3.868 \(\int \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3 (B \cos (c+d x)+C \cos ^2(c+d x)) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.64, antiderivative size = 305, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 42, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3029, 2990, 3033, 3023, 2748, 2635, 2641, 2639} \[ \frac {2 \left (165 a^2 b C+77 a^3 B+165 a b^2 B+45 b^3 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {2 \left (27 a^2 b B+9 a^3 C+21 a b^2 C+7 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 b \left (26 a^2 C+33 a b B+9 b^2 C\right ) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{77 d}+\frac {2 \left (27 a^2 b B+9 a^3 C+21 a b^2 C+7 b^3 B\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {2 \left (165 a^2 b C+77 a^3 B+165 a b^2 B+45 b^3 C\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{231 d}+\frac {2 b^2 (15 a C+11 b B) \sin (c+d x) \cos ^{\frac {7}{2}}(c+d x)}{99 d}+\frac {2 b C \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x) (a+b \cos (c+d x))^2}{11 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2635

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] && Integer
Q[2*n]

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 2748

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 2990

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] /; F
reeQ[{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 3023

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 3029

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

Rule 3033

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 \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3 \left (B \cos (c+d x)+C \cos ^2(c+d x)\right ) \, dx &=\int \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^3 (B+C \cos (c+d x)) \, dx\\ &=\frac {2 b C \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 B+5 b C)+\frac {1}{2} \left (9 b^2 C+11 a (2 b B+a C)\right ) \cos (c+d x)+\frac {1}{2} b (11 b B+15 a C) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 b^2 (11 b B+15 a C) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 b C \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 B+5 b C)+\frac {11}{4} \left (27 a^2 b B+7 b^3 B+9 a^3 C+21 a b^2 C\right ) \cos (c+d x)+\frac {9}{4} b \left (33 a b B+26 a^2 C+9 b^2 C\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 b \left (33 a b B+26 a^2 C+9 b^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac {2 b^2 (11 b B+15 a C) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 b C \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 B+165 a b^2 B+165 a^2 b C+45 b^3 C\right )+\frac {77}{8} \left (27 a^2 b B+7 b^3 B+9 a^3 C+21 a b^2 C\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {2 b \left (33 a b B+26 a^2 C+9 b^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac {2 b^2 (11 b B+15 a C) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 b C \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 b B+7 b^3 B+9 a^3 C+21 a b^2 C\right ) \int \cos ^{\frac {5}{2}}(c+d x) \, dx+\frac {1}{77} \left (77 a^3 B+165 a b^2 B+165 a^2 b C+45 b^3 C\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx\\ &=\frac {2 \left (77 a^3 B+165 a b^2 B+165 a^2 b C+45 b^3 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {2 \left (27 a^2 b B+7 b^3 B+9 a^3 C+21 a b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b \left (33 a b B+26 a^2 C+9 b^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac {2 b^2 (11 b B+15 a C) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 b C \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 b B+7 b^3 B+9 a^3 C+21 a b^2 C\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{231} \left (77 a^3 B+165 a b^2 B+165 a^2 b C+45 b^3 C\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (27 a^2 b B+7 b^3 B+9 a^3 C+21 a b^2 C\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (77 a^3 B+165 a b^2 B+165 a^2 b C+45 b^3 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{231 d}+\frac {2 \left (77 a^3 B+165 a b^2 B+165 a^2 b C+45 b^3 C\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{231 d}+\frac {2 \left (27 a^2 b B+7 b^3 B+9 a^3 C+21 a b^2 C\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b \left (33 a b B+26 a^2 C+9 b^2 C\right ) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{77 d}+\frac {2 b^2 (11 b B+15 a C) \cos ^{\frac {7}{2}}(c+d x) \sin (c+d x)}{99 d}+\frac {2 b C \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 = 2.02, size = 235, normalized size = 0.77 \[ \frac {240 \left (77 a^3 B+165 a^2 b C+165 a b^2 B+45 b^3 C\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+3696 \left (9 a^3 C+27 a^2 b B+21 a b^2 C+7 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+2 \sin (c+d x) \sqrt {\cos (c+d x)} \left (180 b \left (33 a^2 C+33 a b B+16 b^2 C\right ) \cos (2 (c+d x))+154 \left (36 a^3 C+108 a^2 b B+129 a b^2 C+43 b^3 B\right ) \cos (c+d x)+15 \left (616 a^3 B+1716 a^2 b C+1716 a b^2 B+21 b^3 C \cos (4 (c+d x))+531 b^3 C\right )+770 b^2 (3 a C+b B) \cos (3 (c+d x))\right )}{27720 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.65, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (C b^{3} \cos \left (d x + c\right )^{5} + B a^{3} \cos \left (d x + c\right ) + {\left (3 \, C a b^{2} + B b^{3}\right )} \cos \left (d x + c\right )^{4} + 3 \, {\left (C a^{2} b + B a b^{2}\right )} \cos \left (d x + c\right )^{3} + {\left (C a^{3} + 3 \, B a^{2} b\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

integral((C*b^3*cos(d*x + c)^5 + B*a^3*cos(d*x + c) + (3*C*a*b^2 + B*b^3)*cos(d*x + c)^4 + 3*(C*a^2*b + B*a*b^
2)*cos(d*x + c)^3 + (C*a^3 + 3*B*a^2*b)*cos(d*x + c)^2)*sqrt(cos(d*x + c)), x)

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giac [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))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 2.29, size = 825, normalized size = 2.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x)

[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*C*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/
2*c)^12+(-12320*B*b^3-36960*C*a*b^2-50400*C*b^3)*sin(1/2*d*x+1/2*c)^10*cos(1/2*d*x+1/2*c)+(23760*B*a*b^2+24640
*B*b^3+23760*C*a^2*b+73920*C*a*b^2+56880*C*b^3)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-16632*B*a^2*b-35640*
B*a*b^2-22792*B*b^3-5544*C*a^3-35640*C*a^2*b-68376*C*a*b^2-34920*C*b^3)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c
)+(4620*B*a^3+16632*B*a^2*b+27720*B*a*b^2+10472*B*b^3+5544*C*a^3+27720*C*a^2*b+31416*C*a*b^2+13860*C*b^3)*sin(
1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-2310*B*a^3-4158*B*a^2*b-7920*B*a*b^2-1848*B*b^3-1386*C*a^3-7920*C*a^2*b-
5544*C*a*b^2-2790*C*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+1155*a^3*B*(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*B*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*B*(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*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))*b^3+2475*C*a^2*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*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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-2079*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))*a^3-4851*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))*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 \[ \int {\left (C \cos \left (d x + c\right )^{2} + B \cos \left (d x + c\right )\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt {\cos \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*cos(d*x+c))^3*(B*cos(d*x+c)+C*cos(d*x+c)^2)*cos(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 3.03, size = 364, normalized size = 1.19 \[ \frac {B\,a^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}-\frac {2\,C\,a^3\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,b^3\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,b^3\,{\cos \left (c+d\,x\right )}^{13/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {13}{4};\ \frac {17}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{13\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,B\,a^2\,b\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a\,b^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,C\,a^2\,b\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,C\,a\,b^2\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^(1/2)*(B*cos(c + d*x) + C*cos(c + d*x)^2)*(a + b*cos(c + d*x))^3,x)

[Out]

(B*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*C*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*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*C*
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*B*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*B*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*C*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*C*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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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))**3*(B*cos(d*x+c)+C*cos(d*x+c)**2)*cos(d*x+c)**(1/2),x)

[Out]

Timed out

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