3.760 \(\int \frac{(c+d \sin (e+f x))^{5/2}}{(a+b \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=549 \[ \frac{\left (a^2 b^2 d \left (7 c^2-5 d^2\right )+4 a^3 b c d^2+3 a^4 d^3-2 a b^3 c \left (3 c^2+11 d^2\right )+b^4 d \left (11 c^2+8 d^2\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{4 b^3 f \left (a^2-b^2\right )^2 \sqrt{c+d \sin (e+f x)}}+\frac{\left (2 a^2 b^2 \left (4 c^2-3 d^2\right )+4 a^3 b c d+3 a^4 d^2-28 a b^3 c d+b^4 \left (4 c^2+15 d^2\right )\right ) (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{4 b^3 f (a-b)^2 (a+b)^3 \sqrt{c+d \sin (e+f x)}}+\frac{(b c-a d)^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 b f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2}+\frac{3 \left (a^2 d+2 a b c-3 b^2 d\right ) (b c-a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 b f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))}+\frac{3 \left (a^2 d+2 a b c-3 b^2 d\right ) (b c-a d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{4 b^2 f \left (a^2-b^2\right )^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

[Out]

((b*c - a*d)^2*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(2*b*(a^2 - b^2)*f*(a + b*Sin[e + f*x])^2) + (3*(b*c - a
*d)*(2*a*b*c + a^2*d - 3*b^2*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(4*b*(a^2 - b^2)^2*f*(a + b*Sin[e + f*x
])) + (3*(b*c - a*d)*(2*a*b*c + a^2*d - 3*b^2*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e
+ f*x]])/(4*b^2*(a^2 - b^2)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((4*a^3*b*c*d^2 + 3*a^4*d^3 + a^2*b^2*d
*(7*c^2 - 5*d^2) + b^4*d*(11*c^2 + 8*d^2) - 2*a*b^3*c*(3*c^2 + 11*d^2))*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c
+ d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(4*b^3*(a^2 - b^2)^2*f*Sqrt[c + d*Sin[e + f*x]]) + ((b*c - a*d)*(4*
a^3*b*c*d - 28*a*b^3*c*d + 3*a^4*d^2 + 2*a^2*b^2*(4*c^2 - 3*d^2) + b^4*(4*c^2 + 15*d^2))*EllipticPi[(2*b)/(a +
b), (e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(4*(a - b)^2*b^3*(a + b)^3*f*Sqrt[
c + d*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 2.07565, antiderivative size = 549, normalized size of antiderivative = 1., number of steps used = 10, number of rules used = 10, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.37, Rules used = {2792, 3055, 3059, 2655, 2653, 3002, 2663, 2661, 2807, 2805} \[ \frac{\left (a^2 b^2 d \left (7 c^2-5 d^2\right )+4 a^3 b c d^2+3 a^4 d^3-2 a b^3 c \left (3 c^2+11 d^2\right )+b^4 d \left (11 c^2+8 d^2\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{4 b^3 f \left (a^2-b^2\right )^2 \sqrt{c+d \sin (e+f x)}}+\frac{\left (2 a^2 b^2 \left (4 c^2-3 d^2\right )+4 a^3 b c d+3 a^4 d^2-28 a b^3 c d+b^4 \left (4 c^2+15 d^2\right )\right ) (b c-a d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{4 b^3 f (a-b)^2 (a+b)^3 \sqrt{c+d \sin (e+f x)}}+\frac{(b c-a d)^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 b f \left (a^2-b^2\right ) (a+b \sin (e+f x))^2}+\frac{3 \left (a^2 d+2 a b c-3 b^2 d\right ) (b c-a d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 b f \left (a^2-b^2\right )^2 (a+b \sin (e+f x))}+\frac{3 \left (a^2 d+2 a b c-3 b^2 d\right ) (b c-a d) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{4 b^2 f \left (a^2-b^2\right )^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*Sin[e + f*x])^(5/2)/(a + b*Sin[e + f*x])^3,x]

[Out]

((b*c - a*d)^2*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(2*b*(a^2 - b^2)*f*(a + b*Sin[e + f*x])^2) + (3*(b*c - a
*d)*(2*a*b*c + a^2*d - 3*b^2*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(4*b*(a^2 - b^2)^2*f*(a + b*Sin[e + f*x
])) + (3*(b*c - a*d)*(2*a*b*c + a^2*d - 3*b^2*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e
+ f*x]])/(4*b^2*(a^2 - b^2)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((4*a^3*b*c*d^2 + 3*a^4*d^3 + a^2*b^2*d
*(7*c^2 - 5*d^2) + b^4*d*(11*c^2 + 8*d^2) - 2*a*b^3*c*(3*c^2 + 11*d^2))*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c
+ d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(4*b^3*(a^2 - b^2)^2*f*Sqrt[c + d*Sin[e + f*x]]) + ((b*c - a*d)*(4*
a^3*b*c*d - 28*a*b^3*c*d + 3*a^4*d^2 + 2*a^2*b^2*(4*c^2 - 3*d^2) + b^4*(4*c^2 + 15*d^2))*EllipticPi[(2*b)/(a +
b), (e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(4*(a - b)^2*b^3*(a + b)^3*f*Sqrt[
c + d*Sin[e + f*x]])

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 3055

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[((A*b^2 - a*b*B + a^2*C)*Cos[e +
f*x]*(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2)), x] + Dis
t[1/((m + 1)*(b*c - a*d)*(a^2 - b^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[(m + 1)*(b
*c - a*d)*(a*A - b*B + a*C) + d*(A*b^2 - a*b*B + a^2*C)*(m + n + 2) - (c*(A*b^2 - a*b*B + a^2*C) + (m + 1)*(b*
c - a*d)*(A*b - a*B + b*C))*Sin[e + f*x] - d*(A*b^2 - a*b*B + a^2*C)*(m + n + 3)*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] && Lt
Q[m, -1] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2*n] && LtQ[n, -1] && ((IntegerQ[n] &&
!IntegerQ[m]) || EqQ[a, 0])))

Rule 3059

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) +
(f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Dist[C/(b*d), Int[Sqrt[a + b*Sin[e + f*x]]
, x], x] - Dist[1/(b*d), Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[e + f*x], x]/(Sqrt[a + b*Sin[e +
f*x]]*(c + d*Sin[e + f*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]

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]

Rule 3002

Int[(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*sin[
(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[B/d, Int[(a + b*Sin[e + f*x])^m, x], x] - Dist[(B*c - A*d)/d, Int[(a +
b*Sin[e + f*x])^m/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]
&& NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^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 2807

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d*
Sin[e + f*x])/(c + d)]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[c + d]), 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[c + d, 0]

Rubi steps

\begin{align*} \int \frac{(c+d \sin (e+f x))^{5/2}}{(a+b \sin (e+f x))^3} \, dx &=\frac{(b c-a d)^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}-\frac{\int \frac{\frac{1}{2} \left (-4 a b c^3+9 b^2 c^2 d-6 a b c d^2+a^2 d^3\right )-\left (a^2 c d^2+2 a b d \left (2 c^2+d^2\right )-b^2 \left (c^3+6 c d^2\right )\right ) \sin (e+f x)-\frac{1}{2} d \left (2 a b c d+3 a^2 d^2-b^2 \left (c^2+4 d^2\right )\right ) \sin ^2(e+f x)}{(a+b \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}} \, dx}{2 b \left (a^2-b^2\right )}\\ &=\frac{(b c-a d)^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{3 (b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 b \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac{\int \frac{\frac{1}{4} (b c-a d) \left (8 a^2 b c^3+4 b^3 c^3-30 a b^2 c^2 d+9 a^2 b c d^2+15 b^3 c d^2+a^3 d^3-7 a b^2 d^3\right )+\frac{1}{2} d (b c-a d) \left (5 a^2 b c^2+b^3 c^2-a^3 c d-11 a b^2 c d+2 a^2 b d^2+4 b^3 d^2\right ) \sin (e+f x)+\frac{3}{4} d (b c-a d)^2 \left (2 a b c+a^2 d-3 b^2 d\right ) \sin ^2(e+f x)}{(a+b \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{2 b \left (a^2-b^2\right )^2 (b c-a d)}\\ &=\frac{(b c-a d)^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{3 (b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 b \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}-\frac{\int \frac{-\frac{1}{4} d (b c-a d) \left (2 a^2 b^2 c^3+3 a^4 c d^2+a^3 b d \left (3 c^2+d^2\right )-7 a b^3 d \left (3 c^2+d^2\right )+b^4 c \left (4 c^2+15 d^2\right )\right )+\frac{1}{4} d (b c-a d) \left (6 a b^3 c^3-7 a^2 b^2 c^2 d-11 b^4 c^2 d-4 a^3 b c d^2+22 a b^3 c d^2-3 a^4 d^3+5 a^2 b^2 d^3-8 b^4 d^3\right ) \sin (e+f x)}{(a+b \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{2 b^2 \left (a^2-b^2\right )^2 d (b c-a d)}+\frac{\left (3 (b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right )\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{8 b^2 \left (a^2-b^2\right )^2}\\ &=\frac{(b c-a d)^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{3 (b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 b \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac{\left (4 a^3 b c d^2+3 a^4 d^3+a^2 b^2 d \left (7 c^2-5 d^2\right )+b^4 d \left (11 c^2+8 d^2\right )-2 a b^3 c \left (3 c^2+11 d^2\right )\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{8 b^3 \left (a^2-b^2\right )^2}+\frac{\left ((b c-a d) \left (4 a^3 b c d-28 a b^3 c d+3 a^4 d^2+2 a^2 b^2 \left (4 c^2-3 d^2\right )+b^4 \left (4 c^2+15 d^2\right )\right )\right ) \int \frac{1}{(a+b \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{8 b^3 \left (a^2-b^2\right )^2}+\frac{\left (3 (b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right ) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{8 b^2 \left (a^2-b^2\right )^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}\\ &=\frac{(b c-a d)^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{3 (b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 b \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac{3 (b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right ) E\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{c+d \sin (e+f x)}}{4 b^2 \left (a^2-b^2\right )^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (\left (4 a^3 b c d^2+3 a^4 d^3+a^2 b^2 d \left (7 c^2-5 d^2\right )+b^4 d \left (11 c^2+8 d^2\right )-2 a b^3 c \left (3 c^2+11 d^2\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}\right ) \int \frac{1}{\sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}}} \, dx}{8 b^3 \left (a^2-b^2\right )^2 \sqrt{c+d \sin (e+f x)}}+\frac{\left ((b c-a d) \left (4 a^3 b c d-28 a b^3 c d+3 a^4 d^2+2 a^2 b^2 \left (4 c^2-3 d^2\right )+b^4 \left (4 c^2+15 d^2\right )\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}\right ) \int \frac{1}{(a+b \sin (e+f x)) \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}}} \, dx}{8 b^3 \left (a^2-b^2\right )^2 \sqrt{c+d \sin (e+f x)}}\\ &=\frac{(b c-a d)^2 \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{2 b \left (a^2-b^2\right ) f (a+b \sin (e+f x))^2}+\frac{3 (b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{4 b \left (a^2-b^2\right )^2 f (a+b \sin (e+f x))}+\frac{3 (b c-a d) \left (2 a b c+a^2 d-3 b^2 d\right ) E\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{c+d \sin (e+f x)}}{4 b^2 \left (a^2-b^2\right )^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (4 a^3 b c d^2+3 a^4 d^3+a^2 b^2 d \left (7 c^2-5 d^2\right )+b^4 d \left (11 c^2+8 d^2\right )-2 a b^3 c \left (3 c^2+11 d^2\right )\right ) F\left (\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}{4 b^3 \left (a^2-b^2\right )^2 f \sqrt{c+d \sin (e+f x)}}+\frac{(b c-a d) \left (4 a^3 b c d-28 a b^3 c d+3 a^4 d^2+2 a^2 b^2 \left (4 c^2-3 d^2\right )+b^4 \left (4 c^2+15 d^2\right )\right ) \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (e-\frac{\pi }{2}+f x\right )|\frac{2 d}{c+d}\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}{4 (a-b)^2 b^3 (a+b)^3 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 7.99975, size = 1149, normalized size = 2.09 \[ \frac{\sqrt{c+d \sin (e+f x)} \left (\frac{-b^2 \cos (e+f x) c^2+2 a b d \cos (e+f x) c-a^2 d^2 \cos (e+f x)}{2 b \left (b^2-a^2\right ) (a+b \sin (e+f x))^2}-\frac{3 \left (d^2 \cos (e+f x) a^3+b c d \cos (e+f x) a^2-2 b^2 c^2 \cos (e+f x) a-3 b^2 d^2 \cos (e+f x) a+3 b^3 c d \cos (e+f x)\right )}{4 b \left (b^2-a^2\right )^2 (a+b \sin (e+f x))}\right )}{f}-\frac{-\frac{2 \left (-8 c^3 b^3-21 c d^2 b^3+5 a d^3 b^2+54 a c^2 d b^2-16 a^2 c^3 b-15 a^2 c d^2 b+a^3 d^3\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \Pi \left (\frac{2 b}{a+b};\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{(a+b) \sqrt{c+d \sin (e+f x)}}-\frac{2 i \left (4 c d^2 a^3-8 b d^3 a^2-20 b c^2 d a^2+44 b^2 c d^2 a-16 b^3 d^3-4 b^3 c^2 d\right ) \cos (e+f x) \left ((b c-a d) F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )+a d \Pi \left (\frac{b (c+d)}{b c-a d};i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )\right ) \sqrt{\frac{d-d \sin (e+f x)}{c+d}} \sqrt{-\frac{\sin (e+f x) d+d}{c-d}} (-b c+a d+b (c+d \sin (e+f x)))}{b d^2 \sqrt{-\frac{1}{c+d}} (b c-a d) (a+b \sin (e+f x)) \sqrt{1-\sin ^2(e+f x)} \sqrt{-\frac{c^2-2 (c+d \sin (e+f x)) c-d^2+(c+d \sin (e+f x))^2}{d^2}}}-\frac{2 i \left (-9 c d^2 b^3+9 a d^3 b^2+6 a c^2 d b^2-3 a^2 c d^2 b-3 a^3 d^3\right ) \cos (e+f x) \cos (2 (e+f x)) \left (2 b (c-d) (b c-a d) E\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )+d \left (\left (2 a^2-b^2\right ) d \Pi \left (\frac{b (c+d)}{b c-a d};i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )-2 (a+b) (a d-b c) F\left (i \sinh ^{-1}\left (\sqrt{-\frac{1}{c+d}} \sqrt{c+d \sin (e+f x)}\right )|\frac{c+d}{c-d}\right )\right )\right ) \sqrt{\frac{d-d \sin (e+f x)}{c+d}} \sqrt{-\frac{\sin (e+f x) d+d}{c-d}} (-b c+a d+b (c+d \sin (e+f x)))}{b^2 d \sqrt{-\frac{1}{c+d}} (b c-a d) (a+b \sin (e+f x)) \sqrt{1-\sin ^2(e+f x)} \left (-2 c^2+4 (c+d \sin (e+f x)) c+d^2-2 (c+d \sin (e+f x))^2\right ) \sqrt{-\frac{c^2-2 (c+d \sin (e+f x)) c-d^2+(c+d \sin (e+f x))^2}{d^2}}}}{16 (a-b)^2 b (a+b)^2 f} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*Sin[e + f*x])^(5/2)/(a + b*Sin[e + f*x])^3,x]

[Out]

(Sqrt[c + d*Sin[e + f*x]]*((-(b^2*c^2*Cos[e + f*x]) + 2*a*b*c*d*Cos[e + f*x] - a^2*d^2*Cos[e + f*x])/(2*b*(-a^
2 + b^2)*(a + b*Sin[e + f*x])^2) - (3*(-2*a*b^2*c^2*Cos[e + f*x] + a^2*b*c*d*Cos[e + f*x] + 3*b^3*c*d*Cos[e +
f*x] + a^3*d^2*Cos[e + f*x] - 3*a*b^2*d^2*Cos[e + f*x]))/(4*b*(-a^2 + b^2)^2*(a + b*Sin[e + f*x]))))/f - ((-2*
(-16*a^2*b*c^3 - 8*b^3*c^3 + 54*a*b^2*c^2*d - 15*a^2*b*c*d^2 - 21*b^3*c*d^2 + a^3*d^3 + 5*a*b^2*d^3)*EllipticP
i[(2*b)/(a + b), (-e + Pi/2 - f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/((a + b)*Sqrt[c + d*S
in[e + f*x]]) - ((2*I)*(-20*a^2*b*c^2*d - 4*b^3*c^2*d + 4*a^3*c*d^2 + 44*a*b^2*c*d^2 - 8*a^2*b*d^3 - 16*b^3*d^
3)*Cos[e + f*x]*((b*c - a*d)*EllipticF[I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d
)] + a*d*EllipticPi[(b*(c + d))/(b*c - a*d), I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/
(c - d)])*Sqrt[(d - d*Sin[e + f*x])/(c + d)]*Sqrt[-((d + d*Sin[e + f*x])/(c - d))]*(-(b*c) + a*d + b*(c + d*Si
n[e + f*x])))/(b*d^2*Sqrt[-(c + d)^(-1)]*(b*c - a*d)*(a + b*Sin[e + f*x])*Sqrt[1 - Sin[e + f*x]^2]*Sqrt[-((c^2
- d^2 - 2*c*(c + d*Sin[e + f*x]) + (c + d*Sin[e + f*x])^2)/d^2)]) - ((2*I)*(6*a*b^2*c^2*d - 3*a^2*b*c*d^2 - 9
*b^3*c*d^2 - 3*a^3*d^3 + 9*a*b^2*d^3)*Cos[e + f*x]*Cos[2*(e + f*x)]*(2*b*(c - d)*(b*c - a*d)*EllipticE[I*ArcSi
nh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)] + d*(-2*(a + b)*(-(b*c) + a*d)*EllipticF[I*
ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)] + (2*a^2 - b^2)*d*EllipticPi[(b*(c + d
))/(b*c - a*d), I*ArcSinh[Sqrt[-(c + d)^(-1)]*Sqrt[c + d*Sin[e + f*x]]], (c + d)/(c - d)]))*Sqrt[(d - d*Sin[e
+ f*x])/(c + d)]*Sqrt[-((d + d*Sin[e + f*x])/(c - d))]*(-(b*c) + a*d + b*(c + d*Sin[e + f*x])))/(b^2*d*Sqrt[-(
c + d)^(-1)]*(b*c - a*d)*(a + b*Sin[e + f*x])*Sqrt[1 - Sin[e + f*x]^2]*(-2*c^2 + d^2 + 4*c*(c + d*Sin[e + f*x]
) - 2*(c + d*Sin[e + f*x])^2)*Sqrt[-((c^2 - d^2 - 2*c*(c + d*Sin[e + f*x]) + (c + d*Sin[e + f*x])^2)/d^2)]))/(
16*(a - b)^2*b*(a + b)^2*f)

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Maple [B]  time = 8.871, size = 1888, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c+d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))^3,x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*(2*d^3/b^3*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c
+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))
/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-6/b^4*d^2*(a*d-b*c)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e)
)/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(-c/d+a/b)*EllipticPi((
(c+d*sin(f*x+e))/(c-d))^(1/2),(-c/d+1)/(-c/d+a/b),((c-d)/(c+d))^(1/2))+1/b^3*(-a^3*d^3+3*a^2*b*c*d^2-3*a*b^2*c
^2*d+b^3*c^3)*(-1/2*b^2/(a^3*d-a^2*b*c-a*b^2*d+b^3*c)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(a+b*sin(f*x+e))
^2-3/4*b^2*(3*a^2*d-2*a*b*c-b^2*d)/(a^3*d-a^2*b*c-a*b^2*d+b^3*c)^2*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(a+
b*sin(f*x+e))-1/4*d*(7*a^3*d-4*a^2*b*c-a*b^2*d-2*b^3*c)/(a^3*d-a^2*b*c-a*b^2*d+b^3*c)^2*(c/d-1)*((c+d*sin(f*x+
e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)
^2)^(1/2)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-3/4*b*d*(3*a^2*d-2*a*b*c-b^2*d)/(a^3*d
-a^2*b*c-a*b^2*d+b^3*c)^2*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-
1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),(
(c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2)))+1/4*(15*a^4*d^2-20*a^3*b*c*
d+8*a^2*b^2*c^2-6*a^2*b^2*d^2-4*a*b^3*c*d+4*b^4*c^2+3*b^4*d^2)/(a^3*d-a^2*b*c-a*b^2*d+b^3*c)^2/b*(c/d-1)*((c+d
*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*c
os(f*x+e)^2)^(1/2)/(-c/d+a/b)*EllipticPi(((c+d*sin(f*x+e))/(c-d))^(1/2),(-c/d+1)/(-c/d+a/b),((c-d)/(c+d))^(1/2
)))+3/b^3*d*(a^2*d^2-2*a*b*c*d+b^2*c^2)*(-b^2/(a^3*d-a^2*b*c-a*b^2*d+b^3*c)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^
(1/2)/(a+b*sin(f*x+e))-a*d/(a^3*d-a^2*b*c-a*b^2*d+b^3*c)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1/2)*(d*(1-sin(f*x+
e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*EllipticF(((c+d*sin(f
*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-b*d/(a^3*d-a^2*b*c-a*b^2*d+b^3*c)*(c/d-1)*((c+d*sin(f*x+e))/(c-d))^(1
/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*((-
c/d-1)*EllipticE(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),
((c-d)/(c+d))^(1/2)))+(3*a^2*d-2*a*b*c-b^2*d)/(a^3*d-a^2*b*c-a*b^2*d+b^3*c)/b*(c/d-1)*((c+d*sin(f*x+e))/(c-d))
^(1/2)*(d*(1-sin(f*x+e))/(c+d))^(1/2)*((-sin(f*x+e)-1)*d/(c-d))^(1/2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/
(-c/d+a/b)*EllipticPi(((c+d*sin(f*x+e))/(c-d))^(1/2),(-c/d+1)/(-c/d+a/b),((c-d)/(c+d))^(1/2))))/cos(f*x+e)/(c+
d*sin(f*x+e))^(1/2)/f

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Maxima [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((c+d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

Timed out

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Fricas [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((c+d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))^3,x, algorithm="fricas")

[Out]

Timed out

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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((c+d*sin(f*x+e))**(5/2)/(a+b*sin(f*x+e))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{5}{2}}}{{\left (b \sin \left (f x + e\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c+d*sin(f*x+e))^(5/2)/(a+b*sin(f*x+e))^3,x, algorithm="giac")

[Out]

integrate((d*sin(f*x + e) + c)^(5/2)/(b*sin(f*x + e) + a)^3, x)