\(\int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx\) [1370]

Optimal result

Integrand size = 33, antiderivative size = 500 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\frac {a^4 \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} \sqrt [4]{-a^2+b^2} f}-\frac {a^4 \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} \sqrt [4]{-a^2+b^2} f}-\frac {2 a^2 (g \cos (e+f x))^{3/2}}{3 b^3 f g}-\frac {2 (g \cos (e+f x))^{3/2}}{3 b f g}+\frac {2 (g \cos (e+f x))^{7/2}}{7 b f g^3}-\frac {2 a^3 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \sqrt {\cos (e+f x)}}-\frac {4 a \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt {\cos (e+f x)}}+\frac {a^5 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^5 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^5 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} (e+f x),2\right )}{b^5 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {2 a (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f g} \]

[Out]

-2/3*a^2*(g*cos(f*x+e))^(3/2)/b^3/f/g-2/3*(g*cos(f*x+e))^(3/2)/b/f/g+2/7*(g*cos(f*x+e))^(7/2)/b/f/g^3+2/5*a*(g
*cos(f*x+e))^(3/2)*sin(f*x+e)/b^2/f/g+a^4*arctan(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))*g^(1/2
)/b^(9/2)/(-a^2+b^2)^(1/4)/f-a^4*arctanh(b^(1/2)*(g*cos(f*x+e))^(1/2)/(-a^2+b^2)^(1/4)/g^(1/2))*g^(1/2)/b^(9/2
)/(-a^2+b^2)^(1/4)/f+a^5*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)/b^5/f/(b-(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)+a^5*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)/b^5/f/(b+(-a^2+b^2)^(1/2))/(g*cos(f*x+e))^(1/2)-2*a^3*(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^4/f/cos(f*x+e)^(1/2)-4/5*a*(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^2/f/cos(f*x+e)^(1/2)

Rubi [A] (verified)

Time = 0.82 (sec) , antiderivative size = 500, normalized size of antiderivative = 1.00, number of steps used = 21, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {2977, 2721, 2719, 2645, 30, 2648, 14, 2780, 2886, 2884, 335, 304, 211, 214} \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {2 a^3 E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{b^4 f \sqrt {\cos (e+f x)}}-\frac {2 a^2 (g \cos (e+f x))^{3/2}}{3 b^3 f g}+\frac {a^5 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^5 f \left (b-\sqrt {b^2-a^2}\right ) \sqrt {g \cos (e+f x)}}+\frac {a^5 g \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} (e+f x),2\right )}{b^5 f \left (\sqrt {b^2-a^2}+b\right ) \sqrt {g \cos (e+f x)}}+\frac {a^4 \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f \sqrt [4]{b^2-a^2}}-\frac {a^4 \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f \sqrt [4]{b^2-a^2}}+\frac {2 a \sin (e+f x) (g \cos (e+f x))^{3/2}}{5 b^2 f g}-\frac {4 a E\left (\left .\frac {1}{2} (e+f x)\right |2\right ) \sqrt {g \cos (e+f x)}}{5 b^2 f \sqrt {\cos (e+f x)}}+\frac {2 (g \cos (e+f x))^{7/2}}{7 b f g^3}-\frac {2 (g \cos (e+f x))^{3/2}}{3 b f g} \]

[In]

Int[(Sqrt[g*Cos[e + f*x]]*Sin[e + f*x]^4)/(a + b*Sin[e + f*x]),x]

[Out]

(a^4*Sqrt[g]*ArcTan[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(9/2)*(-a^2 + b^2)^(1/4)*
f) - (a^4*Sqrt[g]*ArcTanh[(Sqrt[b]*Sqrt[g*Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*Sqrt[g])])/(b^(9/2)*(-a^2 + b^2)^
(1/4)*f) - (2*a^2*(g*Cos[e + f*x])^(3/2))/(3*b^3*f*g) - (2*(g*Cos[e + f*x])^(3/2))/(3*b*f*g) + (2*(g*Cos[e + f
*x])^(7/2))/(7*b*f*g^3) - (2*a^3*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(b^4*f*Sqrt[Cos[e + f*x]]) -
(4*a*Sqrt[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(5*b^2*f*Sqrt[Cos[e + f*x]]) + (a^5*g*Sqrt[Cos[e + f*x]]*
EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^5*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]])
+ (a^5*g*Sqrt[Cos[e + f*x]]*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (e + f*x)/2, 2])/(b^5*(b + Sqrt[-a^2 + b^
2])*f*Sqrt[g*Cos[e + f*x]]) + (2*a*(g*Cos[e + f*x])^(3/2)*Sin[e + f*x])/(5*b^2*f*g)

Rule 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 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 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 2648

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-a)*(b*Cos[e
+ f*x])^(n + 1)*((a*Sin[e + f*x])^(m - 1)/(b*f*(m + n))), x] + Dist[a^2*((m - 1)/(m + n)), Int[(b*Cos[e + f*x
])^n*(a*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && NeQ[m + n, 0] && IntegersQ[
2*m, 2*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 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 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 {a^3 \sqrt {g \cos (e+f x)}}{b^4}+\frac {a^2 \sqrt {g \cos (e+f x)} \sin (e+f x)}{b^3}-\frac {a \sqrt {g \cos (e+f x)} \sin ^2(e+f x)}{b^2}+\frac {\sqrt {g \cos (e+f x)} \sin ^3(e+f x)}{b}+\frac {a^4 \sqrt {g \cos (e+f x)}}{b^4 (a+b \sin (e+f x))}\right ) \, dx \\ & = -\frac {a^3 \int \sqrt {g \cos (e+f x)} \, dx}{b^4}+\frac {a^4 \int \frac {\sqrt {g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx}{b^4}+\frac {a^2 \int \sqrt {g \cos (e+f x)} \sin (e+f x) \, dx}{b^3}-\frac {a \int \sqrt {g \cos (e+f x)} \sin ^2(e+f x) \, dx}{b^2}+\frac {\int \sqrt {g \cos (e+f x)} \sin ^3(e+f x) \, dx}{b} \\ & = \frac {2 a (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f g}-\frac {(2 a) \int \sqrt {g \cos (e+f x)} \, dx}{5 b^2}-\frac {a^2 \text {Subst}\left (\int \sqrt {x} \, dx,x,g \cos (e+f x)\right )}{b^3 f g}-\frac {\text {Subst}\left (\int \sqrt {x} \left (1-\frac {x^2}{g^2}\right ) \, dx,x,g \cos (e+f x)\right )}{b f g}-\frac {\left (a^5 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 b^5}+\frac {\left (a^5 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 b^5}+\frac {\left (a^4 g\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 )}{b^3 f}-\frac {\left (a^3 \sqrt {g \cos (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \, dx}{b^4 \sqrt {\cos (e+f x)}} \\ & = -\frac {2 a^2 (g \cos (e+f x))^{3/2}}{3 b^3 f g}-\frac {2 a^3 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \sqrt {\cos (e+f x)}}+\frac {2 a (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f g}-\frac {\text {Subst}\left (\int \left (\sqrt {x}-\frac {x^{5/2}}{g^2}\right ) \, dx,x,g \cos (e+f x)\right )}{b f g}+\frac {\left (2 a^4 g\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 )}{b^3 f}-\frac {\left (a^5 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 b^5 \sqrt {g \cos (e+f x)}}+\frac {\left (a^5 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 b^5 \sqrt {g \cos (e+f x)}}-\frac {\left (2 a \sqrt {g \cos (e+f x)}\right ) \int \sqrt {\cos (e+f x)} \, dx}{5 b^2 \sqrt {\cos (e+f x)}} \\ & = -\frac {2 a^2 (g \cos (e+f x))^{3/2}}{3 b^3 f g}-\frac {2 (g \cos (e+f x))^{3/2}}{3 b f g}+\frac {2 (g \cos (e+f x))^{7/2}}{7 b f g^3}-\frac {2 a^3 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \sqrt {\cos (e+f x)}}-\frac {4 a \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt {\cos (e+f x)}}+\frac {a^5 g \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^5 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^5 g \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^5 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {2 a (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f g}-\frac {\left (a^4 g\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 )}{b^4 f}+\frac {\left (a^4 g\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 )}{b^4 f} \\ & = \frac {a^4 \sqrt {g} \arctan \left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} \sqrt [4]{-a^2+b^2} f}-\frac {a^4 \sqrt {g} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt {g}}\right )}{b^{9/2} \sqrt [4]{-a^2+b^2} f}-\frac {2 a^2 (g \cos (e+f x))^{3/2}}{3 b^3 f g}-\frac {2 (g \cos (e+f x))^{3/2}}{3 b f g}+\frac {2 (g \cos (e+f x))^{7/2}}{7 b f g^3}-\frac {2 a^3 \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{b^4 f \sqrt {\cos (e+f x)}}-\frac {4 a \sqrt {g \cos (e+f x)} E\left (\left .\frac {1}{2} (e+f x)\right |2\right )}{5 b^2 f \sqrt {\cos (e+f x)}}+\frac {a^5 g \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^5 \left (b-\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {a^5 g \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^5 \left (b+\sqrt {-a^2+b^2}\right ) f \sqrt {g \cos (e+f x)}}+\frac {2 a (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f g} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

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

Time = 20.70 (sec) , antiderivative size = 816, normalized size of antiderivative = 1.63 \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=-\frac {a \sqrt {g \cos (e+f x)} \left (-\frac {4 a b \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {a \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)}{3 \left (a^2-b^2\right )}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) \left (2 \arctan \left (1-\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2}}\right )-\log \left (\sqrt {-a^2+b^2}-(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )+\log \left (\sqrt {-a^2+b^2}+(1+i) \sqrt {b} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}+i b \cos (e+f x)\right )\right )}{\sqrt {b} \sqrt [4]{-a^2+b^2}}\right ) \sin (e+f x)}{\sqrt {1-\cos ^2(e+f x)} (a+b \sin (e+f x))}-\frac {\left (5 a^2+2 b^2\right ) \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (8 b^{5/2} \operatorname {AppellF1}\left (\frac {3}{4},-\frac {1}{2},1,\frac {7}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \cos ^{\frac {3}{2}}(e+f x)+3 \sqrt {2} a \left (a^2-b^2\right )^{3/4} \left (2 \arctan \left (1-\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \arctan \left (1+\frac {\sqrt {2} \sqrt {b} \sqrt {\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-\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 )+\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 ) \sin ^2(e+f x)}{12 b^{3/2} \left (-a^2+b^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}\right )}{5 b^3 f \sqrt {\cos (e+f x)}}+\frac {\sqrt {g \cos (e+f x)} \left (-\frac {\left (28 a^2+19 b^2\right ) \cos (e+f x)}{42 b^3}+\frac {\cos (3 (e+f x))}{14 b}+\frac {a \sin (2 (e+f x))}{5 b^2}\right )}{f} \]

[In]

Integrate[(Sqrt[g*Cos[e + f*x]]*Sin[e + f*x]^4)/(a + b*Sin[e + f*x]),x]

[Out]

-1/5*(a*Sqrt[g*Cos[e + f*x]]*((-4*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]]))/(Sq
rt[b]*(-a^2 + b^2)^(1/4)))*Sin[e + f*x])/(Sqrt[1 - Cos[e + f*x]^2]*(a + b*Sin[e + f*x])) - ((5*a^2 + 2*b^2)*(a
+ b*Sqrt[1 - Cos[e + f*x]^2])*(8*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) + 3*Sqrt[2]*a*(a^2 - b^2)^(3/4)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]
])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[Cos[e + f*x]])/(a^2 - b^2)^(1/4)] - 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]] + 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]]))*Sin[e + f*x]^2)/(12*b^(3/2)*(-a^2 + b^2)*(1 -
Cos[e + f*x]^2)*(a + b*Sin[e + f*x]))))/(b^3*f*Sqrt[Cos[e + f*x]]) + (Sqrt[g*Cos[e + f*x]]*(-1/42*((28*a^2 +
19*b^2)*Cos[e + f*x])/b^3 + Cos[3*(e + f*x)]/(14*b) + (a*Sin[2*(e + f*x)])/(5*b^2)))/f

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 4.66 (sec) , antiderivative size = 1405, normalized size of antiderivative = 2.81

method	result	size
default	\(\text {Expression too large to display}\)	\(1405\)

[In]

int(sin(f*x+e)^4*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x,method=_RETURNVERBOSE)

[Out]

(1/4/b^5*g*a^4/(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*ln((2*g*cos(1/2*f*x+1/2*e)^2-g-(g^2*(a^2-b^2)/b^2)^(1/4)*(2*g
*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2))/(2*g*cos(1/2*f*x+1/2*e)^2-g+(g^2*(a^2-b^2)/b
^2)^(1/4)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*2^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/2)))+1/2/b^5*g*a^4/(g^2*(a^2-b^2)/
b^2)^(1/4)*2^(1/2)*arctan((2^(1/2)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+(g^2*(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2
)/b^2)^(1/4))+1/2/b^5*g*a^4/(g^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*arctan((2^(1/2)*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/
2)-(g^2*(a^2-b^2)/b^2)^(1/4))/(g^2*(a^2-b^2)/b^2)^(1/4))+2/b^3*a^2*(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)+16/7/b
*cos(1/2*f*x+1/2*e)^6*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)-24/7/b*cos(1/2*f*x+1/2*e)^4*(2*g*cos(1/2*f*x+1/2*e)^2
-g)^(1/2)+8/21/b*cos(1/2*f*x+1/2*e)^2*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)+8/21/b*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(
1/2)-4/3/b^3*cos(1/2*f*x+1/2*e)^2*(2*g*cos(1/2*f*x+1/2*e)^2-g)^(1/2)*a^2-4/3/b^3*(2*g*cos(1/2*f*x+1/2*e)^2-g)^
(1/2)*a^2-1/40*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*a*g*(-128*b^4*cos(1/2*f*x+1/2*e)*sin(
1/2*f*x+1/2*e)^6+128*b^4*sin(1/2*f*x+1/2*e)^4*cos(1/2*f*x+1/2*e)-32*b^4*sin(1/2*f*x+1/2*e)^2*cos(1/2*f*x+1/2*e
)+80*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*a^2*b
^2+32*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2*e)^2-1)^(1/2)*EllipticE(cos(1/2*f*x+1/2*e),2^(1/2))*b^4+
5*a^2*sum(1/_alpha*(8*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(2*sin(1/2*f*x+1/2
*e)^2-1)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),(-4*_alpha^2*b^2+4*b^2)/a^2,2^(1/2))*_alpha^3*b^2-8*b^2*_alpha*(s
in(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)-a^2*2^(1/2)*arctanh(1/2/(g*(2*_alpha^2*b^2+a^2-2*b^2
)/b^2)^(1/2)/(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^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*_alpha^2+3*a^2-3*b^2))*(g*sin(1/2*f*x+1/2*e)^2*(-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)/(g*sin(1/2*f*x+1/2*e)^2*(-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))*(-2*g*sin(1/2*f*x+1/2*e)^4+g*sin(1/2*f*x+1/2*e)^2)^(1/2))/b^6/(-g
*(2*sin(1/2*f*x+1/2*e)^4-sin(1/2*f*x+1/2*e)^2))^(1/2)/sin(1/2*f*x+1/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2))
/f

Fricas [F(-1)]

Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)^4*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="fricas")

[Out]

Timed out

Sympy [F(-1)]

Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)**4*(g*cos(f*x+e))**(1/2)/(a+b*sin(f*x+e)),x)

[Out]

Timed out

Maxima [F]

\[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )^{4}}{b \sin \left (f x + e\right ) + a} \,d x } \]

[In]

integrate(sin(f*x+e)^4*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="maxima")

[Out]

integrate(sqrt(g*cos(f*x + e))*sin(f*x + e)^4/(b*sin(f*x + e) + a), x)

Giac [F]

\[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )} \sin \left (f x + e\right )^{4}}{b \sin \left (f x + e\right ) + a} \,d x } \]

[In]

integrate(sin(f*x+e)^4*(g*cos(f*x+e))^(1/2)/(a+b*sin(f*x+e)),x, algorithm="giac")

[Out]

integrate(sqrt(g*cos(f*x + e))*sin(f*x + e)^4/(b*sin(f*x + e) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx=\int \frac {{\sin \left (e+f\,x\right )}^4\,\sqrt {g\,\cos \left (e+f\,x\right )}}{a+b\,\sin \left (e+f\,x\right )} \,d x \]

[In]

int((sin(e + f*x)^4*(g*cos(e + f*x))^(1/2))/(a + b*sin(e + f*x)),x)

[Out]

int((sin(e + f*x)^4*(g*cos(e + f*x))^(1/2))/(a + b*sin(e + f*x)), x)