3.712 \(\int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{9}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\)

Optimal. Leaf size=270 \[ \frac{2 \left (a^2 (5 A+7 C)+7 A b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 a^3 d}+\frac{2 b \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 d}+\frac{2 b^2 \left (a^2 C+A b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a+b)}+\frac{2 \left (a^2 (5 A+7 C)+7 A b^2\right ) \sin (c+d x)}{21 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{5 a^4 d \sqrt{\cos (c+d x)}}-\frac{2 A b \sin (c+d x)}{5 a^2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A \sin (c+d x)}{7 a d \cos ^{\frac{7}{2}}(c+d x)} \]

[Out]

(2*b*(5*A*b^2 + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*a^4*d) + (2*(7*A*b^2 + a^2*(5*A + 7*C))*Ellipti
cF[(c + d*x)/2, 2])/(21*a^3*d) + (2*b^2*(A*b^2 + a^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a^4*(a + b
)*d) + (2*A*Sin[c + d*x])/(7*a*d*Cos[c + d*x]^(7/2)) - (2*A*b*Sin[c + d*x])/(5*a^2*d*Cos[c + d*x]^(5/2)) + (2*
(7*A*b^2 + a^2*(5*A + 7*C))*Sin[c + d*x])/(21*a^3*d*Cos[c + d*x]^(3/2)) - (2*b*(5*A*b^2 + a^2*(3*A + 5*C))*Sin
[c + d*x])/(5*a^4*d*Sqrt[Cos[c + d*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.49436, antiderivative size = 270, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3056, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac{2 \left (a^2 (5 A+7 C)+7 A b^2\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 a^3 d}+\frac{2 b \left (a^2 (3 A+5 C)+5 A b^2\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 d}+\frac{2 b^2 \left (a^2 C+A b^2\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a+b)}+\frac{2 \left (a^2 (5 A+7 C)+7 A b^2\right ) \sin (c+d x)}{21 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x)}{5 a^4 d \sqrt{\cos (c+d x)}}-\frac{2 A b \sin (c+d x)}{5 a^2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 A \sin (c+d x)}{7 a d \cos ^{\frac{7}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*(a + b*Cos[c + d*x])),x]

[Out]

(2*b*(5*A*b^2 + a^2*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*a^4*d) + (2*(7*A*b^2 + a^2*(5*A + 7*C))*Ellipti
cF[(c + d*x)/2, 2])/(21*a^3*d) + (2*b^2*(A*b^2 + a^2*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a^4*(a + b
)*d) + (2*A*Sin[c + d*x])/(7*a*d*Cos[c + d*x]^(7/2)) - (2*A*b*Sin[c + d*x])/(5*a^2*d*Cos[c + d*x]^(5/2)) + (2*
(7*A*b^2 + a^2*(5*A + 7*C))*Sin[c + d*x])/(21*a^3*d*Cos[c + d*x]^(3/2)) - (2*b*(5*A*b^2 + a^2*(3*A + 5*C))*Sin
[c + d*x])/(5*a^4*d*Sqrt[Cos[c + d*x]])

Rule 3056

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[((A*b^2 + a^2*C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c +
d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), I
nt[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(m + 1)*(b*c - a*d)*(A + C) + d*(A*b^2 + a^2*C)*
(m + n + 2) - (c*(A*b^2 + a^2*C) + b*(m + 1)*(b*c - a*d)*(A + C))*Sin[e + f*x] - d*(A*b^2 + a^2*C)*(m + n + 3)
*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] && LtQ[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ
[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))

Rule 3055

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

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

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 3002

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

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 2805

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

Rubi steps

\begin{align*} \int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{9}{2}}(c+d x) (a+b \cos (c+d x))} \, dx &=\frac{2 A \sin (c+d x)}{7 a d \cos ^{\frac{7}{2}}(c+d x)}+\frac{2 \int \frac{-\frac{7 A b}{2}+\frac{1}{2} a (5 A+7 C) \cos (c+d x)+\frac{5}{2} A b \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{7 a}\\ &=\frac{2 A \sin (c+d x)}{7 a d \cos ^{\frac{7}{2}}(c+d x)}-\frac{2 A b \sin (c+d x)}{5 a^2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{4 \int \frac{\frac{5}{4} \left (7 A b^2+a^2 (5 A+7 C)\right )+a A b \cos (c+d x)-\frac{21}{4} A b^2 \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{35 a^2}\\ &=\frac{2 A \sin (c+d x)}{7 a d \cos ^{\frac{7}{2}}(c+d x)}-\frac{2 A b \sin (c+d x)}{5 a^2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (7 A b^2+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 a^3 d \cos ^{\frac{3}{2}}(c+d x)}+\frac{8 \int \frac{-\frac{21}{8} b \left (5 A b^2+a^2 (3 A+5 C)\right )-\frac{1}{8} a \left (28 A b^2-5 a^2 (5 A+7 C)\right ) \cos (c+d x)+\frac{5}{8} b \left (7 A b^2+a^2 (5 A+7 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{105 a^3}\\ &=\frac{2 A \sin (c+d x)}{7 a d \cos ^{\frac{7}{2}}(c+d x)}-\frac{2 A b \sin (c+d x)}{5 a^2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (7 A b^2+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (5 A b^2+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 a^4 d \sqrt{\cos (c+d x)}}+\frac{16 \int \frac{\frac{5}{16} \left (21 A b^4+7 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)\right )+\frac{1}{4} a b \left (35 A b^2+a^2 (22 A+35 C)\right ) \cos (c+d x)+\frac{21}{16} b^2 \left (5 A b^2+a^2 (3 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{105 a^4}\\ &=\frac{2 A \sin (c+d x)}{7 a d \cos ^{\frac{7}{2}}(c+d x)}-\frac{2 A b \sin (c+d x)}{5 a^2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (7 A b^2+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (5 A b^2+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 a^4 d \sqrt{\cos (c+d x)}}-\frac{16 \int \frac{-\frac{5}{16} b \left (21 A b^4+7 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)\right )-\frac{5}{16} a b^2 \left (7 A b^2+a^2 (5 A+7 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{105 a^4 b}+\frac{\left (b \left (5 A b^2+a^2 (3 A+5 C)\right )\right ) \int \sqrt{\cos (c+d x)} \, dx}{5 a^4}\\ &=\frac{2 b \left (5 A b^2+a^2 (3 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 d}+\frac{2 A \sin (c+d x)}{7 a d \cos ^{\frac{7}{2}}(c+d x)}-\frac{2 A b \sin (c+d x)}{5 a^2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (7 A b^2+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (5 A b^2+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 a^4 d \sqrt{\cos (c+d x)}}+\frac{\left (b^2 \left (A b^2+a^2 C\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{a^4}+\frac{\left (7 A b^2+a^2 (5 A+7 C)\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{21 a^3}\\ &=\frac{2 b \left (5 A b^2+a^2 (3 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 d}+\frac{2 \left (7 A b^2+a^2 (5 A+7 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{21 a^3 d}+\frac{2 b^2 \left (A b^2+a^2 C\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 (a+b) d}+\frac{2 A \sin (c+d x)}{7 a d \cos ^{\frac{7}{2}}(c+d x)}-\frac{2 A b \sin (c+d x)}{5 a^2 d \cos ^{\frac{5}{2}}(c+d x)}+\frac{2 \left (7 A b^2+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 a^3 d \cos ^{\frac{3}{2}}(c+d x)}-\frac{2 b \left (5 A b^2+a^2 (3 A+5 C)\right ) \sin (c+d x)}{5 a^4 d \sqrt{\cos (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 3.92258, size = 342, normalized size = 1.27 \[ \frac{\frac{\left (7 a^2 b^2 (19 A+45 C)+10 a^4 (5 A+7 C)+315 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{8 \left (a^3 (22 A+35 C)+35 a A b^2\right ) \left ((a+b) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )\right )}{a+b}+\frac{5 \left (\left (a^3 (5 A+7 C)+7 a A b^2\right ) \sin (2 (c+d x))+6 a^3 A \tan (c+d x)\right )-42 b \sin (c+d x) \left (\left (a^2 (3 A+5 C)+5 A b^2\right ) \cos ^2(c+d x)+a^2 A\right )}{\cos ^{\frac{5}{2}}(c+d x)}+\frac{21 \left (a^2 (3 A+5 C)+5 A b^2\right ) \sin (c+d x) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a \sqrt{\sin ^2(c+d x)}}}{105 a^4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(9/2)*(a + b*Cos[c + d*x])),x]

[Out]

(((315*A*b^4 + 10*a^4*(5*A + 7*C) + 7*a^2*b^2*(19*A + 45*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b
) + (8*(35*a*A*b^2 + a^3*(22*A + 35*C))*((a + b)*EllipticF[(c + d*x)/2, 2] - a*EllipticPi[(2*b)/(a + b), (c +
d*x)/2, 2]))/(a + b) + (21*(5*A*b^2 + a^2*(3*A + 5*C))*(-2*a*b*EllipticE[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a
*(a + b)*EllipticF[ArcSin[Sqrt[Cos[c + d*x]]], -1] + (2*a^2 - b^2)*EllipticPi[-(b/a), -ArcSin[Sqrt[Cos[c + d*x
]]], -1])*Sin[c + d*x])/(a*Sqrt[Sin[c + d*x]^2]) + (-42*b*(a^2*A + (5*A*b^2 + a^2*(3*A + 5*C))*Cos[c + d*x]^2)
*Sin[c + d*x] + 5*((7*a*A*b^2 + a^3*(5*A + 7*C))*Sin[2*(c + d*x)] + 6*a^3*A*Tan[c + d*x]))/Cos[c + d*x]^(5/2))
/(105*a^4*d)

________________________________________________________________________________________

Maple [B]  time = 2.205, size = 982, normalized size = 3.6 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+b*cos(d*x+c)),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-4*(A*b^2+C*a^2)*b^3/a^4/(-2*a*b+2*b^2)*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli
pticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2/5*A/a^2*b/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*si
n(1/2*d*x+1/2*c)^2-1)/sin(1/2*d*x+1/2*c)^2*(12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1
/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-12*Ellipt
icE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*sin(1/2*d*x+1/2*
c)^2+24*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+3*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(sin(1/2*d*x+1/2*c)^2)
^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)-8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))*(-2*sin(1/2*d*x+1/2*c)^4+si
n(1/2*d*x+1/2*c)^2)^(1/2)+2*(A*b^2+C*a^2)/a^3*(-1/6*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/
2*c)^2)^(1/2)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+1/3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)
/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2)))+2*A/a*(-1/56*cos(
1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1/2+cos(1/2*d*x+1/2*c)^2)^4-5/42*cos(1/2
*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(-1/2+cos(1/2*d*x+1/2*c)^2)^2+5/21*(sin(1/2*d
*x+1/2*c)^2)^(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2)))-2*(A*b^2+C*a^2)/a^4*b*(-(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^
2-1)^(1/2)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+2*(-2*si
n(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*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/(
2*sin(1/2*d*x+1/2*c)^2-1))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

________________________________________________________________________________________

Maxima [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)**2)/cos(d*x+c)**(9/2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac{9}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(9/2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)/((b*cos(d*x + c) + a)*cos(d*x + c)^(9/2)), x)