3.312 \(\int \cos (c+d x) (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx\)

Optimal. Leaf size=372 \[ \frac{2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{315 b d}+\frac{2 \left (45 a^2 A b-10 a^3 B+114 a b^2 B+75 A b^3\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b d}-\frac{2 \left (a^2-b^2\right ) \left (45 a^2 A b-10 a^3 B+114 a b^2 B+75 A b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (45 a^3 A b+279 a^2 b^2 B-10 a^4 B+435 a A b^3+147 b^4 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (9 A b-2 a B) \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{63 b d}+\frac{2 B \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{9 b d} \]

[Out]

(2*(45*a^3*A*b + 435*a*A*b^3 - 10*a^4*B + 279*a^2*b^2*B + 147*b^4*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d
*x)/2, (2*b)/(a + b)])/(315*b^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(45*a^2*A*b + 75*A*b^3
- 10*a^3*B + 114*a*b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(315*b^2*d
*Sqrt[a + b*Cos[c + d*x]]) + (2*(45*a^2*A*b + 75*A*b^3 - 10*a^3*B + 114*a*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[
c + d*x])/(315*b*d) + (2*(45*a*A*b - 10*a^2*B + 49*b^2*B)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(315*b*d) +
(2*(9*A*b - 2*a*B)*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(63*b*d) + (2*B*(a + b*Cos[c + d*x])^(7/2)*Sin[c
+ d*x])/(9*b*d)

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Rubi [A]  time = 0.796367, antiderivative size = 372, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ \frac{2 \left (-10 a^2 B+45 a A b+49 b^2 B\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{315 b d}+\frac{2 \left (45 a^2 A b-10 a^3 B+114 a b^2 B+75 A b^3\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b d}-\frac{2 \left (a^2-b^2\right ) \left (45 a^2 A b-10 a^3 B+114 a b^2 B+75 A b^3\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (45 a^3 A b+279 a^2 b^2 B-10 a^4 B+435 a A b^3+147 b^4 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (9 A b-2 a B) \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{63 b d}+\frac{2 B \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{9 b d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]*(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]),x]

[Out]

(2*(45*a^3*A*b + 435*a*A*b^3 - 10*a^4*B + 279*a^2*b^2*B + 147*b^4*B)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d
*x)/2, (2*b)/(a + b)])/(315*b^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(a^2 - b^2)*(45*a^2*A*b + 75*A*b^3
- 10*a^3*B + 114*a*b^2*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(315*b^2*d
*Sqrt[a + b*Cos[c + d*x]]) + (2*(45*a^2*A*b + 75*A*b^3 - 10*a^3*B + 114*a*b^2*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[
c + d*x])/(315*b*d) + (2*(45*a*A*b - 10*a^2*B + 49*b^2*B)*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(315*b*d) +
(2*(9*A*b - 2*a*B)*(a + b*Cos[c + d*x])^(5/2)*Sin[c + d*x])/(63*b*d) + (2*B*(a + b*Cos[c + d*x])^(7/2)*Sin[c
+ d*x])/(9*b*d)

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]

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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \cos (c+d x) (a+b \cos (c+d x))^{5/2} (A+B \cos (c+d x)) \, dx &=\int (a+b \cos (c+d x))^{5/2} \left (A \cos (c+d x)+B \cos ^2(c+d x)\right ) \, dx\\ &=\frac{2 B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac{2 \int (a+b \cos (c+d x))^{5/2} \left (\frac{7 b B}{2}+\frac{1}{2} (9 A b-2 a B) \cos (c+d x)\right ) \, dx}{9 b}\\ &=\frac{2 (9 A b-2 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac{2 B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac{4 \int (a+b \cos (c+d x))^{3/2} \left (\frac{3}{4} b (15 A b+13 a B)+\frac{1}{4} \left (45 a A b-10 a^2 B+49 b^2 B\right ) \cos (c+d x)\right ) \, dx}{63 b}\\ &=\frac{2 \left (45 a A b-10 a^2 B+49 b^2 B\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}+\frac{2 (9 A b-2 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac{2 B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac{8 \int \sqrt{a+b \cos (c+d x)} \left (\frac{3}{8} b \left (120 a A b+55 a^2 B+49 b^2 B\right )+\frac{3}{8} \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right ) \cos (c+d x)\right ) \, dx}{315 b}\\ &=\frac{2 \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b d}+\frac{2 \left (45 a A b-10 a^2 B+49 b^2 B\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}+\frac{2 (9 A b-2 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac{2 B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac{16 \int \frac{\frac{3}{16} b \left (405 a^2 A b+75 A b^3+155 a^3 B+261 a b^2 B\right )+\frac{3}{16} \left (45 a^3 A b+435 a A b^3-10 a^4 B+279 a^2 b^2 B+147 b^4 B\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{945 b}\\ &=\frac{2 \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b d}+\frac{2 \left (45 a A b-10 a^2 B+49 b^2 B\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}+\frac{2 (9 A b-2 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac{2 B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}-\frac{\left (\left (a^2-b^2\right ) \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{315 b^2}+\frac{\left (45 a^3 A b+435 a A b^3-10 a^4 B+279 a^2 b^2 B+147 b^4 B\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{315 b^2}\\ &=\frac{2 \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b d}+\frac{2 \left (45 a A b-10 a^2 B+49 b^2 B\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}+\frac{2 (9 A b-2 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac{2 B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}+\frac{\left (\left (45 a^3 A b+435 a A b^3-10 a^4 B+279 a^2 b^2 B+147 b^4 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}{315 b^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (\left (a^2-b^2\right ) \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a 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}{315 b^2 \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (45 a^3 A b+435 a A b^3-10 a^4 B+279 a^2 b^2 B+147 b^4 B\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 \left (a^2-b^2\right ) \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a 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 )}{315 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (45 a^2 A b+75 A b^3-10 a^3 B+114 a b^2 B\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b d}+\frac{2 \left (45 a A b-10 a^2 B+49 b^2 B\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}+\frac{2 (9 A b-2 a B) (a+b \cos (c+d x))^{5/2} \sin (c+d x)}{63 b d}+\frac{2 B (a+b \cos (c+d x))^{7/2} \sin (c+d x)}{9 b d}\\ \end{align*}

Mathematica [A]  time = 1.48195, size = 291, normalized size = 0.78 \[ \frac{b (a+b \cos (c+d x)) \left (2 \left (540 a^2 A b+20 a^3 B+747 a b^2 B+345 A b^3\right ) \sin (c+d x)+b \left (\left (300 a^2 B+540 a A b+266 b^2 B\right ) \sin (2 (c+d x))+5 b (2 (19 a B+9 A b) \sin (3 (c+d x))+7 b B \sin (4 (c+d x)))\right )\right )+8 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (b^2 \left (405 a^2 A b+155 a^3 B+261 a b^2 B+75 A b^3\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (45 a^3 A b+279 a^2 b^2 B-10 a^4 B+435 a A b^3+147 b^4 B\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )\right )}{1260 b^2 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]*(a + b*Cos[c + d*x])^(5/2)*(A + B*Cos[c + d*x]),x]

[Out]

(8*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(b^2*(405*a^2*A*b + 75*A*b^3 + 155*a^3*B + 261*a*b^2*B)*EllipticF[(c + d
*x)/2, (2*b)/(a + b)] + (45*a^3*A*b + 435*a*A*b^3 - 10*a^4*B + 279*a^2*b^2*B + 147*b^4*B)*((a + b)*EllipticE[(
c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])) + b*(a + b*Cos[c + d*x])*(2*(540*a^2*A*
b + 345*A*b^3 + 20*a^3*B + 747*a*b^2*B)*Sin[c + d*x] + b*((540*a*A*b + 300*a^2*B + 266*b^2*B)*Sin[2*(c + d*x)]
+ 5*b*(2*(9*A*b + 19*a*B)*Sin[3*(c + d*x)] + 7*b*B*Sin[4*(c + d*x)]))))/(1260*b^2*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 4.468, size = 1635, normalized size = 4.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)*(a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x)

[Out]

-2/315*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*B*b^5*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^10+(720*A*b^5+2080*B*a*b^4+2240*B*b^5)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-1440*A*a*b^4-1080*A*b
^5-1360*B*a^2*b^3-3120*B*a*b^4-2072*B*b^5)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(1080*A*a^2*b^3+1440*A*a*b^
4+840*A*b^5+320*B*a^3*b^2+1360*B*a^2*b^3+2408*B*a*b^4+952*B*b^5)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-270
*A*a^3*b^2-540*A*a^2*b^3-510*A*a*b^4-240*A*b^5-10*B*a^4*b-160*B*a^3*b^2-666*B*a^2*b^3-684*B*a*b^4-168*B*b^5)*s
in(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-45*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^4*b-30*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^2*b^3+75*A*b^
5*(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))+45*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)*El
lipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^4*b-45*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^2+435*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^3-435*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^4+10*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^5-124*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^3*b^2+114*a
*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))*b^4-10*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^5+10*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*b+279*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^2-279*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(co
s(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^2*b^3+147*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^4-147*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))*b^5)/b^
2/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/sin(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., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(5/2)*cos(d*x + c), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (B b^{2} \cos \left (d x + c\right )^{4} + A a^{2} \cos \left (d x + c\right ) +{\left (2 \, B a b + A b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (B a^{2} + 2 \, A a b\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt{b \cos \left (d x + c\right ) + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))**(5/2)*(A+B*cos(d*x+c)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \cos \left (d x + c\right ) + A\right )}{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)*(a+b*cos(d*x+c))^(5/2)*(A+B*cos(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*(b*cos(d*x + c) + a)^(5/2)*cos(d*x + c), x)