\(\int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c+d \sin (e+f x)}} \, dx\) [519]

Optimal result

Integrand size = 27, antiderivative size = 331 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c+d \sin (e+f x)}} \, dx=-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (3+3 \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{45 (c-d)^2 f (3+3 \sin (e+f x))^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d)^3 f (27+27 \sin (e+f x))}-\frac {\left (4 c^2-15 c d+27 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{810 (c-d)^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (4 c^2-11 c d+15 d^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{810 (c-d)^2 f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-1/5*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/(c-d)/f/(a+a*sin(f*x+e))^3-2/15*(c-3*d)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/
2)/a/(c-d)^2/f/(a+a*sin(f*x+e))^2-1/30*(4*c^2-15*c*d+27*d^2)*cos(f*x+e)*(c+d*sin(f*x+e))^(1/2)/(c-d)^3/f/(a^3+
a^3*sin(f*x+e))+1/30*(4*c^2-15*c*d+27*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)*Ellip
ticE(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)^3/f/((c+d*sin(f*x+e))
/(c+d))^(1/2)-1/30*(4*c^2-11*c*d+15*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)*Ellipti
cF(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^3/(c-d)^2/f/(c+d*sin(f*
x+e))^(1/2)

Rubi [A] (verified)

Time = 0.52 (sec) , antiderivative size = 344, 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 = {2845, 3057, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c+d \sin (e+f x)}} \, dx=-\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 f (c-d)^3 \left (a^3 \sin (e+f x)+a^3\right )}+\frac {\left (4 c^2-11 c d+15 d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{30 a^3 f (c-d)^2 \sqrt {c+d \sin (e+f x)}}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{30 a^3 f (c-d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (c-d)^2 (a \sin (e+f x)+a)^2}-\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 f (c-d) (a \sin (e+f x)+a)^3} \]

[In]

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

[Out]

-1/5*(Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/((c - d)*f*(a + a*Sin[e + f*x])^3) - (2*(c - 3*d)*Cos[e + f*x]*Sq
rt[c + d*Sin[e + f*x]])/(15*a*(c - d)^2*f*(a + a*Sin[e + f*x])^2) - ((4*c^2 - 15*c*d + 27*d^2)*Cos[e + f*x]*Sq
rt[c + d*Sin[e + f*x]])/(30*(c - d)^3*f*(a^3 + a^3*Sin[e + f*x])) - ((4*c^2 - 15*c*d + 27*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)^3*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]
) + ((4*c^2 - 11*c*d + 15*d^2)*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)^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 2845

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 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 (c-d) f (a+a \sin (e+f x))^3}-\frac {\int \frac {-\frac {1}{2} a (4 c-9 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 (c-d)} \\ & = -\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}+\frac {\int \frac {\frac {1}{2} a^2 \left (4 c^2-13 c d+21 d^2\right )+a^2 (c-3 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)^2} \\ & = -\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\int \frac {\frac {1}{4} a^3 d^2 (c+15 d)+\frac {1}{4} a^3 d \left (4 c^2-15 c d+27 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 a^6 (c-d)^3} \\ & = -\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {\left (4 c^2-11 c d+15 d^2\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{60 a^3 (c-d)^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{60 a^3 (c-d)^3} \\ & = -\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\left (\left (4 c^2-15 c d+27 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)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (\left (4 c^2-11 c d+15 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)^2 \sqrt {c+d \sin (e+f x)}} \\ & = -\frac {\cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 (c-d) f (a+a \sin (e+f x))^3}-\frac {2 (c-3 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a (c-d)^2 f (a+a \sin (e+f x))^2}-\frac {\left (4 c^2-15 c d+27 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d)^3 f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\left (4 c^2-15 c d+27 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)^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (4 c^2-11 c d+15 d^2\right ) \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)^2 f \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 8.96 (sec) , antiderivative size = 442, normalized size of antiderivative = 1.34 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c+d \sin (e+f x)}} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 \left (-\left (\left (4 c^2-15 c d+27 d^2\right ) (c+d \sin (e+f x))\right )+\frac {2 \left (6 (c-d)^2 \sin \left (\frac {1}{2} (e+f x)\right )-3 (c-d)^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 (c-3 d) (c-d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-2 (c-3 d) (c-d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+\left (4 c^2-15 c d+27 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4\right ) (c+d \sin (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}+d^2 (c+15 d) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}+\left (4 c^2-15 c d+27 d^2\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right )}{810 (c-d)^3 f (1+\sin (e+f x))^3 \sqrt {c+d \sin (e+f x)}} \]

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*(-((4*c^2 - 15*c*d + 27*d^2)*(c + d*Sin[e + f*x])) + (2*(6*(c - d)^2*
Sin[(e + f*x)/2] - 3*(c - d)^2*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2]) + 4*(c - 3*d)*(c - d)*Sin[(e + f*x)/2]*(C
os[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 2*(c - 3*d)*(c - d)*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3 + (4*c^2 -
15*c*d + 27*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 + d^2*(c + 15*d)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e
+ f*x])/(c + d)] + (4*c^2 - 15*c*d + 27*d^2)*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*Ell
ipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*Sqrt[(c + d*Sin[e + f*x])/(c + d)]))/(810*(c - d)^3*f*(1 + Sin[e
+ f*x])^3*Sqrt[c + d*Sin[e + f*x]])

Maple [A] (warning: unable to verify)

Time = 2.93 (sec) , antiderivative size = 593, normalized size of antiderivative = 1.79

method	result	size
default	\(\frac {\sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (-\frac {\sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{5 \left (c -d \right ) \left (\sin \left (f x +e \right )+1\right )^{3}}-\frac {2 \left (c -3 d \right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}{15 \left (c -d \right )^{2} \left (\sin \left (f x +e \right )+1\right )^{2}}-\frac {\left (-d \left (\sin ^{2}\left (f x +e \right )\right )-c \sin \left (f x +e \right )+d \sin \left (f x +e \right )+c \right ) \left (4 c^{2}-15 c d +27 d^{2}\right )}{30 \left (c -d \right )^{3} \sqrt {\left (\sin \left (f x +e \right )+1\right ) \left (\sin \left (f x +e \right )-1\right ) \left (-d \sin \left (f x +e \right )-c \right )}}+\frac {2 \left (-c \,d^{2}-15 d^{3}\right ) \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {\frac {\left (-\sin \left (f x +e \right )-1\right ) d}{c -d}}\, F\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )}{\left (60 c^{3}-180 c^{2} d +180 c \,d^{2}-60 d^{3}\right ) \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}-\frac {d \left (4 c^{2}-15 c d +27 d^{2}\right ) \left (\frac {c}{d}-1\right ) \sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}\, \sqrt {\frac {d \left (1-\sin \left (f x +e \right )\right )}{c +d}}\, \sqrt {\frac {\left (-\sin \left (f x +e \right )-1\right ) d}{c -d}}\, \left (\left (-\frac {c}{d}-1\right ) E\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )+F\left (\sqrt {\frac {c +d \sin \left (f x +e \right )}{c -d}}, \sqrt {\frac {c -d}{c +d}}\right )\right )}{30 \left (c -d \right )^{3} \sqrt {-\left (-d \sin \left (f x +e \right )-c \right ) \left (\cos ^{2}\left (f x +e \right )\right )}}\right )}{a^{3} \cos \left (f x +e \right ) \sqrt {c +d \sin \left (f x +e \right )}\, f}\)	\(593\)

[In]

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

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/a^3*(-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+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)*(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 = 1744, normalized size of antiderivative = 5.27 \[ \int \frac {1}{(3+3 \sin (e+f x))^3 \sqrt {c+d \sin (e+f x)}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

1/180*((sqrt(2)*(8*c^3 - 30*c^2*d + 51*c*d^2 - 45*d^3)*cos(f*x + e)^3 + 3*sqrt(2)*(8*c^3 - 30*c^2*d + 51*c*d^2
- 45*d^3)*cos(f*x + e)^2 - 2*sqrt(2)*(8*c^3 - 30*c^2*d + 51*c*d^2 - 45*d^3)*cos(f*x + e) + (sqrt(2)*(8*c^3 -
30*c^2*d + 51*c*d^2 - 45*d^3)*cos(f*x + e)^2 - 2*sqrt(2)*(8*c^3 - 30*c^2*d + 51*c*d^2 - 45*d^3)*cos(f*x + e) -
4*sqrt(2)*(8*c^3 - 30*c^2*d + 51*c*d^2 - 45*d^3))*sin(f*x + e) - 4*sqrt(2)*(8*c^3 - 30*c^2*d + 51*c*d^2 - 45*
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) + (sqrt(2)*(8*c^3 - 30*c^2*d + 51*c*d^2 - 45*d^3)*cos(f*x + e)^3 + 3*s
qrt(2)*(8*c^3 - 30*c^2*d + 51*c*d^2 - 45*d^3)*cos(f*x + e)^2 - 2*sqrt(2)*(8*c^3 - 30*c^2*d + 51*c*d^2 - 45*d^3
)*cos(f*x + e) + (sqrt(2)*(8*c^3 - 30*c^2*d + 51*c*d^2 - 45*d^3)*cos(f*x + e)^2 - 2*sqrt(2)*(8*c^3 - 30*c^2*d
+ 51*c*d^2 - 45*d^3)*cos(f*x + e) - 4*sqrt(2)*(8*c^3 - 30*c^2*d + 51*c*d^2 - 45*d^3))*sin(f*x + e) - 4*sqrt(2)
*(8*c^3 - 30*c^2*d + 51*c*d^2 - 45*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 - 15*I*c*
d^2 + 27*I*d^3)*cos(f*x + e)^3 + 3*sqrt(2)*(4*I*c^2*d - 15*I*c*d^2 + 27*I*d^3)*cos(f*x + e)^2 + 2*sqrt(2)*(-4*
I*c^2*d + 15*I*c*d^2 - 27*I*d^3)*cos(f*x + e) + (sqrt(2)*(4*I*c^2*d - 15*I*c*d^2 + 27*I*d^3)*cos(f*x + e)^2 +
2*sqrt(2)*(-4*I*c^2*d + 15*I*c*d^2 - 27*I*d^3)*cos(f*x + e) + 4*sqrt(2)*(-4*I*c^2*d + 15*I*c*d^2 - 27*I*d^3))*
sin(f*x + e) + 4*sqrt(2)*(-4*I*c^2*d + 15*I*c*d^2 - 27*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 + 15*I*c*d^2 - 27*I*d^
3)*cos(f*x + e)^3 + 3*sqrt(2)*(-4*I*c^2*d + 15*I*c*d^2 - 27*I*d^3)*cos(f*x + e)^2 + 2*sqrt(2)*(4*I*c^2*d - 15*
I*c*d^2 + 27*I*d^3)*cos(f*x + e) + (sqrt(2)*(-4*I*c^2*d + 15*I*c*d^2 - 27*I*d^3)*cos(f*x + e)^2 + 2*sqrt(2)*(4
*I*c^2*d - 15*I*c*d^2 + 27*I*d^3)*cos(f*x + e) + 4*sqrt(2)*(4*I*c^2*d - 15*I*c*d^2 + 27*I*d^3))*sin(f*x + e) +
4*sqrt(2)*(4*I*c^2*d - 15*I*c*d^2 + 27*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)) - 6*((4*c^2*d - 15*c*d^2 + 27*d^3)*cos(f*x + e)^3 - 6*c^2*d
+ 12*c*d^2 - 6*d^3 - (8*c^2*d - 31*c*d^2 + 39*d^3)*cos(f*x + e)^2 - 2*(9*c^2*d - 29*c*d^2 + 36*d^3)*cos(f*x +
e) + (6*c^2*d - 12*c*d^2 + 6*d^3 - (4*c^2*d - 15*c*d^2 + 27*d^3)*cos(f*x + e)^2 - 2*(6*c^2*d - 23*c*d^2 + 33*
d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(d*sin(f*x + e) + c))/((a^3*c^3*d - 3*a^3*c^2*d^2 + 3*a^3*c*d^3 - a^3*d^4
)*f*cos(f*x + e)^3 + 3*(a^3*c^3*d - 3*a^3*c^2*d^2 + 3*a^3*c*d^3 - a^3*d^4)*f*cos(f*x + e)^2 - 2*(a^3*c^3*d - 3
*a^3*c^2*d^2 + 3*a^3*c*d^3 - a^3*d^4)*f*cos(f*x + e) - 4*(a^3*c^3*d - 3*a^3*c^2*d^2 + 3*a^3*c*d^3 - a^3*d^4)*f
+ ((a^3*c^3*d - 3*a^3*c^2*d^2 + 3*a^3*c*d^3 - a^3*d^4)*f*cos(f*x + e)^2 - 2*(a^3*c^3*d - 3*a^3*c^2*d^2 + 3*a^
3*c*d^3 - a^3*d^4)*f*cos(f*x + e) - 4*(a^3*c^3*d - 3*a^3*c^2*d^2 + 3*a^3*c*d^3 - a^3*d^4)*f)*sin(f*x + e))

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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