3.1400 \(\int \frac{\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=507 \[ -\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{3/2} \left (b^2-a^2\right )^{5/4}}+\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{3/2} \left (b^2-a^2\right )^{5/4}}+\frac{2 b E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{f g^2 \left (a^2-b^2\right ) \sqrt{\cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{a f g \left (a^2-b^2\right ) \sqrt{g \cos (e+f x)}}+\frac{b^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f g \left (a^2-b^2\right ) \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}+\frac{b^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f g \left (a^2-b^2\right ) \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{3/2}}+\frac{2}{a f g \sqrt{g \cos (e+f x)}} \]

[Out]

ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(3/2)) - (b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b
^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^(5/4)*f*g^(3/2)) - ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(3/2)) +
(b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^(5/4)*f*g^(3/2)
) + 2/(a*f*g*Sqrt[g*Cos[e + f*x]]) + (2*b*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/((a^2 - b^2)*f*g^2*S
qrt[Cos[e + f*x]]) + (b^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 -
b^2)*(b - Sqrt[-a^2 + b^2])*f*g*Sqrt[g*Cos[e + f*x]]) + (b^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a
^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b^2)*(b + Sqrt[-a^2 + b^2])*f*g*Sqrt[g*Cos[e + f*x]]) + (2*b*(b - a*Sin[e
+ f*x]))/(a*(a^2 - b^2)*f*g*Sqrt[g*Cos[e + f*x]])

________________________________________________________________________________________

Rubi [A]  time = 1.3614, antiderivative size = 507, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 16, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.516, Rules used = {2898, 2565, 325, 329, 298, 203, 206, 2696, 2867, 2640, 2639, 2701, 2807, 2805, 205, 208} \[ -\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{3/2} \left (b^2-a^2\right )^{5/4}}+\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{3/2} \left (b^2-a^2\right )^{5/4}}+\frac{2 b E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{f g^2 \left (a^2-b^2\right ) \sqrt{\cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{a f g \left (a^2-b^2\right ) \sqrt{g \cos (e+f x)}}+\frac{b^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f g \left (a^2-b^2\right ) \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}+\frac{b^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f g \left (a^2-b^2\right ) \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}+\frac{\tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{3/2}}+\frac{2}{a f g \sqrt{g \cos (e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]

[Out]

ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(3/2)) - (b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b
^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^(5/4)*f*g^(3/2)) - ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(3/2)) +
(b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^(5/4)*f*g^(3/2)
) + 2/(a*f*g*Sqrt[g*Cos[e + f*x]]) + (2*b*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/((a^2 - b^2)*f*g^2*S
qrt[Cos[e + f*x]]) + (b^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 -
b^2)*(b - Sqrt[-a^2 + b^2])*f*g*Sqrt[g*Cos[e + f*x]]) + (b^2*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a
^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b^2)*(b + Sqrt[-a^2 + b^2])*f*g*Sqrt[g*Cos[e + f*x]]) + (2*b*(b - a*Sin[e
+ f*x]))/(a*(a^2 - b^2)*f*g*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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

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 2696

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((g*Co
s[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m + 1)*(b - a*Sin[e + f*x]))/(f*g*(a^2 - b^2)*(p + 1)), x] + Dist[1/
(g^2*(a^2 - b^2)*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m*(a^2*(p + 2) - b^2*(m + p + 2)
+ a*b*(m + p + 3)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] && NeQ[a^2 - b^2, 0] && LtQ[p, -1] &&
IntegersQ[2*m, 2*p]

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 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 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 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{\csc (e+f x)}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx &=\int \left (\frac{\csc (e+f x)}{a (g \cos (e+f x))^{3/2}}-\frac{b}{a (g \cos (e+f x))^{3/2} (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac{\int \frac{\csc (e+f x)}{(g \cos (e+f x))^{3/2}} \, dx}{a}-\frac{b \int \frac{1}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx}{a}\\ &=\frac{2 b (b-a \sin (e+f x))}{a \left (a^2-b^2\right ) f g \sqrt{g \cos (e+f x)}}+\frac{(2 b) \int \frac{\sqrt{g \cos (e+f x)} \left (\frac{a^2}{2}+\frac{b^2}{2}+\frac{1}{2} a b \sin (e+f x)\right )}{a+b \sin (e+f x)} \, dx}{a \left (a^2-b^2\right ) g^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{x^{3/2} \left (1-\frac{x^2}{g^2}\right )} \, dx,x,g \cos (e+f x)\right )}{a f g}\\ &=\frac{2}{a f g \sqrt{g \cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{a \left (a^2-b^2\right ) f g \sqrt{g \cos (e+f x)}}-\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^3}+\frac{b \int \sqrt{g \cos (e+f x)} \, dx}{\left (a^2-b^2\right ) g^2}+\frac{b^3 \int \frac{\sqrt{g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx}{a \left (a^2-b^2\right ) g^2}\\ &=\frac{2}{a f g \sqrt{g \cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{a \left (a^2-b^2\right ) f g \sqrt{g \cos (e+f x)}}-\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^3}-\frac{b^2 \int \frac{1}{\sqrt{g \cos (e+f x)} \left (\sqrt{-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 \left (a^2-b^2\right ) g}+\frac{b^2 \int \frac{1}{\sqrt{g \cos (e+f x)} \left (\sqrt{-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 \left (a^2-b^2\right ) g}+\frac{b^4 \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 \left (a^2-b^2\right ) f g}+\frac{\left (b \sqrt{g \cos (e+f x)}\right ) \int \sqrt{\cos (e+f x)} \, dx}{\left (a^2-b^2\right ) g^2 \sqrt{\cos (e+f x)}}\\ &=\frac{2}{a f g \sqrt{g \cos (e+f x)}}+\frac{2 b \sqrt{g \cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{\left (a^2-b^2\right ) f g^2 \sqrt{\cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{a \left (a^2-b^2\right ) f g \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 g}+\frac{\operatorname{Subst}\left (\int \frac{1}{g+x^2} \, dx,x,\sqrt{g \cos (e+f x)}\right )}{a f g}+\frac{\left (2 b^4\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 \left (a^2-b^2\right ) f g}-\frac{\left (b^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 \left (a^2-b^2\right ) g \sqrt{g \cos (e+f x)}}+\frac{\left (b^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 \left (a^2-b^2\right ) g \sqrt{g \cos (e+f x)}}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{3/2}}+\frac{2}{a f g \sqrt{g \cos (e+f x)}}+\frac{2 b \sqrt{g \cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{\left (a^2-b^2\right ) f g^2 \sqrt{\cos (e+f x)}}+\frac{b^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 )}{\left (a^2-b^2\right ) \left (b-\sqrt{-a^2+b^2}\right ) f g \sqrt{g \cos (e+f x)}}+\frac{b^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 )}{\left (a^2-b^2\right ) \left (b+\sqrt{-a^2+b^2}\right ) f g \sqrt{g \cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{a \left (a^2-b^2\right ) f g \sqrt{g \cos (e+f x)}}-\frac{b^3 \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 \left (a^2-b^2\right ) f g}+\frac{b^3 \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 \left (a^2-b^2\right ) f g}\\ &=\frac{\tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{3/2}}-\frac{b^{5/2} \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 )^{5/4} f g^{3/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{3/2}}+\frac{b^{5/2} \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 )^{5/4} f g^{3/2}}+\frac{2}{a f g \sqrt{g \cos (e+f x)}}+\frac{2 b \sqrt{g \cos (e+f x)} E\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{\left (a^2-b^2\right ) f g^2 \sqrt{\cos (e+f x)}}+\frac{b^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 )}{\left (a^2-b^2\right ) \left (b-\sqrt{-a^2+b^2}\right ) f g \sqrt{g \cos (e+f x)}}+\frac{b^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 )}{\left (a^2-b^2\right ) \left (b+\sqrt{-a^2+b^2}\right ) f g \sqrt{g \cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{a \left (a^2-b^2\right ) f g \sqrt{g \cos (e+f x)}}\\ \end{align*}

Mathematica [C]  time = 27.5923, size = 1587, normalized size = 3.13 \[ \text{result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]/((g*Cos[e + f*x])^(3/2)*(a + b*Sin[e + f*x])),x]

[Out]

-(Cos[e + f*x]^(3/2)*((8*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((a*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*(a^2 - b^2)) + ((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sq
rt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)
^(1/4)] - Log[Sqrt[-a^2 + b^2] - (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]] + L
og[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Cos[e + f*x]] + I*b*Cos[e + f*x]]))/(Sqrt[b]*(-a
^2 + b^2)^(1/4))))/(Sqrt[1 - Cos[e + f*x]^2]*(b + a*Csc[e + f*x])) - ((-2*a^2 + b^2)*(-1 + Cos[e + f*x]^2)*(a
+ b*Sqrt[1 - Cos[e + f*x]^2])*Csc[e + f*x]*(6*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sq
rt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] - 6*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Co
s[e + f*x]])/(a^2 - b^2)^(1/4)] + 12*(a^2 - b^2)*ArcTan[Sqrt[Cos[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) + 6*a^2*Log[1 - Sqrt[Cos[e + f*x]]] - 6
*b^2*Log[1 - Sqrt[Cos[e + f*x]]] - 6*a^2*Log[1 + Sqrt[Cos[e + f*x]]] + 6*b^2*Log[1 + Sqrt[Cos[e + f*x]]] - 3*S
qrt[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]] + 3*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]]))/(12*(a^3 - a*b^2)*(1 - Cos[e + f*x]^2)*(b + a*Csc[e + f*x])) - (Sqrt[b]
*(-1 + Cos[e + f*x]^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e + f*x)]*Csc[e + f*x]*(-42*Sqrt[2]*(a^2 - b^2)
^(3/4)*(2*a^2 - b^2)*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 42*Sqrt[2]*(a^2 - b^
2)^(3/4)*(2*a^2 - b^2)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 84*b^(3/2)*(a^2 -
b^2)*ArcTan[Sqrt[Cos[e + f*x]]] - 56*a*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) + 48*a*b^(5/2)*AppellF1[7/4, 1/2, 1, 11/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]
^2)/(-a^2 + b^2)]*Cos[e + f*x]^(7/2) + 42*b^(3/2)*(a^2 - b^2)*Log[1 - Sqrt[Cos[e + f*x]]] + 42*b^(3/2)*(-a^2 +
b^2)*Log[1 + Sqrt[Cos[e + f*x]]] + 21*Sqrt[2]*(a^2 - b^2)^(3/4)*(2*a^2 - b^2)*Log[Sqrt[a^2 - b^2] - Sqrt[2]*S
qrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Cos[e + f*x]] + b*Cos[e + f*x]] - 21*Sqrt[2]*(a^2 - b^2)^(3/4)*(2*a^2 - b^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]]))/(84*(a^3 - a*b^2)*
(1 - Cos[e + f*x]^2)*(-1 + 2*Cos[e + f*x]^2)*(b + a*Csc[e + f*x]))))/(2*(a - b)*(a + b)*f*(g*Cos[e + f*x])^(3/
2)) + (2*Cos[e + f*x]*(a - b*Sin[e + f*x]))/((a^2 - b^2)*f*(g*Cos[e + f*x])^(3/2))

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Maple [A]  time = 3.954, size = 425, normalized size = 0.8 \begin{align*}{\frac{1}{2\,af} \left ( - \left ( 4\,\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 ){g}^{5/2}+2\,\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 ) \sqrt{-g}{g}^{2}+2\,\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 ) \sqrt{-g}{g}^{2} \right ) \left ( \sin \left ({\frac{fx}{2}}+{\frac{e}{2}} \right ) \right ) ^{2}+2\,\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 ){g}^{5/2}-4\,\sqrt{-2\, \left ( \sin \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}g+g}{g}^{3/2}\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 ) \sqrt{-g}{g}^{2}+\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 ) \sqrt{-g}{g}^{2} \right ){g}^{-{\frac{7}{2}}}{\frac{1}{\sqrt{-g}}} \left ( 2\, \left ( \sin \left ( 1/2\,fx+e/2 \right ) \right ) ^{2}-1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x)

[Out]

1/2*(-(4*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))*g^(5/2)+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))*(-g)^(1/2)*g^2+2*ln(2/(c
os(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))*(-g)^(1/2)*g^2)*s
in(1/2*f*x+1/2*e)^2+2*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))*g^(5/2)-4*(-
2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)*g^(3/2)*(-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))*(-g)^(1/2)*g^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))*(-g)^(1/2)*g^2)/g^(7/2)/(-g)^(1/2)/a/(2*sin(1/2*f*x+1/2*e)^2-
1)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )}{\left (g \cos \left (f x + e\right )\right )^{\frac{3}{2}}{\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)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate(csc(f*x + e)/((g*cos(f*x + e))^(3/2)*(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))^(3/2)/(a+b*sin(f*x+e)),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(csc(f*x+e)/(g*cos(f*x+e))**(3/2)/(a+b*sin(f*x+e)),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )}{\left (g \cos \left (f x + e\right )\right )^{\frac{3}{2}}{\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)/(g*cos(f*x+e))^(3/2)/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)/((g*cos(f*x + e))^(3/2)*(b*sin(f*x + e) + a)), x)