\(\int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^2} \, dx\) [512]

Optimal result

Integrand size = 27, antiderivative size = 224 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^2} \, dx=-\frac {c \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{27 (c-d) f (1+\sin (e+f x))}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (3+3 \sin (e+f x))^2}-\frac {c E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{27 (c-d) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c+d) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{27 f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-1/3*c*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/a^2/(c-d)/f/(1+sin(f*x+e))-1/3*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f/(a
+a*sin(f*x+e))^2+1/3*c*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticE(cos(1/2*e+1/4*P
i+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/a^2/(c-d)/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-1/3*(c+d
)*(sin(1/2*e+1/4*Pi+1/2*f*x)^2)^(1/2)/sin(1/2*e+1/4*Pi+1/2*f*x)*EllipticF(cos(1/2*e+1/4*Pi+1/2*f*x),2^(1/2)*(d
/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/a^2/f/(c+d*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.04, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2843, 3057, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^2} \, dx=-\frac {c \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 a^2 f (c-d) (\sin (e+f x)+1)}+\frac {(c+d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{3 a^2 f \sqrt {c+d \sin (e+f x)}}-\frac {c \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 a^2 f (c-d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (a \sin (e+f x)+a)^2} \]

[In]

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

[Out]

-1/3*(c*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(a^2*(c - d)*f*(1 + Sin[e + f*x])) - (Cos[e + f*x]*Sqrt[c + d*S
in[e + f*x]])/(3*f*(a + a*Sin[e + f*x])^2) - (c*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e
+ f*x]])/(3*a^2*(c - d)*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((c + d)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(
c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*a^2*f*Sqrt[c + d*Sin[e + f*x]])

Rule 2732

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[2*(Sqrt[a + b]/d)*EllipticE[(1/2)*(c - Pi/2
+ d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2734

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 2740

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/(d*Sqrt[a + b]))*EllipticF[(1/2)*(c - P
i/2 + d*x), 2*(b/(a + b))], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2742

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/(a + b))*Sin[c + d*x]], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a
^2 - b^2, 0] &&  !GtQ[a + b, 0]

Rule 2831

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 2843

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 3057

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])

Rubi steps \begin{align*} \text {integral}& = -\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (a+a \sin (e+f x))^2}+\frac {\int \frac {\frac {1}{2} a (2 c+d)+\frac {1}{2} a d \sin (e+f x)}{(a+a \sin (e+f x)) \sqrt {c+d \sin (e+f x)}} \, dx}{3 a^2} \\ & = -\frac {c \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 a^2 (c-d) f (1+\sin (e+f x))}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (a+a \sin (e+f x))^2}-\frac {\int \frac {\frac {a^2 d^2}{2}+\frac {1}{2} a^2 c d \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 a^4 (c-d)} \\ & = -\frac {c \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 a^2 (c-d) f (1+\sin (e+f x))}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (a+a \sin (e+f x))^2}-\frac {c \int \sqrt {c+d \sin (e+f x)} \, dx}{6 a^2 (c-d)}+\frac {(c+d) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{6 a^2} \\ & = -\frac {c \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 a^2 (c-d) f (1+\sin (e+f x))}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (a+a \sin (e+f x))^2}-\frac {\left (c \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{6 a^2 (c-d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left ((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}{6 a^2 \sqrt {c+d \sin (e+f x)}} \\ & = -\frac {c \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 a^2 (c-d) f (1+\sin (e+f x))}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 f (a+a \sin (e+f x))^2}-\frac {c 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)}}{3 a^2 (c-d) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c+d) \operatorname {EllipticF}\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}}}{3 a^2 f \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.81 (sec) , antiderivative size = 253, normalized size of antiderivative = 1.13 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (-c (c+d \sin (e+f x))+\frac {\left (d \cos \left (\frac {1}{2} (e+f x)\right )-c \cos \left (\frac {3}{2} (e+f x)\right )+(3 c-d) \sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}+c (c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}-\left (c^2-d^2\right ) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right )}{27 (c-d) f (1+\sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}} \]

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(-(c*(c + d*Sin[e + f*x])) + ((d*Cos[(e + f*x)/2] - c*Cos[(3*(e + f*x
))/2] + (3*c - d)*Sin[(e + f*x)/2])*(c + d*Sin[e + f*x]))/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + c*(c + d)*
EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)] - (c^2 - d^2)*EllipticF[(-2
*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)]))/(27*(c - d)*f*(1 + Sin[e + f*x])^2*Sqr
t[c + d*Sin[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(905\) vs. \(2(279)=558\).

Time = 2.36 (sec) , antiderivative size = 906, normalized size of antiderivative = 4.04

method	result	size
default	\(\text {Expression too large to display}\)	\(906\)

[In]

int((c+d*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^2,x,method=_RETURNVERBOSE)

[Out]

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

Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.12 (sec) , antiderivative size = 880, normalized size of antiderivative = 3.93 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^2} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/18*((sqrt(2)*(2*c^2 - 3*d^2)*cos(f*x + e)^2 - sqrt(2)*(2*c^2 - 3*d^2)*cos(f*x + e) - (sqrt(2)*(2*c^2 - 3*d^2
)*cos(f*x + e) + 2*sqrt(2)*(2*c^2 - 3*d^2))*sin(f*x + e) - 2*sqrt(2)*(2*c^2 - 3*d^2))*sqrt(I*d)*weierstrassPIn
verse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) -
2*I*c)/d) + (sqrt(2)*(2*c^2 - 3*d^2)*cos(f*x + e)^2 - sqrt(2)*(2*c^2 - 3*d^2)*cos(f*x + e) - (sqrt(2)*(2*c^2 -
3*d^2)*cos(f*x + e) + 2*sqrt(2)*(2*c^2 - 3*d^2))*sin(f*x + e) - 2*sqrt(2)*(2*c^2 - 3*d^2))*sqrt(-I*d)*weierst
rassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x
+ e) + 2*I*c)/d) - 3*(-I*sqrt(2)*c*d*cos(f*x + e)^2 + I*sqrt(2)*c*d*cos(f*x + e) + 2*I*sqrt(2)*c*d + (I*sqrt(
2)*c*d*cos(f*x + e) + 2*I*sqrt(2)*c*d)*sin(f*x + e))*sqrt(I*d)*weierstrassZeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27
*(8*I*c^3 - 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(8*I*c^3 - 9*I*c*d^2)/d^3, 1/3
*(3*d*cos(f*x + e) - 3*I*d*sin(f*x + e) - 2*I*c)/d)) - 3*(I*sqrt(2)*c*d*cos(f*x + e)^2 - I*sqrt(2)*c*d*cos(f*x
+ e) - 2*I*sqrt(2)*c*d + (-I*sqrt(2)*c*d*cos(f*x + e) - 2*I*sqrt(2)*c*d)*sin(f*x + e))*sqrt(-I*d)*weierstrass
Zeta(-4/3*(4*c^2 - 3*d^2)/d^2, -8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, weierstrassPInverse(-4/3*(4*c^2 - 3*d^2)/d^2,
-8/27*(-8*I*c^3 + 9*I*c*d^2)/d^3, 1/3*(3*d*cos(f*x + e) + 3*I*d*sin(f*x + e) + 2*I*c)/d)) + 6*(c*d*cos(f*x +
e)^2 + c*d - d^2 + (2*c*d - d^2)*cos(f*x + e) + (c*d*cos(f*x + e) - c*d + d^2)*sin(f*x + e))*sqrt(d*sin(f*x +
e) + c))/((a^2*c*d - a^2*d^2)*f*cos(f*x + e)^2 - (a^2*c*d - a^2*d^2)*f*cos(f*x + e) - 2*(a^2*c*d - a^2*d^2)*f
- ((a^2*c*d - a^2*d^2)*f*cos(f*x + e) + 2*(a^2*c*d - a^2*d^2)*f)*sin(f*x + e))

Sympy [F]

\[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^2} \, dx=\frac {\int \frac {\sqrt {c + d \sin {\left (e + f x \right )}}}{\sin ^{2}{\left (e + f x \right )} + 2 \sin {\left (e + f x \right )} + 1}\, dx}{a^{2}} \]

[In]

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

[Out]

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

Maxima [F]

\[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^2} \, dx=\int { \frac {\sqrt {d \sin \left (f x + e\right ) + c}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^2} \, dx=\int { \frac {\sqrt {d \sin \left (f x + e\right ) + c}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^2} \, dx=\int \frac {\sqrt {c+d\,\sin \left (e+f\,x\right )}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

[In]

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

[Out]

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