\(\int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx\) [510]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [B] (verified)
   Fricas [C] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [F(-1)]

Optimal result

Integrand size = 27, antiderivative size = 247 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=-\frac {(c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{27 f (1+\sin (e+f x))}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (3+3 \sin (e+f x))^2}-\frac {\left (c^2+5 c d-12 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{27 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c+5 d) \left (c^2-d^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{27 f \sqrt {c+d \sin (e+f x)}} \]

[Out]

-1/3*(c-d)*cos(f*x+e)*(c+d*sin(f*x+e))^(3/2)/f/(a+a*sin(f*x+e))^2-1/3*(c-d)*(c+5*d)*cos(f*x+e)*(c+d*sin(f*x+e)
)^(1/2)/a^2/f/(1+sin(f*x+e))+1/3*(c^2+5*c*d-12*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^2/f/((c+d*sin(f*x+e)
)/(c+d))^(1/2)-1/3*(c+5*d)*(c^2-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)*EllipticF(c
os(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 = 0.36 (sec) , antiderivative size = 256, 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 = {2844, 3056, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\frac {(c+5 d) \left (c^2-d^2\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{3 a^2 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (c^2+5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 a^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {(c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 a^2 f (\sin (e+f x)+1)}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a \sin (e+f x)+a)^2} \]

[In]

Int[(c + d*Sin[e + f*x])^(5/2)/(a + a*Sin[e + f*x])^2,x]

[Out]

-1/3*((c - d)*(c + 5*d)*Cos[e + f*x]*Sqrt[c + d*Sin[e + f*x]])/(a^2*f*(1 + Sin[e + f*x])) - ((c - d)*Cos[e + f
*x]*(c + d*Sin[e + f*x])^(3/2))/(3*f*(a + a*Sin[e + f*x])^2) - ((c^2 + 5*c*d - 12*d^2)*EllipticE[(e - Pi/2 + f
*x)/2, (2*d)/(c + d)]*Sqrt[c + d*Sin[e + f*x]])/(3*a^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + ((c + 5*d)*(c^2
 - d^2)*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*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 3056

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(A*b - a*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*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, 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) (c+d \sin (e+f x))^{3/2}}{3 f (a+a \sin (e+f x))^2}-\frac {\int \frac {\sqrt {c+d \sin (e+f x)} \left (-\frac {1}{2} a \left (2 c^2+7 c d-3 d^2\right )+\frac {1}{2} a (c-7 d) d \sin (e+f x)\right )}{a+a \sin (e+f x)} \, dx}{3 a^2} \\ & = -\frac {(c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 a^2 f (1+\sin (e+f x))}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a+a \sin (e+f x))^2}-\frac {\int \frac {-\frac {1}{2} a^2 (11 c-5 d) d^2+\frac {1}{2} a^2 d \left (c^2+5 c d-12 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 a^4} \\ & = -\frac {(c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 a^2 f (1+\sin (e+f x))}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a+a \sin (e+f x))^2}-\frac {\left (c^2+5 c d-12 d^2\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{6 a^2}+\frac {\left ((c+5 d) \left (c^2-d^2\right )\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{6 a^2} \\ & = -\frac {(c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 a^2 f (1+\sin (e+f x))}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a+a \sin (e+f x))^2}-\frac {\left (\left (c^2+5 c d-12 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}{6 a^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left ((c+5 d) \left (c^2-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}{6 a^2 \sqrt {c+d \sin (e+f x)}} \\ & = -\frac {(c-d) (c+5 d) \cos (e+f x) \sqrt {c+d \sin (e+f x)}}{3 a^2 f (1+\sin (e+f x))}-\frac {(c-d) \cos (e+f x) (c+d \sin (e+f x))^{3/2}}{3 f (a+a \sin (e+f x))^2}-\frac {\left (c^2+5 c d-12 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)}}{3 a^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {(c+5 d) \left (c^2-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}}}{3 a^2 f \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.89 (sec) , antiderivative size = 307, normalized size of antiderivative = 1.24 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\frac {\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^4 \left (-\left (\left (c^2+5 c d-6 d^2\right ) (c+d \sin (e+f x))\right )+\frac {(c-d) \left (7 d \cos \left (\frac {1}{2} (e+f x)\right )-(c+6 d) \cos \left (\frac {3}{2} (e+f x)\right )+(3 c+11 d) \sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))}{\left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3}+d^2 (-11 c+5 d) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}+\left (c^2+5 c d-12 d^2\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}\right )}{27 f (1+\sin (e+f x))^2 \sqrt {c+d \sin (e+f x)}} \]

[In]

Integrate[(c + d*Sin[e + f*x])^(5/2)/(3 + 3*Sin[e + f*x])^2,x]

[Out]

((Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^4*(-((c^2 + 5*c*d - 6*d^2)*(c + d*Sin[e + f*x])) + ((c - d)*(7*d*Cos[(e
 + f*x)/2] - (c + 6*d)*Cos[(3*(e + f*x))/2] + (3*c + 11*d)*Sin[(e + f*x)/2])*(c + d*Sin[e + f*x]))/(Cos[(e + f
*x)/2] + Sin[(e + f*x)/2])^3 + d^2*(-11*c + 5*d)*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]*Sqrt[(c + d*S
in[e + f*x])/(c + d)] + (c^2 + 5*c*d - 12*d^2)*((c + d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*El
lipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)])*Sqrt[(c + d*Sin[e + f*x])/(c + d)]))/(27*f*(1 + Sin[e + f*x])^2
*Sqrt[c + d*Sin[e + f*x]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1371\) vs. \(2(302)=604\).

Time = 21.83 (sec) , antiderivative size = 1372, normalized size of antiderivative = 5.55

method result size
default \(\text {Expression too large to display}\) \(1372\)

[In]

int((c+d*sin(f*x+e))^(5/2)/(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*(-4*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))+2*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)))+6*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)*EllipticF(((c+d*sin(f*x+e))/(c-d))^(1/2),((c-d)/(c+d))^(1/2))+3*d*(c^2-2*c*
d+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-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^3-3*c^2*d+3*c*d^2-d^3)*(-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))+Elliptic
F(((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 = 1152, normalized size of antiderivative = 4.66 \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\text {Too large to display} \]

[In]

integrate((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x, algorithm="fricas")

[Out]

1/18*((sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3)*cos(f*x + e)^2 - sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15
*d^3)*cos(f*x + e) - (sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3)*cos(f*x + e) + 2*sqrt(2)*(2*c^3 + 10*c^2*d
 + 9*c*d^2 - 15*d^3))*sin(f*x + e) - 2*sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3))*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^3 + 10*c^2*d + 9*c*d^2 - 15*d^3)*cos(f*x + e)^2 - sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^
2 - 15*d^3)*cos(f*x + e) - (sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3)*cos(f*x + e) + 2*sqrt(2)*(2*c^3 + 10
*c^2*d + 9*c*d^2 - 15*d^3))*sin(f*x + e) - 2*sqrt(2)*(2*c^3 + 10*c^2*d + 9*c*d^2 - 15*d^3))*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*(sqrt(2)*(I*c^2*d + 5*I*c*d^2 - 12*I*d^3)*cos(f*x + e)^2 + sqrt(2)*(-I*c^2*d - 5*I*c*d^2
 + 12*I*d^3)*cos(f*x + e) + (sqrt(2)*(-I*c^2*d - 5*I*c*d^2 + 12*I*d^3)*cos(f*x + e) + 2*sqrt(2)*(-I*c^2*d - 5*
I*c*d^2 + 12*I*d^3))*sin(f*x + e) + 2*sqrt(2)*(-I*c^2*d - 5*I*c*d^2 + 12*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)*(-I*c^2*d - 5*I*
c*d^2 + 12*I*d^3)*cos(f*x + e)^2 + sqrt(2)*(I*c^2*d + 5*I*c*d^2 - 12*I*d^3)*cos(f*x + e) + (sqrt(2)*(I*c^2*d +
 5*I*c*d^2 - 12*I*d^3)*cos(f*x + e) + 2*sqrt(2)*(I*c^2*d + 5*I*c*d^2 - 12*I*d^3))*sin(f*x + e) + 2*sqrt(2)*(I*
c^2*d + 5*I*c*d^2 - 12*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*(c^2*d - 2*c*d^2 + d^3 + (c^2*d + 5*c*d^2 - 6*d^3)*cos(f*x + e)^2 + (2*c
^2*d + 3*c*d^2 - 5*d^3)*cos(f*x + e) - (c^2*d - 2*c*d^2 + d^3 - (c^2*d + 5*c*d^2 - 6*d^3)*cos(f*x + e))*sin(f*
x + e))*sqrt(d*sin(f*x + e) + c))/(a^2*d*f*cos(f*x + e)^2 - a^2*d*f*cos(f*x + e) - 2*a^2*d*f - (a^2*d*f*cos(f*
x + e) + 2*a^2*d*f)*sin(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\text {Timed out} \]

[In]

integrate((c+d*sin(f*x+e))**(5/2)/(a+a*sin(f*x+e))**2,x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x, algorithm="maxima")

[Out]

integrate((d*sin(f*x + e) + c)^(5/2)/(a*sin(f*x + e) + a)^2, x)

Giac [F]

\[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\int { \frac {{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {5}{2}}}{{\left (a \sin \left (f x + e\right ) + a\right )}^{2}} \,d x } \]

[In]

integrate((c+d*sin(f*x+e))^(5/2)/(a+a*sin(f*x+e))^2,x, algorithm="giac")

[Out]

integrate((d*sin(f*x + e) + c)^(5/2)/(a*sin(f*x + e) + a)^2, x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d \sin (e+f x))^{5/2}}{(3+3 \sin (e+f x))^2} \, dx=\int \frac {{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{5/2}}{{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2} \,d x \]

[In]

int((c + d*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x))^2,x)

[Out]

int((c + d*sin(e + f*x))^(5/2)/(a + a*sin(e + f*x))^2, x)