3.1395 \(\int \frac {\csc ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\)

Optimal. Leaf size=557 \[ -\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f \sqrt {g}}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f \sqrt {g}}-\frac {b^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {b^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {b \csc (e+f x) \sqrt {g \cos (e+f x)}}{a^2 f g}-\frac {b \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \sqrt {g \cos (e+f x)}}+\frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^3 f \sqrt {g} \left (b^2-a^2\right )^{3/4}}+\frac {b^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^3 f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {3 \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f \sqrt {g}}-\frac {\csc ^2(e+f x) \sqrt {g \cos (e+f x)}}{2 a f g}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f \sqrt {g}} \]

[Out]

-3/4*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(1/2)-b^2*arctan((g*cos(f*x+e))^(1/2)/g^(1/2))/a^3/f/g^(1/2)+b
^(7/2)*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^3/(-a^2+b^2)^(3/4)/f/g^(1/2)-3/4*arctan
h((g*cos(f*x+e))^(1/2)/g^(1/2))/a/f/g^(1/2)-b^2*arctanh((g*cos(f*x+e))^(1/2)/g^(1/2))/a^3/f/g^(1/2)+b^(7/2)*ar
ctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))/a^3/(-a^2+b^2)^(3/4)/f/g^(1/2)-b*(cos(1/2*f*x+1/2
*e)^2)^(1/2)/cos(1/2*f*x+1/2*e)*EllipticF(sin(1/2*f*x+1/2*e),2^(1/2))*cos(f*x+e)^(1/2)/a^2/f/(g*cos(f*x+e))^(1
/2)-b^3*(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/(a^2-b*(b-(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)-b^3*(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/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(g*cos(f*x+e))^(1/2)+b*csc(f*x+e)*(g*cos(f*x+e))^(1/2)/a^2/f/g-1/2*csc(f*x+e)^2
*(g*cos(f*x+e))^(1/2)/a/f/g

________________________________________________________________________________________

Rubi [A]  time = 1.03, antiderivative size = 557, 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, 212, 206, 203, 2570, 2642, 2641, 290, 2702, 2807, 2805, 208, 205} \[ \frac {b^{7/2} \tan ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^3 f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f \sqrt {g}}+\frac {b^{7/2} \tanh ^{-1}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a^3 f \sqrt {g} \left (b^2-a^2\right )^{3/4}}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f \sqrt {g}}-\frac {b^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b-\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \left (a^2-b \left (b-\sqrt {b^2-a^2}\right )\right ) \sqrt {g \cos (e+f x)}}-\frac {b^3 \sqrt {\cos (e+f x)} \Pi \left (\frac {2 b}{b+\sqrt {b^2-a^2}};\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \left (a^2-b \left (\sqrt {b^2-a^2}+b\right )\right ) \sqrt {g \cos (e+f x)}}+\frac {b \csc (e+f x) \sqrt {g \cos (e+f x)}}{a^2 f g}-\frac {b \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \sqrt {g \cos (e+f x)}}-\frac {3 \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f \sqrt {g}}-\frac {\csc ^2(e+f x) \sqrt {g \cos (e+f x)}}{2 a f g}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f \sqrt {g}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(4*a*f*Sqrt[g]) - (b^2*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^3*f*
Sqrt[g]) + (b^(7/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^3*(-a^2 + b^2)^(3/
4)*f*Sqrt[g]) - (3*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(4*a*f*Sqrt[g]) - (b^2*ArcTanh[Sqrt[g*Cos[e + f*x]]/
Sqrt[g]])/(a^3*f*Sqrt[g]) + (b^(7/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^
3*(-a^2 + b^2)^(3/4)*f*Sqrt[g]) + (b*Sqrt[g*Cos[e + f*x]]*Csc[e + f*x])/(a^2*f*g) - (Sqrt[g*Cos[e + f*x]]*Csc[
e + f*x]^2)/(2*a*f*g) - (b*Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2])/(a^2*f*Sqrt[g*Cos[e + f*x]]) - (b^3*S
qrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(a^2*(a^2 - b*(b - Sqrt[-a^2 + b^2
]))*f*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*(a^2 - b*(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 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 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 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 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 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 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 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 {\csc ^3(e+f x)}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx &=\int \left (\frac {b^2 \csc (e+f x)}{a^3 \sqrt {g \cos (e+f x)}}-\frac {b \csc ^2(e+f x)}{a^2 \sqrt {g \cos (e+f x)}}+\frac {\csc ^3(e+f x)}{a \sqrt {g \cos (e+f x)}}-\frac {b^3}{a^3 \sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac {\int \frac {\csc ^3(e+f x)}{\sqrt {g \cos (e+f x)}} \, dx}{a}-\frac {b \int \frac {\csc ^2(e+f x)}{\sqrt {g \cos (e+f x)}} \, dx}{a^2}+\frac {b^2 \int \frac {\csc (e+f x)}{\sqrt {g \cos (e+f x)}} \, dx}{a^3}-\frac {b^3 \int \frac {1}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a^3}\\ &=\frac {b \sqrt {g \cos (e+f x)} \csc (e+f x)}{a^2 f g}-\frac {b \int \frac {1}{\sqrt {g \cos (e+f x)}} \, dx}{2 a^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 a^2 \sqrt {-a^2+b^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 a^2 \sqrt {-a^2+b^2}}-\frac {\operatorname {Subst}\left (\int \frac {1}{\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 {1}{\sqrt {x} \left (1-\frac {x^2}{g^2}\right )} \, dx,x,g \cos (e+f x)\right )}{a^3 f g}-\frac {\left (b^4 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^3 f}\\ &=\frac {b \sqrt {g \cos (e+f x)} \csc (e+f x)}{a^2 f g}-\frac {\sqrt {g \cos (e+f x)} \csc ^2(e+f x)}{2 a f g}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{\sqrt {x} \left (1-\frac {x^2}{g^2}\right )} \, dx,x,g \cos (e+f x)\right )}{4 a f g}-\frac {\left (2 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{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 {1}{\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 \sqrt {\cos (e+f x)}\right ) \int \frac {1}{\sqrt {\cos (e+f x)}} \, dx}{2 a^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 a^2 \sqrt {-a^2+b^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 a^2 \sqrt {-a^2+b^2} \sqrt {g \cos (e+f x)}}\\ &=\frac {b \sqrt {g \cos (e+f x)} \csc (e+f x)}{a^2 f g}-\frac {\sqrt {g \cos (e+f x)} \csc ^2(e+f x)}{2 a f g}-\frac {b \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \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 )}{a^2 \sqrt {-a^2+b^2} \left (b-\sqrt {-a^2+b^2}\right ) f \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 )}{a^2 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{g-x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a^3 f}-\frac {b^2 \operatorname {Subst}\left (\int \frac {1}{g+x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a^3 f}+\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^3 \sqrt {-a^2+b^2} f}+\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^3 \sqrt {-a^2+b^2} f}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{1-\frac {x^4}{g^2}} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{2 a f g}\\ &=-\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f \sqrt {g}}+\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^3 \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f \sqrt {g}}+\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^3 \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}+\frac {b \sqrt {g \cos (e+f x)} \csc (e+f x)}{a^2 f g}-\frac {\sqrt {g \cos (e+f x)} \csc ^2(e+f x)}{2 a f g}-\frac {b \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \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 )}{a^2 \sqrt {-a^2+b^2} \left (b-\sqrt {-a^2+b^2}\right ) f \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 )}{a^2 \sqrt {-a^2+b^2} \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{g-x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{4 a f}-\frac {3 \operatorname {Subst}\left (\int \frac {1}{g+x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{4 a f}\\ &=-\frac {3 \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f \sqrt {g}}-\frac {b^2 \tan ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f \sqrt {g}}+\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^3 \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}-\frac {3 \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{4 a f \sqrt {g}}-\frac {b^2 \tanh ^{-1}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a^3 f \sqrt {g}}+\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^3 \left (-a^2+b^2\right )^{3/4} f \sqrt {g}}+\frac {b \sqrt {g \cos (e+f x)} \csc (e+f x)}{a^2 f g}-\frac {\sqrt {g \cos (e+f x)} \csc ^2(e+f x)}{2 a f g}-\frac {b \sqrt {\cos (e+f x)} F\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{a^2 f \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 )}{a^2 \sqrt {-a^2+b^2} \left (b-\sqrt {-a^2+b^2}\right ) f \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 )}{a^2 \sqrt {-a^2+b^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 = 30.85, size = 2129, normalized size = 3.82 \[ \text {Result too large to show} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(Cos[e + f*x]*((b*Csc[e + f*x])/a^2 - Csc[e + f*x]^2/(2*a)))/(f*Sqrt[g*Cos[e + f*x]]) + (Sqrt[Cos[e + f*x]]*((
-2*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)*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))) - ((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]*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]]))/(-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 - C
os[e + f*x]^2])*Cos[2*(e + f*x)]*Csc[e + f*x]*((-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)) + (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*ArcTan[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]*Sqr
t[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*(3*a^2 + 3*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, 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))))/((1 - Cos[e + f*x]^2)*(b
+ a*Csc[e + f*x]))))/(4*a^2*f*Sqrt[g*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/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 3.46, size = 315, normalized size = 0.57 \[ -\frac {3 \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}+4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{8 a \sqrt {g}\, f}+\frac {\sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}}{16 f a g \left (-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}+\frac {3 \ln \left (\frac {-2 g +2 \sqrt {-g}\, \sqrt {2 \left (\cos ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g -g}}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right )}{4 f a \sqrt {-g}}-\frac {3 \ln \left (\frac {2 \sqrt {g}\, \sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}-4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right )}{8 a \sqrt {g}\, f}-\frac {\sqrt {2 \left (\cos ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g -g}}{8 f a g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}-\frac {\sqrt {-2 \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right ) g +g}}{16 f a g \left (\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-3/8/f/a/g^(1/2)*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)))+1/16/f/a/g/(-1+cos(1/2*f*x+1/2*e))*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)+3/4/f/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))-3/8/f/a/g^(1/2)*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))-1/8/f/a/g/cos(1/2*f*x+1/2*e)^2*(2
*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2)-1/16/f/a/g/(cos(1/2*f*x+1/2*e)+1)*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^3/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="maxima")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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