3.84 \(\int \frac{\sqrt{e \sin (c+d x)}}{(a+b \cos (c+d x))^3} \, dx\)

Optimal. Leaf size=529 \[ -\frac{\sqrt{e} \left (3 a^2+2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 \sqrt{b} d \left (b^2-a^2\right )^{9/4}}+\frac{\sqrt{e} \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 \sqrt{b} d \left (b^2-a^2\right )^{9/4}}-\frac{5 a b (e \sin (c+d x))^{3/2}}{4 d e \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{5 a E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{4 d \left (a^2-b^2\right )^2 \sqrt{\sin (c+d x)}}+\frac{a e \left (3 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b d \left (a^2-b^2\right )^2 \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \sin (c+d x)}}+\frac{a e \left (3 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b d \left (a^2-b^2\right )^2 \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \sin (c+d x)}} \]

[Out]

-((3*a^2 + 2*b^2)*Sqrt[e]*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(8*Sqrt[b]*(-a^
2 + b^2)^(9/4)*d) + ((3*a^2 + 2*b^2)*Sqrt[e]*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e
])])/(8*Sqrt[b]*(-a^2 + b^2)^(9/4)*d) + (a*(3*a^2 + 2*b^2)*e*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/
2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b*(a^2 - b^2)^2*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (a*(3*a
^2 + 2*b^2)*e*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b*(a^2 -
b^2)^2*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (5*a*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d
*x]])/(4*(a^2 - b^2)^2*d*Sqrt[Sin[c + d*x]]) - (b*(e*Sin[c + d*x])^(3/2))/(2*(a^2 - b^2)*d*e*(a + b*Cos[c + d*
x])^2) - (5*a*b*(e*Sin[c + d*x])^(3/2))/(4*(a^2 - b^2)^2*d*e*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.21165, antiderivative size = 529, normalized size of antiderivative = 1., number of steps used = 14, number of rules used = 12, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.48, Rules used = {2694, 2864, 2867, 2640, 2639, 2701, 2807, 2805, 329, 298, 205, 208} \[ -\frac{\sqrt{e} \left (3 a^2+2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 \sqrt{b} d \left (b^2-a^2\right )^{9/4}}+\frac{\sqrt{e} \left (3 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 \sqrt{b} d \left (b^2-a^2\right )^{9/4}}-\frac{5 a b (e \sin (c+d x))^{3/2}}{4 d e \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{b (e \sin (c+d x))^{3/2}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}+\frac{5 a E\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{4 d \left (a^2-b^2\right )^2 \sqrt{\sin (c+d x)}}+\frac{a e \left (3 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b d \left (a^2-b^2\right )^2 \left (b-\sqrt{b^2-a^2}\right ) \sqrt{e \sin (c+d x)}}+\frac{a e \left (3 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 b d \left (a^2-b^2\right )^2 \left (\sqrt{b^2-a^2}+b\right ) \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[e*Sin[c + d*x]]/(a + b*Cos[c + d*x])^3,x]

[Out]

-((3*a^2 + 2*b^2)*Sqrt[e]*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(8*Sqrt[b]*(-a^
2 + b^2)^(9/4)*d) + ((3*a^2 + 2*b^2)*Sqrt[e]*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e
])])/(8*Sqrt[b]*(-a^2 + b^2)^(9/4)*d) + (a*(3*a^2 + 2*b^2)*e*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/
2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b*(a^2 - b^2)^2*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (a*(3*a
^2 + 2*b^2)*e*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*b*(a^2 -
b^2)^2*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) + (5*a*EllipticE[(c - Pi/2 + d*x)/2, 2]*Sqrt[e*Sin[c + d
*x]])/(4*(a^2 - b^2)^2*d*Sqrt[Sin[c + d*x]]) - (b*(e*Sin[c + d*x])^(3/2))/(2*(a^2 - b^2)*d*e*(a + b*Cos[c + d*
x])^2) - (5*a*b*(e*Sin[c + d*x])^(3/2))/(4*(a^2 - b^2)^2*d*e*(a + b*Cos[c + d*x]))

Rule 2694

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

Rule 2864

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

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 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 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{e \sin (c+d x)}}{(a+b \cos (c+d x))^3} \, dx &=-\frac{b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{\int \frac{\left (-2 a+\frac{1}{2} b \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{(a+b \cos (c+d x))^2} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac{b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}+\frac{\int \frac{\left (\frac{1}{2} \left (4 a^2+b^2\right )+\frac{5}{4} a b \cos (c+d x)\right ) \sqrt{e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}+\frac{(5 a) \int \sqrt{e \sin (c+d x)} \, dx}{8 \left (a^2-b^2\right )^2}+\frac{\left (3 a^2+2 b^2\right ) \int \frac{\sqrt{e \sin (c+d x)}}{a+b \cos (c+d x)} \, dx}{8 \left (a^2-b^2\right )^2}\\ &=-\frac{b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}-\frac{\left (a \left (3 a^2+2 b^2\right ) e\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{16 b \left (a^2-b^2\right )^2}+\frac{\left (a \left (3 a^2+2 b^2\right ) e\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{16 b \left (a^2-b^2\right )^2}-\frac{\left (b \left (3 a^2+2 b^2\right ) e\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{\left (a^2-b^2\right ) e^2+b^2 x^2} \, dx,x,e \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}+\frac{\left (5 a \sqrt{e \sin (c+d x)}\right ) \int \sqrt{\sin (c+d x)} \, dx}{8 \left (a^2-b^2\right )^2 \sqrt{\sin (c+d x)}}\\ &=\frac{5 a E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt{\sin (c+d x)}}-\frac{b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}-\frac{\left (b \left (3 a^2+2 b^2\right ) e\right ) \operatorname{Subst}\left (\int \frac{x^2}{\left (a^2-b^2\right ) e^2+b^2 x^4} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{4 \left (a^2-b^2\right )^2 d}-\frac{\left (a \left (3 a^2+2 b^2\right ) e \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)} \left (\sqrt{-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{16 b \left (a^2-b^2\right )^2 \sqrt{e \sin (c+d x)}}+\frac{\left (a \left (3 a^2+2 b^2\right ) e \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)} \left (\sqrt{-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{16 b \left (a^2-b^2\right )^2 \sqrt{e \sin (c+d x)}}\\ &=\frac{a \left (3 a^2+2 b^2\right ) e \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{8 b \left (a^2-b^2\right )^2 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{a \left (3 a^2+2 b^2\right ) e \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{8 b \left (a^2-b^2\right )^2 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{5 a E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt{\sin (c+d x)}}-\frac{b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}+\frac{\left (\left (3 a^2+2 b^2\right ) e\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e-b x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{8 \left (a^2-b^2\right )^2 d}-\frac{\left (\left (3 a^2+2 b^2\right ) e\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{-a^2+b^2} e+b x^2} \, dx,x,\sqrt{e \sin (c+d x)}\right )}{8 \left (a^2-b^2\right )^2 d}\\ &=-\frac{\left (3 a^2+2 b^2\right ) \sqrt{e} \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 \sqrt{b} \left (-a^2+b^2\right )^{9/4} d}+\frac{\left (3 a^2+2 b^2\right ) \sqrt{e} \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 \sqrt{b} \left (-a^2+b^2\right )^{9/4} d}+\frac{a \left (3 a^2+2 b^2\right ) e \Pi \left (\frac{2 b}{b-\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{8 b \left (a^2-b^2\right )^2 \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{a \left (3 a^2+2 b^2\right ) e \Pi \left (\frac{2 b}{b+\sqrt{-a^2+b^2}};\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{8 b \left (a^2-b^2\right )^2 \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}+\frac{5 a E\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt{\sin (c+d x)}}-\frac{b (e \sin (c+d x))^{3/2}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{5 a b (e \sin (c+d x))^{3/2}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [C]  time = 14.5601, size = 837, normalized size = 1.58 \[ \frac{\sqrt{e \sin (c+d x)} \left (-\frac{5 a b \sin (c+d x)}{4 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{b \sin (c+d x)}{2 \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\right )}{d}+\frac{\sqrt{e \sin (c+d x)} \left (\frac{5 a \left (8 F_1\left (\frac{3}{4};-\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) \sin ^{\frac{3}{2}}(c+d x) b^{5/2}+3 \sqrt{2} a \left (a^2-b^2\right )^{3/4} \left (2 \tan ^{-1}\left (1-\frac{\sqrt{2} \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}\right )-2 \tan ^{-1}\left (\frac{\sqrt{2} \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{a^2-b^2}}+1\right )-\log \left (b \sin (c+d x)-\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\sin (c+d x)}+\sqrt{a^2-b^2}\right )+\log \left (b \sin (c+d x)+\sqrt{2} \sqrt{b} \sqrt [4]{a^2-b^2} \sqrt{\sin (c+d x)}+\sqrt{a^2-b^2}\right )\right )\right ) \left (a+b \sqrt{1-\sin ^2(c+d x)}\right ) \cos ^2(c+d x)}{12 \sqrt{b} \left (b^2-a^2\right ) (a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}+\frac{2 \left (8 a^2+2 b^2\right ) \left (\frac{a F_1\left (\frac{3}{4};\frac{1}{2},1;\frac{7}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) \sin ^{\frac{3}{2}}(c+d x)}{3 \left (a^2-b^2\right )}+\frac{\left (\frac{1}{8}+\frac{i}{8}\right ) \left (2 \tan ^{-1}\left (1-\frac{(1+i) \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}\right )-2 \tan ^{-1}\left (\frac{(1+i) \sqrt{b} \sqrt{\sin (c+d x)}}{\sqrt [4]{b^2-a^2}}+1\right )-\log \left (i b \sin (c+d x)-(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}\right )+\log \left (i b \sin (c+d x)+(1+i) \sqrt{b} \sqrt [4]{b^2-a^2} \sqrt{\sin (c+d x)}+\sqrt{b^2-a^2}\right )\right )}{\sqrt{b} \sqrt [4]{b^2-a^2}}\right ) \left (a+b \sqrt{1-\sin ^2(c+d x)}\right ) \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt{1-\sin ^2(c+d x)}}\right )}{8 (a-b)^2 (a+b)^2 d \sqrt{\sin (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Sqrt[e*Sin[c + d*x]]/(a + b*Cos[c + d*x])^3,x]

[Out]

(Sqrt[e*Sin[c + d*x]]*(-(b*Sin[c + d*x])/(2*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) - (5*a*b*Sin[c + d*x])/(4*(a^2
- b^2)^2*(a + b*Cos[c + d*x]))))/d + (Sqrt[e*Sin[c + d*x]]*((5*a*Cos[c + d*x]^2*(3*Sqrt[2]*a*(a^2 - b^2)^(3/4
)*(2*ArcTan[1 - (Sqrt[2]*Sqrt[b]*Sqrt[Sin[c + d*x]])/(a^2 - b^2)^(1/4)] - 2*ArcTan[1 + (Sqrt[2]*Sqrt[b]*Sqrt[S
in[c + d*x]])/(a^2 - b^2)^(1/4)] - Log[Sqrt[a^2 - b^2] - Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]]
+ b*Sin[c + d*x]] + Log[Sqrt[a^2 - b^2] + Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(1/4)*Sqrt[Sin[c + d*x]] + b*Sin[c + d*x
]]) + 8*b^(5/2)*AppellF1[3/4, -1/2, 1, 7/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(3
/2))*(a + b*Sqrt[1 - Sin[c + d*x]^2]))/(12*Sqrt[b]*(-a^2 + b^2)*(a + b*Cos[c + d*x])*(1 - Sin[c + d*x]^2)) + (
2*(8*a^2 + 2*b^2)*Cos[c + d*x]*(((1/8 + I/8)*(2*ArcTan[1 - ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(
1/4)] - 2*ArcTan[1 + ((1 + I)*Sqrt[b]*Sqrt[Sin[c + d*x]])/(-a^2 + b^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] - (1 + I)
*Sqrt[b]*(-a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]] + Log[Sqrt[-a^2 + b^2] + (1 + I)*Sqrt[b]*(-
a^2 + b^2)^(1/4)*Sqrt[Sin[c + d*x]] + I*b*Sin[c + d*x]]))/(Sqrt[b]*(-a^2 + b^2)^(1/4)) + (a*AppellF1[3/4, 1/2,
1, 7/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)]*Sin[c + d*x]^(3/2))/(3*(a^2 - b^2)))*(a + b*Sqrt[1
- Sin[c + d*x]^2]))/((a + b*Cos[c + d*x])*Sqrt[1 - Sin[c + d*x]^2])))/(8*(a - b)^2*(a + b)^2*d*Sqrt[Sin[c + d
*x]])

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Maple [B]  time = 11.102, size = 2986, normalized size = 5.6 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*sin(d*x+c))^(1/2)/(a+b*cos(d*x+c))^3,x)

[Out]

3/2/d*e/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/
2)*EllipticE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-3/4/d*e/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)*(1-sin(d*x+
c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-11/4/d*e*a/cos(d
*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticE(
(1-sin(d*x+c))^(1/2),1/2*2^(1/2))+11/8/d*e*a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*(1-sin(d*x+c))^(1/2)*
(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))-3/32/d*e/b/(a^4-2*a^2*b^2+
b^4)/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*ln((e*sin(d*x+c)-(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)
+(e^2*(a^2-b^2)/b^2)^(1/2))/(e*sin(d*x+c)+(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b^2
)/b^2)^(1/2)))*a^2-3/16/d*e/b/(a^4-2*a^2*b^2+b^4)/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2*(a^2-b
^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)+1)*a^2-3/16/d*e/b/(a^4-2*a^2*b^2+b^4)/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*ar
ctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)-1)*a^2+7/4/d*e*a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(
a^2-b^2)^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*EllipticPi((1-s
in(d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-1/2/d*e^3*b^3/(-b^2*cos(d*x+c)^2*e^2+e^2*a^2)^2/(a^2-b^
2)*(e*sin(d*x+c))^(3/2)-1/2/d*e*b^5/(-b^2*cos(d*x+c)^2*e^2+e^2*a^2)^2/(a^4-2*a^2*b^2+b^4)*(e*sin(d*x+c))^(7/2)
-21/16/d*e*a^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2/b^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin
(d*x+c)^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-21/
16/d*e*a^3/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2/b^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x
+c)^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-3/4/d*e
/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*b^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2
)/(1-(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-3/4/d*e/a/cos(d
*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*b^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1+(-a
^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))+9/8/d*e*a/cos(d*x+c)/(e
*sin(d*x+c))^(1/2)/(a^2-b^2)/b^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1-(-a^2+b^2)^(1
/2)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))+9/8/d*e*a/cos(d*x+c)/(e*sin(d*x+c
))^(1/2)/(a^2-b^2)/b^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*Ell
ipticPi((1-sin(d*x+c))^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-1/8/d*e*b/(a^4-2*a^2*b^2+b^4)/(e^2*(a^2-b^2
)/b^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)-1)-1/16/d*e*b/(a^4-2*a^2*b^
2+b^4)/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*ln((e*sin(d*x+c)-(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/
2)+(e^2*(a^2-b^2)/b^2)^(1/2))/(e*sin(d*x+c)+(e^2*(a^2-b^2)/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)*2^(1/2)+(e^2*(a^2-b
^2)/b^2)^(1/2)))-1/8/d*e*b/(a^4-2*a^2*b^2+b^4)/(e^2*(a^2-b^2)/b^2)^(1/4)*2^(1/2)*arctan(2^(1/2)/(e^2*(a^2-b^2)
/b^2)^(1/4)*(e*sin(d*x+c))^(1/2)+1)-3/4/d*e*b^3/(-b^2*cos(d*x+c)^2*e^2+e^2*a^2)^2/(a^4-2*a^2*b^2+b^4)*(e*sin(d
*x+c))^(7/2)*a^2-7/4/d*e^3*b/(-b^2*cos(d*x+c)^2*e^2+e^2*a^2)^2/(a^2-b^2)*(e*sin(d*x+c))^(3/2)*a^2+3/2/d*e/a/co
s(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*Ellipti
cE((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2-3/2/d*e*sin(d*x+c)^2*cos(d*x+c)/a/(e*sin(d*x+c))^(1/2)*b^4/(a^2-b^2)^
2/(-b^2*cos(d*x+c)^2+a^2)-3/2/d*e*sin(d*x+c)^2*cos(d*x+c)/a/(e*sin(d*x+c))^(1/2)*b^2/(a^2-b^2)/(-b^2*cos(d*x+c
)^2+a^2)+1/d*e*sin(d*x+c)^2*cos(d*x+c)*a/(e*sin(d*x+c))^(1/2)*b^2/(a^2-b^2)/(-b^2*cos(d*x+c)^2+a^2)^2+11/4/d*e
*sin(d*x+c)^2*cos(d*x+c)*a/(e*sin(d*x+c))^(1/2)*b^2/(a^2-b^2)^2/(-b^2*cos(d*x+c)^2+a^2)+7/4/d*e*a/cos(d*x+c)/(
e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1+(-a^2+b^2)^(1/
2)/b)*EllipticPi((1-sin(d*x+c))^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-3/4/d*e/a/cos(d*x+c)/(e*sin(d*x+c)
)^(1/2)/(a^2-b^2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1-(-a^2+b^2)^(1/2)/b)*Elliptic
Pi((1-sin(d*x+c))^(1/2),1/(1-(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-3/4/d*e/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b
^2)*(1-sin(d*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)/(1+(-a^2+b^2)^(1/2)/b)*EllipticPi((1-sin(d*x+
c))^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-3/4/d*e/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*(1-sin(d
*x+c))^(1/2)*(2+2*sin(d*x+c))^(1/2)*sin(d*x+c)^(1/2)*EllipticF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2

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Maxima [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((e*sin(d*x+c))^(1/2)/(a+b*cos(d*x+c))^3,x, algorithm="maxima")

[Out]

Timed out

________________________________________________________________________________________

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((e*sin(d*x+c))^(1/2)/(a+b*cos(d*x+c))^3,x, algorithm="fricas")

[Out]

Timed out

________________________________________________________________________________________

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((e*sin(d*x+c))**(1/2)/(a+b*cos(d*x+c))**3,x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{e \sin \left (d x + c\right )}}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*sin(d*x+c))^(1/2)/(a+b*cos(d*x+c))^3,x, algorithm="giac")

[Out]

integrate(sqrt(e*sin(d*x + c))/(b*cos(d*x + c) + a)^3, x)