3.1406 \(\int \frac{\csc (e+f x)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=527 \[ \frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac{b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{5/2} \left (b^2-a^2\right )^{7/4}}-\frac{2 b \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{3 f g^2 \left (a^2-b^2\right ) \sqrt{g \cos (e+f x)}}+\frac{b^3 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{g \cos (e+f x)}}+\frac{b^3 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{g \cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{3 a f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{5/2}}+\frac{2}{3 a f g (g \cos (e+f x))^{3/2}} \]

[Out]

-(ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(5/2))) + (b^(7/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2
+ b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^(7/4)*f*g^(5/2)) - ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(5/2))
+ (b^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^(7/4)*f*g^(5
/2)) + 2/(3*a*f*g*(g*Cos[e + f*x])^(3/2)) - (2*b*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*(a^2 - b^2)*
f*g^2*Sqrt[g*Cos[e + f*x]]) + (b^3*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2]
)/((a^2 - b^2)*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*g^2*Sqrt[g*Cos[e + f*x]]) + (b^3*Sqrt[Cos[e + f*x]]*Elliptic
Pi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b^2)*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*g^2*Sqrt[g*C
os[e + f*x]]) + (2*b*(b - a*Sin[e + f*x]))/(3*a*(a^2 - b^2)*f*g*(g*Cos[e + f*x])^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 1.31351, antiderivative size = 527, 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, 212, 206, 203, 2696, 2867, 2642, 2641, 2702, 2807, 2805, 208, 205} \[ \frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{5/2} \left (b^2-a^2\right )^{7/4}}+\frac{b^{7/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a f g^{5/2} \left (b^2-a^2\right )^{7/4}}-\frac{2 b \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{3 f g^2 \left (a^2-b^2\right ) \sqrt{g \cos (e+f x)}}+\frac{b^3 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{g \cos (e+f x)}}+\frac{b^3 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{f g^2 \left (a^2-b^2\right ) \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{g \cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{3 a f g \left (a^2-b^2\right ) (g \cos (e+f x))^{3/2}}-\frac{\tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{5/2}}+\frac{2}{3 a f g (g \cos (e+f x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]/((g*Cos[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]

[Out]

-(ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(5/2))) + (b^(7/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2
+ b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^(7/4)*f*g^(5/2)) - ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]]/(a*f*g^(5/2))
+ (b^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a*(-a^2 + b^2)^(7/4)*f*g^(5
/2)) + 2/(3*a*f*g*(g*Cos[e + f*x])^(3/2)) - (2*b*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(3*(a^2 - b^2)*
f*g^2*Sqrt[g*Cos[e + f*x]]) + (b^3*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2]
)/((a^2 - b^2)*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*f*g^2*Sqrt[g*Cos[e + f*x]]) + (b^3*Sqrt[Cos[e + f*x]]*Elliptic
Pi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/((a^2 - b^2)*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*g^2*Sqrt[g*C
os[e + f*x]]) + (2*b*(b - a*Sin[e + f*x]))/(3*a*(a^2 - b^2)*f*g*(g*Cos[e + f*x])^(3/2))

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 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 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 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 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)}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx &=\int \left (\frac{\csc (e+f x)}{a (g \cos (e+f x))^{5/2}}-\frac{b}{a (g \cos (e+f x))^{5/2} (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac{\int \frac{\csc (e+f x)}{(g \cos (e+f x))^{5/2}} \, dx}{a}-\frac{b \int \frac{1}{(g \cos (e+f x))^{5/2} (a+b \sin (e+f x))} \, dx}{a}\\ &=\frac{2 b (b-a \sin (e+f x))}{3 a \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}+\frac{(2 b) \int \frac{-\frac{a^2}{2}+\frac{3 b^2}{2}-\frac{1}{2} a b \sin (e+f x)}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{3 a \left (a^2-b^2\right ) g^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{x^{5/2} \left (1-\frac{x^2}{g^2}\right )} \, dx,x,g \cos (e+f x)\right )}{a f g}\\ &=\frac{2}{3 a f g (g \cos (e+f x))^{3/2}}+\frac{2 b (b-a \sin (e+f x))}{3 a \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}-\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^3}-\frac{b \int \frac{1}{\sqrt{g \cos (e+f x)}} \, dx}{3 \left (a^2-b^2\right ) g^2}+\frac{b^3 \int \frac{1}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a \left (a^2-b^2\right ) g^2}\\ &=\frac{2}{3 a f g (g \cos (e+f x))^{3/2}}+\frac{2 b (b-a \sin (e+f x))}{3 a \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}-\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^3}+\frac{b^3 \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 )^{3/2} g^2}+\frac{b^3 \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 )^{3/2} g^2}+\frac{b^4 \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 \left (a^2-b^2\right ) f g}-\frac{\left (b \sqrt{\cos (e+f x)}\right ) \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx}{3 \left (a^2-b^2\right ) g^2 \sqrt{g \cos (e+f x)}}\\ &=\frac{2}{3 a f g (g \cos (e+f x))^{3/2}}-\frac{2 b \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{3 \left (a^2-b^2\right ) f g^2 \sqrt{g \cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{3 a \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{1}{g-x^2} \, dx,x,\sqrt{g \cos (e+f x)}\right )}{a f g^2}-\frac{\operatorname{Subst}\left (\int \frac{1}{g+x^2} \, dx,x,\sqrt{g \cos (e+f x)}\right )}{a f g^2}+\frac{\left (2 b^4\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 \left (a^2-b^2\right ) f g}+\frac{\left (b^3 \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 )^{3/2} g^2 \sqrt{g \cos (e+f x)}}+\frac{\left (b^3 \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 )^{3/2} g^2 \sqrt{g \cos (e+f x)}}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{5/2}}+\frac{2}{3 a f g (g \cos (e+f x))^{3/2}}-\frac{2 b \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{3 \left (a^2-b^2\right ) f g^2 \sqrt{g \cos (e+f x)}}-\frac{b^3 \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 )^{3/2} \left (b-\sqrt{-a^2+b^2}\right ) f g^2 \sqrt{g \cos (e+f x)}}+\frac{b^3 \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 )^{3/2} \left (b+\sqrt{-a^2+b^2}\right ) f g^2 \sqrt{g \cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{3 a \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}+\frac{b^4 \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 )^{3/2} f g^2}+\frac{b^4 \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 )^{3/2} f g^2}\\ &=-\frac{\tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{5/2}}+\frac{b^{7/2} \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 )^{7/4} f g^{5/2}}-\frac{\tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a f g^{5/2}}+\frac{b^{7/2} \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 )^{7/4} f g^{5/2}}+\frac{2}{3 a f g (g \cos (e+f x))^{3/2}}-\frac{2 b \sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{3 \left (a^2-b^2\right ) f g^2 \sqrt{g \cos (e+f x)}}-\frac{b^3 \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 )^{3/2} \left (b-\sqrt{-a^2+b^2}\right ) f g^2 \sqrt{g \cos (e+f x)}}+\frac{b^3 \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 )^{3/2} \left (b+\sqrt{-a^2+b^2}\right ) f g^2 \sqrt{g \cos (e+f x)}}+\frac{2 b (b-a \sin (e+f x))}{3 a \left (a^2-b^2\right ) f g (g \cos (e+f x))^{3/2}}\\ \end{align*}

Mathematica [C]  time = 30.1756, size = 2136, normalized size = 4.05 \[ \text{Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]/((g*Cos[e + f*x])^(5/2)*(a + b*Sin[e + f*x])),x]

[Out]

(Cos[e + f*x]^(5/2)*((-8*a*b*(a + b*Sqrt[1 - Cos[e + f*x]^2])*((5*a*(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[1 - Cos[e + f*x]^2]*(5*(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)] - 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))) - ((1/8 - I/8)*Sqrt
[b]*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Cos[e + f*x]])/(-a^2 + b^2)^(1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqr
t[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]] - 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]]))/(-a^2 + b^2)^(3/4)))/(Sqrt[1 - Cos[e + f*x]^2]*(b + a*Csc[e + f*x])) - (b^2*(-1 + Cos[e + f*x
]^2)*(a + b*Sqrt[1 - Cos[e + f*x]^2])*Cos[2*(e + f*x)]*Csc[e + f*x]*((-10*Sqrt[2]*(2*a^2 - b^2)*ArcTan[1 - (Sq
rt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) + (10*Sqrt[2]*(2*a^2 - b^2
)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) - (20*ArcT
an[Sqrt[Cos[e + f*x]]])/a - (16*b*AppellF1[5/4, 1/2, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)
]*Cos[e + f*x]^(5/2))/(-a^2 + b^2) - (200*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[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))) + (10*Log[1 - Sqrt[Cos[e + f*x]]])/a - (10*
Log[1 + Sqrt[Cos[e + f*x]]])/a - (5*Sqrt[2]*(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]])/(a*Sqrt[b]*(a^2 - b^2)^(3/4)) + (5*Sqrt[2]*(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]])/(a*Sqrt[b]*(a^2 - b^2)^(3/4
))))/(20*(1 - Cos[e + f*x]^2)*(-1 + 2*Cos[e + f*x]^2)*(b + a*Csc[e + f*x])) - (2*(6*a^2 - 7*b^2)*(-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)*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[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, Co
s[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)*ArcT
an[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))))/((
1 - Cos[e + f*x]^2)*(b + a*Csc[e + f*x]))))/(6*(a - b)*(a + b)*f*(g*Cos[e + f*x])^(5/2)) + (2*Cos[e + f*x]*(a
- b*Sin[e + f*x]))/(3*(a^2 - b^2)*f*(g*Cos[e + f*x])^(5/2))

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Maple [A]  time = 4.605, size = 627, normalized size = 1.2 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)/(g*cos(f*x+e))^(5/2)/(a+b*sin(f*x+e)),x)

[Out]

1/6*((24*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^(7/2)-12*(-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^3-12*(-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))*g^3)
*sin(1/2*f*x+1/2*e)^4+(-24*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^(7/2)
+12*(-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^3+12*(-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))*g^3)*sin(1/2*f*x+1/2*e)^2+6*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^(7/2)+4*(-g)^(1/2)*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)*g^(5/2)-3*(-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^3-3*(-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))*g^3)/(-g)^(1/2)/g^(11/2)/a
/(4*sin(1/2*f*x+1/2*e)^4-4*sin(1/2*f*x+1/2*e)^2+1)/f

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Maxima [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))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

Timed out

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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))^(5/2)/(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))**(5/2)/(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{5}{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))^(5/2)/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate(csc(f*x + e)/((g*cos(f*x + e))^(5/2)*(b*sin(f*x + e) + a)), x)