\(\int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx\) [1373]

Optimal result

Integrand size = 31, antiderivative size = 341 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \sqrt [4]{-a^2+b^2} f}+\frac {a \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{3/2} \sqrt [4]{-a^2+b^2} f}+\frac {2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {\cos (e+f x)}}-\frac {a^2 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {a^2 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \]

[Out]

-a*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))*g^(1/2)/b^(3/2)/(-a^2+b^2)^(1/4)/f+a*arctanh(
b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))*g^(1/2)/b^(3/2)/(-a^2+b^2)^(1/4)/f-a^2*g*(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^2/f/(b-(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)-a^2*g*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Ell
ipticPi(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^2/f/(b+(-a^2+b^2)^(1/2))/(g*co
s(f*x+e))^(1/2)+2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))*(g*cos
(f*x+e))^(1/2)/b/f/cos(f*x+e)^(1/2)

Rubi [A] (verified)

Time = 0.50 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {2946, 2721, 2719, 2780, 2886, 2884, 335, 304, 211, 214} \[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt [4]{b^2-a^2}}+\frac {a \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt [4]{b^2-a^2}}-\frac {a^2 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^2 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a^2 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^2 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}+\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}} \]

[In]

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

[Out]

-((a*Sqrt[g]*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(3/2)*(-a^2 + b^2)^(1/4)*
f)) + (a*Sqrt[g]*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(3/2)*(-a^2 + b^2)^(
1/4)*f) + (2*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(b*f*Sqrt[Cos[e + f*x]]) - (a^2*g*Sqrt[Cos[e + f*
x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^2*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x
]]) - (a^2*g*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^2*(b + Sqrt[-a^2
+ b^2])*f*Sqrt[g*Cos[e + f*x]])

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 304

Int[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}
, Dist[s/(2*b), Int[1/(r + s*x^2), x], x] - Dist[s/(2*b), Int[1/(r - s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !
GtQ[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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 2780

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

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

Mathematica [C] (warning: unable to verify)

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

Time = 12.23 (sec) , antiderivative size = 351, normalized size of antiderivative = 1.03 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {\sqrt {g \cos (e+f x)} \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )+\log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{12 b^{3/2} \left (-a^2+b^2\right ) f \sqrt {\cos (e+f x)} (a+b \sin (e+f x))} \]

[In]

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

[Out]

-1/12*(Sqrt[g*Cos[e + f*x]]*(8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2
+ b^2)]*Cos[e + f*x]^(3/2) + 3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(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]*Sqr
t[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]]))*(a + b*Sqrt[Sin[e + f*x]^2]))/(b^(3/2)*(-a^2 + b
^2)*f*Sqrt[Cos[e + f*x]]*(a + b*Sin[e + f*x]))

Maple [C] (warning: unable to verify)

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

Time = 2.37 (sec) , antiderivative size = 839, normalized size of antiderivative = 2.46

method	result	size
default	\(\text {Expression too large to display}\)	\(839\)

[In]

int(sin(f*x+e)*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

(-1/4*a*g/(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*(ln((2*g*cos(1/2*f*x+1/2*e)^2-g-(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g*cos
(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2))/(2*g*cos(1/2*f*x+1/2*e)^2-g+(g^2*(a^2-b^2)/b^2)^
(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2)))+2*arctan((2^(1/2)*(2*g*cos(1/2*f*
x+1/2*e)^2-g)^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2)/b^2)^(1/4))+2*arctan((2^(1/2)*(2*g*cos(1/2*f*x+1
/2*e)^2-g)^(1/2)-(g^2*(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2)/b^2)^(1/4)))/b^2-8*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*si
n(1/2*f*x+1/2*e)^2)^(1/2)*b*g*(1/2*(-1+sin(1/2*f*x+1/2*e)^2)/b^2*(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*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2)
)-1/32/b^4*sum((-2*sin(1/2*f*x+1/2*e)^2*_alpha^2*b^2+sin(1/2*f*x+1/2*e)^2*a^2+2*b^2*_alpha^2-a^2)/_alpha/(2*_a
lpha^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*_a
lpha^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)))/s
in(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. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

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

[In]

integrate(sin(f*x+e)*(g*cos(f*x+e))**(1/2)/(a+b*sin(f*x+e)),x)

[Out]

Timed out

Maxima [F]

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

[In]

integrate(sin(f*x+e)*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

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

Giac [F]

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

[In]

integrate(sin(f*x+e)*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

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

Mupad [F(-1)]

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

[In]

int((sin(e + f*x)*(g*cos(e + f*x))^(1/2))/(a + b*sin(e + f*x)),x)

[Out]

int((sin(e + f*x)*(g*cos(e + f*x))^(1/2))/(a + b*sin(e + f*x)), x)