + d*x])^(3/2))

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fricas [F]  time = 1.53, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (C \cos \left (d x + c\right )^{4} + A \cos \left (d x + c\right )^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

integral((C*cos(d*x + c)^4 + A*cos(d*x + c)^2)*sqrt(b*cos(d*x + c) + a)/(b^3*cos(d*x + c)^3 + 3*a*b^2*cos(d*x
+ c)^2 + 3*a^2*b*cos(d*x + c) + a^3), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^2/(b*cos(d*x + c) + a)^(5/2), x)

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maple [B]  time = 9.93, size = 1323, normalized size = 3.38 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2),x)

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (C \cos \left (d x + c\right )^{2} + A\right )} \cos \left (d x + c\right )^{2}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(A+C*cos(d*x+c)^2)/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)*cos(d*x + c)^2/(b*cos(d*x + c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^2\,\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^2*(A + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(5/2),x)

[Out]

int((cos(c + d*x)^2*(A + C*cos(c + d*x)^2))/(a + b*cos(c + d*x))^(5/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**2*(A+C*cos(d*x+c)**2)/(a+b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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