\(\int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx\) [495]

Optimal result

Integrand size = 27, antiderivative size = 305 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx=\frac {18 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {12 (c+5 d) \cos (e+f x)}{5 d (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {12 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{5 (c-d) d (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {12 \left (c^2+5 c d-12 d^2\right ) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{5 (c-d) d^2 (c+d)^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {12 (c+5 d) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{5 d^2 (c+d)^2 f \sqrt {c+d \sin (e+f x)}} \]

[Out]

2/5*a^2*(c-d)*cos(f*x+e)/d/(c+d)/f/(c+d*sin(f*x+e))^(5/2)-4/15*a^2*(c+5*d)*cos(f*x+e)/d/(c+d)^2/f/(c+d*sin(f*x
+e))^(3/2)-4/15*a^2*(c^2+5*c*d-12*d^2)*cos(f*x+e)/(c-d)/d/(c+d)^3/f/(c+d*sin(f*x+e))^(1/2)+4/15*a^2*(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)/(c-d)/d^2/(c+d)^3/f/((c+d*sin(f*x+e))/(c+d))^(1/2)-4/15*a^2*(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)/d^2/(c+d)^2/f/(c+d*sin(f*x+e))^(1/2)

Rubi [A] (verified)

Time = 0.37 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.05, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2841, 2833, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx=-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{15 d f (c-d) (c+d)^3 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{15 d^2 f (c-d) (c+d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 (c+5 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{15 d^2 f (c+d)^2 \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d f (c+d)^2 (c+d \sin (e+f x))^{3/2}}+\frac {2 a^2 (c-d) \cos (e+f x)}{5 d f (c+d) (c+d \sin (e+f x))^{5/2}} \]

[In]

Int[(a + a*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(2*a^2*(c - d)*Cos[e + f*x])/(5*d*(c + d)*f*(c + d*Sin[e + f*x])^(5/2)) - (4*a^2*(c + 5*d)*Cos[e + f*x])/(15*d
*(c + d)^2*f*(c + d*Sin[e + f*x])^(3/2)) - (4*a^2*(c^2 + 5*c*d - 12*d^2)*Cos[e + f*x])/(15*(c - d)*d*(c + d)^3
*f*Sqrt[c + d*Sin[e + f*x]]) - (4*a^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]])/(15*(c - d)*d^2*(c + d)^3*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (4*a^2*(c + 5*d)*Ellip
ticF[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(15*d^2*(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 2833

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(-(
b*c - a*d))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Dist[1/((m + 1)*(a^2 - b
^2)), Int[(a + b*Sin[e + f*x])^(m + 1)*Simp[(a*c - b*d)*(m + 1) - (b*c - a*d)*(m + 2)*Sin[e + f*x], x], x], x]
/; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && IntegerQ[2*m]

Rule 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c
+ a*d))), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1
)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*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] && GtQ[m, 1] && LtQ[n, -1
] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rubi steps \begin{align*} \text {integral}& = \frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {(2 a) \int \frac {-5 a d-a (c+4 d) \sin (e+f x)}{(c+d \sin (e+f x))^{5/2}} \, dx}{5 d (c+d)} \\ & = \frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d (c+d)^2 f (c+d \sin (e+f x))^{3/2}}+\frac {(4 a) \int \frac {6 a (c-d) d+\frac {1}{2} a (c-d) (c+5 d) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{15 (c-d) d (c+d)^2} \\ & = \frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{15 (c-d) d (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {(8 a) \int \frac {-\frac {1}{4} a (11 c-5 d) (c-d) d+\frac {1}{4} a (c-d) \left (c^2+5 c d-12 d^2\right ) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 (c-d)^2 d (c+d)^3} \\ & = \frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{15 (c-d) d (c+d)^3 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (2 a^2 (c+5 d)\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{15 d^2 (c+d)^2}-\frac {\left (2 a^2 \left (c^2+5 c d-12 d^2\right )\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{15 (c-d) d^2 (c+d)^3} \\ & = \frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{15 (c-d) d (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (2 a^2 \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}{15 (c-d) d^2 (c+d)^3 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (2 a^2 (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}{15 d^2 (c+d)^2 \sqrt {c+d \sin (e+f x)}} \\ & = \frac {2 a^2 (c-d) \cos (e+f x)}{5 d (c+d) f (c+d \sin (e+f x))^{5/2}}-\frac {4 a^2 (c+5 d) \cos (e+f x)}{15 d (c+d)^2 f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 \left (c^2+5 c d-12 d^2\right ) \cos (e+f x)}{15 (c-d) d (c+d)^3 f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 \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)}}{15 (c-d) d^2 (c+d)^3 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 (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}}}{15 d^2 (c+d)^2 f \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.95 (sec) , antiderivative size = 247, normalized size of antiderivative = 0.81 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx=\frac {6 \left (-2 \left ((11 c-5 d) d^2 \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )-\left (c^2+5 c d-12 d^2\right ) \left ((c+d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-c \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right )\right ) (c+d \sin (e+f x))^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}+d \cos (e+f x) \left (3 (c-d)^2 (c+d)^2-2 (c-d) (c+d) (c+5 d) (c+d \sin (e+f x))-2 \left (c^2+5 c d-12 d^2\right ) (c+d \sin (e+f x))^2\right )\right )}{5 (c-d) d^2 (c+d)^3 f (c+d \sin (e+f x))^{5/2}} \]

[In]

Integrate[(3 + 3*Sin[e + f*x])^2/(c + d*Sin[e + f*x])^(7/2),x]

[Out]

(6*(-2*((11*c - 5*d)*d^2*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - (c^2 + 5*c*d - 12*d^2)*((c + d)*Ell
ipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - c*EllipticF[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)]))*(c + d*Sin[
e + f*x])^2*Sqrt[(c + d*Sin[e + f*x])/(c + d)] + d*Cos[e + f*x]*(3*(c - d)^2*(c + d)^2 - 2*(c - d)*(c + d)*(c
+ 5*d)*(c + d*Sin[e + f*x]) - 2*(c^2 + 5*c*d - 12*d^2)*(c + d*Sin[e + f*x])^2)))/(5*(c - d)*d^2*(c + d)^3*f*(c
+ d*Sin[e + f*x])^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1436\) vs. \(2(362)=724\).

Time = 13.32 (sec) , antiderivative size = 1437, normalized size of antiderivative = 4.71

method	result	size
default	\(\text {Expression too large to display}\)	\(1437\)
parts	\(\text {Expression too large to display}\)	\(3056\)

[In]

int((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(7/2),x,method=_RETURNVERBOSE)

[Out]

(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)*a^2*(1/d^2*(2*d*cos(f*x+e)^2/(c^2-d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^
2)^(1/2)+2*c/(c^2-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))+2/(c^2-d^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)*((-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*c+2*d)/d^2*(2/3/(c^2-d^
2)/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^2+8/3*d*cos(f*x+e)^2/(c^2-d^2)^2*c/(-(-d*sin(f*x
+e)-c)*cos(f*x+e)^2)^(1/2)+2*(3*c^2+d^2)/(3*c^4-6*c^2*d^2+3*d^4)*(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))+8/3*c*d/(c^2-d^2)^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)/d^2*(2/5/(c^2-d^2)/d^2*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e
)+c/d)^3+16/15*c/(c^2-d^2)^2/d*(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)/(sin(f*x+e)+c/d)^2+2/15*cos(f*x+e)^2*d/
(c^2-d^2)^3*(23*c^2+9*d^2)/(-(-d*sin(f*x+e)-c)*cos(f*x+e)^2)^(1/2)+2*(15*c^3+17*c*d^2)/(15*c^6-45*c^4*d^2+45*c
^2*d^4-15*d^6)*(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))
+2/15*d*(23*c^2+9*d^2)/(c^2-d^2)^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 = 0.18 (sec) , antiderivative size = 1603, normalized size of antiderivative = 5.26 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx=\text {Too large to display} \]

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(7/2),x, algorithm="fricas")

[Out]

2/45*((3*sqrt(2)*(2*a^2*c^4*d^2 + 10*a^2*c^3*d^3 + 9*a^2*c^2*d^4 - 15*a^2*c*d^5)*cos(f*x + e)^2 + (sqrt(2)*(2*
a^2*c^3*d^3 + 10*a^2*c^2*d^4 + 9*a^2*c*d^5 - 15*a^2*d^6)*cos(f*x + e)^2 - sqrt(2)*(6*a^2*c^5*d + 30*a^2*c^4*d^
2 + 29*a^2*c^3*d^3 - 35*a^2*c^2*d^4 + 9*a^2*c*d^5 - 15*a^2*d^6))*sin(f*x + e) - sqrt(2)*(2*a^2*c^6 + 10*a^2*c^
5*d + 15*a^2*c^4*d^2 + 15*a^2*c^3*d^3 + 27*a^2*c^2*d^4 - 45*a^2*c*d^5))*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)*(2*a^2*c^4*d^2 + 10*a^2*c^3*d^3 + 9*a^2*c^2*d^4 - 15*a^2*c*d^5)*cos(f*x + e)^2 + (sqrt(2)*(2*a^2*c^3*
d^3 + 10*a^2*c^2*d^4 + 9*a^2*c*d^5 - 15*a^2*d^6)*cos(f*x + e)^2 - sqrt(2)*(6*a^2*c^5*d + 30*a^2*c^4*d^2 + 29*a
^2*c^3*d^3 - 35*a^2*c^2*d^4 + 9*a^2*c*d^5 - 15*a^2*d^6))*sin(f*x + e) - sqrt(2)*(2*a^2*c^6 + 10*a^2*c^5*d + 15
*a^2*c^4*d^2 + 15*a^2*c^3*d^3 + 27*a^2*c^2*d^4 - 45*a^2*c*d^5))*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*(3*sqr
t(2)*(I*a^2*c^3*d^3 + 5*I*a^2*c^2*d^4 - 12*I*a^2*c*d^5)*cos(f*x + e)^2 + (sqrt(2)*(I*a^2*c^2*d^4 + 5*I*a^2*c*d
^5 - 12*I*a^2*d^6)*cos(f*x + e)^2 + sqrt(2)*(-3*I*a^2*c^4*d^2 - 15*I*a^2*c^3*d^3 + 35*I*a^2*c^2*d^4 - 5*I*a^2*
c*d^5 + 12*I*a^2*d^6))*sin(f*x + e) + sqrt(2)*(-I*a^2*c^5*d - 5*I*a^2*c^4*d^2 + 9*I*a^2*c^3*d^3 - 15*I*a^2*c^2
*d^4 + 36*I*a^2*c*d^5))*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, w
eierstrassPInverse(-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*si
n(f*x + e) - 2*I*c)/d)) + 3*(3*sqrt(2)*(-I*a^2*c^3*d^3 - 5*I*a^2*c^2*d^4 + 12*I*a^2*c*d^5)*cos(f*x + e)^2 + (s
qrt(2)*(-I*a^2*c^2*d^4 - 5*I*a^2*c*d^5 + 12*I*a^2*d^6)*cos(f*x + e)^2 + sqrt(2)*(3*I*a^2*c^4*d^2 + 15*I*a^2*c^
3*d^3 - 35*I*a^2*c^2*d^4 + 5*I*a^2*c*d^5 - 12*I*a^2*d^6))*sin(f*x + e) + sqrt(2)*(I*a^2*c^5*d + 5*I*a^2*c^4*d^
2 - 9*I*a^2*c^3*d^3 + 15*I*a^2*c^2*d^4 - 36*I*a^2*c*d^5))*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*(2*(a^2*c^2*d^4 + 5*a^2*c*d^5 - 12*a^2*d^6)*c
os(f*x + e)^3 - 2*(3*a^2*c^3*d^3 + 15*a^2*c^2*d^4 - 25*a^2*c*d^5 - 5*a^2*d^6)*cos(f*x + e)*sin(f*x + e) - (a^2
*c^4*d^2 + 20*a^2*c^3*d^3 - 18*a^2*c^2*d^4 - 27*a^2*d^6)*cos(f*x + e))*sqrt(d*sin(f*x + e) + c))/(3*(c^5*d^5 +
2*c^4*d^6 - 2*c^2*d^8 - c*d^9)*f*cos(f*x + e)^2 - (c^7*d^3 + 2*c^6*d^4 + 3*c^5*d^5 + 4*c^4*d^6 - c^3*d^7 - 6*
c^2*d^8 - 3*c*d^9)*f + ((c^4*d^6 + 2*c^3*d^7 - 2*c*d^9 - d^10)*f*cos(f*x + e)^2 - (3*c^6*d^4 + 6*c^5*d^5 + c^4
*d^6 - 4*c^3*d^7 - 3*c^2*d^8 - 2*c*d^9 - d^10)*f)*sin(f*x + e))

Sympy [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx=\text {Timed out} \]

[In]

integrate((a+a*sin(f*x+e))**2/(c+d*sin(f*x+e))**(7/2),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(7/2),x, algorithm="maxima")

[Out]

integrate((a*sin(f*x + e) + a)^2/(d*sin(f*x + e) + c)^(7/2), x)

Giac [F]

\[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx=\int { \frac {{\left (a \sin \left (f x + e\right ) + a\right )}^{2}}{{\left (d \sin \left (f x + e\right ) + c\right )}^{\frac {7}{2}}} \,d x } \]

[In]

integrate((a+a*sin(f*x+e))^2/(c+d*sin(f*x+e))^(7/2),x, algorithm="giac")

[Out]

integrate((a*sin(f*x + e) + a)^2/(d*sin(f*x + e) + c)^(7/2), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{7/2}} \, dx=\int \frac {{\left (a+a\,\sin \left (e+f\,x\right )\right )}^2}{{\left (c+d\,\sin \left (e+f\,x\right )\right )}^{7/2}} \,d x \]

[In]

int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^(7/2),x)

[Out]

int((a + a*sin(e + f*x))^2/(c + d*sin(e + f*x))^(7/2), x)