3.334 \(\int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=413 \[ \frac {2 a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {2 \left (-2 a^2 B+a A b+b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}-\frac {2 a^2 \left (-6 a^3 B+3 a^2 A b+10 a b^2 B-7 A b^3\right ) \sin (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-16 a^4 B+8 a^3 A b+16 a^2 b^2 B-9 a A b^3+b^4 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (-16 a^5 B+8 a^4 A b+28 a^3 b^2 B-15 a^2 A b^3-8 a b^4 B+3 A b^5\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

[Out]

2/3*a*(A*b-B*a)*cos(d*x+c)^2*sin(d*x+c)/b/(a^2-b^2)/d/(a+b*cos(d*x+c))^(3/2)-2/3*a^2*(3*A*a^2*b-7*A*b^3-6*B*a^
3+10*B*a*b^2)*sin(d*x+c)/b^3/(a^2-b^2)^2/d/(a+b*cos(d*x+c))^(1/2)-2/3*(A*a*b-2*B*a^2+B*b^2)*sin(d*x+c)*(a+b*co
s(d*x+c))^(1/2)/b^3/(a^2-b^2)/d+2/3*(8*A*a^4*b-15*A*a^2*b^3+3*A*b^5-16*B*a^5+28*B*a^3*b^2-8*B*a*b^4)*(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^4/(a^2-b^2)^2/d/((a+b*cos(d*x+c))/(a+b))^(1/2)-2/3*(8*A*a^3*b-9*A*a*b^3-16*B*a^4+16*B*a^2*b^2+B*b^4)*(
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^4/(a^2-b^2)/d/(a+b*cos(d*x+c))^(1/2)

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Rubi [A]  time = 0.80, antiderivative size = 413, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.242, Rules used = {2989, 3031, 3023, 2752, 2663, 2661, 2655, 2653} \[ \frac {2 a (A b-a B) \sin (c+d x) \cos ^2(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 A b-6 a^3 B+10 a b^2 B-7 A b^3\right ) \sin (c+d x)}{3 b^3 d \left (a^2-b^2\right )^2 \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (-2 a^2 B+a A b+b^2 B\right ) \sin (c+d x) \sqrt {a+b \cos (c+d x)}}{3 b^3 d \left (a^2-b^2\right )}-\frac {2 \left (8 a^3 A b+16 a^2 b^2 B-16 a^4 B-9 a A b^3+b^4 B\right ) \sqrt {\frac {a+b \cos (c+d x)}{a+b}} F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}+\frac {2 \left (-15 a^2 A b^3+8 a^4 A b+28 a^3 b^2 B-16 a^5 B-8 a b^4 B+3 A b^5\right ) \sqrt {a+b \cos (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )}{3 b^4 d \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

Int[(Cos[c + d*x]^3*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(2*(8*a^4*A*b - 15*a^2*A*b^3 + 3*A*b^5 - 16*a^5*B + 28*a^3*b^2*B - 8*a*b^4*B)*Sqrt[a + b*Cos[c + d*x]]*Ellipti
cE[(c + d*x)/2, (2*b)/(a + b)])/(3*b^4*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*(8*a^3*A*b - 9
*a*A*b^3 - 16*a^4*B + 16*a^2*b^2*B + b^4*B)*Sqrt[(a + b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a
+ b)])/(3*b^4*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x]]) + (2*a*(A*b - a*B)*Cos[c + d*x]^2*Sin[c + d*x])/(3*b*(a
^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) - (2*a^2*(3*a^2*A*b - 7*A*b^3 - 6*a^3*B + 10*a*b^2*B)*Sin[c + d*x])/(3
*b^3*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]) - (2*(a*A*b - 2*a^2*B + b^2*B)*Sqrt[a + b*Cos[c + d*x]]*Sin[c +
d*x])/(3*b^3*(a^2 - b^2)*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 2989

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*c - a*d)*(B*c - A*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 1)
*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(n + 1)*(c^2 - d^2)), x] + Dist[1/(d*(n + 1)*(c^2 - d^2)), Int[(a + b*Sin[
e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1)*Simp[b*(b*c - a*d)*(B*c - A*d)*(m - 1) + a*d*(a*A*c + b*B*c - (
A*b + a*B)*d)*(n + 1) + (b*(b*d*(B*c - A*d) + a*(A*c*d + B*(c^2 - 2*d^2)))*(n + 1) - a*(b*c - a*d)*(B*c - A*d)
*(n + 2))*Sin[e + f*x] + b*(d*(A*b*c + a*B*c - a*A*d)*(m + n + 1) - b*B*(c^2*m + d^2*(n + 1)))*Sin[e + f*x]^2,
x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2,
0] && GtQ[m, 1] && LtQ[n, -1]

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 3031

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[((b*c - a*d)*(A*b^2 - a*b*B + a^2*C)*
Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b^2*f*(m + 1)*(a^2 - b^2)), x] - Dist[1/(b^2*(m + 1)*(a^2 - b^2)),
Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[b*(m + 1)*((b*B - a*C)*(b*c - a*d) - A*b*(a*c - b*d)) + (b*B*(a^2*d + b
^2*d*(m + 1) - a*b*c*(m + 2)) + (b*c - a*d)*(A*b^2*(m + 2) + C*(a^2 + b^2*(m + 1))))*Sin[e + f*x] - b*C*d*(m +
1)*(a^2 - b^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && Ne
Q[a^2 - b^2, 0] && LtQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\cos ^3(c+d x) (A+B \cos (c+d x))}{(a+b \cos (c+d x))^{5/2}} \, dx &=\frac {2 a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 \int \frac {\cos (c+d x) \left (-2 a (A b-a B)+\frac {3}{2} b (A b-a B) \cos (c+d x)+\frac {3}{2} \left (a A b-2 a^2 B+b^2 B\right ) \cos ^2(c+d x)\right )}{(a+b \cos (c+d x))^{3/2}} \, dx}{3 b \left (a^2-b^2\right )}\\ &=\frac {2 a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 A b-7 A b^3-6 a^3 B+10 a b^2 B\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {4 \int \frac {-\frac {1}{4} a b \left (3 a^2 A b-7 A b^3-6 a^3 B+10 a b^2 B\right )-\frac {1}{4} \left (6 a^4 A b-13 a^2 A b^3+3 A b^5-12 a^5 B+22 a^3 b^2 B-6 a b^4 B\right ) \cos (c+d x)+\frac {3}{4} b \left (a^2-b^2\right ) \left (a A b-2 a^2 B+b^2 B\right ) \cos ^2(c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b^3 \left (a^2-b^2\right )^2}\\ &=\frac {2 a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 A b-7 A b^3-6 a^3 B+10 a b^2 B\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (a A b-2 a^2 B+b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {8 \int \frac {-\frac {3}{8} b^2 \left (2 a^3 A b-6 a A b^3-4 a^4 B+7 a^2 b^2 B+b^4 B\right )-\frac {3}{8} b \left (8 a^4 A b-15 a^2 A b^3+3 A b^5-16 a^5 B+28 a^3 b^2 B-8 a b^4 B\right ) \cos (c+d x)}{\sqrt {a+b \cos (c+d x)}} \, dx}{9 b^4 \left (a^2-b^2\right )^2}\\ &=\frac {2 a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 A b-7 A b^3-6 a^3 B+10 a b^2 B\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (a A b-2 a^2 B+b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}-\frac {\left (8 a^3 A b-9 a A b^3-16 a^4 B+16 a^2 b^2 B+b^4 B\right ) \int \frac {1}{\sqrt {a+b \cos (c+d x)}} \, dx}{3 b^4 \left (a^2-b^2\right )}+\frac {\left (8 a^4 A b-15 a^2 A b^3+3 A b^5-16 a^5 B+28 a^3 b^2 B-8 a b^4 B\right ) \int \sqrt {a+b \cos (c+d x)} \, dx}{3 b^4 \left (a^2-b^2\right )^2}\\ &=\frac {2 a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 A b-7 A b^3-6 a^3 B+10 a b^2 B\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (a A b-2 a^2 B+b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}+\frac {\left (\left (8 a^4 A b-15 a^2 A b^3+3 A b^5-16 a^5 B+28 a^3 b^2 B-8 a 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}{3 b^4 \left (a^2-b^2\right )^2 \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {\left (\left (8 a^3 A b-9 a A b^3-16 a^4 B+16 a^2 b^2 B+b^4 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}{3 b^4 \left (a^2-b^2\right ) \sqrt {a+b \cos (c+d x)}}\\ &=\frac {2 \left (8 a^4 A b-15 a^2 A b^3+3 A b^5-16 a^5 B+28 a^3 b^2 B-8 a 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 )}{3 b^4 \left (a^2-b^2\right )^2 d \sqrt {\frac {a+b \cos (c+d x)}{a+b}}}-\frac {2 \left (8 a^3 A b-9 a A b^3-16 a^4 B+16 a^2 b^2 B+b^4 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 )}{3 b^4 \left (a^2-b^2\right ) d \sqrt {a+b \cos (c+d x)}}+\frac {2 a (A b-a B) \cos ^2(c+d x) \sin (c+d x)}{3 b \left (a^2-b^2\right ) d (a+b \cos (c+d x))^{3/2}}-\frac {2 a^2 \left (3 a^2 A b-7 A b^3-6 a^3 B+10 a b^2 B\right ) \sin (c+d x)}{3 b^3 \left (a^2-b^2\right )^2 d \sqrt {a+b \cos (c+d x)}}-\frac {2 \left (a A b-2 a^2 B+b^2 B\right ) \sqrt {a+b \cos (c+d x)} \sin (c+d x)}{3 b^3 \left (a^2-b^2\right ) d}\\ \end {align*}

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Mathematica [A]  time = 2.89, size = 334, normalized size = 0.81 \[ \frac {2 \left (\frac {\left (\frac {a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (b^2 \left (-4 a^4 B+2 a^3 A b+7 a^2 b^2 B-6 a A b^3+b^4 B\right ) F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-\left (16 a^5 B-8 a^4 A b-28 a^3 b^2 B+15 a^2 A b^3+8 a b^4 B-3 A b^5\right ) \left ((a+b) E\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )-a F\left (\frac {1}{2} (c+d x)|\frac {2 b}{a+b}\right )\right )\right )}{(a-b)^2 (a+b)}+\frac {b \sin (c+d x) \left (16 a^6 B-8 a^5 A b-25 a^4 b^2 B+16 a^3 A b^3+B \left (b^3-a^2 b\right )^2 \cos (2 (c+d x))+2 a b \left (10 a^4 B-5 a^3 A b-16 a^2 b^2 B+9 a A b^3+2 b^4 B\right ) \cos (c+d x)+b^6 B\right )}{2 \left (a^2-b^2\right )^2}\right )}{3 b^4 d (a+b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(Cos[c + d*x]^3*(A + B*Cos[c + d*x]))/(a + b*Cos[c + d*x])^(5/2),x]

[Out]

(2*((((a + b*Cos[c + d*x])/(a + b))^(3/2)*(b^2*(2*a^3*A*b - 6*a*A*b^3 - 4*a^4*B + 7*a^2*b^2*B + b^4*B)*Ellipti
cF[(c + d*x)/2, (2*b)/(a + b)] - (-8*a^4*A*b + 15*a^2*A*b^3 - 3*A*b^5 + 16*a^5*B - 28*a^3*b^2*B + 8*a*b^4*B)*(
(a + b)*EllipticE[(c + d*x)/2, (2*b)/(a + b)] - a*EllipticF[(c + d*x)/2, (2*b)/(a + b)])))/((a - b)^2*(a + b))
+ (b*(-8*a^5*A*b + 16*a^3*A*b^3 + 16*a^6*B - 25*a^4*b^2*B + b^6*B + 2*a*b*(-5*a^3*A*b + 9*a*A*b^3 + 10*a^4*B
- 16*a^2*b^2*B + 2*b^4*B)*Cos[c + d*x] + (-(a^2*b) + b^3)^2*B*Cos[2*(c + d*x)])*Sin[c + d*x])/(2*(a^2 - b^2)^2
)))/(3*b^4*d*(a + b*Cos[c + d*x])^(3/2))

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fricas [F]  time = 1.47, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (B \cos \left (d x + c\right )^{4} + A \cos \left (d x + c\right )^{3}\right )} \sqrt {b \cos \left (d x + c\right ) + a}}{b^{3} \cos \left (d x + c\right )^{3} + 3 \, a b^{2} \cos \left (d x + c\right )^{2} + 3 \, a^{2} b \cos \left (d x + c\right ) + a^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^3/(b*cos(d*x + c) + a)^(5/2), x)

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maple [B]  time = 7.26, size = 1389, normalized size = 3.36 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x)

[Out]

-(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/3/b^4*(4*B*b^2*cos(1/2*d*x+1/2*c)*sin(1/2*d*
x+1/2*c)^4+(-2*B*a*b-2*B*b^2)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-9*A*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))+3*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*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^2+17*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^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))-8*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+8*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)*Elliptic
E(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))*a*b)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+2*a
^2/b^4*(3*A*b-4*B*a)/sin(1/2*d*x+1/2*c)^2/(-2*sin(1/2*d*x+1/2*c)^2*b+a+b)/(a^2-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)*((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-(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+2*b*cos(1/2*d*x+1/2*c)*sin(1/2
*d*x+1/2*c)^2)-2*a^3*(A*b-B*a)/b^4*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(
1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2/b*(a-b))^2+8/3*b*sin(1/2*d*x+1/2*c)^2/(a-b)^2/(a+b)^2*cos(1/
2*d*x+1/2*c)*a/(-(-2*cos(1/2*d*x+1/2*c)^2*b-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+(3*a-b)/(3*a^3+3*a^2*b-3*a*b^2-3*
b^3)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b
)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-4/3*a/(a-b)/(a+b)^2*(sin(1/2*d*
x+1/2*c)^2)^(1/2)*((2*cos(1/2*d*x+1/2*c)^2*b+a-b)/(a-b))^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4*b+(a+b)*sin(1/2*d*x+1/
2*c)^2)^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-EllipticE(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{3}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^3*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((B*cos(d*x + c) + A)*cos(d*x + c)^3/(b*cos(d*x + c) + a)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\cos \left (c+d\,x\right )}^3\,\left (A+B\,\cos \left (c+d\,x\right )\right )}{{\left (a+b\,\cos \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((cos(c + d*x)^3*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^(5/2),x)

[Out]

int((cos(c + d*x)^3*(A + B*cos(c + d*x)))/(a + b*cos(c + d*x))^(5/2), 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)**3*(A+B*cos(d*x+c))/(a+b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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