\(\int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx\) [1397]

Optimal result

Integrand size = 35, antiderivative size = 299 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 a \left (5 A b^2+5 a^2 C+3 b^2 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^4 d}+\frac {2 \left (21 a^4 C+7 a^2 b^2 (3 A+C)+b^4 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 b^5 d}-\frac {2 a^3 \left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^5 (a+b) d}+\frac {2 C \sin (c+d x)}{7 b d \sec ^{\frac {5}{2}}(c+d x)}-\frac {2 a C \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 b^3 d \sqrt {\sec (c+d x)}} \]

[Out]

2/7*C*sin(d*x+c)/b/d/sec(d*x+c)^(5/2)-2/5*a*C*sin(d*x+c)/b^2/d/sec(d*x+c)^(3/2)+2/21*(7*a^2*C+b^2*(7*A+5*C))*s
in(d*x+c)/b^3/d/sec(d*x+c)^(1/2)-2/5*a*(5*A*b^2+5*C*a^2+3*C*b^2)*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^4/d+2/21*(21*a^4*C+7*a^2*b^2*(3*A
+C)+b^4*(7*A+5*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))*cos(d
*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^5/d-2*a^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)*Ellipti
cPi(sin(1/2*d*x+1/2*c),2*b/(a+b),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/b^5/(a+b)/d

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 299, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.229, Rules used = {4306, 3129, 3128, 3138, 2719, 3081, 2720, 2884} \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=-\frac {2 a \left (5 a^2 C+5 A b^2+3 b^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 b^4 d}+\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 b^3 d \sqrt {\sec (c+d x)}}+\frac {2 \left (21 a^4 C+7 a^2 b^2 (3 A+C)+b^4 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 b^5 d}-\frac {2 a^3 \left (a^2 C+A b^2\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{b^5 d (a+b)}-\frac {2 a C \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x)}{7 b d \sec ^{\frac {5}{2}}(c+d x)} \]

[In]

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

[Out]

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

Rule 2719

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

Rule 2720

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

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 3081

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 3128

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*
x])^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n +
2) - C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 0])))

Rule 3129

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (C_.)
*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^
(n + 1)/(d*f*(m + n + 2))), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*x])
^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (A*b*d*(m + n + 2) - C*(a*c - b*d*(m + n + 1)))*Sin[e +
f*x] + C*(a*d*m - b*c*(m + 1))*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] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a,
0] && NeQ[c, 0])))

Rule 3138

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 4306

Int[(u_)*((c_.)*sec[(a_.) + (b_.)*(x_)])^(m_.), x_Symbol] :> Dist[(c*Sec[a + b*x])^m*(c*Cos[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Cos[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownSineIntegrandQ[u,
x]

Rubi steps \begin{align*} \text {integral}& = \left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx \\ & = \frac {2 C \sin (c+d x)}{7 b d \sec ^{\frac {5}{2}}(c+d x)}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {3}{2}}(c+d x) \left (\frac {5 a C}{2}+\frac {1}{2} b (7 A+5 C) \cos (c+d x)-\frac {7}{2} a C \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{7 b} \\ & = \frac {2 C \sin (c+d x)}{7 b d \sec ^{\frac {5}{2}}(c+d x)}-\frac {2 a C \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\sqrt {\cos (c+d x)} \left (-\frac {21 a^2 C}{4}+a b C \cos (c+d x)+\frac {5}{4} \left (7 a^2 C+b^2 (7 A+5 C)\right ) \cos ^2(c+d x)\right )}{a+b \cos (c+d x)} \, dx}{35 b^2} \\ & = \frac {2 C \sin (c+d x)}{7 b d \sec ^{\frac {5}{2}}(c+d x)}-\frac {2 a C \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 b^3 d \sqrt {\sec (c+d x)}}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{8} a \left (7 a^2 C+b^2 (7 A+5 C)\right )+\frac {1}{8} b \left (35 A b^2-28 a^2 C+25 b^2 C\right ) \cos (c+d x)-\frac {21}{8} a \left (5 A b^2+5 a^2 C+3 b^2 C\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{105 b^3} \\ & = \frac {2 C \sin (c+d x)}{7 b d \sec ^{\frac {5}{2}}(c+d x)}-\frac {2 a C \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 b^3 d \sqrt {\sec (c+d x)}}-\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {5}{8} a b \left (7 a^2 C+b^2 (7 A+5 C)\right )-\frac {5}{8} \left (21 a^4 C+7 a^2 b^2 (3 A+C)+b^4 (7 A+5 C)\right ) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{105 b^4}-\frac {\left (a \left (5 A b^2+5 a^2 C+3 b^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 b^4} \\ & = -\frac {2 a \left (5 A b^2+5 a^2 C+3 b^2 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^4 d}+\frac {2 C \sin (c+d x)}{7 b d \sec ^{\frac {5}{2}}(c+d x)}-\frac {2 a C \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 b^3 d \sqrt {\sec (c+d x)}}-\frac {\left (a^3 \left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{b^5}+\frac {\left (\left (21 a^4 C+7 a^2 b^2 (3 A+C)+b^4 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{21 b^5} \\ & = -\frac {2 a \left (5 A b^2+5 a^2 C+3 b^2 C\right ) \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 b^4 d}+\frac {2 \left (21 a^4 C+7 a^2 b^2 (3 A+C)+b^4 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{21 b^5 d}-\frac {2 a^3 \left (A b^2+a^2 C\right ) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{b^5 (a+b) d}+\frac {2 C \sin (c+d x)}{7 b d \sec ^{\frac {5}{2}}(c+d x)}-\frac {2 a C \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 a^2 C+b^2 (7 A+5 C)\right ) \sin (c+d x)}{21 b^3 d \sqrt {\sec (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 5.59 (sec) , antiderivative size = 343, normalized size of antiderivative = 1.15 \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=\frac {4 b^2 \left (70 A b^2+70 a^2 C+65 b^2 C-42 a b C \cos (c+d x)+15 b^2 C \cos (2 (c+d x))\right ) \sin (c+d x)-\frac {2 \cos (c+d x) \cot (c+d x) (b+a \sec (c+d x)) \left (-84 a b \left (5 A b^2+5 a^2 C+3 b^2 C\right ) E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-4 b \left (-105 a^3 C+35 a^2 b C-21 a b^2 (5 A+3 C)+5 b^3 (7 A+5 C)\right ) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}-84 a \left (-b \left (5 A b^2+5 a^2 C+3 b^2 C\right ) \tan ^2(c+d x)+5 a \left (A b^2+a^2 C\right ) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right ) \sqrt {\sec (c+d x)} \sqrt {-\tan ^2(c+d x)}\right )\right )}{a+b \cos (c+d x)}}{420 b^5 d \sqrt {\sec (c+d x)}} \]

[In]

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

[Out]

(4*b^2*(70*A*b^2 + 70*a^2*C + 65*b^2*C - 42*a*b*C*Cos[c + d*x] + 15*b^2*C*Cos[2*(c + d*x)])*Sin[c + d*x] - (2*
Cos[c + d*x]*Cot[c + d*x]*(b + a*Sec[c + d*x])*(-84*a*b*(5*A*b^2 + 5*a^2*C + 3*b^2*C)*EllipticE[ArcSin[Sqrt[Se
c[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 4*b*(-105*a^3*C + 35*a^2*b*C - 21*a*b^2*(5*A + 3*
C) + 5*b^3*(7*A + 5*C))*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 8
4*a*(-(b*(5*A*b^2 + 5*a^2*C + 3*b^2*C)*Tan[c + d*x]^2) + 5*a*(A*b^2 + a^2*C)*EllipticPi[-(a/b), ArcSin[Sqrt[Se
c[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2])))/(a + b*Cos[c + d*x]))/(420*b^5*d*Sqrt[Sec[c + d*
x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1243\) vs. \(2(349)=698\).

Time = 4.33 (sec) , antiderivative size = 1244, normalized size of antiderivative = 4.16

method	result	size
default	\(\text {Expression too large to display}\)	\(1244\)

[In]

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

[Out]

-2/105*((-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2)^(1/2)*((240*C*a*b^4-240*C*b^5)*cos(1/2*d*x+1/2*c)*si
n(1/2*d*x+1/2*c)^8+(168*C*a^2*b^3-528*C*a*b^4+360*C*b^5)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(140*A*a*b^4-
140*A*b^5+140*C*a^3*b^2-308*C*a^2*b^3+448*C*a*b^4-280*C*b^5)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-70*A*a*
b^4+70*A*b^5-70*C*a^3*b^2+112*C*a^2*b^3-122*C*a*b^4+80*C*b^5)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+105*A*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^3*b^2-105*
A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^3+
35*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^4
-35*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))*b^5+
105*A*(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))*a^2*
b^3-105*A*(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))*
a*b^4-105*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(
a-b),2^(1/2))*a^3*b^2+105*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))*a^5-105*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*
x+1/2*c),2^(1/2))*a^4*b+35*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d
*x+1/2*c),2^(1/2))*a^3*b^2-35*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/
2*d*x+1/2*c),2^(1/2))*a^2*b^3+25*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos
(1/2*d*x+1/2*c),2^(1/2))*a*b^4-25*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(co
s(1/2*d*x+1/2*c),2^(1/2))*b^5+105*C*(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(co
s(1/2*d*x+1/2*c),2^(1/2))*a^4*b-105*C*(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))*a^3*b^2+63*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Ellipti
cE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b^3-63*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elli
pticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^4-105*C*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*El
lipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2))*a^5)/b^5/(a-b)/(-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)/(-1+2*cos(1/2*d*x+1/2*c)^2)^(1/2)/d

Fricas [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int { \frac {C \cos \left (d x + c\right )^{2} + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {A+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx=\int \frac {C\,{\cos \left (c+d\,x\right )}^2+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]

[In]

int((A + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))),x)

[Out]

int((A + C*cos(c + d*x)^2)/((1/cos(c + d*x))^(5/2)*(a + b*cos(c + d*x))), x)