\(\int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx\) [494]

Optimal result

Integrand size = 27, antiderivative size = 227 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=\frac {6 (c-d) \cos (e+f x)}{d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {12 (c+3 d) \cos (e+f x)}{d (c+d)^2 f \sqrt {c+d \sin (e+f x)}}-\frac {12 (c+3 d) E\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right )|\frac {2 d}{c+d}\right ) \sqrt {c+d \sin (e+f x)}}{d^2 (c+d)^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {12 (c+2 d) \operatorname {EllipticF}\left (\frac {1}{2} \left (e-\frac {\pi }{2}+f x\right ),\frac {2 d}{c+d}\right ) \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}{d^2 (c+d) f \sqrt {c+d \sin (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 247, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {2841, 2833, 2831, 2742, 2740, 2734, 2732} \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=\frac {4 a^2 (c+2 d) \sqrt {\frac {c+d \sin (e+f x)}{c+d}} \operatorname {EllipticF}\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right ),\frac {2 d}{c+d}\right )}{3 d^2 f (c+d) \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 (c+3 d) \sqrt {c+d \sin (e+f x)} E\left (\frac {1}{2} \left (e+f x-\frac {\pi }{2}\right )|\frac {2 d}{c+d}\right )}{3 d^2 f (c+d)^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}-\frac {4 a^2 (c+3 d) \cos (e+f x)}{3 d f (c+d)^2 \sqrt {c+d \sin (e+f x)}}+\frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) (c+d \sin (e+f x))^{3/2}} \]

[In]

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

[Out]

(2*a^2*(c - d)*Cos[e + f*x])/(3*d*(c + d)*f*(c + d*Sin[e + f*x])^(3/2)) - (4*a^2*(c + 3*d)*Cos[e + f*x])/(3*d*
(c + d)^2*f*Sqrt[c + d*Sin[e + f*x]]) - (4*a^2*(c + 3*d)*EllipticE[(e - Pi/2 + f*x)/2, (2*d)/(c + d)]*Sqrt[c +
d*Sin[e + f*x]])/(3*d^2*(c + d)^2*f*Sqrt[(c + d*Sin[e + f*x])/(c + d)]) + (4*a^2*(c + 2*d)*EllipticF[(e - Pi/
2 + f*x)/2, (2*d)/(c + d)]*Sqrt[(c + d*Sin[e + f*x])/(c + d)])/(3*d^2*(c + d)*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)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {(2 a) \int \frac {-3 a d-a (c+2 d) \sin (e+f x)}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 d (c+d)} \\ & = \frac {2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 (c+3 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt {c+d \sin (e+f x)}}+\frac {(4 a) \int \frac {a (c-d) d-\frac {1}{2} a (c-d) (c+3 d) \sin (e+f x)}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 (c-d) d (c+d)^2} \\ & = \frac {2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 (c+3 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt {c+d \sin (e+f x)}}+\frac {\left (2 a^2 (c+2 d)\right ) \int \frac {1}{\sqrt {c+d \sin (e+f x)}} \, dx}{3 d^2 (c+d)}-\frac {\left (2 a^2 (c+3 d)\right ) \int \sqrt {c+d \sin (e+f x)} \, dx}{3 d^2 (c+d)^2} \\ & = \frac {2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 (c+3 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt {c+d \sin (e+f x)}}-\frac {\left (2 a^2 (c+3 d) \sqrt {c+d \sin (e+f x)}\right ) \int \sqrt {\frac {c}{c+d}+\frac {d \sin (e+f x)}{c+d}} \, dx}{3 d^2 (c+d)^2 \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {\left (2 a^2 (c+2 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}{3 d^2 (c+d) \sqrt {c+d \sin (e+f x)}} \\ & = \frac {2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f (c+d \sin (e+f x))^{3/2}}-\frac {4 a^2 (c+3 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt {c+d \sin (e+f x)}}-\frac {4 a^2 (c+3 d) 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 d^2 (c+d)^2 f \sqrt {\frac {c+d \sin (e+f x)}{c+d}}}+\frac {4 a^2 (c+2 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}}}{3 d^2 (c+d) f \sqrt {c+d \sin (e+f x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.62 (sec) , antiderivative size = 147, normalized size of antiderivative = 0.65 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=\frac {6 \left (2 (c+d)^2 \left ((c+3 d) E\left (\frac {1}{4} (-2 e+\pi -2 f x)|\frac {2 d}{c+d}\right )-(c+2 d) \operatorname {EllipticF}\left (\frac {1}{4} (-2 e+\pi -2 f x),\frac {2 d}{c+d}\right )\right ) \left (\frac {c+d \sin (e+f x)}{c+d}\right )^{3/2}-d \cos (e+f x) \left (c^2+6 c d+d^2+2 d (c+3 d) \sin (e+f x)\right )\right )}{d^2 (c+d)^2 f (c+d \sin (e+f x))^{3/2}} \]

[In]

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

[Out]

(6*(2*(c + d)^2*((c + 3*d)*EllipticE[(-2*e + Pi - 2*f*x)/4, (2*d)/(c + d)] - (c + 2*d)*EllipticF[(-2*e + Pi -
2*f*x)/4, (2*d)/(c + d)])*((c + d*Sin[e + f*x])/(c + d))^(3/2) - d*Cos[e + f*x]*(c^2 + 6*c*d + d^2 + 2*d*(c +
3*d)*Sin[e + f*x])))/(d^2*(c + d)^2*f*(c + d*Sin[e + f*x])^(3/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1220\) vs. \(2(293)=586\).

Time = 7.12 (sec) , antiderivative size = 1221, normalized size of antiderivative = 5.38

method	result	size
default	\(\text {Expression too large to display}\)	\(1221\)
parts	\(\text {Expression too large to display}\)	\(2400\)

[In]

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

[Out]

-2/3*a^2*((2*c*d^3+6*d^4)*cos(f*x+e)^2*sin(f*x+e)+(c^2*d^2+6*c*d^3+d^4)*cos(f*x+e)^2+2*(-d/(c+d)*sin(f*x+e)+d/
(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*d*(EllipticF((d/(c-d)*si
n(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d-EllipticF((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+
d))^(1/2))*d^3-EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^3-3*EllipticE((d/(c-d)*si
n(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d+EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+
d))^(1/2))*c*d^2+3*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*d^3)*sin(f*x+e)+2*(d/(c
-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*Ellipt
icF((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^3*d-2*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d
/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticF((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^
(1/2),((c-d)/(c+d))^(1/2))*c*d^3-2*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-
d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^4-6*(d
/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*Ell
ipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c^3*d+2*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*
(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*
c)^(1/2),((c-d)/(c+d))^(1/2))*c^2*d^2+6*(d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2)*(-d/(c+d)*sin(f*x+e)+d/(c+d))^(1/
2)*(-d/(c-d)*sin(f*x+e)-d/(c-d))^(1/2)*EllipticE((d/(c-d)*sin(f*x+e)+1/(c-d)*c)^(1/2),((c-d)/(c+d))^(1/2))*c*d
^3)/(c+d)^2/(c+d*sin(f*x+e))^(3/2)/d^3/cos(f*x+e)/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 = 991, normalized size of antiderivative = 4.37 \[ \int \frac {(3+3 \sin (e+f x))^2}{(c+d \sin (e+f x))^{5/2}} \, dx=\text {Too large to display} \]

[In]

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

[Out]

2/9*(2*(sqrt(2)*(a^2*c^2*d^2 + 3*a^2*c*d^3 + 3*a^2*d^4)*cos(f*x + e)^2 - 2*sqrt(2)*(a^2*c^3*d + 3*a^2*c^2*d^2
+ 3*a^2*c*d^3)*sin(f*x + e) - sqrt(2)*(a^2*c^4 + 3*a^2*c^3*d + 4*a^2*c^2*d^2 + 3*a^2*c*d^3 + 3*a^2*d^4))*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) + 2*(sqrt(2)*(a^2*c^2*d^2 + 3*a^2*c*d^3 + 3*a^2*d^4)*cos(f*x + e)^2 - 2*sqrt(2)*(
a^2*c^3*d + 3*a^2*c^2*d^2 + 3*a^2*c*d^3)*sin(f*x + e) - sqrt(2)*(a^2*c^4 + 3*a^2*c^3*d + 4*a^2*c^2*d^2 + 3*a^2
*c*d^3 + 3*a^2*d^4))*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)*(I*a^2*c*d^3 + 3*I*a^2*d^4)*cos(f*x + e)
^2 + 2*sqrt(2)*(-I*a^2*c^2*d^2 - 3*I*a^2*c*d^3)*sin(f*x + e) + sqrt(2)*(-I*a^2*c^3*d - 3*I*a^2*c^2*d^2 - I*a^2
*c*d^3 - 3*I*a^2*d^4))*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, we
ierstrassPInverse(-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*a^2*c*d^3 - 3*I*a^2*d^4)*cos(f*x + e)^2 + 2*sqrt(2)*(I*a^2*c^2*d^2 + 3
*I*a^2*c*d^3)*sin(f*x + e) + sqrt(2)*(I*a^2*c^3*d + 3*I*a^2*c^2*d^2 + I*a^2*c*d^3 + 3*I*a^2*d^4))*sqrt(-I*d)*w
eierstrassZeta(-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*d^3 + 3*a^2*d^4)*cos(f*x + e)*sin(f*x + e) + (a^2*c^2*d^2 + 6*a^2*c*d^3 + a^2*d^4)*cos(f*x + e))*sqrt(d*s
in(f*x + e) + c))/((c^2*d^5 + 2*c*d^6 + d^7)*f*cos(f*x + e)^2 - 2*(c^3*d^4 + 2*c^2*d^5 + c*d^6)*f*sin(f*x + e)
- (c^4*d^3 + 2*c^3*d^4 + 2*c^2*d^5 + 2*c*d^6 + d^7)*f)

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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