\(\int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx\) [1378]

Optimal result

Integrand size = 33, antiderivative size = 514 \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a^2 \sqrt [4]{-a^2+b^2} g^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}-\frac {a^2 \sqrt [4]{-a^2+b^2} g^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}+\frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g}+\frac {2 a^3 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^4 f \sqrt {g \cos (e+f x)}}-\frac {2 a g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}-\frac {a^3 \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^3 \left (a^2-b^2\right ) g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 a g \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f} \]

[Out]

-a^2*(-a^2+b^2)^(1/4)*g^(3/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(7/2)/f-a^2*(-a^
2+b^2)^(1/4)*g^(3/2)*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/b^(7/2)/f-2/5*(g*cos(f*x+e
))^(5/2)/b/f/g+2*a^3*g^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))
*cos(f*x+e)^(1/2)/b^4/f/(g*cos(f*x+e))^(1/2)-2/3*a*g^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Ellipti
cF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)/b^2/f/(g*cos(f*x+e))^(1/2)-a^3*(a^2-b^2)*g^2*(cos(1/2*f*x+1/2*
e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*cos(f*x+e)^(1/2
)/b^4/f/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)-a^3*(a^2-b^2)*g^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1
/2*f*x+1/2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos(f*x+e)^(1/2)/b^4/f/(a^2-b*(b
+(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)+2*a^2*g*(g*cos(f*x+e))^(1/2)/b^3/f-2/3*a*g*sin(f*x+e)*(g*cos(f*x+e))^
(1/2)/b^2/f

Rubi [A] (verified)

Time = 0.90 (sec) , antiderivative size = 514, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {2977, 2715, 2721, 2720, 2645, 30, 2774, 2946, 2781, 2886, 2884, 335, 218, 214, 211} \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {2 a^3 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^4 f \sqrt {g \cos (e+f x)}}-\frac {a^2 g^{3/2} \sqrt [4]{b^2-a^2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}-\frac {a^2 g^{3/2} \sqrt [4]{b^2-a^2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{7/2} f}+\frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {a^3 g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^4 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {a^3 g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^4 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {2 a g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}-\frac {2 a g \sin (e+f x) \sqrt {g \cos (e+f x)}}{3 b^2 f}-\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g} \]

[In]

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

[Out]

-((a^2*(-a^2 + b^2)^(1/4)*g^(3/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(7/2
)*f)) - (a^2*(-a^2 + b^2)^(1/4)*g^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/
(b^(7/2)*f) + (2*a^2*g*Sqrt[g*Cos[e + f*x]])/(b^3*f) - (2*(g*Cos[e + f*x])^(5/2))/(5*b*f*g) + (2*a^3*g^2*Sqrt[
Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(b^4*f*Sqrt[g*Cos[e + f*x]]) - (2*a*g^2*Sqrt[Cos[e + f*x]]*EllipticF[
(e + f*x)/2, 2])/(3*b^2*f*Sqrt[g*Cos[e + f*x]]) - (a^3*(a^2 - b^2)*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b
- Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^4*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) - (a^3*(a^
2 - b^2)*g^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^4*(a^2 - b*(b + S
qrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) - (2*a*g*Sqrt[g*Cos[e + f*x]]*Sin[e + f*x])/(3*b^2*f)

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 218

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]},
Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !Gt
Q[a/b, 0]

Rule 335

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + b*(x^(k*n)/c^n))^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 2645

Int[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_.)*sin[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> Dist[-(a*f)^(-1), Subst[
Int[x^m*(1 - x^2/a^2)^((n - 1)/2), x], x, a*Cos[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n - 1)/2]
&&  !(IntegerQ[(m - 1)/2] && GtQ[m, 0] && LeQ[m, n])

Rule 2715

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d*x]*((b*Sin[c + d*x])^(n - 1)/(d*n))
, x] + Dist[b^2*((n - 1)/n), Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integ
erQ[2*n]

Rule 2720

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ
[{c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2774

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p))), x] + Dist[g^2*((p - 1)/(b*(m + p))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2781

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[
-a^2 + b^2, 2]}, Dist[-a/(2*q), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Dist[b*(g/f), Sub
st[Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e
+ f*x]]*(q - b*Cos[e + f*x])), x], x])] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a,
b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]

Rule 2886

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist
[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt[c + d*Sin[e + f*x]], Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/
(c + d))*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] && N
eQ[c^2 - d^2, 0] &&  !GtQ[c + d, 0]

Rule 2946

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]))/((a_) + (b_.)*sin[(e_.) + (
f_.)*(x_)]), x_Symbol] :> Dist[d/b, Int[(g*Cos[e + f*x])^p, x], x] + Dist[(b*c - a*d)/b, Int[(g*Cos[e + f*x])^
p/(a + b*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[a^2 - b^2, 0]

Rule 2977

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^(n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])
, x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, sin[e + f*x]^n/(a + b*sin[e + f*x]), x], x] /; FreeQ[{a, b,
e, f, g, p}, x] && NeQ[a^2 - b^2, 0] && IntegerQ[n] && (LtQ[n, 0] || IGtQ[p + 1/2, 0])

Rubi steps \begin{align*} \text {integral}& = \int \left (-\frac {a (g \cos (e+f x))^{3/2}}{b^2}+\frac {(g \cos (e+f x))^{3/2} \sin (e+f x)}{b}+\frac {a^2 (g \cos (e+f x))^{3/2}}{b^2 (a+b \sin (e+f x))}\right ) \, dx \\ & = -\frac {a \int (g \cos (e+f x))^{3/2} \, dx}{b^2}+\frac {a^2 \int \frac {(g \cos (e+f x))^{3/2}}{a+b \sin (e+f x)} \, dx}{b^2}+\frac {\int (g \cos (e+f x))^{3/2} \sin (e+f x) \, dx}{b} \\ & = \frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {2 a g \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f}-\frac {\text {Subst}\left (\int x^{3/2} \, dx,x,g \cos (e+f x)\right )}{b f g}+\frac {\left (a^2 g^2\right ) \int \frac {b+a \sin (e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b^3}-\frac {\left (a g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{3 b^2} \\ & = \frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g}-\frac {2 a g \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f}+\frac {\left (a^3 g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{b^4}-\frac {\left (a^2 \left (a^2-b^2\right ) g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b^4}-\frac {\left (a g^2 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{3 b^2 \sqrt {g \cos (e+f x)}} \\ & = \frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g}-\frac {2 a g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}-\frac {2 a g \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f}-\frac {\left (a^3 \sqrt {-a^2+b^2} g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b^4}-\frac {\left (a^3 \sqrt {-a^2+b^2} g^2\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b^4}-\frac {\left (a^2 \left (a^2-b^2\right ) g^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (\left (a^2-b^2\right ) g^2+b^2 x^2\right )} \, dx,x,g \cos (e+f x)\right )}{b^3 f}+\frac {\left (a^3 g^2 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{b^4 \sqrt {g \cos (e+f x)}} \\ & = \frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g}+\frac {2 a^3 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^4 f \sqrt {g \cos (e+f x)}}-\frac {2 a g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}-\frac {2 a g \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f}-\frac {\left (2 a^2 \left (a^2-b^2\right ) g^3\right ) \text {Subst}\left (\int \frac {1}{\left (a^2-b^2\right ) g^2+b^2 x^4} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b^3 f}-\frac {\left (a^3 \sqrt {-a^2+b^2} g^2 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b^4 \sqrt {g \cos (e+f x)}}-\frac {\left (a^3 \sqrt {-a^2+b^2} g^2 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b^4 \sqrt {g \cos (e+f x)}} \\ & = \frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g}+\frac {2 a^3 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^4 f \sqrt {g \cos (e+f x)}}-\frac {2 a g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {-a^2+b^2} g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {-a^2+b^2} g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 a g \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f}-\frac {\left (a^2 \sqrt {-a^2+b^2} g^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g-b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b^3 f}-\frac {\left (a^2 \sqrt {-a^2+b^2} g^2\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g+b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{b^3 f} \\ & = -\frac {a^2 \sqrt [4]{-a^2+b^2} g^{3/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}-\frac {a^2 \sqrt [4]{-a^2+b^2} g^{3/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{7/2} f}+\frac {2 a^2 g \sqrt {g \cos (e+f x)}}{b^3 f}-\frac {2 (g \cos (e+f x))^{5/2}}{5 b f g}+\frac {2 a^3 g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{b^4 f \sqrt {g \cos (e+f x)}}-\frac {2 a g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticF}\left (\frac {1}{2} (e+f x),2\right )}{3 b^2 f \sqrt {g \cos (e+f x)}}+\frac {a^3 \sqrt {-a^2+b^2} g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^3 \sqrt {-a^2+b^2} g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^4 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {2 a g \sqrt {g \cos (e+f x)} \sin (e+f x)}{3 b^2 f} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 24.59 (sec) , antiderivative size = 1953, normalized size of antiderivative = 3.80 \[ \int \frac {(g \cos (e+f x))^{3/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {(g \cos (e+f x))^{3/2} \sec (e+f x) \left (-\frac {\cos (2 (e+f x))}{5 b}-\frac {2 a \sin (e+f x)}{3 b^2}\right )}{f}+\frac {(g \cos (e+f x))^{3/2} \left (-\frac {2 \left (10 a^2+3 b^2\right ) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \sqrt {\cos (e+f x)}}{\sqrt {1-\cos ^2(e+f x)} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )-2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+\left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )\right ) \cos ^2(e+f x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(e+f x)\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) \sqrt {b} \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )+\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )-\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )}{\left (-a^2+b^2\right )^{3/4}}\right ) \sin (e+f x)}{\sqrt {1-\cos ^2(e+f x)} (a+b \sin (e+f x))}+\frac {\left (30 a^2-3 b^2\right ) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \cos (2 (e+f x)) \left (\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-2 a^2+b^2\right ) \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) \left (-2 a^2+b^2\right ) \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}+\frac {4 \sqrt {\cos (e+f x)}}{b}-\frac {4 a \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {5}{2}}(e+f x)}{5 \left (a^2-b^2\right )}+\frac {10 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \sqrt {\cos (e+f x)}}{\sqrt {1-\cos ^2(e+f x)} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {1}{2},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )-2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},2,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+\left (-a^2+b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{2},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )\right ) \cos ^2(e+f x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(e+f x)\right )\right )}+\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \left (-2 a^2+b^2\right ) \log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}-\frac {\left (\frac {1}{4}-\frac {i}{4}\right ) \left (-2 a^2+b^2\right ) \log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )}{b^{3/2} \left (-a^2+b^2\right )^{3/4}}\right ) \sin (e+f x)}{\sqrt {1-\cos ^2(e+f x)} \left (-1+2 \cos ^2(e+f x)\right ) (a+b \sin (e+f x))}+\frac {28 a b \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {5 b \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \sqrt {\cos (e+f x)} \sqrt {1-\cos ^2(e+f x)}}{\left (-5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+2 \left (2 b^2 \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},2,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+\left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )\right ) \cos ^2(e+f x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(e+f x)\right )\right )}+\frac {a \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )}{4 \sqrt {2} \sqrt {b} \left (a^2-b^2\right )^{3/4}}\right ) \sin ^2(e+f x)}{\left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}\right )}{60 b^2 f \cos ^{\frac {3}{2}}(e+f x)} \]

[In]

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

[Out]

((g*Cos[e + f*x])^(3/2)*Sec[e + f*x]*(-1/5*Cos[2*(e + f*x)]/b - (2*a*Sin[e + f*x])/(3*b^2)))/f + ((g*Cos[e + f
*x])^(3/2)*((-2*(10*a^2 + 3*b^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4,
Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/(Sqrt[1 - Cos[e + f*x]^2]*(5*(a^2 - b^
2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2,
2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Cos[e +
f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*
Sqrt[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]
*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] + Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[
e + f*x]] + I*b*Cos[e + f*x]] - Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] +
I*b*Cos[e + f*x]]))/(-a^2 + b^2)^(3/4))*Sin[e + f*x])/(Sqrt[1 - Cos[e + f*x]^2]*(a + b*Sin[e + f*x])) + ((30*
a^2 - 3*b^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e + f*x)]*(((1/2 - I/2)*(-2*a^2 + b^2)*ArcTan[1 - ((1 + I
)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/2 - I/2)*(-2*a^2 + b^2)*
ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)])/(b^(3/2)*(-a^2 + b^2)^(3/4)) + (4*Sqrt[Co
s[e + f*x]])/b - (4*a*AppellF1[5/4, 1/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Cos[e + f*
x]^(5/2))/(5*(a^2 - b^2)) + (10*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/
(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/(Sqrt[1 - Cos[e + f*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Cos[e +
f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e +
f*x]^2)/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b
^2)])*Cos[e + f*x]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) + ((1/4 - I/4)*(-2*a^2 + b^2)*Log[Sqrt[-a^2 + b^2] -
(1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]])/(b^(3/2)*(-a^2 + b^2)^(3/4)) - ((1/
4 - I/4)*(-2*a^2 + b^2)*Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos
[e + f*x]])/(b^(3/2)*(-a^2 + b^2)^(3/4)))*Sin[e + f*x])/(Sqrt[1 - Cos[e + f*x]^2]*(-1 + 2*Cos[e + f*x]^2)*(a +
b*Sin[e + f*x])) + (28*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Cos
[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]]*Sqrt[1 - Cos[e + f*x]^2])/((-5*(a^2 - b^2)*
AppellF1[1/4, -1/2, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2,
2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Cos[e + f*
x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)*(a^2 + b^2*(-1 + Cos[e + f*x]^2))) + (a*(-2*ArcTan[1
- (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])
/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f
*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]]))/(4*Sqrt[
2]*Sqrt[b]*(a^2 - b^2)^(3/4)))*Sin[e + f*x]^2)/((1 - Cos[e + f*x]^2)*(a + b*Sin[e + f*x]))))/(60*b^2*f*Cos[e +
f*x]^(3/2))

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 5.20 (sec) , antiderivative size = 1054, normalized size of antiderivative = 2.05

method	result	size
default	\(\text {Expression too large to display}\)	\(1054\)

[In]

int((g*cos(f*x+e))^(3/2)*sin(f*x+e)^2/(a+b*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

(-1/20*g/b^3*(32*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*sin(1/2*f*x+1/2*e)^4*b^2+10*a^2*(g^2*(a^2-b^2)/b^2)^(1/4)
*2^(1/2)*arctan((2^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/4))/((a^2*g^2-b^2*g^2)/b^2
)^(1/4))+5*a^2*(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*ln(((g^2*(a^2-b^2)/b^2)^(1/4)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(
1/2)*2^(1/2)-2*g*sin(1/2*f*x+1/2*e)^2+(g^2*(a^2-b^2)/b^2)^(1/2)+g)/(-(g^2*(a^2-b^2)/b^2)^(1/4)*(-2*g*sin(1/2*f
*x+1/2*e)^2+g)^(1/2)*2^(1/2)-2*g*sin(1/2*f*x+1/2*e)^2+(g^2*(a^2-b^2)/b^2)^(1/2)+g))-10*a^2*(g^2*(a^2-b^2)/b^2)
^(1/4)*2^(1/2)*arctan(1/((a^2*g^2-b^2*g^2)/b^2)^(1/4)*(-2^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)+(g^2*(a^2-
b^2)/b^2)^(1/4)))-32*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*sin(1/2*f*x+1/2*e)^2*b^2-40*(-2*g*sin(1/2*f*x+1/2*e)^
2+g)^(1/2)*a^2+8*b^2*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2))-8*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2
)^(1/2)*a*g^2*(-1/12/b^4/(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^2)^(1/2)*(4*cos(1/2*f*x+1/2*e)*sin(1/
2*f*x+1/2*e)^4*b^2-2*cos(1/2*f*x+1/2*e)*sin(1/2*f*x+1/2*e)^2*b^2-3*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(
1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*a^2+EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))*(sin(1/2*f*
x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*b^2)-1/64*a^2*(a^2-b^2)/b^6*sum(1/_alpha/(2*_alpha^2-1)*(2^
(1/2)/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*arctanh(1/2*g*(4*_alpha^2-3)/(4*a^2-3*b^2)*(b^2*_alpha^2+4*a^2*
cos(1/2*f*x+1/2*e)^2-3*b^2*cos(1/2*f*x+1/2*e)^2-3*a^2+2*b^2)*2^(1/2)/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/
(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2))+8*b^2/a^2*_alpha*(_alpha^2-1)*(sin(1/2*f*x+1/2*e)^2)
^(1/2)*(1-2*cos(1/2*f*x+1/2*e)^2)^(1/2)/(-g*sin(1/2*f*x+1/2*e)^2*(2*sin(1/2*f*x+1/2*e)^2-1))^(1/2)*EllipticPi(
cos(1/2*f*x+1/2*e),-4*b^2/a^2*(_alpha^2-1),2^(1/2))),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2)))/sin(1/2*f*x+1/
2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2))/f

Fricas [F(-1)]

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

[In]

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

[Out]

Timed out

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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