3.713 \(\int \frac {A+C \cos ^2(c+d x)}{\cos ^{\frac {11}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\)

Optimal. Leaf size=344 \[ -\frac {2 A b \sin (c+d x)}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}-\frac {2 b^3 \left (a^2 C+A b^2\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^5 d (a+b)}-\frac {2 b \left (a^2 (5 A+7 C)+7 A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 a^4 d}-\frac {2 b \left (a^2 (5 A+7 C)+7 A b^2\right ) \sin (c+d x)}{21 a^4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2 (7 A+9 C)+9 A b^2\right ) \sin (c+d x)}{45 a^3 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2 \left (a^4 (7 A+9 C)+3 a^2 b^2 (3 A+5 C)+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^5 d}+\frac {2 \left (a^4 (7 A+9 C)+3 a^2 b^2 (3 A+5 C)+15 A b^4\right ) \sin (c+d x)}{15 a^5 d \sqrt {\cos (c+d x)}}+\frac {2 A \sin (c+d x)}{9 a d \cos ^{\frac {9}{2}}(c+d x)} \]

[Out]

-2/15*(15*A*b^4+3*a^2*b^2*(3*A+5*C)+a^4*(7*A+9*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^5/d-2/21*b*(7*A*b^2+a^2*(5*A+7*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^4/d-2*b^3*(A*b^2+C*a^2)*(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^5/(a+b)/d+2/9*A*sin(d*x+c)/a/d/cos(d*x+c)^(9/2)-2/7*A*b
*sin(d*x+c)/a^2/d/cos(d*x+c)^(7/2)+2/45*(9*A*b^2+a^2*(7*A+9*C))*sin(d*x+c)/a^3/d/cos(d*x+c)^(5/2)-2/21*b*(7*A*
b^2+a^2*(5*A+7*C))*sin(d*x+c)/a^4/d/cos(d*x+c)^(3/2)+2/15*(15*A*b^4+3*a^2*b^2*(3*A+5*C)+a^4*(7*A+9*C))*sin(d*x
+c)/a^5/d/cos(d*x+c)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.94, antiderivative size = 344, normalized size of antiderivative = 1.00, number of steps used = 10, 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} \[ -\frac {2 b \left (a^2 (5 A+7 C)+7 A b^2\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 a^4 d}-\frac {2 \left (3 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)+15 A b^4\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^5 d}-\frac {2 b^3 \left (a^2 C+A b^2\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^5 d (a+b)}-\frac {2 b \left (a^2 (5 A+7 C)+7 A b^2\right ) \sin (c+d x)}{21 a^4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (a^2 (7 A+9 C)+9 A b^2\right ) \sin (c+d x)}{45 a^3 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \left (3 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)+15 A b^4\right ) \sin (c+d x)}{15 a^5 d \sqrt {\cos (c+d x)}}-\frac {2 A b \sin (c+d x)}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 A \sin (c+d x)}{9 a d \cos ^{\frac {9}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(11/2)*(a + b*Cos[c + d*x])),x]

[Out]

(-2*(15*A*b^4 + 3*a^2*b^2*(3*A + 5*C) + a^4*(7*A + 9*C))*EllipticE[(c + d*x)/2, 2])/(15*a^5*d) - (2*b*(7*A*b^2
+ a^2*(5*A + 7*C))*EllipticF[(c + d*x)/2, 2])/(21*a^4*d) - (2*b^3*(A*b^2 + a^2*C)*EllipticPi[(2*b)/(a + b), (
c + d*x)/2, 2])/(a^5*(a + b)*d) + (2*A*Sin[c + d*x])/(9*a*d*Cos[c + d*x]^(9/2)) - (2*A*b*Sin[c + d*x])/(7*a^2*
d*Cos[c + d*x]^(7/2)) + (2*(9*A*b^2 + a^2*(7*A + 9*C))*Sin[c + d*x])/(45*a^3*d*Cos[c + d*x]^(5/2)) - (2*b*(7*A
*b^2 + a^2*(5*A + 7*C))*Sin[c + d*x])/(21*a^4*d*Cos[c + d*x]^(3/2)) + (2*(15*A*b^4 + 3*a^2*b^2*(3*A + 5*C) + a
^4*(7*A + 9*C))*Sin[c + d*x])/(15*a^5*d*Sqrt[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 {11}{2}}(c+d x) (a+b \cos (c+d x))} \, dx &=\frac {2 A \sin (c+d x)}{9 a d \cos ^{\frac {9}{2}}(c+d x)}+\frac {2 \int \frac {-\frac {9 A b}{2}+\frac {1}{2} a (7 A+9 C) \cos (c+d x)+\frac {7}{2} A b \cos ^2(c+d x)}{\cos ^{\frac {9}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{9 a}\\ &=\frac {2 A \sin (c+d x)}{9 a d \cos ^{\frac {9}{2}}(c+d x)}-\frac {2 A b \sin (c+d x)}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {4 \int \frac {\frac {7}{4} \left (9 A b^2+a^2 (7 A+9 C)\right )+a A b \cos (c+d x)-\frac {45}{4} A b^2 \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{63 a^2}\\ &=\frac {2 A \sin (c+d x)}{9 a d \cos ^{\frac {9}{2}}(c+d x)}-\frac {2 A b \sin (c+d x)}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (9 A b^2+a^2 (7 A+9 C)\right ) \sin (c+d x)}{45 a^3 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8 \int \frac {-\frac {45}{8} b \left (7 A b^2+a^2 (5 A+7 C)\right )-\frac {3}{8} a \left (12 A b^2-7 a^2 (7 A+9 C)\right ) \cos (c+d x)+\frac {21}{8} b \left (9 A b^2+a^2 (7 A+9 C)\right ) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{315 a^3}\\ &=\frac {2 A \sin (c+d x)}{9 a d \cos ^{\frac {9}{2}}(c+d x)}-\frac {2 A b \sin (c+d x)}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (9 A b^2+a^2 (7 A+9 C)\right ) \sin (c+d x)}{45 a^3 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2 b \left (7 A b^2+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 a^4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {16 \int \frac {\frac {63}{16} \left (15 A b^4+3 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)\right )+\frac {9}{4} a b \left (7 A b^2+a^2 (6 A+7 C)\right ) \cos (c+d x)-\frac {45}{16} b^2 \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}{945 a^4}\\ &=\frac {2 A \sin (c+d x)}{9 a d \cos ^{\frac {9}{2}}(c+d x)}-\frac {2 A b \sin (c+d x)}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (9 A b^2+a^2 (7 A+9 C)\right ) \sin (c+d x)}{45 a^3 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2 b \left (7 A b^2+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 a^4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 A b^4+3 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)\right ) \sin (c+d x)}{15 a^5 d \sqrt {\cos (c+d x)}}+\frac {32 \int \frac {-\frac {45}{32} b \left (21 A b^4+7 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)\right )-\frac {9}{32} a \left (140 A b^4+7 a^4 (7 A+9 C)+4 a^2 b^2 (22 A+35 C)\right ) \cos (c+d x)-\frac {63}{32} b \left (15 A b^4+3 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{945 a^5}\\ &=\frac {2 A \sin (c+d x)}{9 a d \cos ^{\frac {9}{2}}(c+d x)}-\frac {2 A b \sin (c+d x)}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (9 A b^2+a^2 (7 A+9 C)\right ) \sin (c+d x)}{45 a^3 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2 b \left (7 A b^2+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 a^4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 A b^4+3 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)\right ) \sin (c+d x)}{15 a^5 d \sqrt {\cos (c+d x)}}-\frac {32 \int \frac {\frac {45}{32} b^2 \left (21 A b^4+7 a^2 b^2 (A+3 C)+a^4 (5 A+7 C)\right )+\frac {45}{32} a b^3 \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}{945 a^5 b}-\frac {\left (15 A b^4+3 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)\right ) \int \sqrt {\cos (c+d x)} \, dx}{15 a^5}\\ &=-\frac {2 \left (15 A b^4+3 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^5 d}+\frac {2 A \sin (c+d x)}{9 a d \cos ^{\frac {9}{2}}(c+d x)}-\frac {2 A b \sin (c+d x)}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (9 A b^2+a^2 (7 A+9 C)\right ) \sin (c+d x)}{45 a^3 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2 b \left (7 A b^2+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 a^4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 A b^4+3 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)\right ) \sin (c+d x)}{15 a^5 d \sqrt {\cos (c+d x)}}-\frac {\left (b^3 \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^5}-\frac {\left (b \left (7 A b^2+a^2 (5 A+7 C)\right )\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 a^4}\\ &=-\frac {2 \left (15 A b^4+3 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 a^5 d}-\frac {2 b \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^4 d}-\frac {2 b^3 \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^5 (a+b) d}+\frac {2 A \sin (c+d x)}{9 a d \cos ^{\frac {9}{2}}(c+d x)}-\frac {2 A b \sin (c+d x)}{7 a^2 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 \left (9 A b^2+a^2 (7 A+9 C)\right ) \sin (c+d x)}{45 a^3 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {2 b \left (7 A b^2+a^2 (5 A+7 C)\right ) \sin (c+d x)}{21 a^4 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (15 A b^4+3 a^2 b^2 (3 A+5 C)+a^4 (7 A+9 C)\right ) \sin (c+d x)}{15 a^5 d \sqrt {\cos (c+d x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 4.29, size = 427, normalized size = 1.24 \[ \frac {\frac {2 \left (70 a^4 A \tan (c+d x)+7 \left (a^4 (7 A+9 C)+9 a^2 A b^2\right ) \sin (2 (c+d x))+6 \sin (c+d x) \left (-15 a^3 A b-5 a b \left (a^2 (5 A+7 C)+7 A b^2\right ) \cos ^2(c+d x)+7 \left (a^4 (7 A+9 C)+3 a^2 b^2 (3 A+5 C)+15 A b^4\right ) \cos ^3(c+d x)\right )\right )}{\cos ^{\frac {7}{2}}(c+d x)}-3 \left (\frac {4 \left (7 a^5 (7 A+9 C)+4 a^3 b^2 (22 A+35 C)+140 a A b^4\right ) \left ((a+b) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )-a \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )\right )}{b (a+b)}+\frac {2 \left (a^4 b (99 A+133 C)+7 a^2 b^3 (19 A+45 C)+315 A b^5\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a+b}+\frac {14 \left (a^4 (7 A+9 C)+3 a^2 b^2 (3 A+5 C)+15 A b^4\right ) \sin (c+d x) \left (\left (b^2-2 a^2\right ) \Pi \left (-\frac {b}{a};\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) F\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )-2 a b E\left (\left .\sin ^{-1}\left (\sqrt {\cos (c+d x)}\right )\right |-1\right )\right )}{a b \sqrt {\sin ^2(c+d x)}}\right )}{630 a^5 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + C*Cos[c + d*x]^2)/(Cos[c + d*x]^(11/2)*(a + b*Cos[c + d*x])),x]

[Out]

(-3*((2*(315*A*b^5 + 7*a^2*b^3*(19*A + 45*C) + a^4*b*(99*A + 133*C))*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]
)/(a + b) + (4*(140*a*A*b^4 + 7*a^5*(7*A + 9*C) + 4*a^3*b^2*(22*A + 35*C))*((a + b)*EllipticF[(c + d*x)/2, 2]
- a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]))/(b*(a + b)) + (14*(15*A*b^4 + 3*a^2*b^2*(3*A + 5*C) + a^4*(7*A
+ 9*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])) + (2*(6*(-15*a^3*A*b - 5*a*b*(7*A*b^2 + a^2*(5*A + 7*C))*Cos[c + d*x]^2 + 7*(15*A*b^4 + 3*a^2*b^2*(3*A +
5*C) + a^4*(7*A + 9*C))*Cos[c + d*x]^3)*Sin[c + d*x] + 7*(9*a^2*A*b^2 + a^4*(7*A + 9*C))*Sin[2*(c + d*x)] + 7
0*a^4*A*Tan[c + d*x]))/Cos[c + d*x]^(7/2))/(630*a^5*d)

________________________________________________________________________________________

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)^(11/2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2)/(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)^(11/2)), x)

________________________________________________________________________________________

maple [B]  time = 12.62, size = 1320, normalized size = 3.84 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2)/(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)*(2/a*A*(-1/144*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)^5-7/180*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)^3-14/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/
(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(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))-7/1
5*(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))-EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))))+4*(A*b^2+C*a^2)*b^4/a^5
/(-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/a^2*b*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*s
in(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)*EllipticF(cos(1/2*d*x
+1/2*c),2^(1/2)))+2*b^2*(A*b^2+C*a^2)/a^5*(-(-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)-2*b*(A*b^2+C*a^2)/a^4*(-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)))-2/5*(A*b^2+C*a^2)/a^3/(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*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)^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))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c
)^2-1)^(1/2)/d

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \cos \left (d x + c\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+C*cos(d*x+c)^2)/cos(d*x+c)^(11/2)/(a+b*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate((C*cos(d*x + c)^2 + A)/((b*cos(d*x + c) + a)*cos(d*x + c)^(11/2)), x)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\cos \left (c+d\,x\right )}^{11/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^(11/2)*(a + b*cos(c + d*x))),x)

[Out]

int((A + C*cos(c + d*x)^2)/(cos(c + d*x)^(11/2)*(a + b*cos(c + d*x))), x)

________________________________________________________________________________________

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)**(11/2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________