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

[Out]

1/5*(35*A*b^4-3*a^2*b^2*(8*A-5*C)-2*a^4*(3*A+5*C))*(cos(1/2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(s
in(1/2*d*x+1/2*c),2^(1/2))/a^4/(a^2-b^2)/d+1/3*b*(7*A*b^2-a^2*(4*A-3*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))/a^3/(a^2-b^2)/d+b*(7*A*b^4-3*a^2*b^2*(3*A-C)-5*a^4*C)*(cos(1/
2*d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))/a^4/(a-b)/(a+b)^2/d-
1/5*(7*A*b^2-a^2*(2*A-5*C))*sin(d*x+c)/a^2/(a^2-b^2)/d/cos(d*x+c)^(5/2)+1/3*b*(7*A*b^2-a^2*(4*A-3*C))*sin(d*x+
c)/a^3/(a^2-b^2)/d/cos(d*x+c)^(3/2)+(A*b^2+C*a^2)*sin(d*x+c)/a/(a^2-b^2)/d/cos(d*x+c)^(5/2)/(a+b*cos(d*x+c))-1
/5*(35*A*b^4-3*a^2*b^2*(8*A-5*C)-2*a^4*(3*A+5*C))*sin(d*x+c)/a^4/(a^2-b^2)/d/cos(d*x+c)^(1/2)

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Rubi [A]  time = 1.81, antiderivative size = 427, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 35, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, 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 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 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]

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 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 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 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]

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*}

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Mathematica [A]  time = 7.01, size = 405, normalized size = 0.95 \[ \frac {4 \sqrt {\cos (c+d x)} \left (2 \tan (c+d x) \left (3 a^2 A \sec ^2(c+d x)+3 a^2 (3 A+5 C)-10 a A b \sec (c+d x)+45 A b^2\right )+\frac {15 \left (a^2 b^3 C+A b^5\right ) \sin (c+d x)}{\left (b^2-a^2\right ) (a+b \cos (c+d x))}\right )-\frac {-\frac {8 \left (3 a^5 (3 A+5 C)+2 a^3 b^2 (23 A-15 C)-70 a A b^4\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{b (a+b)}-\frac {2 \left (2 a^4 b (29 A+75 C)+a^2 b^3 (272 A-135 C)-315 A b^5\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}-\frac {6 \left (2 a^4 (3 A+5 C)+3 a^2 b^2 (8 A-5 C)-35 A b^4\right ) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt {\sin ^2(c+d x)}}}{(b-a) (a+b)}}{60 a^4 d} \]

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*(-315*A*b^5 + a^2*b^3*(272*A - 135*C) + 2*a^4*b*(29*A + 75*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2,
2])/(a + b) - (8*(-70*a*A*b^4 + 2*a^3*b^2*(23*A - 15*C) + 3*a^5*(3*A + 5*C))*((a + b)*EllipticF[(c + d*x)/2, 2
] - a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(b*(a + b)) - (6*(-35*A*b^4 + 3*a^2*b^2*(8*A - 5*C) + 2*a^4*
(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*b*Sqrt[Sin[c +
d*x]^2]))/((-a + b)*(a + b))) + 4*Sqrt[Cos[c + d*x]]*((15*(A*b^5 + a^2*b^3*C)*Sin[c + d*x])/((-a^2 + b^2)*(a +
b*Cos[c + d*x])) + 2*(45*A*b^2 + 3*a^2*(3*A + 5*C) - 10*a*A*b*Sec[c + d*x] + 3*a^2*A*Sec[c + d*x]^2)*Tan[c +
d*x]))/(60*a^4*d)

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fricas [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((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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )}^{2} \cos \left (d x + c\right )^{\frac {7}{2}}}\,{d x} \]

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)

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maple [B]  time = 12.68, size = 1353, normalized size = 3.17 \[ \text {result too large to display} \]

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)*(-2*(A*b^2+C*a^2)*b/a^3*(-b^2/a/(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*b/a/(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*b/a/
(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)*EllipticPi(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)))-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))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*
sin(1/2*d*x+1/2*c)^4-24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(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*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(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)+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)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))+2*(3*A*b^2+C*a^2)/a^4*(-
(-2*sin(1/2*d*x+1/2*c)^4+sin(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)*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))))/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 = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^{7/2}\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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