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

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

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.629096, antiderivative size = 372, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.206, Rules used = {3023, 2753, 2752, 2663, 2661, 2655, 2653} \[ \frac{2 \left (-10 a^2 C+45 a b B+49 b^2 C\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{315 b d}+\frac{2 \left (45 a^2 b B-10 a^3 C+114 a b^2 C+75 b^3 B\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b d}-\frac{2 \left (a^2-b^2\right ) \left (45 a^2 b B-10 a^3 C+114 a b^2 C+75 b^3 B\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (279 a^2 b^2 C+45 a^3 b B-10 a^4 C+435 a b^3 B+147 b^4 C\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 (9 b B-2 a C) \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{63 b d}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{9 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 1.53207, size = 291, normalized size = 0.78 \[ \frac{b (a+b \cos (c+d x)) \left (2 \left (540 a^2 b B+20 a^3 C+747 a b^2 C+345 b^3 B\right ) \sin (c+d x)+b \left (\left (300 a^2 C+540 a b B+266 b^2 C\right ) \sin (2 (c+d x))+5 b (2 (19 a C+9 b B) \sin (3 (c+d x))+7 b C \sin (4 (c+d x)))\right )\right )+8 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (b^2 \left (405 a^2 b B+155 a^3 C+261 a b^2 C+75 b^3 B\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (279 a^2 b^2 C+45 a^3 b B-10 a^4 C+435 a b^3 B+147 b^4 C\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )\right )}{1260 b^2 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [B]  time = 0.863, 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((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x)

[Out]

-2/315*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(-1120*C*b^5*cos(1/2*d*x+1/2*c)*sin(1/2*d*x
+1/2*c)^10+(720*B*b^5+2080*C*a*b^4+2240*C*b^5)*sin(1/2*d*x+1/2*c)^8*cos(1/2*d*x+1/2*c)+(-1440*B*a*b^4-1080*B*b
^5-1360*C*a^2*b^3-3120*C*a*b^4-2072*C*b^5)*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)+(1080*B*a^2*b^3+1440*B*a*b^
4+840*B*b^5+320*C*a^3*b^2+1360*C*a^2*b^3+2408*C*a*b^4+952*C*b^5)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+(-270
*B*a^3*b^2-540*B*a^2*b^3-510*B*a*b^4-240*B*b^5-10*C*a^4*b-160*C*a^3*b^2-666*C*a^2*b^3-684*C*a*b^4-168*C*b^5)*s
in(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+45*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-45*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+435*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^3-435*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^4-45*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*b-30*a^2*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^3+75*b^5*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)*Ellipti
cF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-10*C*(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*C*(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*C*(s
in(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*C*(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+147*C*(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*C*(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+10*C*(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(co
s(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^5-124*C*(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*C*(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)/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

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (C b^{2} \cos \left (d x + c\right )^{4} + B a^{2} \cos \left (d x + c\right ) +{\left (2 \, C a b + B b^{2}\right )} \cos \left (d x + c\right )^{3} +{\left (C a^{2} + 2 \, B 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((a+b*cos(d*x+c))^(5/2)*(B*cos(d*x+c)+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 )}^{\frac{5}{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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