\(\int \frac {x^2}{(a+a \sin (e+f x))^{3/2}} \, dx\) [140]

Optimal result

Integrand size = 18, antiderivative size = 435 \[ \int \frac {x^2}{(a+a \sin (e+f x))^{3/2}} \, dx=-\frac {2 x}{a f^2 \sqrt {a+a \sin (e+f x)}}-\frac {x^2 \cot \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{2 a f \sqrt {a+a \sin (e+f x)}}-\frac {x^2 \text {arctanh}\left (e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f \sqrt {a+a \sin (e+f x)}}-\frac {4 \text {arctanh}\left (\cos \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^3 \sqrt {a+a \sin (e+f x)}}+\frac {2 i x \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^2 \sqrt {a+a \sin (e+f x)}}-\frac {2 i x \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^2 \sqrt {a+a \sin (e+f x)}}-\frac {4 \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^3 \sqrt {a+a \sin (e+f x)}}+\frac {4 \operatorname {PolyLog}\left (3,e^{\frac {1}{4} i (2 e+\pi +2 f x)}\right ) \sin \left (\frac {e}{2}+\frac {\pi }{4}+\frac {f x}{2}\right )}{a f^3 \sqrt {a+a \sin (e+f x)}} \] Output:

-2*x/a/f^2/(a+a*sin(f*x+e))^(1/2)-1/2*x^2*cot(1/2*e+1/4*Pi+1/2*f*x)/a/f/(a 
+a*sin(f*x+e))^(1/2)-x^2*arctanh(exp(1/4*I*(2*f*x+Pi+2*e)))*sin(1/2*e+1/4* 
Pi+1/2*f*x)/a/f/(a+a*sin(f*x+e))^(1/2)-4*arctanh(cos(1/2*e+1/4*Pi+1/2*f*x) 
)*sin(1/2*e+1/4*Pi+1/2*f*x)/a/f^3/(a+a*sin(f*x+e))^(1/2)+2*I*x*polylog(2,- 
exp(1/4*I*(2*f*x+Pi+2*e)))*sin(1/2*e+1/4*Pi+1/2*f*x)/a/f^2/(a+a*sin(f*x+e) 
)^(1/2)-2*I*x*polylog(2,exp(1/4*I*(2*f*x+Pi+2*e)))*sin(1/2*e+1/4*Pi+1/2*f* 
x)/a/f^2/(a+a*sin(f*x+e))^(1/2)-4*polylog(3,-exp(1/4*I*(2*f*x+Pi+2*e)))*si 
n(1/2*e+1/4*Pi+1/2*f*x)/a/f^3/(a+a*sin(f*x+e))^(1/2)+4*polylog(3,exp(1/4*I 
*(2*f*x+Pi+2*e)))*sin(1/2*e+1/4*Pi+1/2*f*x)/a/f^3/(a+a*sin(f*x+e))^(1/2)
 

Mathematica [A] (verified)

Time = 2.47 (sec) , antiderivative size = 352, normalized size of antiderivative = 0.81 \[ \int \frac {x^2}{(a+a \sin (e+f x))^{3/2}} \, dx=\frac {\sqrt [4]{-1} e^{-\frac {3}{2} i (e+f x)} \left (i+e^{i (e+f x)}\right )^3 \left (16 \text {arctanh}\left (\sqrt [4]{-1} e^{\frac {1}{2} i (e+f x)}\right )-f^2 x^2 \log \left (1-\sqrt [4]{-1} e^{\frac {1}{2} i (e+f x)}\right )+f^2 x^2 \log \left (1+\sqrt [4]{-1} e^{\frac {1}{2} i (e+f x)}\right )-4 i f x \operatorname {PolyLog}\left (2,-\sqrt [4]{-1} e^{\frac {1}{2} i (e+f x)}\right )+4 i f x \operatorname {PolyLog}\left (2,\sqrt [4]{-1} e^{\frac {1}{2} i (e+f x)}\right )+8 \operatorname {PolyLog}\left (3,-\sqrt [4]{-1} e^{\frac {1}{2} i (e+f x)}\right )-8 \operatorname {PolyLog}\left (3,\sqrt [4]{-1} e^{\frac {1}{2} i (e+f x)}\right )\right )}{2 \sqrt {2} \left (-i a e^{-i (e+f x)} \left (i+e^{i (e+f x)}\right )^2\right )^{3/2} f^3}-\frac {x \left ((4+f x) \cos \left (\frac {1}{2} (e+f x)\right )+(4-f x) \sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))}}{2 a^2 f^2 \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )^3} \] Input:

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

Output:

((-1)^(1/4)*(I + E^(I*(e + f*x)))^3*(16*ArcTanh[(-1)^(1/4)*E^((I/2)*(e + f 
*x))] - f^2*x^2*Log[1 - (-1)^(1/4)*E^((I/2)*(e + f*x))] + f^2*x^2*Log[1 + 
(-1)^(1/4)*E^((I/2)*(e + f*x))] - (4*I)*f*x*PolyLog[2, -((-1)^(1/4)*E^((I/ 
2)*(e + f*x)))] + (4*I)*f*x*PolyLog[2, (-1)^(1/4)*E^((I/2)*(e + f*x))] + 8 
*PolyLog[3, -((-1)^(1/4)*E^((I/2)*(e + f*x)))] - 8*PolyLog[3, (-1)^(1/4)*E 
^((I/2)*(e + f*x))]))/(2*Sqrt[2]*E^(((3*I)/2)*(e + f*x))*(((-I)*a*(I + E^( 
I*(e + f*x)))^2)/E^(I*(e + f*x)))^(3/2)*f^3) - (x*((4 + f*x)*Cos[(e + f*x) 
/2] + (4 - f*x)*Sin[(e + f*x)/2])*Sqrt[a*(1 + Sin[e + f*x])])/(2*a^2*f^2*( 
Cos[(e + f*x)/2] + Sin[(e + f*x)/2])^3)
 

Rubi [A] (verified)

Time = 0.96 (sec) , antiderivative size = 281, normalized size of antiderivative = 0.65, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.556, Rules used = {3042, 3800, 3042, 4674, 3042, 4257, 4671, 3011, 2720, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^2}{(a \sin (e+f x)+a)^{3/2}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {x^2}{(a \sin (e+f x)+a)^{3/2}}dx\)

\(\Big \downarrow \) 3800

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \int x^2 \csc ^3\left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )dx}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \int x^2 \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )^3dx}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 4674

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (\frac {4 \int \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )dx}{f^2}+\frac {1}{2} \int x^2 \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )dx-\frac {4 x \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (\frac {4 \int \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )dx}{f^2}+\frac {1}{2} \int x^2 \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )dx-\frac {4 x \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 4257

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (\frac {1}{2} \int x^2 \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )dx-\frac {8 \text {arctanh}\left (\cos \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )\right )}{f^3}-\frac {4 x \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 4671

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (\frac {1}{2} \left (-\frac {4 \int x \log \left (1-e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )dx}{f}+\frac {4 \int x \log \left (1+e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )dx}{f}-\frac {4 x^2 \text {arctanh}\left (e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}\right )-\frac {8 \text {arctanh}\left (\cos \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )\right )}{f^3}-\frac {4 x \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 3011

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (\frac {1}{2} \left (\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}-\frac {2 i \int \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )dx}{f}\right )}{f}-\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}-\frac {2 i \int \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )dx}{f}\right )}{f}-\frac {4 x^2 \text {arctanh}\left (e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}\right )-\frac {8 \text {arctanh}\left (\cos \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )\right )}{f^3}-\frac {4 x \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 2720

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (\frac {1}{2} \left (\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}-\frac {4 \int e^{-\frac {1}{4} i (2 e+2 f x+\pi )} \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )de^{\frac {1}{4} i (2 e+2 f x+\pi )}}{f^2}\right )}{f}-\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}-\frac {4 \int e^{-\frac {1}{4} i (2 e+2 f x+\pi )} \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )de^{\frac {1}{4} i (2 e+2 f x+\pi )}}{f^2}\right )}{f}-\frac {4 x^2 \text {arctanh}\left (e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}\right )-\frac {8 \text {arctanh}\left (\cos \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )\right )}{f^3}-\frac {4 x \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {\sin \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \left (-\frac {8 \text {arctanh}\left (\cos \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )\right )}{f^3}+\frac {1}{2} \left (-\frac {4 x^2 \text {arctanh}\left (e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}+\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,-e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}-\frac {4 \operatorname {PolyLog}\left (3,-e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f^2}\right )}{f}-\frac {4 \left (\frac {2 i x \operatorname {PolyLog}\left (2,e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f}-\frac {4 \operatorname {PolyLog}\left (3,e^{\frac {1}{4} i (2 e+2 f x+\pi )}\right )}{f^2}\right )}{f}\right )-\frac {4 x \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f^2}-\frac {x^2 \cot \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right ) \csc \left (\frac {e}{2}+\frac {f x}{2}+\frac {\pi }{4}\right )}{f}\right )}{2 a \sqrt {a \sin (e+f x)+a}}\)

Input:

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

Output:

(((-8*ArcTanh[Cos[e/2 + Pi/4 + (f*x)/2]])/f^3 - (4*x*Csc[e/2 + Pi/4 + (f*x 
)/2])/f^2 - (x^2*Cot[e/2 + Pi/4 + (f*x)/2]*Csc[e/2 + Pi/4 + (f*x)/2])/f + 
((-4*x^2*ArcTanh[E^((I/4)*(2*e + Pi + 2*f*x))])/f + (4*(((2*I)*x*PolyLog[2 
, -E^((I/4)*(2*e + Pi + 2*f*x))])/f - (4*PolyLog[3, -E^((I/4)*(2*e + Pi + 
2*f*x))])/f^2))/f - (4*(((2*I)*x*PolyLog[2, E^((I/4)*(2*e + Pi + 2*f*x))]) 
/f - (4*PolyLog[3, E^((I/4)*(2*e + Pi + 2*f*x))])/f^2))/f)/2)*Sin[e/2 + Pi 
/4 + (f*x)/2])/(2*a*Sqrt[a + a*Sin[e + f*x]])
 

Defintions of rubi rules used

rule 2720

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Simp[v/D[v, x] 
   Subst[Int[FunctionOfExponentialFunction[u, x]/x, x], x, v], x]] /; Funct 
ionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; FreeQ 
[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x)) 
*(F_)[v_] /; FreeQ[{a, b, c}, x] && InverseFunctionQ[F[x]]]
 

rule 3011

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.) 
*(x_))^(m_.), x_Symbol] :> Simp[(-(f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + 
b*x)))^n]/(b*c*n*Log[F])), x] + Simp[g*(m/(b*c*n*Log[F]))   Int[(f + g*x)^( 
m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e 
, f, g, n}, x] && GtQ[m, 0]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3800

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(n_), 
 x_Symbol] :> Simp[(2*a)^IntPart[n]*((a + b*Sin[e + f*x])^FracPart[n]/Sin[e 
/2 + a*(Pi/(4*b)) + f*(x/2)]^(2*FracPart[n]))   Int[(c + d*x)^m*Sin[e/2 + a 
*(Pi/(4*b)) + f*(x/2)]^(2*n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && 
EqQ[a^2 - b^2, 0] && IntegerQ[n + 1/2] && (GtQ[n, 0] || IGtQ[m, 0])
 

rule 4257

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] 
 /; FreeQ[{c, d}, x]
 

rule 4671

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[- 
2*(c + d*x)^m*(ArcTanh[E^(I*(e + f*x))]/f), x] + (-Simp[d*(m/f)   Int[(c + 
d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Simp[d*(m/f)   Int[(c + d*x 
)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IG 
tQ[m, 0]
 

rule 4674

Int[(csc[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((c_.) + (d_.)*(x_))^(m_), x_Symbo 
l] :> Simp[(-b^2)*(c + d*x)^m*Cot[e + f*x]*((b*Csc[e + f*x])^(n - 2)/(f*(n 
- 1))), x] + (-Simp[b^2*d*m*(c + d*x)^(m - 1)*((b*Csc[e + f*x])^(n - 2)/(f^ 
2*(n - 1)*(n - 2))), x] + Simp[b^2*d^2*m*((m - 1)/(f^2*(n - 1)*(n - 2))) 
Int[(c + d*x)^(m - 2)*(b*Csc[e + f*x])^(n - 2), x], x] + Simp[b^2*((n - 2)/ 
(n - 1))   Int[(c + d*x)^m*(b*Csc[e + f*x])^(n - 2), x], x]) /; FreeQ[{b, c 
, d, e, f}, x] && GtQ[n, 1] && NeQ[n, 2] && GtQ[m, 1]
 

rule 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> Simp[PolyLog[n + 1, c*(a + b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d 
, e, n, p}, x] && EqQ[b*d, a*e]
 

Maple [F]

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

Input:

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

Output:

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

Fricas [F]

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

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

Output:

integral(-sqrt(a*sin(f*x + e) + a)*x^2/(a^2*cos(f*x + e)^2 - 2*a^2*sin(f*x 
 + e) - 2*a^2), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

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

Output:

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

Giac [F(-1)]

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

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

Output:

Timed out
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

(sqrt(a)*int((sqrt(sin(e + f*x) + 1)*x**2)/(sin(e + f*x)**2 + 2*sin(e + f* 
x) + 1),x))/a**2