3.720 \(\int \frac{A+C \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\)

Optimal. Leaf size=427 \[ \frac{b \left (7 A b^2-a^2 (4 A-3 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^3 d \left (a^2-b^2\right )}+\frac{\left (-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)+35 A b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 d \left (a^2-b^2\right )}+\frac{b \left (-3 a^2 b^2 (3 A-C)-5 a^4 C+7 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a-b) (a+b)^2}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{b \left (7 A b^2-a^2 (4 A-3 C)\right ) \sin (c+d x)}{3 a^3 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (7 A b^2-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x)}-\frac{\left (-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)+35 A b^4\right ) \sin (c+d x)}{5 a^4 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}} \]

[Out]

((35*A*b^4 - 3*a^2*b^2*(8*A - 5*C) - 2*a^4*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*a^4*(a^2 - b^2)*d) + (b*
(7*A*b^2 - a^2*(4*A - 3*C))*EllipticF[(c + d*x)/2, 2])/(3*a^3*(a^2 - b^2)*d) + (b*(7*A*b^4 - 3*a^2*b^2*(3*A -
C) - 5*a^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a^4*(a - b)*(a + b)^2*d) - ((7*A*b^2 - a^2*(2*A - 5*
C))*Sin[c + d*x])/(5*a^2*(a^2 - b^2)*d*Cos[c + d*x]^(5/2)) + (b*(7*A*b^2 - a^2*(4*A - 3*C))*Sin[c + d*x])/(3*a
^3*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)) - ((35*A*b^4 - 3*a^2*b^2*(8*A - 5*C) - 2*a^4*(3*A + 5*C))*Sin[c + d*x])/(
5*a^4*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]) + ((A*b^2 + a^2*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c + d*x]^(5/2)*(
a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.80726, antiderivative size = 427, 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} \[ \frac{b \left (7 A b^2-a^2 (4 A-3 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^3 d \left (a^2-b^2\right )}+\frac{\left (-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)+35 A b^4\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 d \left (a^2-b^2\right )}+\frac{b \left (-3 a^2 b^2 (3 A-C)-5 a^4 C+7 A b^4\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 d (a-b) (a+b)^2}+\frac{\left (a^2 C+A b^2\right ) \sin (c+d x)}{a d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{b \left (7 A b^2-a^2 (4 A-3 C)\right ) \sin (c+d x)}{3 a^3 d \left (a^2-b^2\right ) \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (7 A b^2-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 d \left (a^2-b^2\right ) \cos ^{\frac{5}{2}}(c+d x)}-\frac{\left (-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)+35 A b^4\right ) \sin (c+d x)}{5 a^4 d \left (a^2-b^2\right ) \sqrt{\cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + b*Cos[c + d*x])^2),x]

[Out]

((35*A*b^4 - 3*a^2*b^2*(8*A - 5*C) - 2*a^4*(3*A + 5*C))*EllipticE[(c + d*x)/2, 2])/(5*a^4*(a^2 - b^2)*d) + (b*
(7*A*b^2 - a^2*(4*A - 3*C))*EllipticF[(c + d*x)/2, 2])/(3*a^3*(a^2 - b^2)*d) + (b*(7*A*b^4 - 3*a^2*b^2*(3*A -
C) - 5*a^4*C)*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a^4*(a - b)*(a + b)^2*d) - ((7*A*b^2 - a^2*(2*A - 5*
C))*Sin[c + d*x])/(5*a^2*(a^2 - b^2)*d*Cos[c + d*x]^(5/2)) + (b*(7*A*b^2 - a^2*(4*A - 3*C))*Sin[c + d*x])/(3*a
^3*(a^2 - b^2)*d*Cos[c + d*x]^(3/2)) - ((35*A*b^4 - 3*a^2*b^2*(8*A - 5*C) - 2*a^4*(3*A + 5*C))*Sin[c + d*x])/(
5*a^4*(a^2 - b^2)*d*Sqrt[Cos[c + d*x]]) + ((A*b^2 + a^2*C)*Sin[c + d*x])/(a*(a^2 - b^2)*d*Cos[c + d*x]^(5/2)*(
a + b*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{7}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx &=\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\int \frac{\frac{1}{2} \left (-7 A b^2+2 a^2 \left (A-\frac{5 C}{2}\right )\right )-a b (A+C) \cos (c+d x)+\frac{5}{2} \left (A b^2+a^2 C\right ) \cos ^2(c+d x)}{\cos ^{\frac{7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{a \left (a^2-b^2\right )}\\ &=-\frac{\left (7 A b^2-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x)}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{2 \int \frac{\frac{5}{4} b \left (7 A b^2-a^2 (4 A-3 C)\right )+\frac{1}{2} a \left (2 A b^2+a^2 (3 A+5 C)\right ) \cos (c+d x)-\frac{3}{4} b \left (7 A b^2-a^2 (2 A-5 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{5 a^2 \left (a^2-b^2\right )}\\ &=-\frac{\left (7 A b^2-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x)}+\frac{b \left (7 A b^2-a^2 (4 A-3 C)\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{4 \int \frac{-\frac{3}{8} \left (35 A b^4-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right )-\frac{1}{4} a b \left (14 A b^2+a^2 (A+15 C)\right ) \cos (c+d x)+\frac{5}{8} b^2 \left (7 A b^2-a^2 (4 A-3 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac{3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{15 a^3 \left (a^2-b^2\right )}\\ &=-\frac{\left (7 A b^2-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x)}+\frac{b \left (7 A b^2-a^2 (4 A-3 C)\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (35 A b^4-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right ) \sin (c+d x)}{5 a^4 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{8 \int \frac{\frac{5}{16} b \left (21 A b^4-a^2 b^2 (20 A-9 C)-4 a^4 (A+3 C)\right )+\frac{1}{8} a \left (70 A b^4-2 a^2 b^2 (23 A-15 C)-3 a^4 (3 A+5 C)\right ) \cos (c+d x)+\frac{3}{16} b \left (35 A b^4-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 a^4 \left (a^2-b^2\right )}\\ &=-\frac{\left (7 A b^2-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x)}+\frac{b \left (7 A b^2-a^2 (4 A-3 C)\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (35 A b^4-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right ) \sin (c+d x)}{5 a^4 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))}-\frac{8 \int \frac{-\frac{5}{16} b^2 \left (21 A b^4-a^2 b^2 (20 A-9 C)-4 a^4 (A+3 C)\right )-\frac{5}{16} a b^3 \left (7 A b^2-a^2 (4 A-3 C)\right ) \cos (c+d x)}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 a^4 b \left (a^2-b^2\right )}+\frac{\left (35 A b^4-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right ) \int \sqrt{\cos (c+d x)} \, dx}{10 a^4 \left (a^2-b^2\right )}\\ &=\frac{\left (35 A b^4-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 \left (a^2-b^2\right ) d}-\frac{\left (7 A b^2-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x)}+\frac{b \left (7 A b^2-a^2 (4 A-3 C)\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (35 A b^4-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right ) \sin (c+d x)}{5 a^4 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))}+\frac{\left (b \left (7 A b^2-a^2 (4 A-3 C)\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)}} \, dx}{6 a^3 \left (a^2-b^2\right )}+\frac{\left (b \left (7 A b^4-3 a^2 b^2 (3 A-C)-5 a^4 C\right )\right ) \int \frac{1}{\sqrt{\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{2 a^4 \left (a^2-b^2\right )}\\ &=\frac{\left (35 A b^4-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right ) E\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{5 a^4 \left (a^2-b^2\right ) d}+\frac{b \left (7 A b^2-a^2 (4 A-3 C)\right ) F\left (\left .\frac{1}{2} (c+d x)\right |2\right )}{3 a^3 \left (a^2-b^2\right ) d}+\frac{b \left (7 A b^4-3 a^2 b^2 (3 A-C)-5 a^4 C\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a^4 (a-b) (a+b)^2 d}-\frac{\left (7 A b^2-a^2 (2 A-5 C)\right ) \sin (c+d x)}{5 a^2 \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x)}+\frac{b \left (7 A b^2-a^2 (4 A-3 C)\right ) \sin (c+d x)}{3 a^3 \left (a^2-b^2\right ) d \cos ^{\frac{3}{2}}(c+d x)}-\frac{\left (35 A b^4-3 a^2 b^2 (8 A-5 C)-2 a^4 (3 A+5 C)\right ) \sin (c+d x)}{5 a^4 \left (a^2-b^2\right ) d \sqrt{\cos (c+d x)}}+\frac{\left (A b^2+a^2 C\right ) \sin (c+d x)}{a \left (a^2-b^2\right ) d \cos ^{\frac{5}{2}}(c+d x) (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [A]  time = 7.01756, size = 496, normalized size = 1.16 \[ \frac{\sqrt{\cos (c+d x)} \left (\frac{-a^2 b^3 C \sin (c+d x)-A b^5 \sin (c+d x)}{a^4 \left (a^2-b^2\right ) (a+b \cos (c+d x))}+\frac{2 \sec (c+d x) \left (3 a^2 A \sin (c+d x)+5 a^2 C \sin (c+d x)+15 A b^2 \sin (c+d x)\right )}{5 a^4}-\frac{4 A b \tan (c+d x) \sec (c+d x)}{3 a^3}+\frac{2 A \tan (c+d x) \sec ^2(c+d x)}{5 a^2}\right )}{d}-\frac{\frac{2 \left (-272 a^2 A b^3-58 a^4 A b+135 a^2 b^3 C-150 a^4 b C+315 A b^5\right ) \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}+\frac{\left (-184 a^3 A b^2-36 a^5 A+120 a^3 b^2 C-60 a^5 C+280 a A b^4\right ) \left (2 F\left (\left .\frac{1}{2} (c+d x)\right |2\right )-\frac{2 a \Pi \left (\frac{2 b}{a+b};\left .\frac{1}{2} (c+d x)\right |2\right )}{a+b}\right )}{b}+\frac{2 \left (-72 a^2 A b^3-18 a^4 A b+45 a^2 b^3 C-30 a^4 b C+105 A b^5\right ) \sin (c+d x) \cos (2 (c+d x)) \left (\left (2 a^2-b^2\right ) \Pi \left (-\frac{b}{a};\left .-\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt{\cos (c+d x)}\right )\right |-1\right )\right )}{a b^2 \sqrt{1-\cos ^2(c+d x)} \left (2 \cos ^2(c+d x)-1\right )}}{60 a^4 d (b-a) (a+b)} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(7/2)*(a + b*Cos[c + d*x])^2),x]

[Out]

-((2*(-58*a^4*A*b - 272*a^2*A*b^3 + 315*A*b^5 - 150*a^4*b*C + 135*a^2*b^3*C)*EllipticPi[(2*b)/(a + b), (c + d*
x)/2, 2])/(a + b) + ((-36*a^5*A - 184*a^3*A*b^2 + 280*a*A*b^4 - 60*a^5*C + 120*a^3*b^2*C)*(2*EllipticF[(c + d*
x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b)))/b + (2*(-18*a^4*A*b - 72*a^2*A*b^3 + 105*
A*b^5 - 30*a^4*b*C + 45*a^2*b^3*C)*Cos[2*(c + d*x)]*(-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*b^2*Sqrt[1 - Cos[c + d*x]^2]*(-1 + 2*Cos[c + d*x]^2)))/(60*a^4*(-a + b)*(a + b)*d) + (
Sqrt[Cos[c + d*x]]*((2*Sec[c + d*x]*(3*a^2*A*Sin[c + d*x] + 15*A*b^2*Sin[c + d*x] + 5*a^2*C*Sin[c + d*x]))/(5*
a^4) + (-(A*b^5*Sin[c + d*x]) - a^2*b^3*C*Sin[c + d*x])/(a^4*(a^2 - b^2)*(a + b*Cos[c + d*x])) - (4*A*b*Sec[c
+ d*x]*Tan[c + d*x])/(3*a^3) + (2*A*Sec[c + d*x]^2*Tan[c + d*x])/(5*a^2)))/d

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Maple [B]  time = 2.956, size = 1353, normalized size = 3.2 \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)^(7/2)/(a+b*cos(d*x+c))^2,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*b^2*(3*A*b^2+C*a^2)/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)*Ell
ipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))-2/5*A/a^2/(8*sin(1/2*d*x+1/2*c)^6-12*sin(1/2*d*x+1/2*c)^4+6*sin
(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*Ellipti
cE(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+sin
(1/2*d*x+1/2*c)^2)^(1/2)+2*(3*A*b^2+C*a^2)/a^4*(-(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*sin(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)-4*A/a^3*b*(-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*b^2+C*a^2)*b/a^3*(-1/a*b^2/(a^2-
b^2)*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)/(2*cos(1/2*d*x+1/2*c)^2*b+a-b)-1/
2/(a+b)/a*(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))-1/2/a*b/(a^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(
1/2))+1/2/a*b/(a^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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))-3*a/(a^2-b^2)/(-2*a*b+2*b^2)*b*(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)*Ellip
ticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+1/a/(a^2-b^2)/(-2*a*b+2*b^2)*b^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)*EllipticPi(cos(1/2*d*x+1/2
*c),-2*b/(a-b),2^(1/2))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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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)^(7/2)/(a+b*cos(d*x+c))^2,x, algorithm="maxima")

[Out]

Timed out

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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)^(7/2)/(a+b*cos(d*x+c))^2,x, algorithm="fricas")

[Out]

Timed out

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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)**(7/2)/(a+b*cos(d*x+c))**2,x)

[Out]

Timed out

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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 )}^{2} \cos \left (d x + c\right )^{\frac{7}{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)^(7/2)/(a+b*cos(d*x+c))^2,x, algorithm="giac")

[Out]

integrate((C*cos(d*x + c)^2 + A)/((b*cos(d*x + c) + a)^2*cos(d*x + c)^(7/2)), x)