3.1394 \(\int \frac{\csc ^2(e+f x)}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\)
Optimal. Leaf size=448 \[ -\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f \sqrt{g} \left (b^2-a^2\right )^{3/4}}-\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f \sqrt{g} \left (b^2-a^2\right )^{3/4}}+\frac{b^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{a f \left (b \sqrt{b^2-a^2}+a^2-b^2\right ) \sqrt{g \cos (e+f x)}}+\frac{b^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{a f \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{g \cos (e+f x)}}+\frac{b \tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a^2 f \sqrt{g}}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a^2 f \sqrt{g}}-\frac{\csc (e+f x) \sqrt{g \cos (e+f x)}}{a f g}+\frac{\sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{a f \sqrt{g \cos (e+f x)}} \]
[Out]
(b*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2*f*Sqrt[g]) - (b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a
^2 + b^2)^(1/4)*Sqrt[g])])/(a^2*(-a^2 + b^2)^(3/4)*f*Sqrt[g]) + (b*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2
*f*Sqrt[g]) - (b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^2*(-a^2 + b^2)
^(3/4)*f*Sqrt[g]) - (Sqrt[g*Cos[e + f*x]]*Csc[e + f*x])/(a*f*g) + (Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2
])/(a*f*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*(a^2 - b^2 + b*Sqrt[-a^2 + b^2])*f*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*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*Sqrt[g*Cos[e + f*x]])
________________________________________________________________________________________
Rubi [A] time = 0.975404, antiderivative size = 448, normalized size of antiderivative = 1.,
number of steps used = 19, number of rules used = 14, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.424, Rules
used = {2898, 2565, 329, 212, 206, 203, 2570, 2642, 2641, 2702, 2807, 2805, 208, 205}
\[ -\frac{b^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f \sqrt{g} \left (b^2-a^2\right )^{3/4}}-\frac{b^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{a^2 f \sqrt{g} \left (b^2-a^2\right )^{3/4}}+\frac{b^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{a f \left (b \sqrt{b^2-a^2}+a^2-b^2\right ) \sqrt{g \cos (e+f x)}}+\frac{b^2 \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{a f \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{g \cos (e+f x)}}+\frac{b \tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a^2 f \sqrt{g}}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a^2 f \sqrt{g}}-\frac{\csc (e+f x) \sqrt{g \cos (e+f x)}}{a f g}+\frac{\sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{a f \sqrt{g \cos (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[Csc[e + f*x]^2/(Sqrt[g*Cos[e + f*x]]*(a + b*Sin[e + f*x])),x]
[Out]
(b*ArcTan[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2*f*Sqrt[g]) - (b^(5/2)*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a
^2 + b^2)^(1/4)*Sqrt[g])])/(a^2*(-a^2 + b^2)^(3/4)*f*Sqrt[g]) + (b*ArcTanh[Sqrt[g*Cos[e + f*x]]/Sqrt[g]])/(a^2
*f*Sqrt[g]) - (b^(5/2)*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(a^2*(-a^2 + b^2)
^(3/4)*f*Sqrt[g]) - (Sqrt[g*Cos[e + f*x]]*Csc[e + f*x])/(a*f*g) + (Sqrt[Cos[e + f*x]]*EllipticF[(e + f*x)/2, 2
])/(a*f*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*(a^2 - b^2 + b*Sqrt[-a^2 + b^2])*f*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*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*f*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 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 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 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 ^2(e+f x)}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx &=\int \left (-\frac{b \csc (e+f x)}{a^2 \sqrt{g \cos (e+f x)}}+\frac{\csc ^2(e+f x)}{a \sqrt{g \cos (e+f x)}}+\frac{b^2}{a^2 \sqrt{g \cos (e+f x)} (a+b \sin (e+f x))}\right ) \, dx\\ &=\frac{\int \frac{\csc ^2(e+f x)}{\sqrt{g \cos (e+f x)}} \, dx}{a}-\frac{b \int \frac{\csc (e+f x)}{\sqrt{g \cos (e+f x)}} \, dx}{a^2}+\frac{b^2 \int \frac{1}{\sqrt{g \cos (e+f x)} (a+b \sin (e+f x))} \, dx}{a^2}\\ &=-\frac{\sqrt{g \cos (e+f x)} \csc (e+f x)}{a f g}+\frac{\int \frac{1}{\sqrt{g \cos (e+f x)}} \, dx}{2 a}-\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 a \sqrt{-a^2+b^2}}-\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 a \sqrt{-a^2+b^2}}+\frac{b \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^2 f g}+\frac{\left (b^3 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^2 f}\\ &=-\frac{\sqrt{g \cos (e+f x)} \csc (e+f x)}{a f g}+\frac{(2 b) \operatorname{Subst}\left (\int \frac{1}{1-\frac{x^4}{g^2}} \, dx,x,\sqrt{g \cos (e+f x)}\right )}{a^2 f g}+\frac{\left (2 b^3 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^2 f}+\frac{\sqrt{\cos (e+f x)} \int \frac{1}{\sqrt{\cos (e+f x)}} \, dx}{2 a \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 a \sqrt{-a^2+b^2} \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 a \sqrt{-a^2+b^2} \sqrt{g \cos (e+f x)}}\\ &=-\frac{\sqrt{g \cos (e+f x)} \csc (e+f x)}{a f g}+\frac{\sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{a f \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 )}{a \left (a^2-b^2+b \sqrt{-a^2+b^2}\right ) f \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 )}{a \left (a^2-b \left (b+\sqrt{-a^2+b^2}\right )\right ) f \sqrt{g \cos (e+f x)}}+\frac{b \operatorname{Subst}\left (\int \frac{1}{g-x^2} \, dx,x,\sqrt{g \cos (e+f x)}\right )}{a^2 f}+\frac{b \operatorname{Subst}\left (\int \frac{1}{g+x^2} \, dx,x,\sqrt{g \cos (e+f x)}\right )}{a^2 f}-\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^2 \sqrt{-a^2+b^2} f}-\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^2 \sqrt{-a^2+b^2} f}\\ &=\frac{b \tan ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a^2 f \sqrt{g}}-\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^2 \left (-a^2+b^2\right )^{3/4} f \sqrt{g}}+\frac{b \tanh ^{-1}\left (\frac{\sqrt{g \cos (e+f x)}}{\sqrt{g}}\right )}{a^2 f \sqrt{g}}-\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^2 \left (-a^2+b^2\right )^{3/4} f \sqrt{g}}-\frac{\sqrt{g \cos (e+f x)} \csc (e+f x)}{a f g}+\frac{\sqrt{\cos (e+f x)} F\left (\left .\frac{1}{2} (e+f x)\right |2\right )}{a f \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 )}{a \left (a^2-b^2+b \sqrt{-a^2+b^2}\right ) f \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 )}{a \left (a^2-b \left (b+\sqrt{-a^2+b^2}\right )\right ) f \sqrt{g \cos (e+f x)}}\\ \end{align*}
Mathematica [C] time = 30.1144, size = 2093, normalized size = 4.67 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Csc[e + f*x]^2/(Sqrt[g*Cos[e + f*x]]*(a + b*Sin[e + f*x])),x]
[Out]
-(Cot[e + f*x]/(a*f*Sqrt[g*Cos[e + f*x]])) - (Sqrt[Cos[e + f*x]]*((4*a*(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*(-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 - (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*Appell
F1[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*Co
s[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])) - (6*b*(-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*App
ellF1[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)*A
rcTan[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]]] - Sqr
t[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]] + Sq
rt[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*f*Sqrt[g*Cos[e + f*x]])
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Maple [C] time = 10.609, size = 1217, normalized size = 2.7 \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)^2/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x)
[Out]
1/8*(-4*(-2*sin(1/2*f*x+1/2*e)^4*g+sin(1/2*f*x+1/2*e)^2*g)^(3/2)*g^(3/2)*(-g)^(1/2)*a+4*cos(1/2*f*x+1/2*e)*sin
(1/2*f*x+1/2*e)*(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)*(-2*sin(1/2*f*x+1/2*e)^4*g+sin(1/2*f*x+1/2*e)^2*g)^(3/2)*b
*(4*g^(5/2)*(-g)^(1/2)*sum(1/(_R^6*b^2-3*_R^4*b^2*g+8*_R^2*a^2*g^2-5*_R^2*b^2*g^2-b^2*g^3)*_R*ln((-2*sin(1/2*f
*x+1/2*e)^2*g+g)^(1/2)-cos(1/2*f*x+1/2*e)*g^(1/2)*2^(1/2)-_R)*(_R^2+g),_R=RootOf(b^2*_Z^8-4*b^2*g*_Z^6+(16*a^2
*g^2-10*b^2*g^2)*_Z^4-4*b^2*g^3*_Z^2+b^2*g^4))*b^2-2*g^(3/2)*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))+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+(-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)+(-8*(-2*sin(1/2*f*x+1/2*e)^4*g+sin(1/2*f*x+1/2*e)^2*g)^(3/2)*EllipticF(cos(
1/2*f*x+1/2*e),2^(1/2))*g^(3/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*(-g)^(1/2)*a-g^(
7/2)*sin(1/2*f*x+1/2*e)^4*(2*sin(1/2*f*x+1/2*e)^2-1)^2/a*sum(1/_alpha/(2*_alpha^2-1)*(8*(sin(1/2*f*x+1/2*e)^2)
^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),(-4*_alpha^2*b^2+4*b^2)/a^2,2^(1/2))*(g*
(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*_alpha^3*b^2-8*b^2*_alpha*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/
2*e)^2-1)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),(-4*_alpha^2*b^2+4*b^2)/a^2,2^(1/2))*(g*(2*_alpha^2*b^2+a^2-2*b^
2)/b^2)^(1/2)+2^(1/2)*a^2*arctanh(1/2/(-2*sin(1/2*f*x+1/2*e)^4*g+sin(1/2*f*x+1/2*e)^2*g)^(1/2)/(g*(2*_alpha^2*
b^2+a^2-2*b^2)/b^2)^(1/2)/(4*a^2-3*b^2)*g*2^(1/2)*(-16*sin(1/2*f*x+1/2*e)^2*_alpha^2*a^2+12*sin(1/2*f*x+1/2*e)
^2*_alpha^2*b^2+4*_alpha^4*b^2+12*sin(1/2*f*x+1/2*e)^2*a^2-9*sin(1/2*f*x+1/2*e)^2*b^2+4*_alpha^2*a^2-7*b^2*_al
pha^2-3*a^2+3*b^2))*(sin(1/2*f*x+1/2*e)^2*g*(-2*sin(1/2*f*x+1/2*e)^2+1))^(1/2))/(g*(2*_alpha^2*b^2+a^2-2*b^2)/
b^2)^(1/2)/(sin(1/2*f*x+1/2*e)^2*g*(-2*sin(1/2*f*x+1/2*e)^2+1))^(1/2),_alpha=RootOf(4*_Z^4*b^2-4*_Z^2*b^2+a^2)
)*(-g)^(1/2))*cos(1/2*f*x+1/2*e)+8*(-2*sin(1/2*f*x+1/2*e)^4*g+sin(1/2*f*x+1/2*e)^2*g)^(3/2)*g^(3/2)*(-g)^(1/2)
*a*sin(1/2*f*x+1/2*e)^2)/a^2/g^(3/2)/(-g)^(1/2)/cos(1/2*f*x+1/2*e)/(-2*sin(1/2*f*x+1/2*e)^4*g+sin(1/2*f*x+1/2*
e)^2*g)^(3/2)/sin(1/2*f*x+1/2*e)/(-2*sin(1/2*f*x+1/2*e)^2*g+g)^(1/2)/f
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )^{2}}{\sqrt{g \cos \left (f x + e\right )}{\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)^2/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="maxima")
[Out]
integrate(csc(f*x + e)^2/(sqrt(g*cos(f*x + e))*(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)^2/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc ^{2}{\left (e + f x \right )}}{\sqrt{g \cos{\left (e + f x \right )}} \left (a + b \sin{\left (e + f x \right )}\right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(csc(f*x+e)**2/(a+b*sin(f*x+e))/(g*cos(f*x+e))**(1/2),x)
[Out]
Integral(csc(e + f*x)**2/(sqrt(g*cos(e + f*x))*(a + b*sin(e + f*x))), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\csc \left (f x + e\right )^{2}}{\sqrt{g \cos \left (f x + e\right )}{\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)^2/(a+b*sin(f*x+e))/(g*cos(f*x+e))^(1/2),x, algorithm="giac")
[Out]
integrate(csc(f*x + e)^2/(sqrt(g*cos(f*x + e))*(b*sin(f*x + e) + a)), x)