3.1384 \(\int \frac{(g \cos (e+f x))^{5/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx\)

Optimal. Leaf size=501 \[ \frac{a^2 g^{5/2} \left (b^2-a^2\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f}-\frac{a^2 g^{5/2} \left (b^2-a^2\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f}+\frac{2 a^3 g^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{b^4 f \sqrt{\cos (e+f x)}}-\frac{a^3 g^3 \left (a^2-b^2\right ) \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^5 f \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}-\frac{a^3 g^3 \left (a^2-b^2\right ) \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^5 f \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}+\frac{2 a^2 g (g \cos (e+f x))^{3/2}}{3 b^3 f}-\frac{6 a g^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 b^2 f \sqrt{\cos (e+f x)}}-\frac{2 a g \sin (e+f x) (g \cos (e+f x))^{3/2}}{5 b^2 f}-\frac{2 (g \cos (e+f x))^{7/2}}{7 b f g} \]

[Out]

(a^2*(-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])])/(b^(9/2)*
f) - (a^2*(-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])])/(b^
(9/2)*f) + (2*a^2*g*(g*Cos[e + f*x])^(3/2))/(3*b^3*f) - (2*(g*Cos[e + f*x])^(7/2))/(7*b*f*g) + (2*a^3*g^2*Sqrt
[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(b^4*f*Sqrt[Cos[e + f*x]]) - (6*a*g^2*Sqrt[g*Cos[e + f*x]]*Ellipti
cE[(e + f*x)/2, 2])/(5*b^2*f*Sqrt[Cos[e + f*x]]) - (a^3*(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^5*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - (a^3*(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^5*(b + Sqrt[-a^2 + b^2])*f
*Sqrt[g*Cos[e + f*x]]) - (2*a*g*(g*Cos[e + f*x])^(3/2)*Sin[e + f*x])/(5*b^2*f)

________________________________________________________________________________________

Rubi [A]  time = 1.19368, antiderivative size = 501, normalized size of antiderivative = 1., number of steps used = 20, number of rules used = 15, integrand size = 33, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.454, Rules used = {2898, 2635, 2640, 2639, 2565, 30, 2695, 2867, 2701, 2807, 2805, 329, 298, 205, 208} \[ \frac{a^2 g^{5/2} \left (b^2-a^2\right )^{3/4} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f}-\frac{a^2 g^{5/2} \left (b^2-a^2\right )^{3/4} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt{g} \sqrt [4]{b^2-a^2}}\right )}{b^{9/2} f}+\frac{2 a^3 g^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{b^4 f \sqrt{\cos (e+f x)}}-\frac{a^3 g^3 \left (a^2-b^2\right ) \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^5 f \left (b-\sqrt{b^2-a^2}\right ) \sqrt{g \cos (e+f x)}}-\frac{a^3 g^3 \left (a^2-b^2\right ) \sqrt{\cos (e+f x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} (e+f x)\right |2\right )}{b^5 f \left (\sqrt{b^2-a^2}+b\right ) \sqrt{g \cos (e+f x)}}+\frac{2 a^2 g (g \cos (e+f x))^{3/2}}{3 b^3 f}-\frac{6 a g^2 E\left (\left .\frac{1}{2} (e+f x)\right |2\right ) \sqrt{g \cos (e+f x)}}{5 b^2 f \sqrt{\cos (e+f x)}}-\frac{2 a g \sin (e+f x) (g \cos (e+f x))^{3/2}}{5 b^2 f}-\frac{2 (g \cos (e+f x))^{7/2}}{7 b f g} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(-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])])/(b^(9/2)*
f) - (a^2*(-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])])/(b^
(9/2)*f) + (2*a^2*g*(g*Cos[e + f*x])^(3/2))/(3*b^3*f) - (2*(g*Cos[e + f*x])^(7/2))/(7*b*f*g) + (2*a^3*g^2*Sqrt
[g*Cos[e + f*x]]*EllipticE[(e + f*x)/2, 2])/(b^4*f*Sqrt[Cos[e + f*x]]) - (6*a*g^2*Sqrt[g*Cos[e + f*x]]*Ellipti
cE[(e + f*x)/2, 2])/(5*b^2*f*Sqrt[Cos[e + f*x]]) - (a^3*(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^5*(b - Sqrt[-a^2 + b^2])*f*Sqrt[g*Cos[e + f*x]]) - (a^3*(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^5*(b + Sqrt[-a^2 + b^2])*f
*Sqrt[g*Cos[e + f*x]]) - (2*a*g*(g*Cos[e + f*x])^(3/2)*Sin[e + f*x])/(5*b^2*f)

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 2635

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

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 2695

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(g*(g*
Cos[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 2867

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 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{(g \cos (e+f x))^{5/2} \sin ^2(e+f x)}{a+b \sin (e+f x)} \, dx &=\int \left (-\frac{a (g \cos (e+f x))^{5/2}}{b^2}+\frac{(g \cos (e+f x))^{5/2} \sin (e+f x)}{b}+\frac{a^2 (g \cos (e+f x))^{5/2}}{b^2 (a+b \sin (e+f x))}\right ) \, dx\\ &=-\frac{a \int (g \cos (e+f x))^{5/2} \, dx}{b^2}+\frac{a^2 \int \frac{(g \cos (e+f x))^{5/2}}{a+b \sin (e+f x)} \, dx}{b^2}+\frac{\int (g \cos (e+f x))^{5/2} \sin (e+f x) \, dx}{b}\\ &=\frac{2 a^2 g (g \cos (e+f x))^{3/2}}{3 b^3 f}-\frac{2 a g (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f}-\frac{\operatorname{Subst}\left (\int x^{5/2} \, dx,x,g \cos (e+f x)\right )}{b f g}+\frac{\left (a^2 g^2\right ) \int \frac{\sqrt{g \cos (e+f x)} (b+a \sin (e+f x))}{a+b \sin (e+f x)} \, dx}{b^3}-\frac{\left (3 a g^2\right ) \int \sqrt{g \cos (e+f x)} \, dx}{5 b^2}\\ &=\frac{2 a^2 g (g \cos (e+f x))^{3/2}}{3 b^3 f}-\frac{2 (g \cos (e+f x))^{7/2}}{7 b f g}-\frac{2 a g (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f}+\frac{\left (a^3 g^2\right ) \int \sqrt{g \cos (e+f x)} \, dx}{b^4}-\frac{\left (a^2 \left (a^2-b^2\right ) g^2\right ) \int \frac{\sqrt{g \cos (e+f x)}}{a+b \sin (e+f x)} \, dx}{b^4}-\frac{\left (3 a g^2 \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 (g \cos (e+f x))^{3/2}}{3 b^3 f}-\frac{2 (g \cos (e+f x))^{7/2}}{7 b f g}-\frac{6 a g^2 \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{2 a g (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f}+\frac{\left (a^3 \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^5}-\frac{\left (a^3 \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^5}-\frac{\left (a^2 \left (a^2-b^2\right ) g^3\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 g^2 \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 (g \cos (e+f x))^{3/2}}{3 b^3 f}-\frac{2 (g \cos (e+f x))^{7/2}}{7 b f g}+\frac{2 a^3 g^2 \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{6 a g^2 \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{2 a g (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f}-\frac{\left (2 a^2 \left (a^2-b^2\right ) g^3\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^3 \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^5 \sqrt{g \cos (e+f x)}}-\frac{\left (a^3 \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^5 \sqrt{g \cos (e+f x)}}\\ &=\frac{2 a^2 g (g \cos (e+f x))^{3/2}}{3 b^3 f}-\frac{2 (g \cos (e+f x))^{7/2}}{7 b f g}+\frac{2 a^3 g^2 \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{6 a g^2 \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^3 \left (a^2-b^2\right ) g^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 )}{b^5 \left (b-\sqrt{-a^2+b^2}\right ) f \sqrt{g \cos (e+f x)}}-\frac{a^3 \left (a^2-b^2\right ) g^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 )}{b^5 \left (b+\sqrt{-a^2+b^2}\right ) f \sqrt{g \cos (e+f x)}}-\frac{2 a g (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f}+\frac{\left (a^2 \left (a^2-b^2\right ) g^3\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^2 \left (a^2-b^2\right ) g^3\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^2 \left (-a^2+b^2\right )^{3/4} g^{5/2} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt{g}}\right )}{b^{9/2} f}-\frac{a^2 \left (-a^2+b^2\right )^{3/4} g^{5/2} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{g \cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt{g}}\right )}{b^{9/2} f}+\frac{2 a^2 g (g \cos (e+f x))^{3/2}}{3 b^3 f}-\frac{2 (g \cos (e+f x))^{7/2}}{7 b f g}+\frac{2 a^3 g^2 \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{6 a g^2 \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^3 \left (a^2-b^2\right ) g^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 )}{b^5 \left (b-\sqrt{-a^2+b^2}\right ) f \sqrt{g \cos (e+f x)}}-\frac{a^3 \left (a^2-b^2\right ) g^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 )}{b^5 \left (b+\sqrt{-a^2+b^2}\right ) f \sqrt{g \cos (e+f x)}}-\frac{2 a g (g \cos (e+f x))^{3/2} \sin (e+f x)}{5 b^2 f}\\ \end{align*}

Mathematica [C]  time = 27.1328, size = 824, normalized size = 1.64 \[ \frac{a \left (-\frac{\left (5 a^2-3 b^2\right ) \left (a+b \sqrt{1-\cos ^2(e+f x)}\right ) \left (8 F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right ) \cos ^{\frac{3}{2}}(e+f x) b^{5/2}+3 \sqrt{2} a \left (a^2-b^2\right )^{3/4} \left (2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (b \cos (e+f x)-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}\right )+\log \left (b \cos (e+f x)+\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\cos (e+f x)}+\sqrt{a^2-b^2}\right )\right )\right ) \sin ^2(e+f x)}{12 b^{3/2} \left (b^2-a^2\right ) \left (1-\cos ^2(e+f x)\right ) (a+b \sin (e+f x))}-\frac{4 a b \left (a+b \sqrt{1-\cos ^2(e+f x)}\right ) \left (\frac{a F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\cos ^2(e+f x),\frac{b^2 \cos ^2(e+f x)}{b^2-a^2}\right ) \cos ^{\frac{3}{2}}(e+f x)}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac{(1+i) \sqrt{b} \sqrt{\cos (e+f x)}}{\sqrt [4]{b^2-a^2}}+1\right )-\log \left (i b \cos (e+f x)-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (e+f x)}+\sqrt{b^2-a^2}\right )+\log \left (i b \cos (e+f x)+(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\cos (e+f x)}+\sqrt{b^2-a^2}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right ) \sin (e+f x)}{\sqrt{1-\cos ^2(e+f x)} (a+b \sin (e+f x))}\right ) (g \cos (e+f x))^{5/2}}{5 b^3 f \cos ^{\frac{5}{2}}(e+f x)}+\frac{\sec ^2(e+f x) \left (-\frac{\left (9 b^2-28 a^2\right ) \cos (e+f x)}{42 b^3}-\frac{\cos (3 (e+f x))}{14 b}-\frac{a \sin (2 (e+f x))}{5 b^2}\right ) (g \cos (e+f x))^{5/2}}{f} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*(g*Cos[e + f*x])^(5/2)*((-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 - 3*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]*Sqr
t[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 - Co
s[e + f*x]^2)*(a + b*Sin[e + f*x]))))/(5*b^3*f*Cos[e + f*x]^(5/2)) + ((g*Cos[e + f*x])^(5/2)*Sec[e + f*x]^2*(-
((-28*a^2 + 9*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 = 6.252, size = 1937, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/7/f*g^2/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*g^2/b*cos(1/2*f*x+1/2*e)^4*(2*cos
(1/2*f*x+1/2*e)^2*g-g)^(1/2)-12/7/f*g^2/b*cos(1/2*f*x+1/2*e)^2*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2)-12/7/f*g^2/b
*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2)+4/3/f*g^2/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*g^2/b^3*(2*cos(1/2*f*x+1/2*e)^2*g-g)^(1/2)*a^2-2/f*g^2/b^3*(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*a^2+2/f*
g^2/b*(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)-1/2/f*g^3/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))+1/2/f*g^3
/b*a^2*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
)*g^3*a/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)^7-32/5/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*g^3*a/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)^5+4/f*(g*(2*cos(1/2*f*x+1/2*e)^2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*g^3*a/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)^3+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)*g^3*a^3/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)*(sin(1/2*f*x+1/2*e)^2)
^(1/2)*(-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))-6/5/f*(g*(2*cos(1/2*f*x+1/2*e)^
2-1)*sin(1/2*f*x+1/2*e)^2)^(1/2)*g^3*a/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*cos(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)*g^3*a/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)+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)*g^3*a/b^6/sin(1/2*f*x+1
/2*e)/(g*(2*cos(1/2*f*x+1/2*e)^2-1))^(1/2)*sum((a^2-b^2)/_alpha*(8*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)*(s
in(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(1/2)*EllipticPi(cos(1/2*f*x+1/2*e),-4*b^2/a^2*(_alpha^
2-1),2^(1/2))*_alpha^3*b^2-8*b^2*_alpha*(sin(1/2*f*x+1/2*e)^2)^(1/2)*(-2*cos(1/2*f*x+1/2*e)^2+1)^(1/2)*Ellipti
cPi(cos(1/2*f*x+1/2*e),-4*b^2/a^2*(_alpha^2-1),2^(1/2))*(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)+2^(1/2)*a^2*a
rctanh(1/2*g*(4*_alpha^2-3)/(4*a^2-3*b^2)*(4*cos(1/2*f*x+1/2*e)^2*a^2-3*b^2*cos(1/2*f*x+1/2*e)^2+b^2*_alpha^2-
3*a^2+2*b^2)*2^(1/2)/(g*(2*_alpha^2*b^2+a^2-2*b^2)/b^2)^(1/2)/(-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)^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))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (g \cos \left (f x + e\right )\right )^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{b \sin \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((g*cos(f*x + e))^(5/2)*sin(f*x + e)^2/(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((g*cos(f*x+e))^(5/2)*sin(f*x+e)^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((g*cos(f*x+e))**(5/2)*sin(f*x+e)**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{\left (g \cos \left (f x + e\right )\right )^{\frac{5}{2}} \sin \left (f x + e\right )^{2}}{b \sin \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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