3.520 \(\int \frac{1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=423 \[ -\frac{d \left (-21 c^2 d+4 c^3+62 c d^2+147 d^3\right ) \cos (e+f x)}{30 a^3 f (c-d)^4 (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{\left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x)}{30 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right ) \sqrt{c+d \sin (e+f x)}}+\frac{\left (4 c^2-21 c d+65 d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{30 a^3 f (c-d)^3 \sqrt{c+d \sin (e+f x)}}-\frac{\left (-21 c^2 d+4 c^3+62 c d^2+147 d^3\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{30 a^3 f (c-d)^4 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 (c-4 d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2 \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 \sqrt{c+d \sin (e+f x)}} \]

[Out]

-(d*(4*c^3 - 21*c^2*d + 62*c*d^2 + 147*d^3)*Cos[e + f*x])/(30*a^3*(c - d)^4*(c + d)*f*Sqrt[c + d*Sin[e + f*x]]
) - Cos[e + f*x]/(5*(c - d)*f*(a + a*Sin[e + f*x])^3*Sqrt[c + d*Sin[e + f*x]]) - (2*(c - 4*d)*Cos[e + f*x])/(1
5*a*(c - d)^2*f*(a + a*Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]]) - ((4*c^2 - 21*c*d + 65*d^2)*Cos[e + f*x])/(3
0*(c - d)^3*f*(a^3 + a^3*Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]) - ((4*c^3 - 21*c^2*d + 62*c*d^2 + 147*d^3)*El
lipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(30*a^3*(c - d)^4*(c + d)*f*Sqrt[(c + d*S
in[e + f*x])/(c + d)]) + ((4*c^2 - 21*c*d + 65*d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*S
in[e + f*x])/(c + d)])/(30*a^3*(c - d)^3*f*Sqrt[c + d*Sin[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.02531, antiderivative size = 423, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.296, Rules used = {2766, 2978, 2754, 2752, 2663, 2661, 2655, 2653} \[ -\frac{d \left (-21 c^2 d+4 c^3+62 c d^2+147 d^3\right ) \cos (e+f x)}{30 a^3 f (c-d)^4 (c+d) \sqrt{c+d \sin (e+f x)}}-\frac{\left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x)}{30 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right ) \sqrt{c+d \sin (e+f x)}}+\frac{\left (4 c^2-21 c d+65 d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{30 a^3 f (c-d)^3 \sqrt{c+d \sin (e+f x)}}-\frac{\left (-21 c^2 d+4 c^3+62 c d^2+147 d^3\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{30 a^3 f (c-d)^4 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}-\frac{2 (c-4 d) \cos (e+f x)}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2 \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{5 f (c-d) (a \sin (e+f x)+a)^3 \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^(3/2)),x]

[Out]

-(d*(4*c^3 - 21*c^2*d + 62*c*d^2 + 147*d^3)*Cos[e + f*x])/(30*a^3*(c - d)^4*(c + d)*f*Sqrt[c + d*Sin[e + f*x]]
) - Cos[e + f*x]/(5*(c - d)*f*(a + a*Sin[e + f*x])^3*Sqrt[c + d*Sin[e + f*x]]) - (2*(c - 4*d)*Cos[e + f*x])/(1
5*a*(c - d)^2*f*(a + a*Sin[e + f*x])^2*Sqrt[c + d*Sin[e + f*x]]) - ((4*c^2 - 21*c*d + 65*d^2)*Cos[e + f*x])/(3
0*(c - d)^3*f*(a^3 + a^3*Sin[e + f*x])*Sqrt[c + d*Sin[e + f*x]]) - ((4*c^3 - 21*c^2*d + 62*c*d^2 + 147*d^3)*El
lipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(30*a^3*(c - d)^4*(c + d)*f*Sqrt[(c + d*S
in[e + f*x])/(c + d)]) + ((4*c^2 - 21*c*d + 65*d^2)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*S
in[e + f*x])/(c + d)])/(30*a^3*(c - d)^3*f*Sqrt[c + d*Sin[e + f*x]])

Rule 2766

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

Rule 2978

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

Rule 2754

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> -Simp[((
b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 - b^2)), x] + Dist[1/((m + 1)*(a^2 - b^2
)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*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] && LtQ[m, -1] && 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 \frac{1}{(a+a \sin (e+f x))^3 (c+d \sin (e+f x))^{3/2}} \, dx &=-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 \sqrt{c+d \sin (e+f x)}}-\frac{\int \frac{-\frac{1}{2} a (4 c-11 d)-\frac{5}{2} a d \sin (e+f x)}{(a+a \sin (e+f x))^2 (c+d \sin (e+f x))^{3/2}} \, dx}{5 a^2 (c-d)}\\ &=-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 \sqrt{c+d \sin (e+f x)}}-\frac{2 (c-4 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}+\frac{\int \frac{\frac{1}{2} a^2 \left (4 c^2-15 c d+41 d^2\right )+3 a^2 (c-4 d) d \sin (e+f x)}{(a+a \sin (e+f x)) (c+d \sin (e+f x))^{3/2}} \, dx}{15 a^4 (c-d)^2}\\ &=-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 \sqrt{c+d \sin (e+f x)}}-\frac{2 (c-4 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}-\frac{\left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x)}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) \sqrt{c+d \sin (e+f x)}}-\frac{\int \frac{-\frac{3}{4} a^3 (c-49 d) d^2-\frac{1}{4} a^3 d \left (4 c^2-21 c d+65 d^2\right ) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 a^6 (c-d)^3}\\ &=-\frac{d \left (4 c^3-21 c^2 d+62 c d^2+147 d^3\right ) \cos (e+f x)}{30 a^3 (c-d)^4 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 \sqrt{c+d \sin (e+f x)}}-\frac{2 (c-4 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}-\frac{\left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x)}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) \sqrt{c+d \sin (e+f x)}}+\frac{2 \int \frac{-\frac{1}{8} a^3 d^2 \left (c^2+126 c d+65 d^2\right )-\frac{1}{8} a^3 d \left (4 c^3-21 c^2 d+62 c d^2+147 d^3\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{15 a^6 (c-d)^4 (c+d)}\\ &=-\frac{d \left (4 c^3-21 c^2 d+62 c d^2+147 d^3\right ) \cos (e+f x)}{30 a^3 (c-d)^4 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 \sqrt{c+d \sin (e+f x)}}-\frac{2 (c-4 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}-\frac{\left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x)}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) \sqrt{c+d \sin (e+f x)}}+\frac{\left (4 c^2-21 c d+65 d^2\right ) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{60 a^3 (c-d)^3}-\frac{\left (4 c^3-21 c^2 d+62 c d^2+147 d^3\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{60 a^3 (c-d)^4 (c+d)}\\ &=-\frac{d \left (4 c^3-21 c^2 d+62 c d^2+147 d^3\right ) \cos (e+f x)}{30 a^3 (c-d)^4 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 \sqrt{c+d \sin (e+f x)}}-\frac{2 (c-4 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}-\frac{\left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x)}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) \sqrt{c+d \sin (e+f x)}}-\frac{\left (\left (4 c^3-21 c^2 d+62 c d^2+147 d^3\right ) \sqrt{c+d \sin (e+f x)}\right ) \int \sqrt{\frac{c}{c+d}+\frac{d \sin (e+f x)}{c+d}} \, dx}{60 a^3 (c-d)^4 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (\left (4 c^2-21 c d+65 d^2\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}{60 a^3 (c-d)^3 \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{d \left (4 c^3-21 c^2 d+62 c d^2+147 d^3\right ) \cos (e+f x)}{30 a^3 (c-d)^4 (c+d) f \sqrt{c+d \sin (e+f x)}}-\frac{\cos (e+f x)}{5 (c-d) f (a+a \sin (e+f x))^3 \sqrt{c+d \sin (e+f x)}}-\frac{2 (c-4 d) \cos (e+f x)}{15 a (c-d)^2 f (a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}}-\frac{\left (4 c^2-21 c d+65 d^2\right ) \cos (e+f x)}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right ) \sqrt{c+d \sin (e+f x)}}-\frac{\left (4 c^3-21 c^2 d+62 c d^2+147 d^3\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)}}{30 a^3 (c-d)^4 (c+d) f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left (4 c^2-21 c d+65 d^2\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}}}{30 a^3 (c-d)^3 f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 6.54493, size = 745, normalized size = 1.76 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6 \sqrt{c+d \sin (e+f x)} \left (\frac{4 c^2 \sin \left (\frac{1}{2} (e+f x)\right )-25 c d \sin \left (\frac{1}{2} (e+f x)\right )+87 d^2 \sin \left (\frac{1}{2} (e+f x)\right )}{15 (c-d)^4 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )}-\frac{-21 c^2 d+4 c^3+62 c d^2+117 d^3}{15 (c-d)^4 (c+d)}-\frac{2 d^4 \cos (e+f x)}{(c-d)^4 (c+d) (c+d \sin (e+f x))}+\frac{2 \left (2 c \sin \left (\frac{1}{2} (e+f x)\right )-11 d \sin \left (\frac{1}{2} (e+f x)\right )\right )}{15 (c-d)^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3}+\frac{11 d-2 c}{15 (c-d)^3 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2}-\frac{1}{5 (c-d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4}+\frac{2 \sin \left (\frac{1}{2} (e+f x)\right )}{5 (c-d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5}\right )}{f (a \sin (e+f x)+a)^3}+\frac{d \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6 \left (\frac{2 \left (-21 c^2 d+4 c^3+62 c d^2+147 d^3\right ) \cos ^2(e+f x) \sqrt{c+d \sin (e+f x)}}{d \left (1-\sin ^2(e+f x)\right )}-\frac{2 \left (-c^2 d-126 c d^2-65 d^3\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 )}{\sqrt{c+d \sin (e+f x)}}-\frac{\left (21 c^2 d-4 c^3-62 c d^2-147 d^3\right ) \left (\frac{2 (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} E\left (\frac{1}{2} \left (-e-f x+\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{\sqrt{c+d \sin (e+f x)}}-\frac{2 c \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 )}{\sqrt{c+d \sin (e+f x)}}\right )}{d}\right )}{60 f (c-d)^4 (c+d) (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((a + a*Sin[e + f*x])^3*(c + d*Sin[e + f*x])^(3/2)),x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*Sqrt[c + d*Sin[e + f*x]]*(-(4*c^3 - 21*c^2*d + 62*c*d^2 + 117*d^3)/(1
5*(c - d)^4*(c + d)) + (2*Sin[(e + f*x)/2])/(5*(c - d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5) - 1/(5*(c -
d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4) + (-2*c + 11*d)/(15*(c - d)^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/
2])^2) + (2*(2*c*Sin[(e + f*x)/2] - 11*d*Sin[(e + f*x)/2]))/(15*(c - d)^3*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]
)^3) + (4*c^2*Sin[(e + f*x)/2] - 25*c*d*Sin[(e + f*x)/2] + 87*d^2*Sin[(e + f*x)/2])/(15*(c - d)^4*(Cos[(e + f*
x)/2] + Sin[(e + f*x)/2])) - (2*d^4*Cos[e + f*x])/((c - d)^4*(c + d)*(c + d*Sin[e + f*x]))))/(f*(a + a*Sin[e +
f*x])^3) + (d*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*((-2*(-(c^2*d) - 126*c*d^2 - 65*d^3)*EllipticF[(-e + Pi
/2 - f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/Sqrt[c + d*Sin[e + f*x]] + (2*(4*c^3 - 21*c^2*
d + 62*c*d^2 + 147*d^3)*Cos[e + f*x]^2*Sqrt[c + d*Sin[e + f*x]])/(d*(1 - Sin[e + f*x]^2)) - ((-4*c^3 + 21*c^2*
d - 62*c*d^2 - 147*d^3)*((2*(c + d)*EllipticE[(-e + Pi/2 - f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c
+ d)])/Sqrt[c + d*Sin[e + f*x]] - (2*c*EllipticF[(-e + Pi/2 - f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x]
)/(c + d)])/Sqrt[c + d*Sin[e + f*x]]))/d))/(60*(c - d)^4*(c + d)*f*(a + a*Sin[e + f*x])^3)

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Maple [B]  time = 7.014, size = 1851, 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(1/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(3/2),x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/a^3*(1/(c-d)*(-1/5/(c-d)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(1+si
n(f*x+e))^3-2/15*(c-3*d)/(c-d)^2*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(1+sin(f*x+e))^2-1/30*(-sin(f*x+e)^2*
d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)^3*(4*c^2-15*c*d+27*d^2)/((-d*sin(f*x+e)-c)*(-1+sin(f*x+e))*(1+sin(f*x+e))
)^(1/2)+2*(-c*d^2-15*d^3)/(60*c^3-180*c^2*d+180*c*d^2-60*d^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))-1/30*d*(4*c^2-15*c*d+27*d^2)/(c-d)^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)*((-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))))-d/(c-d)^2*(-1/3/(c-d)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(1+sin(f*x+e))^2-
1/3*(-sin(f*x+e)^2*d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)^2*(c-3*d)/((-d*sin(f*x+e)-c)*(-1+sin(f*x+e))*(1+sin(f*
x+e)))^(1/2)+2*d^2/(3*c^2-6*c*d+3*d^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))-1/3*d*(c-3*d)/(c-d)^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))))-d^3/(c-
d)^3*(2*d*cos(f*x+e)^2/(c^2-d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*c/(c^2-d^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))+2/(c^2-d^2)*d*(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))))+d^2/(c-d)^3*(-(-sin(f*x+e)^2*d-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)/((-d*sin(f*
x+e)-c)*(-1+sin(f*x+e))*(1+sin(f*x+e)))^(1/2)-2*d/(2*c-2*d)*(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*si
n(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))-d/(c-d)*(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)))))
/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(1/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \sin \left (f x + e\right ) + c}}{4 \, a^{3} c^{2} + 8 \, a^{3} c d + 4 \, a^{3} d^{2} +{\left (2 \, a^{3} c d + 3 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{4} -{\left (3 \, a^{3} c^{2} + 10 \, a^{3} c d + 7 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{2} +{\left (a^{3} d^{2} \cos \left (f x + e\right )^{4} + 4 \, a^{3} c^{2} + 8 \, a^{3} c d + 4 \, a^{3} d^{2} -{\left (a^{3} c^{2} + 6 \, a^{3} c d + 5 \, a^{3} d^{2}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*sin(f*x + e) + c)/(4*a^3*c^2 + 8*a^3*c*d + 4*a^3*d^2 + (2*a^3*c*d + 3*a^3*d^2)*cos(f*x + e)^4
- (3*a^3*c^2 + 10*a^3*c*d + 7*a^3*d^2)*cos(f*x + e)^2 + (a^3*d^2*cos(f*x + e)^4 + 4*a^3*c^2 + 8*a^3*c*d + 4*a^
3*d^2 - (a^3*c^2 + 6*a^3*c*d + 5*a^3*d^2)*cos(f*x + e)^2)*sin(f*x + e)), 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(1/(a+a*sin(f*x+e))**3/(c+d*sin(f*x+e))**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (a \sin \left (f x + e\right ) + a\right )}^{3}{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+a*sin(f*x+e))^3/(c+d*sin(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((a*sin(f*x + e) + a)^3*(d*sin(f*x + e) + c)^(3/2)), x)