\(\int \frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx\) [1386]

Optimal result

Integrand size = 31, antiderivative size = 425 \[ \int \frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=\frac {g^{5/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a b^{3/2} f}-\frac {g^{5/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a b^{3/2} f}-\frac {2 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {\cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 425, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.516, Rules used = {2977, 2645, 327, 335, 304, 209, 212, 2774, 2946, 2721, 2719, 2780, 2886, 2884, 211, 214} \[ \int \frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {g^{5/2} \left (b^2-a^2\right )^{3/4} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a b^{3/2} f}+\frac {g^{5/2} \left (b^2-a^2\right )^{3/4} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{a b^{3/2} f}+\frac {g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^2 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {g^3 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^2 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}+\frac {g^{5/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {g^{5/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {2 g^2 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b f \sqrt {\cos (e+f x)}} \]

[In]

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

[Out]

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

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 304

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 327

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

Rule 335

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 2645

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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)*(c - Pi/2 + d*x), 2], x] /; FreeQ[{
c, d}, x]

Rule 2721

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[(b*Sin[c + d*x])^n/Sin[c + d*x]^n, Int[Sin[c + d*x]
^n, x], x] /; FreeQ[{b, c, d}, x] && LtQ[-1, n, 1] && IntegerQ[2*n]

Rule 2774

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[g*(g*C
os[e + f*x])^(p - 1)*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + p))), x] + Dist[g^2*((p - 1)/(b*(m + p))), Int[(g
*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^m*(b + a*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f, g, m}, x] &&
NeQ[a^2 - b^2, 0] && GtQ[p, 1] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2780

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 2884

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp
[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(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 2886

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/
(c + d))*Sin[e + f*x]]), 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 2946

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 2977

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*} \text {integral}& = \int \left (\frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a}-\frac {b (g \cos (e+f x))^{5/2}}{a (a+b \sin (e+f x))}\right ) \, dx \\ & = \frac {\int (g \cos (e+f x))^{5/2} \csc (e+f x) \, dx}{a}-\frac {b \int \frac {(g \cos (e+f x))^{5/2}}{a+b \sin (e+f x)} \, dx}{a} \\ & = -\frac {2 g (g \cos (e+f x))^{3/2}}{3 a f}-\frac {\text {Subst}\left (\int \frac {x^{5/2}}{1-\frac {x^2}{g^2}} \, dx,x,g \cos (e+f x)\right )}{a f g}-\frac {g^2 \int \frac {\sqrt {g \cos (e+f x)} (b+a \sin (e+f x))}{a+b \sin (e+f x)} \, dx}{a} \\ & = -\frac {g \text {Subst}\left (\int \frac {\sqrt {x}}{1-\frac {x^2}{g^2}} \, dx,x,g \cos (e+f x)\right )}{a f}-\frac {g^2 \int \sqrt {g \cos (e+f x)} \, dx}{b}-\frac {\left (\left (-a^2+b^2\right ) g^2\right ) \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx}{a b} \\ & = -\frac {(2 g) \text {Subst}\left (\int \frac {x^2}{1-\frac {x^4}{g^2}} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}-\frac {\left (\left (a^2-b^2\right ) g^3\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}-b \cos (e+f x)\right )} \, dx}{2 b^2}+\frac {\left (\left (a^2-b^2\right ) g^3\right ) \int \frac {1}{\sqrt {g \cos (e+f x)} \left (\sqrt {-a^2+b^2}+b \cos (e+f x)\right )} \, dx}{2 b^2}+\frac {\left (\left (a^2-b^2\right ) g^3\right ) \text {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 f}-\frac {\left (g^2 \sqrt {g \cos (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \, dx}{b \sqrt {\cos (e+f x)}} \\ & = -\frac {2 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {\cos (e+f x)}}-\frac {g^3 \text {Subst}\left (\int \frac {1}{g-x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}+\frac {g^3 \text {Subst}\left (\int \frac {1}{g+x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a f}+\frac {\left (2 \left (a^2-b^2\right ) g^3\right ) \text {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 f}-\frac {\left (\left (a^2-b^2\right ) g^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 b^2 \sqrt {g \cos (e+f x)}}+\frac {\left (\left (a^2-b^2\right ) g^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 b^2 \sqrt {g \cos (e+f x)}} \\ & = \frac {g^{5/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {g^{5/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {2 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {\cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}-\frac {\left (\left (a^2-b^2\right ) g^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g-b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a b f}+\frac {\left (\left (a^2-b^2\right ) g^3\right ) \text {Subst}\left (\int \frac {1}{\sqrt {-a^2+b^2} g+b x^2} \, dx,x,\sqrt {g \cos (e+f x)}\right )}{a b f} \\ & = \frac {g^{5/2} \arctan \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}-\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a b^{3/2} f}-\frac {g^{5/2} \text {arctanh}\left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g}}\right )}{a f}+\frac {\left (-a^2+b^2\right )^{3/4} g^{5/2} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{a b^{3/2} f}-\frac {2 g^2 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b f \sqrt {\cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {\left (a^2-b^2\right ) g^3 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^2 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 20.33 (sec) , antiderivative size = 484, normalized size of antiderivative = 1.14 \[ \int \frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=\frac {(g \cos (e+f x))^{5/2} \csc (e+f x) \left (8 a b^{5/2} \operatorname {AppellF1}\left (\frac {7}{4},\frac {1}{2},1,\frac {11}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {7}{2}}(e+f x)+7 \left (a^2-b^2\right ) \left (-2 \sqrt {2} \left (a^2-b^2\right )^{3/4} \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+2 \sqrt {2} \left (a^2-b^2\right )^{3/4} \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )+4 b^{3/2} \arctan \left (\sqrt {\cos (e+f x)}\right )+2 b^{3/2} \log \left (1-\sqrt {\cos (e+f x)}\right )-2 b^{3/2} \log \left (1+\sqrt {\cos (e+f x)}\right )+\sqrt {2} \left (a^2-b^2\right )^{3/4} \log \left (\sqrt {a^2-b^2}-\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )-\sqrt {2} \left (a^2-b^2\right )^{3/4} \log \left (\sqrt {a^2-b^2}+\sqrt {2} \sqrt {b} \sqrt [4]{a^2-b^2} \sqrt {\cos (e+f x)}+b \cos (e+f x)\right )\right )\right ) \left (a+b \sqrt {\sin ^2(e+f x)}\right )}{28 a b^{3/2} \left (a^2-b^2\right ) f \cos ^{\frac {5}{2}}(e+f x) (b+a \csc (e+f x))} \]

[In]

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

[Out]

((g*Cos[e + f*x])^(5/2)*Csc[e + f*x]*(8*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) + 7*(a^2 - b^2)*(-2*Sqrt[2]*(a^2 - b^2)^(3/4)*ArcTan[1 - (Sqrt[2]*Sqrt[b
]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] + 2*Sqrt[2]*(a^2 - b^2)^(3/4)*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e
+ f*x]])/(a^2 - b^2)^(1/4)] + 4*b^(3/2)*ArcTan[Sqrt[Cos[e + f*x]]] + 2*b^(3/2)*Log[1 - Sqrt[Cos[e + f*x]]] - 2
*b^(3/2)*Log[1 + Sqrt[Cos[e + f*x]]] + Sqrt[2]*(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]] - Sqrt[2]*(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]]))*(a + b*Sqrt[Sin[e + f*x]^2]))/(28*a*b^(3/2)*(a^2
- b^2)*f*Cos[e + f*x]^(5/2)*(b + a*Csc[e + f*x]))

Maple [A] (verified)

Time = 5.50 (sec) , antiderivative size = 255, normalized size of antiderivative = 0.60

method	result	size
default	\(-\frac {g^{\frac {3}{2}} \left (8 \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}\, \sqrt {-g}\, \sqrt {g}\, \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+6 \ln \left (\frac {2 \sqrt {-g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}-2 g}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) g^{\frac {3}{2}}-4 \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}\, \sqrt {-g}\, \sqrt {g}+3 \ln \left (\frac {4 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )+2 \sqrt {g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}-2 g}{-1+\cos \left (\frac {f x}{2}+\frac {e}{2}\right )}\right ) g \sqrt {-g}+3 \ln \left (-\frac {2 \left (2 g \cos \left (\frac {f x}{2}+\frac {e}{2}\right )-\sqrt {g}\, \sqrt {-2 g \left (\sin ^{2}\left (\frac {f x}{2}+\frac {e}{2}\right )\right )+g}+g \right )}{\cos \left (\frac {f x}{2}+\frac {e}{2}\right )+1}\right ) g \sqrt {-g}\right )}{6 a \sqrt {-g}\, f}\)	\(255\)

[In]

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

[Out]

-1/6*g^(3/2)/a/(-g)^(1/2)*(8*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)*(-g)^(1/2)*g^(1/2)*sin(1/2*f*x+1/2*e)^2+6*ln(
2/cos(1/2*f*x+1/2*e)*((-g)^(1/2)*(-2*g*sin(1/2*f*x+1/2*e)^2+g)^(1/2)-g))*g^(3/2)-4*(-2*g*sin(1/2*f*x+1/2*e)^2+
g)^(1/2)*(-g)^(1/2)*g^(1/2)+3*ln(2/(-1+cos(1/2*f*x+1/2*e))*(2*g*cos(1/2*f*x+1/2*e)+g^(1/2)*(-2*g*sin(1/2*f*x+1
/2*e)^2+g)^(1/2)-g))*g*(-g)^(1/2)+3*ln(-2/(cos(1/2*f*x+1/2*e)+1)*(2*g*cos(1/2*f*x+1/2*e)-g^(1/2)*(-2*g*sin(1/2
*f*x+1/2*e)^2+g)^(1/2)+g))*g*(-g)^(1/2))/f

Fricas [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

\[ \int \frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \csc \left (f x + e\right )}{b \sin \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {5}{2}} \csc \left (f x + e\right )}{b \sin \left (f x + e\right ) + a} \,d x } \]

[In]

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

[Out]

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

Mupad [F(-1)]

Timed out. \[ \int \frac {(g \cos (e+f x))^{5/2} \csc (e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{5/2}}{\sin \left (e+f\,x\right )\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]

[In]

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

[Out]

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