\(\int \frac {(a+a \cos (c+d x))^2 (A+B \cos (c+d x)+C \cos ^2(c+d x))}{\cos ^{\frac {7}{2}}(c+d x)} \, dx\) [444]

Optimal result

Integrand size = 43, antiderivative size = 174 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {4 a^2 (4 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (A+2 B+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (17 A+25 B+15 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \]

[Out]

-4/5*a^2*(4*A+5*B)*(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))/d+4/3
*a^2*(A+2*B+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))/d+2/5*A
*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(5/2)+2/15*(4*A+5*B)*(a^2+a^2*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^
(3/2)+2/15*a^2*(17*A+25*B+15*C)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.163, Rules used = {3122, 3054, 3047, 3100, 2827, 2720, 2719} \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {4 a^2 (A+2 B+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (17 A+25 B+15 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}-\frac {4 a^2 (4 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (4 A+5 B) \sin (c+d x) \left (a^2 \cos (c+d x)+a^2\right )}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 A \sin (c+d x) (a \cos (c+d x)+a)^2}{5 d \cos ^{\frac {5}{2}}(c+d x)} \]

[In]

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

[Out]

(-4*a^2*(4*A + 5*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^2*(A + 2*B + 3*C)*EllipticF[(c + d*x)/2, 2])/(3*d)
+ (2*a^2*(17*A + 25*B + 15*C)*Sin[c + d*x])/(15*d*Sqrt[Cos[c + d*x]]) + (2*A*(a + a*Cos[c + d*x])^2*Sin[c + d
*x])/(5*d*Cos[c + d*x]^(5/2)) + (2*(4*A + 5*B)*(a^2 + a^2*Cos[c + d*x])*Sin[c + d*x])/(15*d*Cos[c + d*x]^(3/2)
)

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 2827

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3054

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

Rule 3100

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(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)/(b*f*(m
+ 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(a*A - b*B +
a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A*b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b,
e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]

Rule 3122

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[(-(c^2*C - B*c*d + A*d^2))*Cos[e
+ f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(c^2 - d^2))), x] + Dist[1/(b*d*(n +
1)*(c^2 - d^2)), Int[(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(a*d*m + b*c*(n + 1)) + (c*C
- B*d)*(a*c*m + b*d*(n + 1)) + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)))*Sin[e + f*x], x]
, x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] &&  !LtQ[m, -2^(-1)] && (LtQ[n, -1] || EqQ[m + n + 2, 0])

Rubi steps \begin{align*} \text {integral}& = \frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 \int \frac {(a+a \cos (c+d x))^2 \left (\frac {1}{2} a (4 A+5 B)-\frac {1}{2} a (A-5 C) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx}{5 a} \\ & = \frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{4} a^2 (17 A+25 B+15 C)-\frac {1}{4} a^2 (7 A+5 B-15 C) \cos (c+d x)\right )}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a} \\ & = \frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 \int \frac {\frac {1}{4} a^3 (17 A+25 B+15 C)+\left (-\frac {1}{4} a^3 (7 A+5 B-15 C)+\frac {1}{4} a^3 (17 A+25 B+15 C)\right ) \cos (c+d x)-\frac {1}{4} a^3 (7 A+5 B-15 C) \cos ^2(c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx}{15 a} \\ & = \frac {2 a^2 (17 A+25 B+15 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {8 \int \frac {\frac {5}{4} a^3 (A+2 B+3 C)-\frac {3}{4} a^3 (4 A+5 B) \cos (c+d x)}{\sqrt {\cos (c+d x)}} \, dx}{15 a} \\ & = \frac {2 a^2 (17 A+25 B+15 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)}-\frac {1}{5} \left (2 a^2 (4 A+5 B)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (2 a^2 (A+2 B+3 C)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = -\frac {4 a^2 (4 A+5 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^2 (A+2 B+3 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 a^2 (17 A+25 B+15 C) \sin (c+d x)}{15 d \sqrt {\cos (c+d x)}}+\frac {2 A (a+a \cos (c+d x))^2 \sin (c+d x)}{5 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {2 (4 A+5 B) \left (a^2+a^2 \cos (c+d x)\right ) \sin (c+d x)}{15 d \cos ^{\frac {3}{2}}(c+d x)} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 9.22 (sec) , antiderivative size = 1041, normalized size of antiderivative = 5.98 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^2 \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (-\frac {(-16 A-20 B-5 C+5 C \cos (2 c)) \csc (c) \sec (c)}{20 d}+\frac {A \sec (c) \sec ^3(c+d x) \sin (d x)}{10 d}+\frac {\sec (c) \sec ^2(c+d x) (3 A \sin (c)+10 A \sin (d x)+5 B \sin (d x))}{30 d}+\frac {\sec (c) \sec (c+d x) (10 A \sin (c)+5 B \sin (c)+24 A \sin (d x)+30 B \sin (d x)+15 C \sin (d x))}{30 d}\right )-\frac {A (a+a \cos (c+d x))^2 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d \sqrt {1+\cot ^2(c)}}-\frac {2 B (a+a \cos (c+d x))^2 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{3 d \sqrt {1+\cot ^2(c)}}-\frac {C (a+a \cos (c+d x))^2 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\arctan (\cot (c))) \sqrt {1-\sin (d x-\arctan (\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\arctan (\cot (c)))} \sqrt {1+\sin (d x-\arctan (\cot (c)))}}{d \sqrt {1+\cot ^2(c)}}+\frac {2 A (a+a \cos (c+d x))^2 \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{5 d}+\frac {B (a+a \cos (c+d x))^2 \csc (c) \sec ^4\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {\, _2F_1\left (-\frac {1}{2},-\frac {1}{4};\frac {3}{4};\cos ^2(d x+\arctan (\tan (c)))\right ) \sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\arctan (\tan (c)))} \sqrt {1+\cos (d x+\arctan (\tan (c)))} \sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\arctan (\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\arctan (\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{2 d} \]

[In]

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

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^2*Sec[c/2 + (d*x)/2]^4*(-1/20*((-16*A - 20*B - 5*C + 5*C*Cos[2*c])*Csc
[c]*Sec[c])/d + (A*Sec[c]*Sec[c + d*x]^3*Sin[d*x])/(10*d) + (Sec[c]*Sec[c + d*x]^2*(3*A*Sin[c] + 10*A*Sin[d*x]
+ 5*B*Sin[d*x]))/(30*d) + (Sec[c]*Sec[c + d*x]*(10*A*Sin[c] + 5*B*Sin[c] + 24*A*Sin[d*x] + 30*B*Sin[d*x] + 15
*C*Sin[d*x]))/(30*d)) - (A*(a + a*Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan
[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1
+ Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) -
(2*B*(a + a*Cos[c + d*x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2
+ (d*x)/2]^4*Sec[d*x - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Si
n[d*x - ArcTan[Cot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(3*d*Sqrt[1 + Cot[c]^2]) - (C*(a + a*Cos[c + d*
x])^2*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^4*Sec[d*x -
ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[Cot[c]]]
)]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(d*Sqrt[1 + Cot[c]^2]) + (2*A*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 +
(d*x)/2]^4*((HypergeometricPFQ[{-1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan
[c])/(Sqrt[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan
[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1 + Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Co
s[c]^2*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c
]]]*Sqrt[1 + Tan[c]^2]]))/(5*d) + (B*(a + a*Cos[c + d*x])^2*Csc[c]*Sec[c/2 + (d*x)/2]^4*((HypergeometricPFQ[{-
1/2, -1/4}, {3/4}, Cos[d*x + ArcTan[Tan[c]]]^2]*Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/(Sqrt[1 - Cos[d*x + ArcTan[T
an[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]*Sqrt[1
+ Tan[c]^2]) - ((Sin[d*x + ArcTan[Tan[c]]]*Tan[c])/Sqrt[1 + Tan[c]^2] + (2*Cos[c]^2*Cos[d*x + ArcTan[Tan[c]]]*
Sqrt[1 + Tan[c]^2])/(Cos[c]^2 + Sin[c]^2))/Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqrt[1 + Tan[c]^2]]))/(2*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(878\) vs. \(2(210)=420\).

Time = 19.16 (sec) , antiderivative size = 879, normalized size of antiderivative = 5.05

method	result	size
default	\(\text {Expression too large to display}\)	\(879\)
parts	\(\text {Expression too large to display}\)	\(950\)

[In]

int((a+cos(d*x+c)*a)^2*(A+B*cos(d*x+c)+C*cos(d*x+c)^2)/cos(d*x+c)^(7/2),x,method=_RETURNVERBOSE)

[Out]

-8*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^2*(1/4*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(
1/2))+1/4*C*(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))+1/4*C*(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))-Ell
ipticE(cos(1/2*d*x+1/2*c),2^(1/2)))+1/20*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*(24*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6-12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(s
in(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*sin(1/2*d*x+1/2*c)^4-24*sin(1/2*d*x+1/2*c)^4*
cos(1/2*d*x+1/2*c)+12*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(1/2*d*x+1/2*
c),2^(1/2))*sin(1/2*d*x+1/2*c)^2+8*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*s
in(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2)))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c
)^2)^(1/2)+(1/2*A+1/4*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)/(cos(1/
2*d*x+1/2*c)^2-1/2)^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)))+(1/4*A+1/2*B+1/4*C)/sin(1/2*d*x+1/2*c)^
2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(
1/2*d*x+1/2*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))))/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.43 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (A + 2 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (A + 2 \, B + 3 \, C\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 3 i \, \sqrt {2} {\left (4 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 3 i \, \sqrt {2} {\left (4 \, A + 5 \, B\right )} a^{2} \cos \left (d x + c\right )^{3} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (3 \, {\left (8 \, A + 10 \, B + 5 \, C\right )} a^{2} \cos \left (d x + c\right )^{2} + 5 \, {\left (2 \, A + B\right )} a^{2} \cos \left (d x + c\right ) + 3 \, A a^{2}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{15 \, d \cos \left (d x + c\right )^{3}} \]

[In]

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

[Out]

-2/15*(5*I*sqrt(2)*(A + 2*B + 3*C)*a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)
) - 5*I*sqrt(2)*(A + 2*B + 3*C)*a^2*cos(d*x + c)^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) +
3*I*sqrt(2)*(4*A + 5*B)*a^2*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I
*sin(d*x + c))) - 3*I*sqrt(2)*(4*A + 5*B)*a^2*cos(d*x + c)^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0,
cos(d*x + c) - I*sin(d*x + c))) - (3*(8*A + 10*B + 5*C)*a^2*cos(d*x + c)^2 + 5*(2*A + B)*a^2*cos(d*x + c) + 3
*A*a^2)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\int { \frac {{\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{2}}{\cos \left (d x + c\right )^{\frac {7}{2}}} \,d x } \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 3.98 (sec) , antiderivative size = 313, normalized size of antiderivative = 1.80 \[ \int \frac {(a+a \cos (c+d x))^2 \left (A+B \cos (c+d x)+C \cos ^2(c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx=\frac {6\,A\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+20\,A\,a^2\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+30\,A\,a^2\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{15\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {2\,C\,a^2\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,C\,a^2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {4\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,C\,a^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

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

[Out]

(6*A*a^2*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2) + 20*A*a^2*cos(c + d*x)*sin(c + d*x)*hyperg
eom([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 30*A*a^2*cos(c + d*x)^2*sin(c + d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c
+ d*x)^2))/(15*d*cos(c + d*x)^(5/2)*(1 - cos(c + d*x)^2)^(1/2)) + (2*B*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (
2*C*a^2*ellipticE(c/2 + (d*x)/2, 2))/d + (4*C*a^2*ellipticF(c/2 + (d*x)/2, 2))/d + (4*B*a^2*sin(c + d*x)*hyper
geom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2)) + (2*B*a^2*sin(c + d*x)*
hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(3*d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*C*a^2*sin(c
+ d*x)*hypergeom([-1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(1/2)*(sin(c + d*x)^2)^(1/2))