3.2.37 \(\int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx\) [137]

Optimal. Leaf size=237 \[ \frac {4 a^3 (17 A+15 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^3 (121 A+105 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {4 a^3 (121 A+105 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {4 a^3 (17 A+15 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {20 a^3 (22 A+21 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 a B \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (11 A+15 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{99 d} \]

[Out]

4/15*a^3*(17*A+15*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
/231*a^3*(121*A+105*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
+4/45*a^3*(17*A+15*B)*cos(d*x+c)^(3/2)*sin(d*x+c)/d+20/693*a^3*(22*A+21*B)*cos(d*x+c)^(5/2)*sin(d*x+c)/d+2/11*
a*B*cos(d*x+c)^(5/2)*(a+a*cos(d*x+c))^2*sin(d*x+c)/d+2/99*(11*A+15*B)*cos(d*x+c)^(5/2)*(a^3+a^3*cos(d*x+c))*si
n(d*x+c)/d+4/231*a^3*(121*A+105*B)*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.32, antiderivative size = 237, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3055, 3047, 3102, 2827, 2715, 2720, 2719} \begin {gather*} \frac {4 a^3 (121 A+105 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {4 a^3 (17 A+15 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {20 a^3 (22 A+21 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{693 d}+\frac {4 a^3 (17 A+15 B) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {2 (11 A+15 B) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) \left (a^3 \cos (c+d x)+a^3\right )}{99 d}+\frac {4 a^3 (121 A+105 B) \sin (c+d x) \sqrt {\cos (c+d x)}}{231 d}+\frac {2 a B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a \cos (c+d x)+a)^2}{11 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^3*(17*A + 15*B)*EllipticE[(c + d*x)/2, 2])/(15*d) + (4*a^3*(121*A + 105*B)*EllipticF[(c + d*x)/2, 2])/(23
1*d) + (4*a^3*(121*A + 105*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (4*a^3*(17*A + 15*B)*Cos[c + d*x]^(3/
2)*Sin[c + d*x])/(45*d) + (20*a^3*(22*A + 21*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(693*d) + (2*a*B*Cos[c + d*x]
^(5/2)*(a + a*Cos[c + d*x])^2*Sin[c + d*x])/(11*d) + (2*(11*A + 15*B)*Cos[c + d*x]^(5/2)*(a^3 + a^3*Cos[c + d*
x])*Sin[c + d*x])/(99*d)

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[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 3055

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)*B*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)*((c + d*Sin[e + f*x
])^(n + 1)/(d*f*(m + n + 1))), x] + Dist[1/(d*(m + n + 1)), Int[(a + b*Sin[e + f*x])^(m - 1)*(c + d*Sin[e + f*
x])^n*Simp[a*A*d*(m + n + 1) + B*(a*c*(m - 1) + b*d*(n + 1)) + (A*b*d*(m + n + 1) - B*(b*c*m - a*d*(2*m + n)))
*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, 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 3102

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

Rubi steps

\begin {align*} \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx &=\frac {2 a B \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2}{11} \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x))^2 \left (\frac {1}{2} a (11 A+5 B)+\frac {1}{2} a (11 A+15 B) \cos (c+d x)\right ) \, dx\\ &=\frac {2 a B \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (11 A+15 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{99 d}+\frac {4}{99} \int \cos ^{\frac {3}{2}}(c+d x) (a+a \cos (c+d x)) \left (\frac {1}{2} a^2 (77 A+60 B)+\frac {5}{2} a^2 (22 A+21 B) \cos (c+d x)\right ) \, dx\\ &=\frac {2 a B \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (11 A+15 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{99 d}+\frac {4}{99} \int \cos ^{\frac {3}{2}}(c+d x) \left (\frac {1}{2} a^3 (77 A+60 B)+\left (\frac {5}{2} a^3 (22 A+21 B)+\frac {1}{2} a^3 (77 A+60 B)\right ) \cos (c+d x)+\frac {5}{2} a^3 (22 A+21 B) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {20 a^3 (22 A+21 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 a B \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (11 A+15 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{99 d}+\frac {8}{693} \int \cos ^{\frac {3}{2}}(c+d x) \left (\frac {9}{4} a^3 (121 A+105 B)+\frac {77}{4} a^3 (17 A+15 B) \cos (c+d x)\right ) \, dx\\ &=\frac {20 a^3 (22 A+21 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 a B \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (11 A+15 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{99 d}+\frac {1}{9} \left (2 a^3 (17 A+15 B)\right ) \int \cos ^{\frac {5}{2}}(c+d x) \, dx+\frac {1}{77} \left (2 a^3 (121 A+105 B)\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {4 a^3 (121 A+105 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {4 a^3 (17 A+15 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {20 a^3 (22 A+21 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 a B \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (11 A+15 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{99 d}+\frac {1}{15} \left (2 a^3 (17 A+15 B)\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{231} \left (2 a^3 (121 A+105 B)\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {4 a^3 (17 A+15 B) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {4 a^3 (121 A+105 B) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {4 a^3 (121 A+105 B) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {4 a^3 (17 A+15 B) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {20 a^3 (22 A+21 B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{693 d}+\frac {2 a B \cos ^{\frac {5}{2}}(c+d x) (a+a \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2 (11 A+15 B) \cos ^{\frac {5}{2}}(c+d x) \left (a^3+a^3 \cos (c+d x)\right ) \sin (c+d x)}{99 d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 6.35, size = 990, normalized size = 4.18 \begin {gather*} \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 {(17 A+15 B) \cot (c)}{30 d}+\frac {(2134 A+1953 B) \cos (d x) \sin (c)}{7392 d}+\frac {(73 A+75 B) \cos (2 d x) \sin (2 c)}{720 d}+\frac {3 (44 A+63 B) \cos (3 d x) \sin (3 c)}{4928 d}+\frac {(A+3 B) \cos (4 d x) \sin (4 c)}{288 d}+\frac {B \cos (5 d x) \sin (5 c)}{704 d}+\frac {(2134 A+1953 B) \cos (c) \sin (d x)}{7392 d}+\frac {(73 A+75 B) \cos (2 c) \sin (2 d x)}{720 d}+\frac {3 (44 A+63 B) \cos (3 c) \sin (3 d x)}{4928 d}+\frac {(A+3 B) \cos (4 c) \sin (4 d x)}{288 d}+\frac {B \cos (5 c) \sin (5 d x)}{704 d}\right )-\frac {11 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-\text {ArcTan}(\cot (c)))\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{42 d \sqrt {1+\cot ^2(c)}}-\frac {5 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-\text {ArcTan}(\cot (c)))\right ) \sec ^6\left (\frac {c}{2}+\frac {d x}{2}\right ) \sec (d x-\text {ArcTan}(\cot (c))) \sqrt {1-\sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {-\sqrt {1+\cot ^2(c)} \sin (c) \sin (d x-\text {ArcTan}(\cot (c)))} \sqrt {1+\sin (d x-\text {ArcTan}(\cot (c)))}}{22 d \sqrt {1+\cot ^2(c)}}-\frac {17 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+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{60 d}-\frac {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+\text {ArcTan}(\tan (c)))\right ) \sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1-\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {1+\cos (d x+\text {ArcTan}(\tan (c)))} \sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}} \sqrt {1+\tan ^2(c)}}-\frac {\frac {\sin (d x+\text {ArcTan}(\tan (c))) \tan (c)}{\sqrt {1+\tan ^2(c)}}+\frac {2 \cos ^2(c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}{\cos ^2(c)+\sin ^2(c)}}{\sqrt {\cos (c) \cos (d x+\text {ArcTan}(\tan (c))) \sqrt {1+\tan ^2(c)}}}\right )}{4 d} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

Sqrt[Cos[c + d*x]]*(a + a*Cos[c + d*x])^3*Sec[c/2 + (d*x)/2]^6*(-1/30*((17*A + 15*B)*Cot[c])/d + ((2134*A + 19
53*B)*Cos[d*x]*Sin[c])/(7392*d) + ((73*A + 75*B)*Cos[2*d*x]*Sin[2*c])/(720*d) + (3*(44*A + 63*B)*Cos[3*d*x]*Si
n[3*c])/(4928*d) + ((A + 3*B)*Cos[4*d*x]*Sin[4*c])/(288*d) + (B*Cos[5*d*x]*Sin[5*c])/(704*d) + ((2134*A + 1953
*B)*Cos[c]*Sin[d*x])/(7392*d) + ((73*A + 75*B)*Cos[2*c]*Sin[2*d*x])/(720*d) + (3*(44*A + 63*B)*Cos[3*c]*Sin[3*
d*x])/(4928*d) + ((A + 3*B)*Cos[4*c]*Sin[4*d*x])/(288*d) + (B*Cos[5*c]*Sin[5*d*x])/(704*d)) - (11*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 - ArcTan[Cot[c]]]*Sqrt[1 - Sin[d*x - ArcTan[Cot[c]]]]*Sqrt[-(Sqrt[1 + Cot[c]^2]*Sin[c]*Sin[d*x - ArcTan[C
ot[c]]])]*Sqrt[1 + Sin[d*x - ArcTan[Cot[c]]]])/(42*d*Sqrt[1 + Cot[c]^2]) - (5*B*(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 - 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 + S
in[d*x - ArcTan[Cot[c]]]])/(22*d*Sqrt[1 + Cot[c]^2]) - (17*A*(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])/(Sqr
t[1 - Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[1 + Cos[d*x + ArcTan[Tan[c]]]]*Sqrt[Cos[c]*Cos[d*x + ArcTan[Tan[c]]]*Sqr
t[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*Co
s[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]]))/(60*d) - (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]) - ((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]]))/(4*d)

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Maple [A]
time = 0.30, size = 441, normalized size = 1.86

method	result	size
default	\(-\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{3} \left (10080 B \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{12}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\left (-6160 A -43680 B \right ) \left (\sin ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (24200 A +77280 B \right ) \left (\sin ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-37532 A -72240 B \right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (29722 A +39270 B \right ) \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-8118 A -8820 B \right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+1815 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3927 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+1575 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-3465 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{3465 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(441\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-4/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(10080*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-6160*A-43680*B)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(24200*A+77280*B)*sin(1/2*d*x+1/2*c)^8*cos(
1/2*d*x+1/2*c)+(-37532*A-72240*B)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(29722*A+39270*B)*sin(1/2*d*x+1/2*c)
^4*cos(1/2*d*x+1/2*c)+(-8118*A-8820*B)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+1815*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))-3927*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))+1575*B*(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))-3465*B*(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)/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.14, size = 243, normalized size = 1.03 \begin {gather*} -\frac {2 \, {\left (15 i \, \sqrt {2} {\left (121 \, A + 105 \, B\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 i \, \sqrt {2} {\left (121 \, A + 105 \, B\right )} a^{3} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 231 i \, \sqrt {2} {\left (17 \, A + 15 \, B\right )} a^{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 ) + 231 i \, \sqrt {2} {\left (17 \, A + 15 \, B\right )} a^{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 (315 \, B a^{3} \cos \left (d x + c\right )^{4} + 385 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{3} + 135 \, {\left (11 \, A + 14 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 154 \, {\left (17 \, A + 15 \, B\right )} a^{3} \cos \left (d x + c\right ) + 30 \, {\left (121 \, A + 105 \, B\right )} a^{3}\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right )\right )}}{3465 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*(15*I*sqrt(2)*(121*A + 105*B)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 15*I*sqr
t(2)*(121*A + 105*B)*a^3*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 231*I*sqrt(2)*(17*A + 15*
B)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 231*I*sqrt(2)*(17*A
+ 15*B)*a^3*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c))) - (315*B*a^3*co
s(d*x + c)^4 + 385*(A + 3*B)*a^3*cos(d*x + c)^3 + 135*(11*A + 14*B)*a^3*cos(d*x + c)^2 + 154*(17*A + 15*B)*a^3
*cos(d*x + c) + 30*(121*A + 105*B)*a^3)*sqrt(cos(d*x + c))*sin(d*x + c))/d

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 3880 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 1.31, size = 360, normalized size = 1.52 \begin {gather*} \frac {A\,a^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}-\frac {6\,A\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,a^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,A\,a^3\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,B\,a^3\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a^3\,{\cos \left (c+d\,x\right )}^{13/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {13}{4};\ \frac {17}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{13\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(A*a^3*((2*cos(c + d*x)^(1/2)*sin(c + d*x))/3 + (2*ellipticF(c/2 + (d*x)/2, 2))/3))/d - (6*A*a^3*cos(c + d*x)^
(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^(1/2)) - (2*A*a^3*cos(c
+ d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(3*d*(sin(c + d*x)^2)^(1/2)) - (2*A*a^3
*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*(sin(c + d*x)^2)^(1/2))
- (2*B*a^3*cos(c + d*x)^(7/2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + d*x)^2)^
(1/2)) - (2*B*a^3*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(3*d*(sin(c + d
*x)^2)^(1/2)) - (6*B*a^3*cos(c + d*x)^(11/2)*sin(c + d*x)*hypergeom([1/2, 11/4], 15/4, cos(c + d*x)^2))/(11*d*
(sin(c + d*x)^2)^(1/2)) - (2*B*a^3*cos(c + d*x)^(13/2)*sin(c + d*x)*hypergeom([1/2, 13/4], 17/4, cos(c + d*x)^
2))/(13*d*(sin(c + d*x)^2)^(1/2))

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