3.1376 \(\int \frac {\sqrt {g \cos (e+f x)} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=544 \[ \frac {b^2 \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}-\frac {b^2 \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}-\frac {b^2 g \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {b^2 g \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}+\frac {b \csc (e+f x) (g \cos (e+f x))^{3/2}}{a^2 f g}+\frac {b E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{a^2 f \sqrt {\cos (e+f x)}}-\frac {b^{5/2} \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^3 f \sqrt [4]{b^2-a^2}}+\frac {b^{5/2} \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^3 f \sqrt [4]{b^2-a^2}}+\frac {\sqrt {g} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f}-\frac {\csc ^2(e+f x) (g \cos (e+f x))^{3/2}}{2 a f g}-\frac {\sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f} \]

[Out]

b*(g*cos(f*x+e))^(3/2)*csc(f*x+e)/a^2/f/g-1/2*(g*cos(f*x+e))^(3/2)*csc(f*x+e)^2/a/f/g+1/4*arctan((g*cos(f*x+e)
)^(1/2)/g^(1/2))*g^(1/2)/a/f+b^2*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))*g^(1/2)/a^3/f-b^(5/2)*arctan(b^(1/2)*(g*
cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))*g^(1/2)/a^3/(-a^2+b^2)^(1/4)/f-1/4*arctanh((g*cos(f*x+e))^(1/2)/g^
(1/2))*g^(1/2)/a/f-b^2*arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))*g^(1/2)/a^3/f+b^(5/2)*arctanh(b^(1/2)*(g*cos(f*x+
e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))*g^(1/2)/a^3/(-a^2+b^2)^(1/4)/f-b^2*g*(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/f/(b-(-a^2+b^2
)^(1/2))/(g*cos(f*x+e))^(1/2)-b^2*g*(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/f/(b+(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)+b*(cos(
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/f/co
s(f*x+e)^(1/2)

________________________________________________________________________________________

Rubi [A]  time = 1.06, antiderivative size = 544, normalized size of antiderivative = 1.00, number of steps used = 25, number of rules used = 15, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.454, Rules used = {2898, 2565, 329, 298, 203, 206, 2570, 2640, 2639, 290, 2701, 2807, 2805, 205, 208} \[ -\frac {b^{5/2} \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^3 f \sqrt [4]{b^2-a^2}}+\frac {b^2 \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}+\frac {b^{5/2} \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^3 f \sqrt [4]{b^2-a^2}}-\frac {b^2 \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}-\frac {b^2 g \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}-\frac {b^2 g \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}+\frac {b \csc (e+f x) (g \cos (e+f x))^{3/2}}{a^2 f g}+\frac {b E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{a^2 f \sqrt {\cos (e+f x)}}+\frac {\sqrt {g} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f}-\frac {\csc ^2(e+f x) (g \cos (e+f x))^{3/2}}{2 a f g}-\frac {\sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[g]*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(4*a*f) + (b^2*Sqrt[g]*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a
^3*f) - (b^(5/2)*Sqrt[g]*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^3*(-a^2 + b^2
)^(1/4)*f) - (Sqrt[g]*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(4*a*f) - (b^2*Sqrt[g]*ArcTanh[Sqrt[g*Cos[e + f*x
]]/Sqrt[g]])/(a^3*f) + (b^(5/2)*Sqrt[g]*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/
(a^3*(-a^2 + b^2)^(1/4)*f) + (b*(g*Cos[e + f*x])^(3/2)*Csc[e + f*x])/(a^2*f*g) - ((g*Cos[e + f*x])^(3/2)*Csc[e
+ f*x]^2)/(2*a*f*g) + (b*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(a^2*f*Sqrt[Cos[e + f*x]]) - (b^2*g*
Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a^2*(b - Sqrt[-a^2 + b^2])*f*Sqr
t[g*Cos[e + f*x]]) - (b^2*g*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a^2*
(b + Sqrt[-a^2 + b^2])*f*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 290

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(
a*c*n*(p + 1)), x] + Dist[(m + n*(p + 1) + 1)/(a*n*(p + 1)), Int[(c*x)^m*(a + b*x^n)^(p + 1), x], x] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 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 2570

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[((b*Cos[e + f
*x])^(n + 1)*(a*Sin[e + f*x])^(m + 1))/(a*b*f*(m + 1)), x] + Dist[(m + n + 2)/(a^2*(m + 1)), Int[(b*Cos[e + f*
x])^n*(a*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && IntegersQ[2*m, 2*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 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 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 {\sqrt {g \cos (e+f x)} \csc ^3(e+f x)}{a+b \sin (e+f x)} \, dx &=\int \left (\frac {b^2 \sqrt {g \cos (e+f x)} \csc (e+f x)}{a^3}-\frac {b \sqrt {g \cos (e+f x)} \csc ^2(e+f x)}{a^2}+\frac {\sqrt {g \cos (e+f x)} \csc ^3(e+f x)}{a}-\frac {b^3 \sqrt {g \cos (e+f x)}}{a^3 (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac {\int \sqrt {g \cos (e+f x)} \csc ^3(e+f x) \, dx}{a}-\frac {b \int \sqrt {g \cos (e+f x)} \csc ^2(e+f x) \, dx}{a^2}+\frac {b^2 \int \sqrt {g \cos (e+f x)} \csc (e+f x) \, dx}{a^3}-\frac {b^3 \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx}{a^3}\\ &=\frac {b (g \cos (e+f x))^{3/2} \csc (e+f x)}{a^2 f g}+\frac {b \int \sqrt {g \cos (e+f x)} \, dx}{2 a^2}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {x}}{\left (1-\frac {x^2}{g^2}\right )^2} \, dx,x,g \cos (e+f x)\right )}{a f g}-\frac {b^2 \operatorname {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{g^2}} \, dx,x,g \cos (e+f x)\right )}{a^3 f g}+\frac {\left (b^2 g\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 a^2}-\frac {\left (b^2 g\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 a^2}-\frac {\left (b^4 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^3 f}\\ &=\frac {b (g \cos (e+f x))^{3/2} \csc (e+f x)}{a^2 f g}-\frac {(g \cos (e+f x))^{3/2} \csc ^2(e+f x)}{2 a f g}-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{g^2}} \, dx,x,g \cos (e+f x)\right )}{4 a f g}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{g^2}} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a^3 f g}-\frac {\left (2 b^4 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^3 f}+\frac {\left (b^2 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 a^2 \sqrt {g \cos (e+f x)}}-\frac {\left (b^2 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 a^2 \sqrt {g \cos (e+f x)}}+\frac {\left (b \sqrt {g \cos (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \, dx}{2 a^2 \sqrt {\cos (e+f x)}}\\ &=\frac {b (g \cos (e+f x))^{3/2} \csc (e+f x)}{a^2 f g}-\frac {(g \cos (e+f x))^{3/2} \csc ^2(e+f x)}{2 a f g}+\frac {b \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \sqrt {\cos (e+f x)}}-\frac {b^2 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 )}{a^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {b^2 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 )}{a^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {\operatorname {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{g^2}} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{2 a f g}-\frac {\left (b^2 g\right ) \operatorname {Subst}\left (\int \frac {1}{g-x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a^3 f}+\frac {\left (b^2 g\right ) \operatorname {Subst}\left (\int \frac {1}{g+x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a^3 f}+\frac {\left (b^3 g\right ) \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^3 f}-\frac {\left (b^3 g\right ) \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^3 f}\\ &=\frac {b^2 \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}-\frac {b^{5/2} \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^3 \sqrt [4]{-a^2+b^2} f}-\frac {b^2 \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}+\frac {b^{5/2} \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^3 \sqrt [4]{-a^2+b^2} f}+\frac {b (g \cos (e+f x))^{3/2} \csc (e+f x)}{a^2 f g}-\frac {(g \cos (e+f x))^{3/2} \csc ^2(e+f x)}{2 a f g}+\frac {b \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \sqrt {\cos (e+f x)}}-\frac {b^2 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 )}{a^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {b^2 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 )}{a^2 \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 )}{4 a f}+\frac {g \operatorname {Subst}\left (\int \frac {1}{g+x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{4 a f}\\ &=\frac {\sqrt {g} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f}+\frac {b^2 \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}-\frac {b^{5/2} \sqrt {g} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^3 \sqrt [4]{-a^2+b^2} f}-\frac {\sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f}-\frac {b^2 \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f}+\frac {b^{5/2} \sqrt {g} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a^3 \sqrt [4]{-a^2+b^2} f}+\frac {b (g \cos (e+f x))^{3/2} \csc (e+f x)}{a^2 f g}-\frac {(g \cos (e+f x))^{3/2} \csc ^2(e+f x)}{2 a f g}+\frac {b \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \sqrt {\cos (e+f x)}}-\frac {b^2 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 )}{a^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {b^2 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 )}{a^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}\\ \end {align*}

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Mathematica [C]  time = 29.19, size = 1582, normalized size = 2.91 \[ \text {result too large to display} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Sqrt[g*Cos[e + f*x]]*((b*Cot[e + f*x])/a^2 - (Cot[e + f*x]*Csc[e + f*x])/(2*a)))/f - (Sqrt[g*Cos[e + f*x]]*((
6*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))))/(Sqr
t[1 - Cos[e + f*x]^2]*(b + a*Csc[e + f*x])) - ((-a^2 - 5*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]*Sqrt[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*C
os[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*Sqr
t[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)*Ar
cTan[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]))))/(4*a^2*f*Sqrt[Cos[e + f*x]])

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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)^3*(g*cos(f*x+e))^(1/2)/(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 \[ \int \frac {\sqrt {g \cos \left (f x + e\right )} \csc \left (f x + e\right )^{3}}{b \sin \left (f x + e\right ) + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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maple [A]  time = 3.25, size = 307, normalized size = 0.56 \[ \frac {\sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}}{16 f a \left (-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}-\frac {\sqrt {g}\, \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}+4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{8 f a}+\frac {\sqrt {2 \left (\cos ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g -g}}{8 f a \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}-\frac {g \ln \left (\frac {-2 g +2 \sqrt {-g}\, \sqrt {2 \left (\cos ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g -g}}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{4 f a \sqrt {-g}}-\frac {\sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}}{16 f a \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {\sqrt {g}\, \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{8 a f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16/f/a/(-1+cos(1/2*f*x+1/2*e))*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-1/8/f*g^(1/2)/a*ln((4*g*cos(1/2*f*x+1/2*e
)+2*g^(1/2)*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-2*g)/(-1+cos(1/2*f*x+1/2*e)))+1/8/f/a/cos(1/2*f*x+1/2*e)^2*(2*
cos(1/2*f*x+1/2*e)^2*g-g)^(1/2)-1/4/f*g/a/(-g)^(1/2)*ln((-2*g+2*(-g)^(1/2)*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2))
/cos(1/2*f*x+1/2*e))-1/16/f/a/(cos(1/2*f*x+1/2*e)+1)*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)-1/8/f*g^(1/2)/a*ln((-
4*g*cos(1/2*f*x+1/2*e)+2*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)+1))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**3*(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)**3/(a + b*sin(e + f*x)), x)

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