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

Optimal. Leaf size=436 \[ -\frac{2 a^2 \sin (c+d x) \cos ^3(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac{8 a^2 \left (2 a^2-3 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-71 a^2 b^2+48 a^4+3 b^4\right ) \sin (c+d x) \cos (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b^3 d \left (a^2-b^2\right )^2}-\frac{4 a \left (-49 a^2 b^2+32 a^4+7 b^4\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b^4 d \left (a^2-b^2\right )^2}-\frac{2 a \left (-116 a^2 b^2+128 a^4-17 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^5 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-212 a^4 b^2+55 a^2 b^4+128 a^6+9 b^6\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^5 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

[Out]

(2*(128*a^6 - 212*a^4*b^2 + 55*a^2*b^4 + 9*b^6)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)]
)/(15*b^5*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*a*(128*a^4 - 116*a^2*b^2 - 17*b^4)*Sqrt[(a
+ b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(15*b^5*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x
]]) - (2*a^2*Cos[c + d*x]^3*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) - (8*a^2*(2*a^2 - 3*b
^2)*Cos[c + d*x]^2*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]) - (4*a*(32*a^4 - 49*a^2*b^2
+ 7*b^4)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*b^4*(a^2 - b^2)^2*d) + (2*(48*a^4 - 71*a^2*b^2 + 3*b^4)*Co
s[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*b^3*(a^2 - b^2)^2*d)

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Rubi [A]  time = 0.862177, antiderivative size = 436, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 9, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.391, Rules used = {2792, 3047, 3049, 3023, 2752, 2663, 2661, 2655, 2653} \[ -\frac{2 a^2 \sin (c+d x) \cos ^3(c+d x)}{3 b d \left (a^2-b^2\right ) (a+b \cos (c+d x))^{3/2}}-\frac{8 a^2 \left (2 a^2-3 b^2\right ) \sin (c+d x) \cos ^2(c+d x)}{3 b^2 d \left (a^2-b^2\right )^2 \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-71 a^2 b^2+48 a^4+3 b^4\right ) \sin (c+d x) \cos (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b^3 d \left (a^2-b^2\right )^2}-\frac{4 a \left (-49 a^2 b^2+32 a^4+7 b^4\right ) \sin (c+d x) \sqrt{a+b \cos (c+d x)}}{15 b^4 d \left (a^2-b^2\right )^2}-\frac{2 a \left (-116 a^2 b^2+128 a^4-17 b^4\right ) \sqrt{\frac{a+b \cos (c+d x)}{a+b}} F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^5 d \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (-212 a^4 b^2+55 a^2 b^4+128 a^6+9 b^6\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^5 d \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(128*a^6 - 212*a^4*b^2 + 55*a^2*b^4 + 9*b^6)*Sqrt[a + b*Cos[c + d*x]]*EllipticE[(c + d*x)/2, (2*b)/(a + b)]
)/(15*b^5*(a^2 - b^2)^2*d*Sqrt[(a + b*Cos[c + d*x])/(a + b)]) - (2*a*(128*a^4 - 116*a^2*b^2 - 17*b^4)*Sqrt[(a
+ b*Cos[c + d*x])/(a + b)]*EllipticF[(c + d*x)/2, (2*b)/(a + b)])/(15*b^5*(a^2 - b^2)*d*Sqrt[a + b*Cos[c + d*x
]]) - (2*a^2*Cos[c + d*x]^3*Sin[c + d*x])/(3*b*(a^2 - b^2)*d*(a + b*Cos[c + d*x])^(3/2)) - (8*a^2*(2*a^2 - 3*b
^2)*Cos[c + d*x]^2*Sin[c + d*x])/(3*b^2*(a^2 - b^2)^2*d*Sqrt[a + b*Cos[c + d*x]]) - (4*a*(32*a^4 - 49*a^2*b^2
+ 7*b^4)*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*b^4*(a^2 - b^2)^2*d) + (2*(48*a^4 - 71*a^2*b^2 + 3*b^4)*Co
s[c + d*x]*Sqrt[a + b*Cos[c + d*x]]*Sin[c + d*x])/(15*b^3*(a^2 - b^2)^2*d)

Rule 2792

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -S
imp[((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*(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 - 3)*(c + d*Sin[e +
f*x])^(n + 1)*Simp[b*(m - 2)*(b*c - a*d)^2 + a*d*(n + 1)*(c*(a^2 + b^2) - 2*a*b*d) + (b*(n + 1)*(a*b*c^2 + c*
d*(a^2 + b^2) - 3*a*b*d^2) - a*(n + 2)*(b*c - a*d)^2)*Sin[e + f*x] + b*(b^2*(c^2 - d^2) - m*(b*c - a*d)^2 + d*
n*(2*a*b*c - d*(a^2 + b^2)))*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] &
& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[m, 2] && LtQ[n, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n])

Rule 3047

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*s
in[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Simp[((c^2*C - B*c*d + A*d^2)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^m*(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 - 1)*(c + d*Sin[e + f*x])^(n + 1)*Simp[A*d*(b*d*m + a*c*(n + 1)) + (c
*C - B*d)*(b*c*m + a*d*(n + 1)) - (d*(A*(a*d*(n + 2) - b*c*(n + 1)) + B*(b*d*(n + 1) - a*c*(n + 2))) - C*(b*c*
d*(n + 1) - a*(c^2 + d^2*(n + 1))))*Sin[e + f*x] + b*(d*(B*c - A*d)*(m + n + 2) - C*(c^2*(m + 1) + d^2*(n + 1)
))*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] && NeQ[a^2 - b^2,
0] && NeQ[c^2 - d^2, 0] && GtQ[m, 0] && LtQ[n, -1]

Rule 3049

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((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*(c + d*Sin[e + f*x])^(n + 1))/(d*f*(m + n + 2)), x] + Dist[1/(d*(m + n + 2)), Int[(a + b*Sin[e + f*x]
)^(m - 1)*(c + d*Sin[e + f*x])^n*Simp[a*A*d*(m + n + 2) + C*(b*c*m + a*d*(n + 1)) + (d*(A*b + a*B)*(m + n + 2)
- C*(a*c - b*d*(m + n + 1)))*Sin[e + f*x] + (C*(a*d*m - b*c*(m + 1)) + b*B*d*(m + n + 2))*Sin[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2
, 0] && GtQ[m, 0] &&  !(IGtQ[n, 0] && ( !IntegerQ[m] || (EqQ[a, 0] && NeQ[c, 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 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 \frac{\cos ^5(c+d x)}{(a+b \cos (c+d x))^{5/2}} \, dx &=-\frac{2 a^2 \cos ^3(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 ^2(c+d x) \left (3 a^2-\frac{3}{2} a b \cos (c+d x)-\frac{1}{2} \left (8 a^2-3 b^2\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^2 \cos ^3(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{8 a^2 \left (2 a^2-3 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{4 \int \frac{\cos (c+d x) \left (-4 a^2 \left (2 a^2-3 b^2\right )+\frac{1}{2} a b \left (a^2-3 b^2\right ) \cos (c+d x)+\frac{1}{4} \left (48 a^4-71 a^2 b^2+3 b^4\right ) \cos ^2(c+d x)\right )}{\sqrt{a+b \cos (c+d x)}} \, dx}{3 b^2 \left (a^2-b^2\right )^2}\\ &=-\frac{2 a^2 \cos ^3(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{8 a^2 \left (2 a^2-3 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}+\frac{2 \left (48 a^4-71 a^2 b^2+3 b^4\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right )^2 d}+\frac{8 \int \frac{\frac{1}{4} a \left (48 a^4-71 a^2 b^2+3 b^4\right )-\frac{1}{8} b \left (16 a^4-27 a^2 b^2-9 b^4\right ) \cos (c+d x)-\frac{3}{4} a \left (32 a^4-49 a^2 b^2+7 b^4\right ) \cos ^2(c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b^3 \left (a^2-b^2\right )^2}\\ &=-\frac{2 a^2 \cos ^3(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{8 a^2 \left (2 a^2-3 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}-\frac{4 a \left (32 a^4-49 a^2 b^2+7 b^4\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^4 \left (a^2-b^2\right )^2 d}+\frac{2 \left (48 a^4-71 a^2 b^2+3 b^4\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right )^2 d}+\frac{16 \int \frac{\frac{3}{4} a b \left (8 a^4-11 a^2 b^2-2 b^4\right )+\frac{3}{16} \left (128 a^6-212 a^4 b^2+55 a^2 b^4+9 b^6\right ) \cos (c+d x)}{\sqrt{a+b \cos (c+d x)}} \, dx}{45 b^4 \left (a^2-b^2\right )^2}\\ &=-\frac{2 a^2 \cos ^3(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{8 a^2 \left (2 a^2-3 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}-\frac{4 a \left (32 a^4-49 a^2 b^2+7 b^4\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^4 \left (a^2-b^2\right )^2 d}+\frac{2 \left (48 a^4-71 a^2 b^2+3 b^4\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right )^2 d}-\frac{\left (a \left (128 a^4-116 a^2 b^2-17 b^4\right )\right ) \int \frac{1}{\sqrt{a+b \cos (c+d x)}} \, dx}{15 b^5 \left (a^2-b^2\right )}+\frac{\left (128 a^6-212 a^4 b^2+55 a^2 b^4+9 b^6\right ) \int \sqrt{a+b \cos (c+d x)} \, dx}{15 b^5 \left (a^2-b^2\right )^2}\\ &=-\frac{2 a^2 \cos ^3(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{8 a^2 \left (2 a^2-3 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}-\frac{4 a \left (32 a^4-49 a^2 b^2+7 b^4\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^4 \left (a^2-b^2\right )^2 d}+\frac{2 \left (48 a^4-71 a^2 b^2+3 b^4\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right )^2 d}+\frac{\left (\left (128 a^6-212 a^4 b^2+55 a^2 b^4+9 b^6\right ) \sqrt{a+b \cos (c+d x)}\right ) \int \sqrt{\frac{a}{a+b}+\frac{b \cos (c+d x)}{a+b}} \, dx}{15 b^5 \left (a^2-b^2\right )^2 \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{\left (a \left (128 a^4-116 a^2 b^2-17 b^4\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}{15 b^5 \left (a^2-b^2\right ) \sqrt{a+b \cos (c+d x)}}\\ &=\frac{2 \left (128 a^6-212 a^4 b^2+55 a^2 b^4+9 b^6\right ) \sqrt{a+b \cos (c+d x)} E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )}{15 b^5 \left (a^2-b^2\right )^2 d \sqrt{\frac{a+b \cos (c+d x)}{a+b}}}-\frac{2 a \left (128 a^4-116 a^2 b^2-17 b^4\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 )}{15 b^5 \left (a^2-b^2\right ) d \sqrt{a+b \cos (c+d x)}}-\frac{2 a^2 \cos ^3(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{8 a^2 \left (2 a^2-3 b^2\right ) \cos ^2(c+d x) \sin (c+d x)}{3 b^2 \left (a^2-b^2\right )^2 d \sqrt{a+b \cos (c+d x)}}-\frac{4 a \left (32 a^4-49 a^2 b^2+7 b^4\right ) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^4 \left (a^2-b^2\right )^2 d}+\frac{2 \left (48 a^4-71 a^2 b^2+3 b^4\right ) \cos (c+d x) \sqrt{a+b \cos (c+d x)} \sin (c+d x)}{15 b^3 \left (a^2-b^2\right )^2 d}\\ \end{align*}

Mathematica [A]  time = 1.90669, size = 272, normalized size = 0.62 \[ \frac{b \left (\frac{10 a^5 \sin (c+d x)}{a^2-b^2}-\frac{10 a^4 \left (11 a^2-15 b^2\right ) \sin (c+d x) (a+b \cos (c+d x))}{\left (a^2-b^2\right )^2}-28 a \sin (c+d x) (a+b \cos (c+d x))^2+3 b \sin (2 (c+d x)) (a+b \cos (c+d x))^2\right )+\frac{2 \left (\frac{a+b \cos (c+d x)}{a+b}\right )^{3/2} \left (a \left (116 a^3 b^2-116 a^2 b^3+128 a^4 b-128 a^5+17 a b^4-17 b^5\right ) F\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )+\left (-212 a^4 b^2+55 a^2 b^4+128 a^6+9 b^6\right ) E\left (\frac{1}{2} (c+d x)|\frac{2 b}{a+b}\right )\right )}{(a-b)^2}}{15 b^5 d (a+b \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((2*((a + b*Cos[c + d*x])/(a + b))^(3/2)*((128*a^6 - 212*a^4*b^2 + 55*a^2*b^4 + 9*b^6)*EllipticE[(c + d*x)/2,
(2*b)/(a + b)] + a*(-128*a^5 + 128*a^4*b + 116*a^3*b^2 - 116*a^2*b^3 + 17*a*b^4 - 17*b^5)*EllipticF[(c + d*x)/
2, (2*b)/(a + b)]))/(a - b)^2 + b*((10*a^5*Sin[c + d*x])/(a^2 - b^2) - (10*a^4*(11*a^2 - 15*b^2)*(a + b*Cos[c
+ d*x])*Sin[c + d*x])/(a^2 - b^2)^2 - 28*a*(a + b*Cos[c + d*x])^2*Sin[c + d*x] + 3*b*(a + b*Cos[c + d*x])^2*Si
n[2*(c + d*x)]))/(15*b^5*d*(a + b*Cos[c + d*x])^(3/2))

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Maple [B]  time = 17.893, size = 1684, normalized size = 3.9 \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)^5/(a+b*cos(d*x+c))^(5/2),x)

[Out]

-(-(-2*b*cos(1/2*d*x+1/2*c)^2-a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(16/b^2*(-1/10/b*cos(1/2*d*x+1/2*c)^3*(-2*b*sin
(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)-1/60/b^2*(-4*a+12*b)*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+
1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/60/b^2*(-4*a+12*b)*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1
/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1
/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/60*(4*a^2-15*a*b+27*b^2)/b^3*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(
1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(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))))-8/b^3*(2*a+3*b)*(-1/6/b
*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)+1/6*(a-b)/b*(sin(1/2*d*x+1/2*
c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2
)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))-1/12/b^2*(-2*a+6*b)*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2
)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(E
llipticF(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))))-2/b^5*(3*a^
2+4*a*b+3*b^2)*(a-b)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d
*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(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)))-2*(4*a^3+3*a^2*b+2*a*b^2+b^3)/b^5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*((2*b*cos(1
/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1
/2*d*x+1/2*c),(-2*b/(a-b))^(1/2))+10/b^5*a^4/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*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)^(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*co
s(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2)-2/b^5*a^5*(1/6/b/(a-b)/(a+b)*cos(1/2*d*x+1/2*c)*(-2*b*sin(1/2*d*x+1/2*c
)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)/(cos(1/2*d*x+1/2*c)^2+1/2*(a-b)/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*b*cos(1/2*d*x+1/2*c)^2-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*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x
+1/2*c)^4+(a+b)*sin(1/2*d*x+1/2*c)^2)^(1/2)*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*b*cos(1/2*d*x+1/2*c)^2+a-b)/(a-b))^(1/2)/(-2*b*sin(1/2*d*x+1/2*c)^4+(a+b)*
sin(1/2*d*x+1/2*c)^2)^(1/2)*(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., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{5}}{{\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(cos(d*x+c)^5/(a+b*cos(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate(cos(d*x + c)^5/(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 (\frac{\sqrt{b \cos \left (d x + c\right ) + a} \cos \left (d x + c\right )^{5}}{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 ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*cos(d*x + c) + a)*cos(d*x + c)^5/(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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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)**5/(a+b*cos(d*x+c))**(5/2),x)

[Out]

Timed out

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

[Out]

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