3.1373 \(\int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=341 \[ -\frac {a^2 g \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a^2 g \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}-\frac {a \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt [4]{b^2-a^2}}+\frac {a \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt [4]{b^2-a^2}}+\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}} \]

[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]  time = 0.73, antiderivative size = 341, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 10, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.323, Rules used = {2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ -\frac {a \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt [4]{b^2-a^2}}+\frac {a \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{3/2} f \sqrt [4]{b^2-a^2}}-\frac {a^2 g \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {a^2 g \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}+\frac {2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}} \]

Antiderivative was successfully verified.

[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 205

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

Rule 208

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

Rule 298

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 329

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 2639

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

Rule 2640

Int[Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[b*Sin[c + d*x]]/Sqrt[Sin[c + d*x]], Int[Sqrt[Si
n[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2701

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 2805

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2*EllipticPi[(2*b)/(a + b), (1*(e - Pi/2 + f*x))/2, (2*d)/(c + d)])/(f*(a + b)*Sqrt[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 2807

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*
Sin[e + f*x])/(c + d)]), 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 2867

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*} \int \frac {\sqrt {g \cos (e+f x)} \sin (e+f x)}{a+b \sin (e+f x)} \, dx &=\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) \operatorname {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) \operatorname {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)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |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)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {(a g) \operatorname {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) \operatorname {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} \tan ^{-1}\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} \tanh ^{-1}\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)} \Pi \left (\frac {2 b}{b-\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |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)} \Pi \left (\frac {2 b}{b+\sqrt {-a^2+b^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}\\ \end {align*}

________________________________________________________________________________________

Mathematica [C]  time = 19.80, size = 351, normalized size = 1.03 \[ -\frac {\sqrt {g \cos (e+f x)} \left (a+b \sqrt {\sin ^2(e+f x)}\right ) \left (8 b^{5/2} \cos ^{\frac {3}{2}}(e+f x) F_1\left (\frac {3}{4};-\frac {1}{2},1;\frac {7}{4};\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{b^2-a^2}\right )+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (-\log \left (-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+\sqrt {a^2-b^2}+b \cos (e+f x)\right )+\log \left (\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+\sqrt {a^2-b^2}+b \cos (e+f x)\right )+2 \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )\right )\right )}{12 b^{3/2} f \left (b^2-a^2\right ) \sqrt {\cos (e+f x)} (a+b \sin (e+f x))} \]

Warning: Unable to verify antiderivative.

[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]))

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[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

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )}{b \sin \left (f x + e\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

maple [C]  time = 6.87, size = 884, normalized size = 2.59 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/f*g*a*sum((_R^6-_R^4*g-_R^2*g^2+g^3)/(_R^7*b^2-3*_R^5*b^2*g+8*_R^3*a^2*g^2-5*_R^3*b^2*g^2-_R*b^2*g^3)*ln(
(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-cos(1/2*f*x+1/2*e)*g^(1/2)*2^(1/2)-_R),_R=RootOf(b^2*_Z^8-4*b^2*g*_Z^6+(16
*a^2*g^2-10*b^2*g^2)*_Z^4-4*b^2*g^3*_Z^2+b^2*g^4))-4/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/
2)*g/b/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)*sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-
1))^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(1/2)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))
+4/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*g/b/(-g*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2
*e)^2))^(1/2)/sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(1/2
*f*x+1/2*e)^2+1)^(1/2)*EllipticF(cos(1/2*f*x+1/2*e),2^(1/2))+1/4/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1
/2*e)^2)^(1/2)*g/b^3/sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*sum((-2*sin(1/2*f*x+1/2*e)^2*_alp
ha^2*b^2+sin(1/2*f*x+1/2*e)^2*a^2+2*b^2*_alpha^2-a^2)/_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)*(4*cos(1/2*f*x+1/2*e)^2*a^2-3*b^2*cos(1/2*f*x+1/2*e
)^2+b^2*_alpha^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)*(-2*cos(1/2*f*x+1/2*e)^2
+1)^(1/2)/(-sin(1/2*f*x+1/2*e)^2*g*(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))

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )}{b \sin \left (f x + e\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {\sin \left (e+f\,x\right )\,\sqrt {g\,\cos \left (e+f\,x\right )}}{a+b\,\sin \left (e+f\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________