\(\int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx\) [143]

Optimal result

Integrand size = 33, antiderivative size = 204 \[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {4 a^3 (7 A+9 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (13 A+21 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (7 A+9 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (11 A+7 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)} \]

[Out]

-4/5*a^3*(7*A+9*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+4/2
1*a^3*(13*A+21*B)*(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+4/10
5*a^3*(41*A+42*B)*sin(d*x+c)/d/cos(d*x+c)^(3/2)+2/7*a*A*(a+a*cos(d*x+c))^2*sin(d*x+c)/d/cos(d*x+c)^(7/2)+2/35*
(11*A+7*B)*(a^3+a^3*cos(d*x+c))*sin(d*x+c)/d/cos(d*x+c)^(5/2)+4/5*a^3*(7*A+9*B)*sin(d*x+c)/d/cos(d*x+c)^(1/2)

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 204, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3054, 3047, 3100, 2827, 2716, 2719, 2720} \[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {4 a^3 (13 A+21 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}-\frac {4 a^3 (7 A+9 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 (11 A+7 B) \sin (c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4 a^3 (7 A+9 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a A \sin (c+d x) (a \cos (c+d x)+a)^2}{7 d \cos ^{\frac {7}{2}}(c+d x)} \]

[In]

Int[((a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(9/2),x]

[Out]

(-4*a^3*(7*A + 9*B)*EllipticE[(c + d*x)/2, 2])/(5*d) + (4*a^3*(13*A + 21*B)*EllipticF[(c + d*x)/2, 2])/(21*d)
+ (4*a^3*(41*A + 42*B)*Sin[c + d*x])/(105*d*Cos[c + d*x]^(3/2)) + (4*a^3*(7*A + 9*B)*Sin[c + d*x])/(5*d*Sqrt[C
os[c + d*x]]) + (2*a*A*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(7*d*Cos[c + d*x]^(7/2)) + (2*(11*A + 7*B)*(a^3 +
a^3*Cos[c + d*x])*Sin[c + d*x])/(35*d*Cos[c + d*x]^(5/2))

Rule 2716

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*((b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1
))), x] + Dist[(n + 2)/(b^2*(n + 1)), Int[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1
] && IntegerQ[2*n]

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]

Rubi steps \begin{align*} \text {integral}& = \frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2}{7} \int \frac {(a+a \cos (c+d x))^2 \left (\frac {1}{2} a (11 A+7 B)+\frac {1}{2} a (A+7 B) \cos (c+d x)\right )}{\cos ^{\frac {7}{2}}(c+d x)} \, dx \\ & = \frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (11 A+7 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4}{35} \int \frac {(a+a \cos (c+d x)) \left (\frac {1}{2} a^2 (41 A+42 B)+\frac {1}{2} a^2 (8 A+21 B) \cos (c+d x)\right )}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (11 A+7 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {4}{35} \int \frac {\frac {1}{2} a^3 (41 A+42 B)+\left (\frac {1}{2} a^3 (8 A+21 B)+\frac {1}{2} a^3 (41 A+42 B)\right ) \cos (c+d x)+\frac {1}{2} a^3 (8 A+21 B) \cos ^2(c+d x)}{\cos ^{\frac {5}{2}}(c+d x)} \, dx \\ & = \frac {4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (11 A+7 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {8}{105} \int \frac {\frac {21}{4} a^3 (7 A+9 B)+\frac {5}{4} a^3 (13 A+21 B) \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (11 A+7 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}+\frac {1}{5} \left (2 a^3 (7 A+9 B)\right ) \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x)} \, dx+\frac {1}{21} \left (2 a^3 (13 A+21 B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx \\ & = \frac {4 a^3 (13 A+21 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (7 A+9 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (11 A+7 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{2}}(c+d x)}-\frac {1}{5} \left (2 a^3 (7 A+9 B)\right ) \int \sqrt {\cos (c+d x)} \, dx \\ & = -\frac {4 a^3 (7 A+9 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {4 a^3 (13 A+21 B) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {4 a^3 (41 A+42 B) \sin (c+d x)}{105 d \cos ^{\frac {3}{2}}(c+d x)}+\frac {4 a^3 (7 A+9 B) \sin (c+d x)}{5 d \sqrt {\cos (c+d x)}}+\frac {2 a A (a+a \cos (c+d x))^2 \sin (c+d x)}{7 d \cos ^{\frac {7}{2}}(c+d x)}+\frac {2 (11 A+7 B) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{35 d \cos ^{\frac {5}{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 = 8.88 (sec) , antiderivative size = 925, normalized size of antiderivative = 4.53 \[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\sqrt {\cos (c+d x)} (a+a \cos (c+d x))^3 \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \left (\frac {(7 A+9 B) \csc (c) \sec (c)}{10 d}+\frac {A \sec (c) \sec ^4(c+d x) \sin (d x)}{28 d}+\frac {\sec (c) \sec ^3(c+d x) (5 A \sin (c)+21 A \sin (d x)+7 B \sin (d x))}{140 d}+\frac {\sec (c) \sec ^2(c+d x) (63 A \sin (c)+21 B \sin (c)+130 A \sin (d x)+105 B \sin (d x))}{420 d}+\frac {\sec (c) \sec (c+d x) (130 A \sin (c)+105 B \sin (c)+294 A \sin (d x)+378 B \sin (d x))}{420 d}\right )-\frac {13 A (a+a \cos (c+d x))^3 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^6\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)))}}{42 d \sqrt {1+\cot ^2(c)}}-\frac {B (a+a \cos (c+d x))^3 \csc (c) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};\sin ^2(d x-\arctan (\cot (c)))\right ) \sec ^6\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)))}}{2 d \sqrt {1+\cot ^2(c)}}+\frac {7 A (a+a \cos (c+d x))^3 \csc (c) \sec ^6\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 )}{20 d}+\frac {9 B (a+a \cos (c+d x))^3 \csc (c) \sec ^6\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 )}{20 d} \]

[In]

Integrate[((a + a*Cos[c + d*x])^3*(A + B*Cos[c + d*x]))/Cos[c + d*x]^(9/2),x]

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(((7*A + 9*B)*Csc[c]*Sec[c])/(10*d) + (A*Sec[c]
*Sec[c + d*x]^4*Sin[d*x])/(28*d) + (Sec[c]*Sec[c + d*x]^3*(5*A*Sin[c] + 21*A*Sin[d*x] + 7*B*Sin[d*x]))/(140*d)
+ (Sec[c]*Sec[c + d*x]^2*(63*A*Sin[c] + 21*B*Sin[c] + 130*A*Sin[d*x] + 105*B*Sin[d*x]))/(420*d) + (Sec[c]*Sec
[c + d*x]*(130*A*Sin[c] + 105*B*Sin[c] + 294*A*Sin[d*x] + 378*B*Sin[d*x]))/(420*d)) - (13*A*(a + a*Cos[c + d*x
])^3*Csc[c]*HypergeometricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*Sec[d*x - A
rcTan[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]]]])/(42*d*Sqrt[1 + Cot[c]^2]) - (B*(a + a*Cos[c + d*x])^3*Csc[c]*Hypergeome
tricPFQ[{1/4, 1/2}, {5/4}, Sin[d*x - ArcTan[Cot[c]]]^2]*Sec[c/2 + (d*x)/2]^6*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 - A
rcTan[Cot[c]]]])/(2*d*Sqrt[1 + Cot[c]^2]) + (7*A*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((Hypergeo
metricPFQ[{-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*Cos[c]^2*Cos[d*x + ArcT
an[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
]]))/(20*d) + (9*B*(a + a*Cos[c + d*x])^3*Csc[c]*Sec[c/2 + (d*x)/2]^6*((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]) - ((S
in[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]]))/(20*d)

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(901\) vs. \(2(236)=472\).

Time = 13.76 (sec) , antiderivative size = 902, normalized size of antiderivative = 4.42

method	result	size
default	\(\text {Expression too large to display}\)	\(902\)
parts	\(\text {Expression too large to display}\)	\(1056\)

[In]

int((a+cos(d*x+c)*a)^3*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x,method=_RETURNVERBOSE)

[Out]

-16*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(1/8*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))+1/8*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)/(cos(1/2*d*x+1/2*c
)^2-1/2)^4-5/42*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+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)))+1/5*(1/8*B+3/8*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)*(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)^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*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*sin(1/
2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)+(3/8*A+3/8*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)))+(1/8*A+
3/8*B)/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)*Ellipti
cE(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 = 0.13 (sec) , antiderivative size = 263, normalized size of antiderivative = 1.29 \[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=-\frac {2 \, {\left (5 i \, \sqrt {2} {\left (13 \, A + 21 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 i \, \sqrt {2} {\left (13 \, A + 21 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 21 i \, \sqrt {2} {\left (7 \, A + 9 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 21 i \, \sqrt {2} {\left (7 \, A + 9 \, B\right )} a^{3} \cos \left (d x + c\right )^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - {\left (42 \, {\left (7 \, A + 9 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 5 \, {\left (26 \, A + 21 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 21 \, {\left (3 \, A + B\right )} a^{3} \cos \left (d x + c\right ) + 15 \, A a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{105 \, d \cos \left (d x + c\right )^{4}} \]

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x, algorithm="fricas")

[Out]

-2/105*(5*I*sqrt(2)*(13*A + 21*B)*a^3*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))
- 5*I*sqrt(2)*(13*A + 21*B)*a^3*cos(d*x + c)^4*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 21
*I*sqrt(2)*(7*A + 9*B)*a^3*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*s
in(d*x + c))) - 21*I*sqrt(2)*(7*A + 9*B)*a^3*cos(d*x + c)^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0,
cos(d*x + c) - I*sin(d*x + c))) - (42*(7*A + 9*B)*a^3*cos(d*x + c)^3 + 5*(26*A + 21*B)*a^3*cos(d*x + c)^2 + 21
*(3*A + B)*a^3*cos(d*x + c) + 15*A*a^3)*sqrt(cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^4)

Sympy [F(-1)]

Timed out. \[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\text {Timed out} \]

[In]

integrate((a+a*cos(d*x+c))**3*(A+B*cos(d*x+c))/cos(d*x+c)**(9/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(9/2), x)

Giac [F]

\[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} {\left (a \cos \left (d x + c\right ) + a\right )}^{3}}{\cos \left (d x + c\right )^{\frac {9}{2}}} \,d x } \]

[In]

integrate((a+a*cos(d*x+c))^3*(A+B*cos(d*x+c))/cos(d*x+c)^(9/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(a*cos(d*x + c) + a)^3/cos(d*x + c)^(9/2), x)

Mupad [B] (verification not implemented)

Time = 2.77 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.50 \[ \int \frac {(a+a \cos (c+d x))^3 (A+B \cos (c+d x))}{\cos ^{\frac {9}{2}}(c+d x)} \, dx=\frac {2\,B\,a^3\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{d}+\frac {\frac {2\,A\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {7}{4},\frac {1}{2};\ -\frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7}+\frac {6\,A\,a^3\,\cos \left (c+d\,x\right )\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5}+2\,A\,a^3\,{\cos \left (c+d\,x\right )}^2\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )+2\,A\,a^3\,{\cos \left (c+d\,x\right )}^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{7/2}\,\sqrt {1-{\cos \left (c+d\,x\right )}^2}}+\frac {6\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {1}{4},\frac {1}{2};\ \frac {3}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,\sqrt {\cos \left (c+d\,x\right )}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{4},\frac {1}{2};\ \frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{d\,{\cos \left (c+d\,x\right )}^{3/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}}+\frac {2\,B\,a^3\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (-\frac {5}{4},\frac {1}{2};\ -\frac {1}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{5\,d\,{\cos \left (c+d\,x\right )}^{5/2}\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

[In]

int(((A + B*cos(c + d*x))*(a + a*cos(c + d*x))^3)/cos(c + d*x)^(9/2),x)

[Out]

(2*B*a^3*ellipticF(c/2 + (d*x)/2, 2))/d + ((2*A*a^3*sin(c + d*x)*hypergeom([-7/4, 1/2], -3/4, cos(c + d*x)^2))
/7 + (6*A*a^3*cos(c + d*x)*sin(c + d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/5 + 2*A*a^3*cos(c + d*x)
^2*sin(c + d*x)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2) + 2*A*a^3*cos(c + d*x)^3*sin(c + d*x)*hypergeom([-
1/4, 1/2], 3/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(7/2)*(1 - cos(c + d*x)^2)^(1/2)) + (6*B*a^3*sin(c + d*x)*hyp
ergeom([-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^3*sin(c + d*x
)*hypergeom([-3/4, 1/2], 1/4, cos(c + d*x)^2))/(d*cos(c + d*x)^(3/2)*(sin(c + d*x)^2)^(1/2)) + (2*B*a^3*sin(c
+ d*x)*hypergeom([-5/4, 1/2], -1/4, cos(c + d*x)^2))/(5*d*cos(c + d*x)^(5/2)*(sin(c + d*x)^2)^(1/2))