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

Optimal. Leaf size=350 \[ -\frac{2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{315 b d}+\frac{4 a \left (-5 a^2 C+84 A b^2+57 b^2 C\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b d}-\frac{4 a \left (a^2-b^2\right ) \left (-5 a^2 C+84 A b^2+57 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (-3 a^2 b^2 (161 A+93 C)+10 a^4 C-21 b^4 (9 A+7 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{9 b d}-\frac{4 a C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{63 b d} \]

[Out]

(-2*(10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*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)]) - (4*a*(a^2 - b^2)*(84*A*b^2 - 5*a^2*C + 57*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]]) + (4*a*(84*A*b^2 - 5*a^2*C + 57*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(315*b*d) - (2*(10*a^2*C -
7*b^2*(9*A + 7*C))*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(315*b*d) - (4*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)

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Rubi [A]  time = 0.648788, antiderivative size = 350, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {3024, 2753, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \sin (c+d x) (a+b \cos (c+d x))^{3/2}}{315 b d}+\frac{4 a \left (-5 a^2 C+84 A b^2+57 b^2 C\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{315 b d}-\frac{4 a \left (a^2-b^2\right ) \left (-5 a^2 C+84 A b^2+57 b^2 C\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{a+b \cos (c+d x)}}-\frac{2 \left (-3 a^2 b^2 (161 A+93 C)+10 a^4 C-21 b^4 (9 A+7 C)\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{315 b^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}+\frac{2 C \sin (c+d x) (a+b \cos (c+d x))^{7/2}}{9 b d}-\frac{4 a C \sin (c+d x) (a+b \cos (c+d x))^{5/2}}{63 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(10*a^4*C - 21*b^4*(9*A + 7*C) - 3*a^2*b^2*(161*A + 93*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)]) - (4*a*(a^2 - b^2)*(84*A*b^2 - 5*a^2*C + 57*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]]) + (4*a*(84*A*b^2 - 5*a^2*C + 57*b^2*C)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(315*b*d) - (2*(10*a^2*C -
7*b^2*(9*A + 7*C))*(a + b*Cos[c + d*x])^(3/2)*Sin[c + d*x])/(315*b*d) - (4*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 3024

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (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) - a*C*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, 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 (A+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{1}{2} b (9 A+7 C)-a C \cos (c+d x)\right ) \, dx}{9 b}\\ &=-\frac{4 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} a b (21 A+13 C)-\frac{1}{4} \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) \cos (c+d x)\right ) \, dx}{63 b}\\ &=-\frac{2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}-\frac{4 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 (7 b^2 (9 A+7 C)+5 a^2 (21 A+11 C)\right )+\frac{3}{4} a \left (84 A b^2-5 a^2 C+57 b^2 C\right ) \cos (c+d x)\right ) \, dx}{315 b}\\ &=\frac{4 a \left (84 A b^2-5 a^2 C+57 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b d}-\frac{2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}-\frac{4 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} a b \left (5 a^2 (63 A+31 C)+3 b^2 (119 A+87 C)\right )-\frac{3}{16} \left (10 a^4 C-21 b^4 (9 A+7 C)-3 a^2 b^2 (161 A+93 C)\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{945 b}\\ &=\frac{4 a \left (84 A b^2-5 a^2 C+57 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b d}-\frac{2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}-\frac{4 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 (2 a \left (a^2-b^2\right ) \left (84 A b^2-5 a^2 C+57 b^2 C\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{315 b^2}-\frac{\left (10 a^4 C-21 b^4 (9 A+7 C)-3 a^2 b^2 (161 A+93 C)\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{315 b^2}\\ &=\frac{4 a \left (84 A b^2-5 a^2 C+57 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b d}-\frac{2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}-\frac{4 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 (10 a^4 C-21 b^4 (9 A+7 C)-3 a^2 b^2 (161 A+93 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 (2 a \left (a^2-b^2\right ) \left (84 A b^2-5 a^2 C+57 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 (10 a^4 C-21 b^4 (9 A+7 C)-3 a^2 b^2 (161 A+93 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{4 a \left (a^2-b^2\right ) \left (84 A b^2-5 a^2 C+57 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{4 a \left (84 A b^2-5 a^2 C+57 b^2 C\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{315 b d}-\frac{2 \left (10 a^2 C-7 b^2 (9 A+7 C)\right ) (a+b \cos (c+d x))^{3/2} \sin (c+d x)}{315 b d}-\frac{4 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 = 1.3186, size = 274, normalized size = 0.78 \[ \frac{b (a+b \cos (c+d x)) \left (2 a \left (20 a^2 C+924 A b^2+747 b^2 C\right ) \sin (c+d x)+b \left (\left (300 a^2 C+252 A b^2+266 b^2 C\right ) \sin (2 (c+d x))+5 b C (38 a \sin (3 (c+d x))+7 b \sin (4 (c+d x)))\right )\right )+8 \sqrt{\frac{a+b \cos (c+d x)}{a+b}} \left (a b^2 \left (5 a^2 (63 A+31 C)+3 b^2 (119 A+87 C)\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (3 a^2 b^2 (161 A+93 C)-10 a^4 C+21 b^4 (9 A+7 C)\right ) \left ((a+b) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )-a F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )\right )}{1260 b^2 d \sqrt{a+b \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(8*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*(a*b^2*(5*a^2*(63*A + 31*C) + 3*b^2*(119*A + 87*C))*EllipticF[(c + d*x)/
2, (2*b)/(a + b)] + (-10*a^4*C + 21*b^4*(9*A + 7*C) + 3*a^2*b^2*(161*A + 93*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*a*(924*A*b^2 + 20*a^2
*C + 747*b^2*C)*Sin[c + d*x] + b*((252*A*b^2 + 300*a^2*C + 266*b^2*C)*Sin[2*(c + d*x)] + 5*b*C*(38*a*Sin[3*(c
+ d*x)] + 7*b*Sin[4*(c + d*x)]))))/(1260*b^2*d*Sqrt[a + b*Cos[c + d*x]])

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Maple [B]  time = 0.492, size = 1527, 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)*(A+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+(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)+(-504*A*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)+(1176*A*a*b^4+504*A*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)+(-462*A*a^2*b^3-588*A*a*b^4-126*A*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)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-168*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^2+168*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)*E
llipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*b^4+483*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^2-483*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+189*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^4-189*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))*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(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a^5-124*C*(si
n(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^4-10*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*(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*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)/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 (C \cos \left (d x + c\right )^{2} + A\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)*(A+C*cos(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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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} + 2 \, C a b \cos \left (d x + c\right )^{3} + 2 \, A a b \cos \left (d x + c\right ) + A a^{2} +{\left (C a^{2} + A b^{2}\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)*(A+C*cos(d*x+c)^2),x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C \cos \left (d x + c\right )^{2} + A\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)*(A+C*cos(d*x+c)^2),x, algorithm="giac")

[Out]

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