3.305 \(\int \cos (c+d x) (a+b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=297 \[ \frac {2 \left (-6 a^2 B+21 a A b+25 b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 b d}-\frac {2 \left (a^2-b^2\right ) \left (-6 a^2 B+21 a A b+25 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (-6 a^3 B+21 a^2 A b+82 a b^2 B+63 A b^3\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (7 A b-2 a B) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b d}+\frac {2 B \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d} \]

[Out]

2/35*(7*A*b-2*B*a)*(a+b*cos(d*x+c))^(3/2)*sin(d*x+c)/b/d+2/7*B*(a+b*cos(d*x+c))^(5/2)*sin(d*x+c)/b/d+2/105*(21
*A*a*b-6*B*a^2+25*B*b^2)*sin(d*x+c)*(a+b*cos(d*x+c))^(1/2)/b/d+2/105*(21*A*a^2*b+63*A*b^3-6*B*a^3+82*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)*(b/(a+b))^(1/2))*(a+b*cos
(d*x+c))^(1/2)/b^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/105*(a^2-b^2)*(21*A*a*b-6*B*a^2+25*B*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)*(b/(a+b))^(1/2))*((a+b*cos(d*x+c))/(a+b
))^(1/2)/b^2/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.53, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {2968, 3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 \left (-6 a^2 B+21 a A b+25 b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{105 b d}-\frac {2 \left (a^2-b^2\right ) \left (-6 a^2 B+21 a A b+25 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (21 a^2 A b-6 a^3 B+82 a b^2 B+63 A b^3\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}+\frac {2 (7 A b-2 a B) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{35 b d}+\frac {2 B \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{7 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(21*a^2*A*b + 63*A*b^3 - 6*a^3*B + 82*a*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b
)])/(105*b^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(21*a*A*b - 6*a^2*B + 25*b^2*B)*Sqrt[(a +
b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(105*b^2*d*Sqrt[a + b*Cos[c + d*x]]) + (2*(21*
a*A*b - 6*a^2*B + 25*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(105*b*d) + (2*(7*A*b - 2*a*B)*(a + b*Cos[c
+ d*x])^(3/2)*Sin[c + d*x])/(35*b*d) + (2*B*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(7*b*d)

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2661

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, (2*b)
/(a + b)])/(d*Sqrt[a + b]), x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2663

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[(a + b*Sin[c + d*x])/(a + b)]/Sqrt[a
+ b*Sin[c + d*x]], Int[1/Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2752

Int[((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(b*c
- a*d)/b, Int[1/Sqrt[a + b*Sin[e + f*x]], x], x] + Dist[d/b, Int[Sqrt[a + b*Sin[e + f*x]], x], x] /; FreeQ[{a
, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0]

Rule 2753

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[(d
*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(f*(m + 1)), x] + Dist[1/(m + 1), Int[(a + b*Sin[e + f*x])^(m - 1)*Simp[
b*d*m + a*c*(m + 1) + (a*d*m + b*c*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*
c - a*d, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0] && IntegerQ[2*m]

Rule 2968

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 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]

Rubi steps

\begin {align*} \int \cos (c+d x) (a+b \cos (c+d x))^{3/2} (A+B \cos (c+d x)) \, dx &=\int (a+b \cos (c+d x))^{3/2} \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx\\ &=\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {2 \int (a+b \cos (c+d x))^{3/2} \left (\frac {5 b B}{2}+\frac {1}{2} (7 A b-2 a B) \cos (c+d x)\right ) \, dx}{7 b}\\ &=\frac {2 (7 A b-2 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {4 \int \sqrt {a+b \cos (c+d x)} \left (\frac {1}{4} b (21 A b+19 a B)+\frac {1}{4} \left (21 a A b-6 a^2 B+25 b^2 B\right ) \cos (c+d x)\right ) \, dx}{35 b}\\ &=\frac {2 \left (21 a A b-6 a^2 B+25 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}+\frac {2 (7 A b-2 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {8 \int \frac {\frac {1}{8} b \left (84 a A b+51 a^2 B+25 b^2 B\right )+\frac {1}{8} \left (21 a^2 A b+63 A b^3-6 a^3 B+82 a b^2 B\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b}\\ &=\frac {2 \left (21 a A b-6 a^2 B+25 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}+\frac {2 (7 A b-2 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}-\frac {\left (\left (a^2-b^2\right ) \left (21 a A b-6 a^2 B+25 b^2 B\right )\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{105 b^2}+\frac {\left (21 a^2 A b+63 A b^3-6 a^3 B+82 a b^2 B\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{105 b^2}\\ &=\frac {2 \left (21 a A b-6 a^2 B+25 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}+\frac {2 (7 A b-2 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}+\frac {\left (\left (21 a^2 A b+63 A b^3-6 a^3 B+82 a b^2 B\right ) \sqrt {a+b \cos (c+d x)}\right ) \int \sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}} \, dx}{105 b^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (a^2-b^2\right ) \left (21 a A b-6 a^2 B+25 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}}\right ) \int \frac {1}{\sqrt {\frac {a}{a+b}+\frac {b \cos (c+d x)}{a+b}}} \, dx}{105 b^2 \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (21 a^2 A b+63 A b^3-6 a^3 B+82 a b^2 B\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (a^2-b^2\right ) \left (21 a A b-6 a^2 B+25 b^2 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{105 b^2 d \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (21 a A b-6 a^2 B+25 b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{105 b d}+\frac {2 (7 A b-2 a B) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{35 b d}+\frac {2 B (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{7 b d}\\ \end {align*}

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Mathematica [A]  time = 1.07, size = 233, normalized size = 0.78 \[ \frac {b (a+b \cos (c+d x)) \left (\left (12 a^2 B+168 a A b+115 b^2 B\right ) \sin (c+d x)+3 b (2 (8 a B+7 A b) \sin (2 (c+d x))+5 b B \sin (3 (c+d x)))\right )+4 \sqrt {\frac {a+b \cos (c+d x)}{a+b}} \left (b^2 \left (51 a^2 B+84 a A b+25 b^2 B\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )+\left (-6 a^3 B+21 a^2 A b+82 a b^2 B+63 A b^3\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )\right )}{210 b^2 d \sqrt {a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(84*a*A*b + 51*a^2*B + 25*b^2*B)*EllipticF[(c + d*x)/2, (2*b)/(a +
b)] + (21*a^2*A*b + 63*A*b^3 - 6*a^3*B + 82*a*b^2*B)*((a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*Ellipt
icF[(c + d*x)/2, (2*b)/(a + b)])) + b*(a + b*Cos[c + d*x])*((168*a*A*b + 12*a^2*B + 115*b^2*B)*Sin[c + d*x] +
3*b*(2*(7*A*b + 8*a*B)*Sin[2*(c + d*x)] + 5*b*B*Sin[3*(c + d*x)])))/(210*b^2*d*Sqrt[a + b*Cos[c + d*x]])

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fricas [F]  time = 0.88, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (B b \cos \left (d x + c\right )^{3} + A a \cos \left (d x + c\right ) + {\left (B a + A b\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {b \cos \left (d x + c\right ) + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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maple [B]  time = 1.58, size = 1305, normalized size = 4.39 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/105*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*B*b^4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1
/2*c)^8+(-168*A*b^4-312*B*a*b^3-360*B*b^4)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(252*A*a*b^3+168*A*b^4+108*
B*a^2*b^2+312*B*a*b^3+280*B*b^4)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-84*A*a^2*b^2-126*A*a*b^3-42*A*b^4-6
*B*a^3*b-54*B*a^2*b^2-128*B*a*b^3-80*B*b^4)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-21*A*(sin(1/2*d*x+1/2*c)^2
)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a
^3*b+21*a*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2
*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^3+21*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(
a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b-21*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-
b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2+63*A*(sin(
1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b
/(a-b))^(1/2))*a*b^3-63*A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*Ell
ipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^4+6*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2
*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4-31*B*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^2
+25*B*b^4*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticF(cos(1/2*d
*x+1/2*c),(-2*b/(a-b))^(1/2))-6*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(
1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4+6*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2
*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^3*b+82*B*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2
))*a^2*b^2-82*B*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*b/(a-b)*sin(1/2*d*x+1/2*c)^2+(a+b)/(a-b))^(1/2)*EllipticE(cos
(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b^3)/b^2/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/si
n(1/2*d*x+1/2*c)/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (B \cos \left (d x + c\right ) + A\right )} {\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \cos \left (c+d\,x\right )\,\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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