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

[Out]

arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(3/2)-b^(5/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/
g^(1/2))/a/(-a^2+b^2)^(5/4)/f/g^(3/2)-arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(3/2)+b^(5/2)*arctanh(b^(1/2
)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a/(-a^2+b^2)^(5/4)/f/g^(3/2)+2/a/f/g/(g*cos(f*x+e))^(1/2)+2*b
*(b-a*sin(f*x+e))/a/(a^2-b^2)/f/g/(g*cos(f*x+e))^(1/2)+b^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*Ell
ipticPi(sin(1/2*f*x+1/2*e),2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*cos(f*x+e)^(1/2)/(a^2-b^2)/f/g/(b-(-a^2+b^2)^(1/2
))/(g*cos(f*x+e))^(1/2)+b^2*(cos(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticPi(sin(1/2*f*x+1/2*e),2*b/
(b+(-a^2+b^2)^(1/2)),2^(1/2))*cos(f*x+e)^(1/2)/(a^2-b^2)/f/g/(b+(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)+2*b*(co
s(1/2*f*x+1/2*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticE(sin(1/2*f*x+1/2*e),2^(1/2))*(g*cos(f*x+e))^(1/2)/(a^2-b
^2)/f/g^2/cos(f*x+e)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.36, antiderivative size = 507, normalized size of antiderivative = 1.00, 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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 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])

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*}

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Mathematica [C]  time = 27.98, 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]

-1/2*(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
)*Sqrt[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]]
+ 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]]))/(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
]*Sqrt[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]*Sqr
t[Cos[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*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]] + 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])) - (Sqr
t[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]*Sqrt[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]))))/((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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fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]

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)

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maple [A]  time = 4.58, size = 425, normalized size = 0.84 \[ \frac {-\left (4 \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) g^{\frac {5}{2}}+2 \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}+4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) \sqrt {-g}\, g^{2}+2 \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}\, g^{2}\right ) \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+2 \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) g^{\frac {5}{2}}-4 \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}\, g^{\frac {3}{2}} \sqrt {-g}+\ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}+4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) \sqrt {-g}\, g^{2}+\ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) \sqrt {-g}\, g^{2}}{2 g^{\frac {7}{2}} \sqrt {-g}\, a \left (2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )-1\right ) f} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc \left (f x + e\right )}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{\sin \left (e+f\,x\right )\,{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(sin(e + f*x)*(g*cos(e + f*x))^(3/2)*(a + b*sin(e + f*x))),x)

[Out]

int(1/(sin(e + f*x)*(g*cos(e + f*x))^(3/2)*(a + b*sin(e + f*x))), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\csc {\left (e + f x \right )}}{\left (g \cos {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]

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]

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

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