3.1374 \(\int \frac{\sqrt{g \cos (e+f x)} \csc (e+f x)}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=355 \[ -\frac{\sqrt{b} \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt [4]{b^2-a^2}}+\frac{\sqrt{b} \sqrt{g} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt [4]{b^2-a^2}}-\frac{g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}-\frac{g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}+\frac{\sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f}-\frac{\sqrt{g} \tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f} \]

[Out]

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

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Rubi [A]  time = 0.832532, antiderivative size = 355, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 11, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.355, Rules used = {2898, 2565, 329, 298, 203, 206, 2701, 2807, 2805, 205, 208} \[ -\frac{\sqrt{b} \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt [4]{b^2-a^2}}+\frac{\sqrt{b} \sqrt{g} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt [4]{b^2-a^2}}-\frac{g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}-\frac{g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}+\frac{\sqrt{g} \tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f}-\frac{\sqrt{g} \tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2898

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])

Rule 2565

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 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 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 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 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 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]

Rubi steps

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

Mathematica [C]  time = 14.9523, size = 534, normalized size = 1.5 \[ \frac{\csc (e+f x) \sqrt{g \cos (e+f x)} \left (a+b \sqrt{\sin ^2(e+f x)}\right ) \left (8 a b \cos ^{\frac{3}{2}}(e+f x) F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )+3 \left (-\sqrt{2} \sqrt{b} \left (a^2-b^2\right )^{3/4} \log \left (-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}+b \cos (e+f x)\right )+\sqrt{2} \sqrt{b} \left (a^2-b^2\right )^{3/4} \log \left (\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}+b \cos (e+f x)\right )+2 \sqrt{2} \sqrt{b} \left (a^2-b^2\right )^{3/4} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \sqrt{2} \sqrt{b} \left (a^2-b^2\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )+2 a^2 \log \left (1-\sqrt{\cos (e+f x)}\right )-2 a^2 \log \left (\sqrt{\cos (e+f x)}+1\right )+4 a^2 \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )-2 b^2 \log \left (1-\sqrt{\cos (e+f x)}\right )+2 b^2 \log \left (\sqrt{\cos (e+f x)}+1\right )-4 b^2 \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )\right )\right )}{12 a f \left (a^2-b^2\right ) \sqrt{\cos (e+f x)} (a \csc (e+f x)+b)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[g*Cos[e + f*x]]*Csc[e + f*x]*(8*a*b*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*(2*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e +
f*x]])/(a^2 - b^2)^(1/4)] - 2*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]
])/(a^2 - b^2)^(1/4)] + 4*a^2*ArcTan[Sqrt[Cos[e + f*x]]] - 4*b^2*ArcTan[Sqrt[Cos[e + f*x]]] + 2*a^2*Log[1 - Sq
rt[Cos[e + f*x]]] - 2*b^2*Log[1 - Sqrt[Cos[e + f*x]]] - 2*a^2*Log[1 + Sqrt[Cos[e + f*x]]] + 2*b^2*Log[1 + Sqrt
[Cos[e + f*x]]] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sq
rt[Cos[e + f*x]] + b*Cos[e + f*x]] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*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]]))*(a + b*Sqrt[Sin[e + f*x]^2]))/(12*a*(a^2 - b^2)*f*Sqrt
[Cos[e + f*x]]*(b + a*Csc[e + f*x]))

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Maple [A]  time = 2.449, size = 188, normalized size = 0.5 \begin{align*} -{\frac{1}{2\,af}\sqrt{g}\ln \left ( 2\,{\frac{\sqrt{g}\sqrt{-2\, \left ( \sin \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}g+g}+2\,g\cos \left ( 1/2\,fx+e/2 \right ) -g}{-1+\cos \left ( 1/2\,fx+e/2 \right ) }} \right ) }-{\frac{1}{2\,af}\sqrt{g}\ln \left ( 2\,{\frac{\sqrt{g}\sqrt{-2\, \left ( \sin \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}g+g}-2\,g\cos \left ( 1/2\,fx+e/2 \right ) -g}{\cos \left ( 1/2\,fx+e/2 \right ) +1}} \right ) }-{\frac{g}{af}\ln \left ( 2\,{\frac{\sqrt{-g}\sqrt{-2\, \left ( \sin \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}g+g}-g}{\cos \left ( 1/2\,fx+e/2 \right ) }} \right ){\frac{1}{\sqrt{-g}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/a/f*g^(1/2)*ln(2/(-1+cos(1/2*f*x+1/2*e))*(g^(1/2)*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)+2*g*cos(1/2*f*x+1/2
*e)-g))-1/2/a/f*g^(1/2)*ln(2/(cos(1/2*f*x+1/2*e)+1)*(g^(1/2)*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-2*g*cos(1/2*f
*x+1/2*e)-g))-1/a/(-g)^(1/2)/f*g*ln(2/cos(1/2*f*x+1/2*e)*((-g)^(1/2)*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-g))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(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))*csc(f*x + e)/(b*sin(f*x + e) + a), x)