3.4.59 \(\int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx\) [359]

Optimal. Leaf size=305 \[ \frac {2 \left (27 a^2 A b+7 A b^3+9 a^3 B+21 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (77 a^3 A+165 a A b^2+165 a^2 b B+45 b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {2 \left (77 a^3 A+165 a A b^2+165 a^2 b B+45 b^3 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {2 \left (27 a^2 A b+7 A b^3+9 a^3 B+21 a b^2 B\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b \left (33 a A b+26 a^2 B+9 b^2 B\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac {2 b^2 (11 A b+15 a B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 b B \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{11 d} \]

[Out]

2/15*(27*A*a^2*b+7*A*b^3+9*B*a^3+21*B*a*b^2)*(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+2/231*(77*A*a^3+165*A*a*b^2+165*B*a^2*b+45*B*b^3)*(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+2/45*(27*A*a^2*b+7*A*b^3+9*B*a^3+21*B*a*b^2)*cos(d*x+c)^(3/
2)*sin(d*x+c)/d+2/77*b*(33*A*a*b+26*B*a^2+9*B*b^2)*cos(d*x+c)^(5/2)*sin(d*x+c)/d+2/99*b^2*(11*A*b+15*B*a)*cos(
d*x+c)^(7/2)*sin(d*x+c)/d+2/11*b*B*cos(d*x+c)^(5/2)*(a+b*cos(d*x+c))^2*sin(d*x+c)/d+2/231*(77*A*a^3+165*A*a*b^
2+165*B*a^2*b+45*B*b^3)*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]
time = 0.35, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3069, 3112, 3102, 2827, 2715, 2720, 2719} \begin {gather*} \frac {2 b \left (26 a^2 B+33 a A b+9 b^2 B\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{77 d}+\frac {2 \left (77 a^3 A+165 a^2 b B+165 a A b^2+45 b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {2 \left (9 a^3 B+27 a^2 A b+21 a b^2 B+7 A b^3\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (9 a^3 B+27 a^2 A b+21 a b^2 B+7 A b^3\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {2 \left (77 a^3 A+165 a^2 b B+165 a A b^2+45 b^3 B\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{231 d}+\frac {2 b^2 (15 a B+11 A b) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{99 d}+\frac {2 b B \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(27*a^2*A*b + 7*A*b^3 + 9*a^3*B + 21*a*b^2*B)*EllipticE[(c + d*x)/2, 2])/(15*d) + (2*(77*a^3*A + 165*a*A*b^
2 + 165*a^2*b*B + 45*b^3*B)*EllipticF[(c + d*x)/2, 2])/(231*d) + (2*(77*a^3*A + 165*a*A*b^2 + 165*a^2*b*B + 45
*b^3*B)*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(231*d) + (2*(27*a^2*A*b + 7*A*b^3 + 9*a^3*B + 21*a*b^2*B)*Cos[c + d*
x]^(3/2)*Sin[c + d*x])/(45*d) + (2*b*(33*a*A*b + 26*a^2*B + 9*b^2*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(77*d) +
(2*b^2*(11*A*b + 15*a*B)*Cos[c + d*x]^(7/2)*Sin[c + d*x])/(99*d) + (2*b*B*Cos[c + d*x]^(5/2)*(a + b*Cos[c + d
*x])^2*Sin[c + d*x])/(11*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 3069

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

Rule 3112

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])*((A_.) + (B_.)*sin[(e
_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-C)*d*Cos[e + f*x]*Sin[e + f*x]*((a +
b*Sin[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Dist[1/(b*(m + 3)), Int[(a + b*Sin[e + f*x])^m*Simp[a*C*d + A*b*
c*(m + 3) + b*(B*c*(m + 3) + d*(C*(m + 2) + A*(m + 3)))*Sin[e + f*x] - (2*a*C*d - b*(c*C + B*d)*(m + 3))*Sin[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] &&
!LtQ[m, -1]

Rubi steps

\begin {align*} \int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx &=\frac {2 b B \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {2}{11} \int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x)) \left (\frac {1}{2} a (11 a A+5 b B)+\frac {1}{2} \left (9 b^2 B+11 a (2 A b+a B)\right ) \cos (c+d x)+\frac {1}{2} b (11 A b+15 a B) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 b^2 (11 A b+15 a B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 b B \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {4}{99} \int \cos ^{\frac {3}{2}}(c+d x) \left (\frac {9}{4} a^2 (11 a A+5 b B)+\frac {11}{4} \left (27 a^2 A b+7 A b^3+9 a^3 B+21 a b^2 B\right ) \cos (c+d x)+\frac {9}{4} b \left (33 a A b+26 a^2 B+9 b^2 B\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 b \left (33 a A b+26 a^2 B+9 b^2 B\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac {2 b^2 (11 A b+15 a B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 b B \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {8}{693} \int \cos ^{\frac {3}{2}}(c+d x) \left (\frac {9}{8} \left (77 a^3 A+165 a A b^2+165 a^2 b B+45 b^3 B\right )+\frac {77}{8} \left (27 a^2 A b+7 A b^3+9 a^3 B+21 a b^2 B\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {2 b \left (33 a A b+26 a^2 B+9 b^2 B\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac {2 b^2 (11 A b+15 a B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 b B \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {1}{9} \left (27 a^2 A b+7 A b^3+9 a^3 B+21 a b^2 B\right ) \int \cos ^{\frac {5}{2}}(c+d x) \, dx+\frac {1}{77} \left (77 a^3 A+165 a A b^2+165 a^2 b B+45 b^3 B\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 \left (77 a^3 A+165 a A b^2+165 a^2 b B+45 b^3 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {2 \left (27 a^2 A b+7 A b^3+9 a^3 B+21 a b^2 B\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b \left (33 a A b+26 a^2 B+9 b^2 B\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac {2 b^2 (11 A b+15 a B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 b B \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{11 d}+\frac {1}{15} \left (27 a^2 A b+7 A b^3+9 a^3 B+21 a b^2 B\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{231} \left (77 a^3 A+165 a A b^2+165 a^2 b B+45 b^3 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (27 a^2 A b+7 A b^3+9 a^3 B+21 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (77 a^3 A+165 a A b^2+165 a^2 b B+45 b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {2 \left (77 a^3 A+165 a A b^2+165 a^2 b B+45 b^3 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {2 \left (27 a^2 A b+7 A b^3+9 a^3 B+21 a b^2 B\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b \left (33 a A b+26 a^2 B+9 b^2 B\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac {2 b^2 (11 A b+15 a B) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 b B \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{11 d}\\ \end {align*}

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Mathematica [A]
time = 2.04, size = 235, normalized size = 0.77 \begin {gather*} \frac {3696 \left (27 a^2 A b+7 A b^3+9 a^3 B+21 a b^2 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+240 \left (77 a^3 A+165 a A b^2+165 a^2 b B+45 b^3 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \sqrt {\cos (c+d x)} \left (154 \left (108 a^2 A b+43 A b^3+36 a^3 B+129 a b^2 B\right ) \cos (c+d x)+180 b \left (33 a A b+33 a^2 B+16 b^2 B\right ) \cos (2 (c+d x))+770 b^2 (A b+3 a B) \cos (3 (c+d x))+15 \left (616 a^3 A+1716 a A b^2+1716 a^2 b B+531 b^3 B+21 b^3 B \cos (4 (c+d x))\right )\right ) \sin (c+d x)}{27720 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3696*(27*a^2*A*b + 7*A*b^3 + 9*a^3*B + 21*a*b^2*B)*EllipticE[(c + d*x)/2, 2] + 240*(77*a^3*A + 165*a*A*b^2 +
165*a^2*b*B + 45*b^3*B)*EllipticF[(c + d*x)/2, 2] + 2*Sqrt[Cos[c + d*x]]*(154*(108*a^2*A*b + 43*A*b^3 + 36*a^3
*B + 129*a*b^2*B)*Cos[c + d*x] + 180*b*(33*a*A*b + 33*a^2*B + 16*b^2*B)*Cos[2*(c + d*x)] + 770*b^2*(A*b + 3*a*
B)*Cos[3*(c + d*x)] + 15*(616*a^3*A + 1716*a*A*b^2 + 1716*a^2*b*B + 531*b^3*B + 21*b^3*B*Cos[4*(c + d*x)]))*Si
n[c + d*x])/(27720*d)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(824\) vs. \(2(333)=666\).
time = 0.37, size = 825, normalized size = 2.70

method	result	size
default	\(\text {Expression too large to display}\)	\(825\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3465*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(20160*B*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)
^12*b^3+(-12320*A*b^3-36960*B*a*b^2-50400*B*b^3)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(23760*A*a*b^2+24640
*A*b^3+23760*B*a^2*b+73920*B*a*b^2+56880*B*b^3)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-16632*A*a^2*b-35640*
A*a*b^2-22792*A*b^3-5544*B*a^3-35640*B*a^2*b-68376*B*a*b^2-34920*B*b^3)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c
)+(4620*A*a^3+16632*A*a^2*b+27720*A*a*b^2+10472*A*b^3+5544*B*a^3+27720*B*a^2*b+31416*B*a*b^2+13860*B*b^3)*sin(
1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-2310*A*a^3-4158*A*a^2*b-7920*A*a*b^2-1848*A*b^3-1386*B*a^3-7920*B*a^2*b-
5544*B*a*b^2-2790*B*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+1155*A*a^3*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+2475*A*a*b^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))-6237*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*si
n(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^2*b-1617*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))*b^3+2475*a^2*b*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))+675*b^3*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))-2079*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a^3-4851*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))*a*b^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+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*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 = 0.16, size = 358, normalized size = 1.17 \begin {gather*} \frac {2 \, {\left (315 \, B b^{3} \cos \left (d x + c\right )^{4} + 1155 \, A a^{3} + 2475 \, B a^{2} b + 2475 \, A a b^{2} + 675 \, B b^{3} + 385 \, {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + 135 \, {\left (11 \, B a^{2} b + 11 \, A a b^{2} + 3 \, B b^{3}\right )} \cos \left (d x + c\right )^{2} + 77 \, {\left (9 \, B a^{3} + 27 \, A a^{2} b + 21 \, B a b^{2} + 7 \, A b^{3}\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 15 \, \sqrt {2} {\left (77 i \, A a^{3} + 165 i \, B a^{2} b + 165 i \, A a b^{2} + 45 i \, B b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 15 \, \sqrt {2} {\left (-77 i \, A a^{3} - 165 i \, B a^{2} b - 165 i \, A a b^{2} - 45 i \, B b^{3}\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 231 \, \sqrt {2} {\left (-9 i \, B a^{3} - 27 i \, A a^{2} b - 21 i \, B a b^{2} - 7 i \, A b^{3}\right )} {\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 \, \sqrt {2} {\left (9 i \, B a^{3} + 27 i \, A a^{2} b + 21 i \, B a b^{2} + 7 i \, A b^{3}\right )} {\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 )}{3465 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3465*(2*(315*B*b^3*cos(d*x + c)^4 + 1155*A*a^3 + 2475*B*a^2*b + 2475*A*a*b^2 + 675*B*b^3 + 385*(3*B*a*b^2 +
A*b^3)*cos(d*x + c)^3 + 135*(11*B*a^2*b + 11*A*a*b^2 + 3*B*b^3)*cos(d*x + c)^2 + 77*(9*B*a^3 + 27*A*a^2*b + 21
*B*a*b^2 + 7*A*b^3)*cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) - 15*sqrt(2)*(77*I*A*a^3 + 165*I*B*a^2*b + 1
65*I*A*a*b^2 + 45*I*B*b^3)*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 15*sqrt(2)*(-77*I*A*a^3
- 165*I*B*a^2*b - 165*I*A*a*b^2 - 45*I*B*b^3)*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 231
*sqrt(2)*(-9*I*B*a^3 - 27*I*A*a^2*b - 21*I*B*a*b^2 - 7*I*A*b^3)*weierstrassZeta(-4, 0, weierstrassPInverse(-4,
0, cos(d*x + c) + I*sin(d*x + c))) - 231*sqrt(2)*(9*I*B*a^3 + 27*I*A*a^2*b + 21*I*B*a*b^2 + 7*I*A*b^3)*weiers
trassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) - I*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+b*cos(d*x+c))**3*(A+B*cos(d*x+c)),x)

[Out]

Exception raised: SystemError >> excessive stack use: stack is 7319 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+b*cos(d*x+c))^3*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.74, size = 364, normalized size = 1.19 \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 {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\,A\,b^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\,b^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}}-\frac {6\,A\,a^2\,b\,{\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\,b^2\,{\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\,B\,a^2\,b\,{\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\,b^2\,{\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}} \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 + b*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 - (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*A*b^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*
b^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
)) - (6*A*a^2*b*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*b^2*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 13/4, cos(c + d*x)^2))/(3*d*(si
n(c + d*x)^2)^(1/2)) - (2*B*a^2*b*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*b^2*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))

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