\(\int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx\) [518]

Optimal result

Integrand size = 27, antiderivative size = 321 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 f (3+3 \sin (e+f x))^3}-\frac {(2 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{45 (c-d) f (3+3 \sin (e+f x))^2}-\frac {\left (4 c^2-5 c d-3 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d)^2 f (27+27 \sin (e+f x))}-\frac {\left (4 c^2-5 c d-3 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{810 (c-d)^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(4 c-5 d) (c+d) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{810 (c-d) f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-1/5*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/f/(a+a*sin(f*x+e))^3-1/15*(2*c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/a/(
c-d)/f/(a+a*sin(f*x+e))^2-1/30*(4*c^2-5*c*d-3*d^2)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/(c-d)^2/f/(a^3+a^3*sin(f*
x+e))+1/30*(4*c^2-5*c*d-3*d^2)*(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*Pi+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*(c+d*sin(f*x+e))^(1/2)/a^3/(c-d)^2/f/((c+d*sin(f*x+e))/(c+d))^(1/2
)-1/30*(4*c-5*d)*(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*P
i+1/2*f*x),2^(1/2)*(d/(c+d))^(1/2))*((c+d*sin(f*x+e))/(c+d))^(1/2)/a^3/(c-d)/f/(c+d*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.53 (sec) , antiderivative size = 334, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2843, 3057, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx=-\frac {\left (4 c^2-5 c d-3 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 f (c-d)^2 \left (a^3 \sin (e+f x)+a^3\right )}-\frac {\left (4 c^2-5 c d-3 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{30 a^3 f (c-d)^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(4 c-5 d) (c+d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{30 a^3 f (c-d) \sqrt {c+d \sin (e+f x)}}-\frac {(2 c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (c-d) (a \sin (e+f x)+a)^2}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 f (a \sin (e+f x)+a)^3} \]

[In]

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

[Out]

-1/5*(Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f*(a + a*Sin[e + f*x])^3) - ((2*c - d)*Cos[e + f*x]*Sqrt[c + d*S
in[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*Sq
rt[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)}}{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) \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}}}{30 a^3 (c-d) f \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.22 (sec) , antiderivative size = 446, normalized size of antiderivative = 1.39 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 \left (-\left (\left (4 c^2-5 c d-3 d^2\right ) (c+d \sin (e+f x))\right )+\frac {2 \left (6 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 (c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+2 (c-d) (2 c-d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-(c-d) (2 c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+\left (4 c^2-5 c d-3 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right ) (c+d \sin (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}+(c-5 d) d^2 \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}+\left (4 c^2-5 c d-3 d^2\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right )}{810 (c-d)^2 f (1+\sin (e+f x))^3 \sqrt {c+d \sin (e+f x)}} \]

[In]

Integrate[Sqrt[c + d*Sin[e + f*x]]/(3 + 3*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)]))/(810*(c - d)^2*f*(1 + Sin[e + f*x])^
3*Sqrt[c + d*Sin[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1055\) vs. \(2(376)=752\).

Time = 3.92 (sec) , antiderivative size = 1056, normalized size of antiderivative = 3.29

method	result	size
default	\(\text {Expression too large to display}\)	\(1056\)

[In]

int((c+d*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^3,x,method=_RETURNVERBOSE)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/a^3*(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)))
)+(c-d)*(-1/5/(c-d)*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+1)^3-2/15*(c-3*d)/(c-d)^2*(-(-d*sin(f*
x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+1)^2-1/30*(-d*sin(f*x+e)^2-c*sin(f*x+e)+d*sin(f*x+e)+c)/(c-d)^3*(4*c^2
-15*c*d+27*d^2)/((sin(f*x+e)+1)*(sin(f*x+e)-1)*(-d*sin(f*x+e)-c))^(1/2)+2*(-c*d^2-15*d^3)/(60*c^3-180*c^2*d+18
0*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)*(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)
)-1/30*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)*(
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 = 0.16 (sec) , antiderivative size = 1658, normalized size of antiderivative = 5.17 \[ \int \frac {\sqrt {c+d \sin (e+f x)}}{(3+3 \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/180*((sqrt(2)*(8*c^3 - 10*c^2*d - 9*c*d^2 + 15*d^3)*cos(f*x + e)^3 + 3*sqrt(2)*(8*c^3 - 10*c^2*d - 9*c*d^2 +
15*d^3)*cos(f*x + e)^2 - 2*sqrt(2)*(8*c^3 - 10*c^2*d - 9*c*d^2 + 15*d^3)*cos(f*x + e) + (sqrt(2)*(8*c^3 - 10*
c^2*d - 9*c*d^2 + 15*d^3)*cos(f*x + e)^2 - 2*sqrt(2)*(8*c^3 - 10*c^2*d - 9*c*d^2 + 15*d^3)*cos(f*x + e) - 4*sq
rt(2)*(8*c^3 - 10*c^2*d - 9*c*d^2 + 15*d^3))*sin(f*x + e) - 4*sqrt(2)*(8*c^3 - 10*c^2*d - 9*c*d^2 + 15*d^3))*s
qrt(I*d)*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) + (sqrt(2)*(8*c^3 - 10*c^2*d - 9*c*d^2 + 15*d^3)*cos(f*x + e)^3 + 3*sqrt(2)*(
8*c^3 - 10*c^2*d - 9*c*d^2 + 15*d^3)*cos(f*x + e)^2 - 2*sqrt(2)*(8*c^3 - 10*c^2*d - 9*c*d^2 + 15*d^3)*cos(f*x
+ e) + (sqrt(2)*(8*c^3 - 10*c^2*d - 9*c*d^2 + 15*d^3)*cos(f*x + e)^2 - 2*sqrt(2)*(8*c^3 - 10*c^2*d - 9*c*d^2 +
15*d^3)*cos(f*x + e) - 4*sqrt(2)*(8*c^3 - 10*c^2*d - 9*c*d^2 + 15*d^3))*sin(f*x + e) - 4*sqrt(2)*(8*c^3 - 10*
c^2*d - 9*c*d^2 + 15*d^3))*sqrt(-I*d)*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*(sqrt(2)*(-4*I*c^2*d + 5*I*c*d^2 + 3*I*d^3)
*cos(f*x + e)^3 + 3*sqrt(2)*(-4*I*c^2*d + 5*I*c*d^2 + 3*I*d^3)*cos(f*x + e)^2 + 2*sqrt(2)*(4*I*c^2*d - 5*I*c*d
^2 - 3*I*d^3)*cos(f*x + e) + (sqrt(2)*(-4*I*c^2*d + 5*I*c*d^2 + 3*I*d^3)*cos(f*x + e)^2 + 2*sqrt(2)*(4*I*c^2*d
- 5*I*c*d^2 - 3*I*d^3)*cos(f*x + e) + 4*sqrt(2)*(4*I*c^2*d - 5*I*c*d^2 - 3*I*d^3))*sin(f*x + e) + 4*sqrt(2)*(
4*I*c^2*d - 5*I*c*d^2 - 3*I*d^3))*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*(sqrt(2)*(4*I*c^2*d - 5*I*c*d^2 - 3*I*d^3)*cos(f*x + e)^3 + 3*sqrt(2)*(4
*I*c^2*d - 5*I*c*d^2 - 3*I*d^3)*cos(f*x + e)^2 + 2*sqrt(2)*(-4*I*c^2*d + 5*I*c*d^2 + 3*I*d^3)*cos(f*x + e) + (
sqrt(2)*(4*I*c^2*d - 5*I*c*d^2 - 3*I*d^3)*cos(f*x + e)^2 + 2*sqrt(2)*(-4*I*c^2*d + 5*I*c*d^2 + 3*I*d^3)*cos(f*
x + e) + 4*sqrt(2)*(-4*I*c^2*d + 5*I*c*d^2 + 3*I*d^3))*sin(f*x + e) + 4*sqrt(2)*(-4*I*c^2*d + 5*I*c*d^2 + 3*I*
d^3))*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, weierstrassPInver
se(-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*((4*c^2*d - 5*c*d^2 - 3*d^3)*cos(f*x + e)^3 - 6*c^2*d + 12*c*d^2 - 6*d^3 - (8*c^2*d - 11*c*d^2 -
d^3)*cos(f*x + e)^2 - 2*(9*c^2*d - 14*c*d^2 + d^3)*cos(f*x + e) + (6*c^2*d - 12*c*d^2 + 6*d^3 - (4*c^2*d - 5*c
*d^2 - 3*d^3)*cos(f*x + e)^2 - 4*(3*c^2*d - 4*c*d^2 - d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(d*sin(f*x + e) + c
))/((a^3*c^2*d - 2*a^3*c*d^2 + a^3*d^3)*f*cos(f*x + e)^3 + 3*(a^3*c^2*d - 2*a^3*c*d^2 + a^3*d^3)*f*cos(f*x + e
)^2 - 2*(a^3*c^2*d - 2*a^3*c*d^2 + a^3*d^3)*f*cos(f*x + e) - 4*(a^3*c^2*d - 2*a^3*c*d^2 + a^3*d^3)*f + ((a^3*c
^2*d - 2*a^3*c*d^2 + a^3*d^3)*f*cos(f*x + e)^2 - 2*(a^3*c^2*d - 2*a^3*c*d^2 + a^3*d^3)*f*cos(f*x + e) - 4*(a^3
*c^2*d - 2*a^3*c*d^2 + a^3*d^3)*f)*sin(f*x + e))

Sympy [F]

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

[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

Maxima [F]

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

[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)

Giac [F]

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

[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)

Mupad [F(-1)]

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

[In]

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

[Out]

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