3.360 \(\int \sqrt {\cos (c+d x)} (a+b \cos (c+d x))^3 (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=255 \[ \frac {2 b \left (22 a^2 B+27 a A b+7 b^2 B\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {2 \left (7 a^3 B+21 a^2 A b+15 a b^2 B+5 A b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (7 a^3 B+21 a^2 A b+15 a b^2 B+5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}+\frac {2 b^2 (13 a B+9 A b) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{63 d}+\frac {2 b B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2}{9 d} \]

[Out]

2/15*(15*A*a^3+27*A*a*b^2+27*B*a^2*b+7*B*b^3)*(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+2/21*(21*A*a^2*b+5*A*b^3+7*B*a^3+15*B*a*b^2)*(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*b*(27*A*a*b+22*B*a^2+7*B*b^2)*cos(d*x+c)^(3/2)*sin(d*x+c)/
d+2/63*b^2*(9*A*b+13*B*a)*cos(d*x+c)^(5/2)*sin(d*x+c)/d+2/9*b*B*cos(d*x+c)^(3/2)*(a+b*cos(d*x+c))^2*sin(d*x+c)
/d+2/21*(21*A*a^2*b+5*A*b^3+7*B*a^3+15*B*a*b^2)*sin(d*x+c)*cos(d*x+c)^(1/2)/d

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Rubi [A]  time = 0.50, antiderivative size = 255, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 7, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {2990, 3033, 3023, 2748, 2639, 2635, 2641} \[ \frac {2 \left (21 a^2 A b+7 a^3 B+15 a b^2 B+5 A b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 b \left (22 a^2 B+27 a A b+7 b^2 B\right ) \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{45 d}+\frac {2 \left (21 a^2 A b+7 a^3 B+15 a b^2 B+5 A b^3\right ) \sin (c+d x) \sqrt {\cos (c+d x)}}{21 d}+\frac {2 b^2 (13 a B+9 A b) \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{63 d}+\frac {2 b B \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2}{9 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(15*a^3*A + 27*a*A*b^2 + 27*a^2*b*B + 7*b^3*B)*EllipticE[(c + d*x)/2, 2])/(15*d) + (2*(21*a^2*A*b + 5*A*b^3
+ 7*a^3*B + 15*a*b^2*B)*EllipticF[(c + d*x)/2, 2])/(21*d) + (2*(21*a^2*A*b + 5*A*b^3 + 7*a^3*B + 15*a*b^2*B)*
Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(21*d) + (2*b*(27*a*A*b + 22*a^2*B + 7*b^2*B)*Cos[c + d*x]^(3/2)*Sin[c + d*x]
)/(45*d) + (2*b^2*(9*A*b + 13*a*B)*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(63*d) + (2*b*B*Cos[c + d*x]^(3/2)*(a + b*
Cos[c + d*x])^2*Sin[c + d*x])/(9*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 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 (A+B \cos (c+d x)) \, dx &=\frac {2 b B \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac {2}{9} \int \sqrt {\cos (c+d x)} (a+b \cos (c+d x)) \left (\frac {3}{2} a (3 a A+b B)+\frac {1}{2} \left (7 b^2 B+9 a (2 A b+a B)\right ) \cos (c+d x)+\frac {1}{2} b (9 A b+13 a B) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 b^2 (9 A b+13 a B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b B \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac {4}{63} \int \sqrt {\cos (c+d x)} \left (\frac {21}{4} a^2 (3 a A+b B)+\frac {9}{4} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \cos (c+d x)+\frac {7}{4} b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b^2 (9 A b+13 a B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b B \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac {8}{315} \int \sqrt {\cos (c+d x)} \left (\frac {21}{8} \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right )+\frac {45}{8} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \cos (c+d x)\right ) \, dx\\ &=\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b^2 (9 A b+13 a B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b B \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac {1}{7} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \int \cos ^{\frac {3}{2}}(c+d x) \, dx+\frac {1}{15} \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {2 \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b^2 (9 A b+13 a B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b B \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d}+\frac {1}{21} \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {2 \left (15 a^3 A+27 a A b^2+27 a^2 b B+7 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{21 d}+\frac {2 \left (21 a^2 A b+5 A b^3+7 a^3 B+15 a b^2 B\right ) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 b \left (27 a A b+22 a^2 B+7 b^2 B\right ) \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{45 d}+\frac {2 b^2 (9 A b+13 a B) \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{63 d}+\frac {2 b B \cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2 \sin (c+d x)}{9 d}\\ \end {align*}

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Mathematica [A]  time = 1.23, size = 197, normalized size = 0.77 \[ \frac {60 \left (7 a^3 B+21 a^2 A b+15 a b^2 B+5 A b^3\right ) F\left (\left .\frac {1}{2} (c+d x)\right |2\right )+84 \left (15 a^3 A+27 a^2 b B+27 a A b^2+7 b^3 B\right ) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+\sin (c+d x) \sqrt {\cos (c+d x)} \left (7 b \left (108 a^2 B+108 a A b+43 b^2 B\right ) \cos (c+d x)+5 \left (84 a^3 B+252 a^2 A b+18 b^2 (3 a B+A b) \cos (2 (c+d x))+234 a b^2 B+78 A b^3+7 b^3 B \cos (3 (c+d x))\right )\right )}{630 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(84*(15*a^3*A + 27*a*A*b^2 + 27*a^2*b*B + 7*b^3*B)*EllipticE[(c + d*x)/2, 2] + 60*(21*a^2*A*b + 5*A*b^3 + 7*a^
3*B + 15*a*b^2*B)*EllipticF[(c + d*x)/2, 2] + Sqrt[Cos[c + d*x]]*(7*b*(108*a*A*b + 108*a^2*B + 43*b^2*B)*Cos[c
+ d*x] + 5*(252*a^2*A*b + 78*A*b^3 + 84*a^3*B + 234*a*b^2*B + 18*b^2*(A*b + 3*a*B)*Cos[2*(c + d*x)] + 7*b^3*B
*Cos[3*(c + d*x)]))*Sin[c + d*x])/(630*d)

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b^{3} \cos \left (d x + c\right )^{4} + A a^{3} + {\left (3 \, B a b^{2} + A b^{3}\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (B a^{2} b + A a b^{2}\right )} \cos \left (d x + c\right )^{2} + {\left (B a^{3} + 3 \, A a^{2} b\right )} \cos \left (d x + c\right )\right )} \sqrt {\cos \left (d x + c\right )}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt {\cos \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(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*sqrt(cos(d*x + c)), x)

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maple [B]  time = 1.58, size = 745, normalized size = 2.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*B*b^3*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2
*c)^10+(720*A*b^3+2160*B*a*b^2+2240*B*b^3)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-1512*A*a*b^2-1080*A*b^3-1
512*B*a^2*b-3240*B*a*b^2-2072*B*b^3)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(1260*A*a^2*b+1512*A*a*b^2+840*A*
b^3+420*B*a^3+1512*B*a^2*b+2520*B*a*b^2+952*B*b^3)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-630*A*a^2*b-378*A
*a*b^2-240*A*b^3-210*B*a^3-378*B*a^2*b-720*B*a*b^2-168*B*b^3)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-315*A*(s
in(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^3-567*A*(s
in(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+315*A*
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))+75*A
*b^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)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-567*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^2*b-147
*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+105
*a^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))+225
*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)))/
(-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 \[ \int {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt {\cos \left (d x + c\right )}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^(1/2)*(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*sqrt(cos(d*x + c)), x)

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mupad [B]  time = 1.54, size = 328, normalized size = 1.29 \[ \frac {2\,\left (A\,a^3\,\mathrm {E}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+A\,a^2\,b\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )+A\,a^2\,b\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )\right )}{d}+\frac {B\,a^3\,\left (\frac {2\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3}+\frac {2\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3}\right )}{d}-\frac {2\,A\,b^3\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,b^3\,{\cos \left (c+d\,x\right )}^{11/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {11}{4};\ \frac {15}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{11\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,A\,a\,b^2\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {6\,B\,a^2\,b\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,a\,b^2\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{3\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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