3.518 \(\int \frac{\sqrt{c+d \sin (e+f x)}}{(a+a \sin (e+f x))^3} \, dx\)

Optimal. Leaf size=334 \[ -\frac{\left (4 c^2-5 c d-3 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{30 f (c-d)^2 \left (a^3 \sin (e+f x)+a^3\right )}-\frac{\left (4 c^2-5 c d-3 d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{30 a^3 f (c-d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{(4 c-5 d) (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{30 a^3 f (c-d) \sqrt{c+d \sin (e+f x)}}-\frac{(2 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 a f (c-d) (a \sin (e+f x)+a)^2}-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{5 f (a \sin (e+f x)+a)^3} \]

[Out]

-(Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(5*f*(a + a*Sin[e + f*x])^3) - ((2*c - d)*Cos[e + f*x]*Sqrt[c + d*Sin
[e + f*x]])/(15*a*(c - d)*f*(a + a*Sin[e + f*x])^2) - ((4*c^2 - 5*c*d - 3*d^2)*Cos[e + f*x]*Sqrt[c + d*Sin[e +
f*x]])/(30*(c - d)^2*f*(a^3 + a^3*Sin[e + f*x])) - ((4*c^2 - 5*c*d - 3*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*
d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(30*a^3*(c - d)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((4*c - 5*d)*(
c + d)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(30*a^3*(c - d)*f*Sqrt
[c + d*Sin[e + f*x]])

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Rubi [A]  time = 0.767721, antiderivative size = 334, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2764, 2978, 2752, 2663, 2661, 2655, 2653} \[ -\frac{\left (4 c^2-5 c d-3 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{30 f (c-d)^2 \left (a^3 \sin (e+f x)+a^3\right )}-\frac{\left (4 c^2-5 c d-3 d^2\right ) \sqrt{c+d \sin (e+f x)} E\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{30 a^3 f (c-d)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{(4 c-5 d) (c+d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{2} \left (e+f x-\frac{\pi }{2}\right )|\frac{2 d}{c+d}\right )}{30 a^3 f (c-d) \sqrt{c+d \sin (e+f x)}}-\frac{(2 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 a f (c-d) (a \sin (e+f x)+a)^2}-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{5 f (a \sin (e+f x)+a)^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(5*f*(a + a*Sin[e + f*x])^3) - ((2*c - d)*Cos[e + f*x]*Sqrt[c + d*Sin
[e + f*x]])/(15*a*(c - d)*f*(a + a*Sin[e + f*x])^2) - ((4*c^2 - 5*c*d - 3*d^2)*Cos[e + f*x]*Sqrt[c + d*Sin[e +
f*x]])/(30*(c - d)^2*f*(a^3 + a^3*Sin[e + f*x])) - ((4*c^2 - 5*c*d - 3*d^2)*EllipticE[(e - Pi/2 + f*x)/2, (2*
d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(30*a^3*(c - d)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((4*c - 5*d)*(
c + d)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(30*a^3*(c - d)*f*Sqrt
[c + d*Sin[e + f*x]])

Rule 2764

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)),
Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 1)*Simp[a*d*n - b*c*(m + 1) - b*d*(m + n + 1)*Sin[
e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^
2, 0] && LtQ[m, -1] && LtQ[0, n, 1] && (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 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{\sqrt{c+d \sin (e+f x)}}{(a+a \sin (e+f x))^3} \, dx &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{5 f (a+a \sin (e+f x))^3}+\frac{\int \frac{\frac{1}{2} a (4 c+d)+\frac{3}{2} a d \sin (e+f x)}{(a+a \sin (e+f x))^2 \sqrt{c+d \sin (e+f x)}} \, dx}{5 a^2}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{5 f (a+a \sin (e+f x))^3}-\frac{(2 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 a (c-d) f (a+a \sin (e+f x))^2}-\frac{\int \frac{-\frac{1}{2} a^2 \left (4 c^2-3 c d-4 d^2\right )-\frac{1}{2} a^2 (2 c-d) d \sin (e+f x)}{(a+a \sin (e+f x)) \sqrt{c+d \sin (e+f x)}} \, dx}{15 a^4 (c-d)}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{5 f (a+a \sin (e+f x))^3}-\frac{(2 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 a (c-d) f (a+a \sin (e+f x))^2}-\frac{\left (4 c^2-5 c d-3 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{30 (c-d)^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac{\int \frac{-\frac{1}{4} a^3 (c-5 d) d^2-\frac{1}{4} a^3 d \left (4 c^2-5 c d-3 d^2\right ) \sin (e+f x)}{\sqrt{c+d \sin (e+f x)}} \, dx}{15 a^6 (c-d)^2}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{5 f (a+a \sin (e+f x))^3}-\frac{(2 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 a (c-d) f (a+a \sin (e+f x))^2}-\frac{\left (4 c^2-5 c d-3 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{30 (c-d)^2 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac{((4 c-5 d) (c+d)) \int \frac{1}{\sqrt{c+d \sin (e+f x)}} \, dx}{60 a^3 (c-d)}-\frac{\left (4 c^2-5 c d-3 d^2\right ) \int \sqrt{c+d \sin (e+f x)} \, dx}{60 a^3 (c-d)^2}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{5 f (a+a \sin (e+f x))^3}-\frac{(2 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 a (c-d) f (a+a \sin (e+f x))^2}-\frac{\left (4 c^2-5 c d-3 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{30 (c-d)^2 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac{\left (\left (4 c^2-5 c d-3 d^2\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)^2 \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{\left ((4 c-5 d) (c+d) \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) \sqrt{c+d \sin (e+f x)}}\\ &=-\frac{\cos (e+f x) \sqrt{c+d \sin (e+f x)}}{5 f (a+a \sin (e+f x))^3}-\frac{(2 c-d) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{15 a (c-d) f (a+a \sin (e+f x))^2}-\frac{\left (4 c^2-5 c d-3 d^2\right ) \cos (e+f x) \sqrt{c+d \sin (e+f x)}}{30 (c-d)^2 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac{\left (4 c^2-5 c d-3 d^2\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)^2 f \sqrt{\frac{c+d \sin (e+f x)}{c+d}}}+\frac{(4 c-5 d) (c+d) 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) f \sqrt{c+d \sin (e+f x)}}\\ \end{align*}

Mathematica [A]  time = 5.95959, size = 449, normalized size = 1.34 \[ \frac{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^6 \left (-\left (4 c^2-5 c d-3 d^2\right ) (c+d \sin (e+f x))+\frac{2 (c+d \sin (e+f x)) \left (\left (4 c^2-5 c d-3 d^2\right ) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^4+6 (c-d)^2 \sin \left (\frac{1}{2} (e+f x)\right )-(c-d) (2 c-d) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^3+2 (c-d) (2 c-d) \sin \left (\frac{1}{2} (e+f x)\right ) \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^2-3 (c-d)^2 \left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )\right )}{\left (\sin \left (\frac{1}{2} (e+f x)\right )+\cos \left (\frac{1}{2} (e+f x)\right )\right )^5}+\left (4 c^2-5 c d-3 d^2\right ) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} \left ((c+d) E\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )-c F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )+d^2 (c-5 d) \sqrt{\frac{c+d \sin (e+f x)}{c+d}} F\left (\frac{1}{4} (-2 e-2 f x+\pi )|\frac{2 d}{c+d}\right )\right )}{30 a^3 f (c-d)^2 (\sin (e+f x)+1)^3 \sqrt{c+d \sin (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(-((4*c^2 - 5*c*d - 3*d^2)*(c + d*Sin[e + f*x])) + (2*(6*(c - d)^2*Si
n[(e + f*x)/2] - 3*(c - d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 2*(c - d)*(2*c - d)*Sin[(e + f*x)/2]*(Cos
[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - (c - d)*(2*c - d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (4*c^2 - 5*c
*d - 3*d^2)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4)*(c + d*Sin[e + f*x]))/(Cos[(e + f*x)/2]
+ Sin[(e + f*x)/2])^5 + (c - 5*d)*d^2*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x]
)/(c + d)] + (4*c^2 - 5*c*d - 3*d^2)*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-
2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*Sqrt[(c + d*Sin[e + f*x])/(c + d)]))/(30*a^3*(c - d)^2*f*(1 + Sin[e + f*x
])^3*Sqrt[c + d*Sin[e + f*x]])

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Maple [B]  time = 5.708, size = 1056, normalized size = 3.2 \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))^(1/2)/(a+a*sin(f*x+e))^3,x)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/a^3*((c-d)*(-1/5/(c-d)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(1+sin(
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*si
n(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*(-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)*((-si
n(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]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{d \sin \left (f x + e\right ) + c}}{{\left (a \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))^(1/2)/(a+a*sin(f*x+e))^3,x, algorithm="maxima")

[Out]

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

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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}}{3 \, a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3} +{\left (a^{3} \cos \left (f x + e\right )^{2} - 4 \, a^{3}\right )} \sin \left (f x + e\right )}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(d*sin(f*x + e) + c)/(3*a^3*cos(f*x + e)^2 - 4*a^3 + (a^3*cos(f*x + e)^2 - 4*a^3)*sin(f*x + e)),
x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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