\(\int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^3} \, dx\) [517]

Optimal result

Integrand size = 27, antiderivative size = 310 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^3} \, dx=-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 f (3+3 \sin (e+f x))^3}-\frac {2 (c+2 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{45 f (3+3 \sin (e+f x))^2}-\frac {\left (4 c^2+5 c d-3 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 (c-d) f (27+27 \sin (e+f x))}-\frac {\left (4 c^2+5 c d-3 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{810 (c-d) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c+d) (4 c+5 d) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{810 f \sqrt {c+d \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.61 (sec) , antiderivative size = 323, 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 = {2844, 3057, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^3} \, dx=-\frac {\left (4 c^2+5 c d-3 d^2\right ) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{30 f (c-d) \left (a^3 \sin (e+f x)+a^3\right )}-\frac {\left (4 c^2+5 c d-3 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{30 a^3 f (c-d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c+d) (4 c+5 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{30 a^3 f \sqrt {c+d \sin (e+f x)}}-\frac {2 (c+2 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (a \sin (e+f x)+a)^2}-\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 f (a \sin (e+f x)+a)^3} \]

[In]

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

[Out]

-1/5*((c - d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(f*(a + a*Sin[e + f*x])^3) - (2*(c + 2*d)*Cos[e + f*x]*Sq
rt[c + d*Sin[e + f*x]])/(15*a*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)*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)*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((c + d)*(4*c
+ 5*d)*EllipticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(30*a^3*f*Sqrt[c + d*S
in[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 2844

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^(n - 1)/(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 - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*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] && GtQ[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 {(c-d) \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 \left (4 c^2+7 c d-d^2\right )-\frac {1}{2} a d (3 c+7 d) \sin (e+f x)}{(a+a \sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}} \, dx}{5 a^2} \\ & = -\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 f (a+a \sin (e+f x))^3}-\frac {2 (c+2 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a f (a+a \sin (e+f x))^2}+\frac {\int \frac {\frac {1}{2} a^2 (c-d) \left (4 c^2+7 c d+d^2\right )+a^2 (c-d) d (c+2 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 {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 f (a+a \sin (e+f x))^3}-\frac {2 (c+2 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a 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) f \left (a^3+a^3 \sin (e+f x)\right )}-\frac {\int \frac {\frac {1}{4} a^3 (c-d) d^2 (c+5 d)+\frac {1}{4} a^3 (c-d) 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 {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 f (a+a \sin (e+f x))^3}-\frac {2 (c+2 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a 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) f \left (a^3+a^3 \sin (e+f x)\right )}+\frac {((c+d) (4 c+5 d)) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{60 a^3}-\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)} \\ & = -\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 f (a+a \sin (e+f x))^3}-\frac {2 (c+2 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a 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) 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) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left ((c+d) (4 c+5 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 \sqrt {c+d \sin (e+f x)}} \\ & = -\frac {(c-d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{5 f (a+a \sin (e+f x))^3}-\frac {2 (c+2 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{15 a 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) 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) f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c+d) (4 c+5 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 f \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 4.29 (sec) , antiderivative size = 438, normalized size of antiderivative = 1.41 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^3} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^6 \left (\frac {2 \left (6 (c-d) \sin \left (\frac {1}{2} (e+f x)\right )+3 (-c+d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )+4 (c+2 d) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^2-2 (c+2 d) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3+\frac {\left (4 c^2+5 c d-3 d^2\right ) \sin \left (\frac {1}{2} (e+f x)\right ) \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4}{c-d}\right ) (c+d \sin (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^5}-\frac {\left (4 c^2+5 c d-3 d^2\right ) (c+d \sin (e+f x))-d^2 (c+5 d) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}+\left (-4 c^2-5 c d+3 d^2\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{c-d}\right )}{810 f (1+\sin (e+f x))^3 \sqrt {c+d \sin (e+f x)}} \]

[In]

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

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^6*((2*(6*(c - d)*Sin[(e + f*x)/2] + 3*(-c + d)*(Cos[(e + f*x)/2] + Sin[
(e + f*x)/2]) + 4*(c + 2*d)*Sin[(e + f*x)/2]*(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^2 - 2*(c + 2*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))*(c + d*Sin[e + f*x]))/(Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^5 - ((4*c^2 + 5*c*d - 3*d^2)*(c + d*S
in[e + f*x]) - d^2*(c + 5*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 - 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)])/(c - d)))/(810*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. \(1461\) vs. \(2(365)=730\).

Time = 4.43 (sec) , antiderivative size = 1462, normalized size of antiderivative = 4.72

method	result	size
default	\(\text {Expression too large to display}\)	\(1462\)

[In]

int((c+d*sin(f*x+e))^(3/2)/(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^2*(-(-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-si
n(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))))+2*d*(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*s
in(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^2-2*c*d+d^2)*(-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)/((si
n(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 = 0.14 (sec) , antiderivative size = 1589, normalized size of antiderivative = 5.13 \[ \int \frac {(c+d \sin (e+f x))^{3/2}}{(3+3 \sin (e+f x))^3} \, dx=\text {Too large to display} \]

[In]

integrate((c+d*sin(f*x+e))^(3/2)/(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 + 9*c*d^2 - 1
1*d^3)*cos(f*x + e)^2 - 2*(9*c^2*d + c*d^2 - 4*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 - 2*(6*c^2*d + 7*c*d^2 - 7*d^3)*cos(f*x + e))*sin(f*x + e))*sqrt(d*sin(f*x + e)
+ c))/((a^3*c*d - a^3*d^2)*f*cos(f*x + e)^3 + 3*(a^3*c*d - a^3*d^2)*f*cos(f*x + e)^2 - 2*(a^3*c*d - a^3*d^2)*f
*cos(f*x + e) - 4*(a^3*c*d - a^3*d^2)*f + ((a^3*c*d - a^3*d^2)*f*cos(f*x + e)^2 - 2*(a^3*c*d - a^3*d^2)*f*cos(
f*x + e) - 4*(a^3*c*d - a^3*d^2)*f)*sin(f*x + e))

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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