3.1393 \(\int \frac{\csc (e+f x)}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=369 \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt{g} \left (b^2-a^2\right )^{3/4}}+\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt{g} \left (b^2-a^2\right )^{3/4}}-\frac{b \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{g \cos (e+f x)}}-\frac{b \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{g \cos (e+f x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f \sqrt{g}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f \sqrt{g}} \]

[Out]

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

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Rubi [A]  time = 0.818003, antiderivative size = 369, 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, 212, 206, 203, 2702, 2807, 2805, 208, 205} \[ \frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt{g} \left (b^2-a^2\right )^{3/4}}+\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f \sqrt{g} \left (b^2-a^2\right )^{3/4}}-\frac{b \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{g \cos (e+f x)}}-\frac{b \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{g \cos (e+f x)}}-\frac{\tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f \sqrt{g}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f \sqrt{g}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*Sqrt[g])) + (b^(3/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2
+ b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^(3/4)*f*Sqrt[g]) - ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*Sqrt[g])
+ (b^(3/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^(3/4)*f*Sqrt
[g]) - (b*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b*(b - Sqrt[-a^
2 + b^2]))*f*Sqrt[g*Cos[e + f*x]]) - (b*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/
2, 2])/((a^2 - b*(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 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 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 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 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 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{\csc (e+f x)}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx &=\int \left (\frac{\csc (e+f x)}{a \sqrt{g \cos (e+f x)}}-\frac{b}{a \sqrt{g \cos (e+f x)} (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac{\int \frac{\csc (e+f x)}{\sqrt{g \cos (e+f x)}} \, dx}{a}-\frac{b \int \frac{1}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a}\\ &=\frac{b \int \frac{1}{\sqrt{g \cos (e+f x)} \left (\sqrt{-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 \sqrt{-a^2+b^2}}+\frac{b \int \frac{1}{\sqrt{g \cos (e+f x)} \left (\sqrt{-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 \sqrt{-a^2+b^2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1-\frac{x^2}{g^2}\right )} \, dx,x,g \cos (e+f x)\right )}{a f g}-\frac{\left (b^2 g\right ) \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 )}{a f}\\ &=-\frac{2 \operatorname{Subst}\left (\int \frac{1}{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{1}{\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 (b \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{-a^2+b^2} \sqrt{g \cos (e+f x)}}+\frac{\left (b \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{-a^2+b^2} \sqrt{g \cos (e+f x)}}\\ &=-\frac{b \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 (a^2-b \left (b-\sqrt{-a^2+b^2}\right )\right ) f \sqrt{g \cos (e+f x)}}-\frac{b \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 (a^2-b \left (b+\sqrt{-a^2+b^2}\right )\right ) f \sqrt{g \cos (e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{1}{g-x^2} \, dx,x,\sqrt{g \cos (e+f x)}\right )}{a f}-\frac{\operatorname{Subst}\left (\int \frac{1}{g+x^2} \, dx,x,\sqrt{g \cos (e+f x)}\right )}{a f}+\frac{b^2 \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 \sqrt{-a^2+b^2} f}+\frac{b^2 \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 \sqrt{-a^2+b^2} f}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f \sqrt{g}}+\frac{b^{3/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt{g}}\right )}{a \left (-a^2+b^2\right )^{3/4} f \sqrt{g}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f \sqrt{g}}+\frac{b^{3/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt{g}}\right )}{a \left (-a^2+b^2\right )^{3/4} f \sqrt{g}}-\frac{b \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 (a^2-b \left (b-\sqrt{-a^2+b^2}\right )\right ) f \sqrt{g \cos (e+f x)}}-\frac{b \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 (a^2-b \left (b+\sqrt{-a^2+b^2}\right )\right ) f \sqrt{g \cos (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 20.9325, size = 698, normalized size = 1.89 \[ -\frac{2 \sqrt{\cos (e+f x)} \left (\cos ^2(e+f x)-1\right ) \csc (e+f x) \left (a+b \sqrt{1-\cos ^2(e+f x)}\right ) \left (\frac{5 b \left (a^2-b^2\right ) \sqrt{\cos (e+f x)} F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )}{\sqrt{1-\cos ^2(e+f x)} \left (a^2+b^2 \left (\cos ^2(e+f x)-1\right )\right ) \left (5 \left (a^2-b^2\right ) F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )-2 \cos ^2(e+f x) \left (2 b^2 F_1\left (\frac{5}{4};\frac{1}{2},2;\frac{9}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )+\left (b^2-a^2\right ) F_1\left (\frac{5}{4};\frac{3}{2},1;\frac{9}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right )\right )\right )}-\frac{-\sqrt{2} b^{3/2} \log \left (-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}+b \cos (e+f x)\right )+\sqrt{2} b^{3/2} \log \left (\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}+b \cos (e+f x)\right )-2 \left (a^2-b^2\right )^{3/4} \log \left (1-\sqrt{\cos (e+f x)}\right )+2 \left (a^2-b^2\right )^{3/4} \log \left (\sqrt{\cos (e+f x)}+1\right )-2 \sqrt{2} b^{3/2} \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \sqrt{2} b^{3/2} \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )+4 \left (a^2-b^2\right )^{3/4} \tan ^{-1}\left (\sqrt{\cos (e+f x)}\right )}{8 a \left (a^2-b^2\right )^{3/4}}\right )}{f \left (1-\cos ^2(e+f x)\right ) \sqrt{g \cos (e+f x)} (a \csc (e+f x)+b)} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(-2*Sqrt[Cos[e + f*x]]*(-1 + Cos[e + f*x]^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Csc[e + f*x]*((5*b*(a^2 - b^2)*A
ppellF1[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))) - (-2*Sqrt[2]*b^(3/2)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 2*Sqrt[2]
*b^(3/2)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 4*(a^2 - b^2)^(3/4)*ArcTan[Sqrt[
Cos[e + f*x]]] - 2*(a^2 - b^2)^(3/4)*Log[1 - Sqrt[Cos[e + f*x]]] + 2*(a^2 - b^2)^(3/4)*Log[1 + Sqrt[Cos[e + f*
x]]] - Sqrt[2]*b^(3/2)*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]] + Sqrt[2]*b^(3/2)*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]])/(8*a*(a^2 - b^2)^(3/4))))/(f*Sqrt[g*Cos[e + f*x]]*(1 - Cos[e + f*x]^2)*(b + a*Csc[e + f*x]))

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Maple [A]  time = 2.322, size = 186, normalized size = 0.5 \begin{align*}{\frac{1}{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}}}}-{\frac{1}{2\,af}\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{1}{\sqrt{g}}}}-{\frac{1}{2\,af}\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}{\sqrt{g}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/a/(-g)^(1/2)/f*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))-1/2/a/g^(1/2)/f*l
n(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))-1/2/a/g^(1/
2)/f*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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \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(csc(f*x+e)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(csc(e + f*x)/(sqrt(g*cos(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{\csc \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(csc(f*x+e)/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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