3.1370 \(\int \frac{\sqrt{g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=500 \[ -\frac{2 a^2 (g \cos (e+f x))^{3/2}}{3 b^3 f g}+\frac{a^4 \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f \sqrt [4]{b^2-a^2}}-\frac{a^4 \sqrt{g} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f \sqrt [4]{b^2-a^2}}-\frac{2 a^3 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{b^4 f \sqrt{\cos (e+f x)}}+\frac{a^5 g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^5 f \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}+\frac{a^5 g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^5 f \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}+\frac{2 a \sin (e+f x) (g \cos (e+f x))^{3/2}}{5 b^2 f g}-\frac{4 a E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 b^2 f \sqrt{\cos (e+f x)}}+\frac{2 (g \cos (e+f x))^{7/2}}{7 b f g^3}-\frac{2 (g \cos (e+f x))^{3/2}}{3 b f g} \]

[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)

________________________________________________________________________________________

Rubi [A]  time = 1.23742, antiderivative size = 500, normalized size of antiderivative = 1., number of steps used = 21, number of rules used = 14, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.424, Rules used = {2898, 2640, 2639, 2565, 30, 2568, 14, 2701, 2807, 2805, 329, 298, 205, 208} \[ -\frac{2 a^2 (g \cos (e+f x))^{3/2}}{3 b^3 f g}+\frac{a^4 \sqrt{g} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f \sqrt [4]{b^2-a^2}}-\frac{a^4 \sqrt{g} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f \sqrt [4]{b^2-a^2}}-\frac{2 a^3 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{b^4 f \sqrt{\cos (e+f x)}}+\frac{a^5 g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^5 f \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}+\frac{a^5 g \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^5 f \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}+\frac{2 a \sin (e+f x) (g \cos (e+f x))^{3/2}}{5 b^2 f g}-\frac{4 a E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 b^2 f \sqrt{\cos (e+f x)}}+\frac{2 (g \cos (e+f x))^{7/2}}{7 b f g^3}-\frac{2 (g \cos (e+f x))^{3/2}}{3 b f g} \]

Antiderivative was successfully verified.

[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 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 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 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 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 30

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

Rule 2568

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 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 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 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 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 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 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 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]

Rubi steps

\begin{align*} \int \frac{\sqrt{g \cos (e+f x)} \sin ^4(e+f x)}{a+b \sin (e+f x)} \, dx &=\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 \operatorname{Subst}\left (\int \sqrt{x} \, dx,x,g \cos (e+f x)\right )}{b^3 f g}-\frac{\operatorname{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 ) \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 )}{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{\operatorname{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 ) \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 )}{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)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (e+f x)\right |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)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (e+f x)\right |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 ) \operatorname{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 ) \operatorname{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} \tan ^{-1}\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} \tanh ^{-1}\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)} \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} (e+f x)\right |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)} \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} (e+f x)\right |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]  time = 26.9448, size = 816, normalized size = 1.63 \[ \frac{\sqrt{g \cos (e+f x)} \left (-\frac{\left (28 a^2+19 b^2\right ) \cos (e+f x)}{42 b^3}+\frac{\cos (3 (e+f x))}{14 b}+\frac{a \sin (2 (e+f x))}{5 b^2}\right )}{f}-\frac{a \sqrt{g \cos (e+f x)} \left (-\frac{\left (5 a^2+2 b^2\right ) \left (a+b \sqrt{1-\cos ^2(e+f x)}\right ) \left (8 F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right ) \cos ^{\frac{3}{2}}(e+f x) b^{5/2}+3 \sqrt{2} a \left (a^2-b^2\right )^{3/4} \left (2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (b \cos (e+f x)-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}\right )+\log \left (b \cos (e+f x)+\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}\right )\right )\right ) \sin ^2(e+f x)}{12 b^{3/2} \left (b^2-a^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}-\frac{4 a b \left (a+b \sqrt{1-\cos ^2(e+f x)}\right ) \left (\frac{a F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right ) \cos ^{\frac{3}{2}}(e+f x)}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac{(1+i) \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{b^2-a^2}}+1\right )-\log \left (i b \cos (e+f x)-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (e+f x)}+\sqrt{b^2-a^2}\right )+\log \left (i b \cos (e+f x)+(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (e+f x)}+\sqrt{b^2-a^2}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right ) \sin (e+f x)}{\sqrt{1-\cos ^2(e+f x)} (a+b \sin (e+f x))}\right )}{5 b^3 f \sqrt{\cos (e+f x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-(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]]))/(Sqrt[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]))))/(5*b^3*f*Sqrt[Cos[e + f*x]]) + (Sqrt[g*Cos[e + f*x]]*(-((28*a^2 + 19*b^2)
*Cos[e + f*x])/(42*b^3) + Cos[3*(e + f*x)]/(14*b) + (a*Sin[2*(e + f*x)])/(5*b^2)))/f

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Maple [C]  time = 5.607, size = 1674, normalized size = 3.4 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

16/7/f/b*cos(1/2*f*x+1/2*e)^6*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2)-24/7/f/b*cos(1/2*f*x+1/2*e)^4*(2*cos(1/2*f*x+
1/2*e)^2*g-g)^(1/2)+8/21/f/b*cos(1/2*f*x+1/2*e)^2*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2)+8/21/f/b*(2*cos(1/2*f*x+1
/2*e)^2*g-g)^(1/2)-4/3/f/b^3*cos(1/2*f*x+1/2*e)^2*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2)*a^2-4/3/f/b^3*(2*cos(1/2*
f*x+1/2*e)^2*g-g)^(1/2)*a^2+2/f/b^3*a^2*(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)+1/2/f*g/b^3*a^4*sum((_R^6-_R^4*g-
_R^2*g^2+g^3)/(_R^7*b^2-3*_R^5*b^2*g+8*_R^3*a^2*g^2-5*_R^3*b^2*g^2-_R*b^2*g^3)*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=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))+16/5/f*(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/b^2/(-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)^5/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*cos(1/2*f
*x+1/2*e)-16/5/f*(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/b^2/(-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)^3/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*cos(1/2*f*x+1/2*e)+4/5
/f*(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/b^2/(-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)*cos(1/2*f*x+1/2*e)-2/f*(g*(2*cos(1/2*f
*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*a^3*g/b^4/(-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)*EllipticE(cos(1/2*f*x+1/2*e),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)-4/5/f*(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/b^2/(-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)*EllipticE(cos(1/2*f*x+1/2*e),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)
-1/8/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*a^3*g/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)*sum(1/_alpha*(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*_al
pha^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*_alpha^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))*(-2*sin(1/2*f*x+1/2*e)^4*g+sin(1/2*f*x+1/2*e)^2*g)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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

[Out]

Timed out

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)