3.1392 \(\int \frac{\sin (e+f x)}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=352 \[ \frac{a \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} f \sqrt{g} \left (b^2-a^2\right )^{3/4}}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} f \sqrt{g} \left (b^2-a^2\right )^{3/4}}-\frac{a^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b f \left (b \sqrt{b^2-a^2}+a^2-b^2\right ) \sqrt{g \cos (e+f x)}}-\frac{a^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b f \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{g \cos (e+f x)}}+\frac{2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{b f \sqrt{g \cos (e+f x)}} \]

[Out]

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

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Rubi [A]  time = 0.72542, antiderivative size = 352, normalized size of antiderivative = 1., 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, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ \frac{a \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} f \sqrt{g} \left (b^2-a^2\right )^{3/4}}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{\sqrt{b} f \sqrt{g} \left (b^2-a^2\right )^{3/4}}-\frac{a^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b f \left (b \sqrt{b^2-a^2}+a^2-b^2\right ) \sqrt{g \cos (e+f x)}}-\frac{a^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b f \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{g \cos (e+f x)}}+\frac{2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{b f \sqrt{g \cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

Rule 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

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

Rule 2702

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 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 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 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 212

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] &&
!GtQ[a/b, 0]

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 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]

Rubi steps

\begin{align*} \int \frac{\sin (e+f x)}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx &=\frac{\int \frac{1}{\sqrt{g \cos (e+f x)}} \, dx}{b}-\frac{a \int \frac{1}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{b}\\ &=\frac{a^2 \int \frac{1}{\sqrt{g \cos (e+f x)} \left (\sqrt{-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b \sqrt{-a^2+b^2}}+\frac{a^2 \int \frac{1}{\sqrt{g \cos (e+f x)} \left (\sqrt{-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b \sqrt{-a^2+b^2}}-\frac{(a g) \operatorname{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 )}{f}+\frac{\sqrt{\cos (e+f x)} \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx}{b \sqrt{g \cos (e+f x)}}\\ &=\frac{2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{b f \sqrt{g \cos (e+f x)}}-\frac{(2 a g) \operatorname{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 )}{f}+\frac{\left (a^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 \sqrt{-a^2+b^2} \sqrt{g \cos (e+f x)}}+\frac{\left (a^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 \sqrt{-a^2+b^2} \sqrt{g \cos (e+f x)}}\\ &=\frac{2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{b f \sqrt{g \cos (e+f x)}}-\frac{a^2 \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 \left (a^2-b^2+b \sqrt{-a^2+b^2}\right ) f \sqrt{g \cos (e+f x)}}+\frac{a^2 \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 \sqrt{-a^2+b^2} \left (b+\sqrt{-a^2+b^2}\right ) f \sqrt{g \cos (e+f x)}}+\frac{a \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} g-b x^2} \, dx,x,\sqrt{g \cos (e+f x)}\right )}{\sqrt{-a^2+b^2} f}+\frac{a \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} g+b x^2} \, dx,x,\sqrt{g \cos (e+f x)}\right )}{\sqrt{-a^2+b^2} f}\\ &=\frac{a \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt{g}}\right )}{\sqrt{b} \left (-a^2+b^2\right )^{3/4} f \sqrt{g}}+\frac{a \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt{g}}\right )}{\sqrt{b} \left (-a^2+b^2\right )^{3/4} f \sqrt{g}}+\frac{2 \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{b f \sqrt{g \cos (e+f x)}}-\frac{a^2 \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 \left (a^2-b^2+b \sqrt{-a^2+b^2}\right ) f \sqrt{g \cos (e+f x)}}+\frac{a^2 \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 \sqrt{-a^2+b^2} \left (b+\sqrt{-a^2+b^2}\right ) f \sqrt{g \cos (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 6.28664, size = 546, normalized size = 1.55 \[ -\frac{2 \sqrt{\cos (e+f x)} \left (a+b \sqrt{\sin ^2(e+f x)}\right ) \left (\frac{5 b \left (a^2-b^2\right ) \sqrt{\sin ^2(e+f x)} \sqrt{\cos (e+f x)} F_1\left (\frac{1}{4};-\frac{1}{2},1;\frac{5}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}{\left (a^2+b^2 \cos ^2(e+f x)-b^2\right ) \left (2 \cos ^2(e+f x) \left (2 b^2 F_1\left (\frac{5}{4};-\frac{1}{2},2;\frac{9}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )+\left (a^2-b^2\right ) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )\right )-5 \left (a^2-b^2\right ) F_1\left (\frac{1}{4};-\frac{1}{2},1;\frac{5}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )\right )}+\frac{a \left (-\log \left (-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}+b \cos (e+f x)\right )+\log \left (\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}+b \cos (e+f x)\right )-2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )\right )}{4 \sqrt{2} \sqrt{b} \left (a^2-b^2\right )^{3/4}}\right )}{f \sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Sqrt[Cos[e + f*x]]*(a + b*Sqrt[Sin[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)) + (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[
Sin[e + f*x]^2])/((a^2 - b^2 + b^2*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)])*C
os[e + f*x]^2))))/(f*Sqrt[g*Cos[e + f*x]]*(a + b*Sin[e + f*x]))

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Maple [C]  time = 5.057, size = 1181, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/f*a*g*sum((_R^4+_R^2*g)/(_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))-1/2/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)/b*sum(_alp
ha/(2*_alpha^2-1)*(8*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(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)*EllipticPi(cos(1/2*f*x+1/2*e),-4*b^2/a^2*(_alpha^2-1),2^(1/2))*_alpha^3*b^2-8*b^2*_alpha*(sin(1
/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(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))*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)+2^(1/2)*a^2*arctanh(1/2*g*(4*_alpha^2-3)/(4*a^2-3*b^2)*(4*c
os(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))*(-sin(1/2*f*x+1/2*e)^2*g*(2*sin(1/2*
f*x+1/2*e)^2-1))^(1/2))/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(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),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^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)+1/2/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)/b*sum(_alpha/(2*_alpha^2-1)*(8*(g*(2*_al
pha^2*b^2+a^2-2*b^2)/b^2)^(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)*EllipticPi(cos(
1/2*f*x+1/2*e),-4*b^2/a^2*(_alpha^2-1),2^(1/2))*_alpha^3*b^2-8*b^2*_alpha*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos
(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))*(g*(2*_alpha^2*b^2+a
^2-2*b^2)/b^2)^(1/2)+2^(1/2)*a^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))*(-sin(1/2*f*x+1/2*e)^2*g*(2*sin(1/2*f*x+1/2*e)^2-1))^(1/2))/(g*(2*
_alpha^2*b^2+a^2-2*b^2)/b^2)^(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),_alpha=RootOf(4*
_Z^4*b^2-4*_Z^2*b^2+a^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)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )}{\sqrt{g \cos \left (f x + e\right )}{\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )}{\sqrt{g \cos \left (f x + e\right )}{\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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