3.1488 \(\int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=345 \[ -\frac {2 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a+b)}+\frac {2 \sin (c+d x) \left (7 a^2 C-7 a b B+7 A b^2+5 b^2 C\right )}{21 b^3 d \sqrt {\sec (c+d x)}}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-5 a^3 C+5 a^2 b B-a b^2 (5 A+3 C)+3 b^3 B\right )}{5 b^4 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-21 a^4 C+21 a^3 b B-7 a^2 b^2 (3 A+C)+7 a b^3 B-b^4 (7 A+5 C)\right )}{21 b^5 d}+\frac {2 (b B-a C) \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x)}{7 b d \sec ^{\frac {5}{2}}(c+d x)} \]

[Out]

2/7*C*sin(d*x+c)/b/d/sec(d*x+c)^(5/2)+2/5*(B*b-C*a)*sin(d*x+c)/b^2/d/sec(d*x+c)^(3/2)+2/21*(7*A*b^2-7*B*a*b+7*
C*a^2+5*C*b^2)*sin(d*x+c)/b^3/d/sec(d*x+c)^(1/2)+2/5*(5*a^2*b*B+3*b^3*B-5*a^3*C-a*b^2*(5*A+3*C))*(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^3*b*B+7*a*b^3*B-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-a*(B*b-C*
a))*(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))*cos(d*x+c
)^(1/2)*sec(d*x+c)^(1/2)/b^5/(a+b)/d

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Rubi [A]  time = 1.48, antiderivative size = 345, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 43, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {4221, 3049, 3059, 2639, 3002, 2641, 2805} \[ \frac {2 \sin (c+d x) \left (7 a^2 C-7 a b B+7 A b^2+5 b^2 C\right )}{21 b^3 d \sqrt {\sec (c+d x)}}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (-7 a^2 b^2 (3 A+C)+21 a^3 b B-21 a^4 C+7 a b^3 B-b^4 (7 A+5 C)\right )}{21 b^5 d}+\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (5 a^2 b B-5 a^3 C-a b^2 (5 A+3 C)+3 b^3 B\right )}{5 b^4 d}-\frac {2 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{b^5 d (a+b)}+\frac {2 (b B-a C) \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 C \sin (c+d x)}{7 b d \sec ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(5*a^2*b*B + 3*b^3*B - 5*a^3*C - a*b^2*(5*A + 3*C))*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c
+ d*x]])/(5*b^4*d) - (2*(21*a^3*b*B + 7*a*b^3*B - 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*(b*B - a*C))*Sqrt[Cos[c
+ d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2]*Sqrt[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*(b*B - a*C)*Sin[c + d*x])/(5*b^2*d*Sec[c + d*x]^(3/2)) + (2*(7*A*b^2 - 7*a*b*B +
7*a^2*C + 5*b^2*C)*Sin[c + d*x])/(21*b^3*d*Sqrt[Sec[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 3049

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 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 4221

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*} \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{(a+b \cos (c+d x)) \sec ^{\frac {5}{2}}(c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\cos ^{\frac {5}{2}}(c+d x) \left (A+B \cos (c+d x)+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} (b B-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 (b B-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}{4} a (b B-a C)+\frac {1}{4} b (21 b B+4 a C) \cos (c+d x)+\frac {5}{4} \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 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 (b B-a C) \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 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 b^2-7 a b B+7 a^2 C+5 b^2 C\right )+\frac {1}{8} b \left (35 A b^2+28 a b B-28 a^2 C+25 b^2 C\right ) \cos (c+d x)+\frac {21}{8} \left (5 a^2 b B+3 b^3 B-5 a^3 C-a b^2 (5 A+3 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 (b B-a C) \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 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 b^2-7 a b B+7 a^2 C+5 b^2 C\right )+\frac {5}{8} \left (21 a^3 b B+7 a b^3 B-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 (\left (5 a^2 b B+3 b^3 B-5 a^3 C-a b^2 (5 A+3 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 \left (5 a^2 b B+3 b^3 B-5 a^3 C-a b^2 (5 A+3 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 (b B-a C) \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 C\right ) \sin (c+d x)}{21 b^3 d \sqrt {\sec (c+d x)}}-\frac {\left (\left (21 a^3 b B+7 a b^3 B-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 {\left (a^3 \left (A b^2-a (b B-a 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 {2 \left (5 a^2 b B+3 b^3 B-5 a^3 C-a b^2 (5 A+3 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^3 b B+7 a b^3 B-21 a^4 C-7 a^2 b^2 (3 A+C)-b^4 (7 A+5 C)\right ) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{21 b^5 d}-\frac {2 a^3 \left (A b^2-a (b B-a C)\right ) \sqrt {\cos (c+d x)} \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |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 (b B-a C) \sin (c+d x)}{5 b^2 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 \left (7 A b^2-7 a b B+7 a^2 C+5 b^2 C\right ) \sin (c+d x)}{21 b^3 d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 6.67, size = 532, normalized size = 1.54 \[ \frac {4 b^2 \sin (c+d x) \left (70 a^2 C+42 b (b B-a C) \cos (c+d x)-70 a b B+70 A b^2+15 b^2 C \cos (2 (c+d x))+65 b^2 C\right )-\frac {2 \cos (c+d x) \cot (c+d x) (a \sec (c+d x)+b) \left (-4 a b^2 \sqrt {-\tan ^2(c+d x)} \sqrt {\sec (c+d x)} \left (-28 a^2 C+28 a b B+35 A b^2+25 b^2 C\right ) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-2 b^2 \sqrt {-\tan ^2(c+d x)} \sqrt {\sec (c+d x)} \left (-35 a^3 C+35 a^2 b B-a b^2 (35 A+13 C)+63 b^3 B\right ) \left (F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-\Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )\right )-21 \left (5 a^3 C-5 a^2 b B+a b^2 (5 A+3 C)-3 b^3 B\right ) \left (4 a^2 \sqrt {-\tan ^2(c+d x)} \sqrt {\sec (c+d x)} \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-2 b^2 \sqrt {-\tan ^2(c+d x)} \sqrt {\sec (c+d x)} \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b \sec ^2(c+d x)-2 b (2 a-b) \sqrt {-\tan ^2(c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+4 a b \sqrt {-\tan ^2(c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+4 a b\right )\right )}{a (a+b \cos (c+d x))}}{420 b^5 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Cos[c + d*x] + 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*b*B + 70*a^2*C + 65*b^2*C + 42*b*(b*B - a*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])*(-2*b^2*(35*a^2*b*B + 63*b^3*B - 35*a^3*C -
a*b^2*(35*A + 13*C))*(EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] - EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]
], -1])*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 4*a*b^2*(35*A*b^2 + 28*a*b*B - 28*a^2*C + 25*b^2*C)*Ellipti
cPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 21*(-5*a^2*b*B - 3*b^3*
B + 5*a^3*C + a*b^2*(5*A + 3*C))*(4*a*b - 4*a*b*Sec[c + d*x]^2 + 4*a*b*EllipticE[ArcSin[Sqrt[Sec[c + d*x]]], -
1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan[c + d*x]^2] - 2*(2*a - b)*b*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec
[c + d*x]]*Sqrt[-Tan[c + d*x]^2] + 4*a^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]
*Sqrt[-Tan[c + d*x]^2] - 2*b^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[-Tan
[c + d*x]^2])))/(a*(a + b*Cos[c + d*x])))/(420*b^5*d*Sqrt[Sec[c + d*x]])

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

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [B]  time = 9.50, size = 1097, normalized size = 3.18 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(8/105*C/b*(60*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c
)^8-258*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+448*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+85*(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))-168*(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))-167*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)+4/5/b^2*(B*b-C*a-4*C*b)*(-4*cos(1/2*d*x+1/2
*c)*sin(1/2*d*x+1/2*c)^6+14*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(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))-9*(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))-6*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)+4/3/b^3*(A*b^2-B*a*b-3*B*b^2+C*a^2+3*C*a*b+6*C*b^2)*(2*sin(1/2*d*x+1/2*c
)^4*cos(1/2*d*x+1/2*c)+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)*EllipticF(cos(1/2*d*x+1
/2*c),2^(1/2))-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))-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/b^4*(A*
a*b^2+2*A*b^3-B*a^2*b-2*B*a*b^2-3*B*b^3+C*a^3+2*C*a^2*b+3*C*a*b^2+4*C*b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*co
s(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)))+2*(A*a^2*b^2+A*a*b^3+A*b^4-B*a^3*b-B*a^2*b^2-B*a*b^3-B*b^4+C*
a^4+C*a^3*b+C*a^2*b^2+C*a*b^3+C*b^4)/b^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*si
n(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))+4*a^3*(A*b^2-B*a*b+C*a^2)
/b^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+si
n(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]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A}{{\left (b \cos \left (d x + c\right ) + a\right )} \sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {C\,{\cos \left (c+d\,x\right )}^2+B\,\cos \left (c+d\,x\right )+A}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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