3.1482 \(\int \frac {(A+B \cos (c+d x)+C \cos ^2(c+d x)) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx\)

Optimal. Leaf size=294 \[ -\frac {2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d (a+b)}-\frac {2 (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a^2 d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 a d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 1.42, antiderivative size = 294, 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, 3055, 3059, 2639, 3002, 2641, 2805} \[ \frac {2 \sin (c+d x) \sqrt {\sec (c+d x)} \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d}-\frac {2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \left (a^2 (3 A+5 C)-5 a b B+5 A b^2\right )}{5 a^3 d}-\frac {2 b \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \left (A b^2-a (b B-a C)\right ) \Pi \left (\frac {2 b}{a+b};\left .\frac {1}{2} (c+d x)\right |2\right )}{a^3 d (a+b)}-\frac {2 (A b-a B) \sin (c+d x) \sec ^{\frac {3}{2}}(c+d x)}{3 a^2 d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 a^2 d}+\frac {2 A \sin (c+d x) \sec ^{\frac {5}{2}}(c+d x)}{5 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 {\left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right ) \sec ^{\frac {7}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx &=\left (\sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {A+B \cos (c+d x)+C \cos ^2(c+d x)}{\cos ^{\frac {7}{2}}(c+d x) (a+b \cos (c+d x))} \, dx\\ &=\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {\left (2 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {5}{2} (A b-a B)+\frac {1}{2} a (3 A+5 C) \cos (c+d x)+\frac {3}{2} A b \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{5 a}\\ &=-\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {\left (4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {3}{4} \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right )+\frac {1}{4} a (4 A b+5 a B) \cos (c+d x)-\frac {5}{4} b (A b-a B) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))} \, dx}{15 a^2}\\ &=\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a^3 d}-\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}+\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {-\frac {5}{8} \left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (A+3 C)\right )-\frac {1}{8} a \left (20 A b^2-20 a b B+3 a^2 (3 A+5 C)\right ) \cos (c+d x)-\frac {3}{8} b \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \cos ^2(c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 a^3}\\ &=\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a^3 d}-\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {\left (8 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {\frac {5}{8} b \left (3 A b^3-a^3 B-3 a b^2 B+a^2 b (A+3 C)\right )+\frac {5}{8} a b^2 (A b-a B) \cos (c+d x)}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))} \, dx}{15 a^3 b}-\frac {\left (\left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx}{5 a^3}\\ &=-\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 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 a^3 d}+\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a^3 d}-\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}-\frac {\left ((A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx}{3 a^2}-\frac {\left (b \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}{a^3}\\ &=-\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 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 a^3 d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 a^2 d}-\frac {2 b \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)}}{a^3 (a+b) d}+\frac {2 \left (5 A b^2-5 a b B+a^2 (3 A+5 C)\right ) \sqrt {\sec (c+d x)} \sin (c+d x)}{5 a^3 d}-\frac {2 (A b-a B) \sec ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{3 a^2 d}+\frac {2 A \sec ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{5 a d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [B]  time = 7.05, size = 692, normalized size = 2.35 \[ \frac {\sqrt {\sec (c+d x)} \left (\frac {2 \sec (c+d x) (a B \sin (c+d x)-A b \sin (c+d x))}{3 a^2}+\frac {2 \sin (c+d x) \left (3 a^2 A+5 a^2 C-5 a b B+5 A b^2\right )}{5 a^3}+\frac {2 A \tan (c+d x) \sec (c+d x)}{5 a}\right )}{d}-\frac {\frac {\sin (c+d x) \cos (2 (c+d x)) \left (9 a^2 A b+15 a^2 b C-15 a b^2 B+15 A b^3\right ) (a \sec (c+d x)+b) \left (-4 a^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+2 b^2 \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )+4 a b \sec ^2(c+d x)+2 b (2 a-b) \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b \sqrt {\sec (c+d x)} \sqrt {1-\sec ^2(c+d x)} E\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-4 a b\right )}{a b^2 \left (1-\cos ^2(c+d x)\right ) \sqrt {\sec (c+d x)} \left (2-\sec ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac {2 \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} \left (18 a^3 A+30 a^3 C-40 a^2 b B+40 a A b^2\right ) (a \sec (c+d x)+b) \Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )}{b \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}+\frac {2 \sin (c+d x) \cos ^2(c+d x) \sqrt {1-\sec ^2(c+d x)} \left (-10 a^3 B+19 a^2 A b+45 a^2 b C-45 a b^2 B+45 A b^3\right ) (a \sec (c+d x)+b) \left (F\left (\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-\Pi \left (-\frac {a}{b};\left .\sin ^{-1}\left (\sqrt {\sec (c+d x)}\right )\right |-1\right )\right )}{a \left (1-\cos ^2(c+d x)\right ) (a+b \cos (c+d x))}}{30 a^3 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/30*((2*(19*a^2*A*b + 45*A*b^3 - 10*a^3*B - 45*a*b^2*B + 45*a^2*b*C)*Cos[c + d*x]^2*(EllipticF[ArcSin[Sqrt[S
ec[c + d*x]]], -1] - EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1])*(b + a*Sec[c + d*x])*Sqrt[1 - Sec[c +
d*x]^2]*Sin[c + d*x])/(a*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + (2*(18*a^3*A + 40*a*A*b^2 - 40*a^2*b*B
+ 30*a^3*C)*Cos[c + d*x]^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1]*(b + a*Sec[c + d*x])*Sqrt[1 - Se
c[c + d*x]^2]*Sin[c + d*x])/(b*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)) + ((9*a^2*A*b + 15*A*b^3 - 15*a*b^2*
B + 15*a^2*b*C)*Cos[2*(c + d*x)]*(b + a*Sec[c + d*x])*(-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[1 - Sec[c + d*x]^2] + 2*(2*a - b)*b*EllipticF[ArcSin[Sqrt[Sec
[c + d*x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] - 4*a^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*
x]]], -1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2] + 2*b^2*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -
1]*Sqrt[Sec[c + d*x]]*Sqrt[1 - Sec[c + d*x]^2])*Sin[c + d*x])/(a*b^2*(a + b*Cos[c + d*x])*(1 - Cos[c + d*x]^2)
*Sqrt[Sec[c + d*x]]*(2 - Sec[c + d*x]^2)))/(a^3*d) + (Sqrt[Sec[c + d*x]]*((2*(3*a^2*A + 5*A*b^2 - 5*a*b*B + 5*
a^2*C)*Sin[c + d*x])/(5*a^3) + (2*Sec[c + d*x]*(-(A*b*Sin[c + d*x]) + a*B*Sin[c + d*x]))/(3*a^2) + (2*A*Sec[c
+ d*x]*Tan[c + d*x])/(5*a)))/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+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(a+b*cos(d*x+c)),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

maple [B]  time = 11.13, size = 802, normalized size = 2.73 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*cos(d*x+c)+C*cos(d*x+c)^2)*sec(d*x+c)^(7/2)/(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)*(-2/5*A/a/(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)+4*(A*b^2-B*a*b+C*a^2)*b^2/a^3/(-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*b^2-B*a*b+C*a^2)/a^3*(-(-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*s
in(1/2*d*x+1/2*c)^2-1)+2*(-A*b+B*a)/a^2*(-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*s
in(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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((1/cos(c + d*x))^(7/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^2))/(a + b*cos(c + d*x)),x)

[Out]

int(((1/cos(c + d*x))^(7/2)*(A + B*cos(c + d*x) + C*cos(c + d*x)^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+B*cos(d*x+c)+C*cos(d*x+c)**2)*sec(d*x+c)**(7/2)/(a+b*cos(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________