3.85 \(\int \frac{1}{(a+b \cos (c+d x))^3 \sqrt{e \sin (c+d x)}} \, dx\)

Optimal. Leaf size=535 \[ \frac{3 \sqrt{b} \left (5 a^2+2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 d \sqrt{e} \left (b^2-a^2\right )^{11/4}}+\frac{3 \sqrt{b} \left (5 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 d \sqrt{e} \left (b^2-a^2\right )^{11/4}}-\frac{7 a b \sqrt{e \sin (c+d x)}}{4 d e \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{b \sqrt{e \sin (c+d x)}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{7 a \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{4 d \left (a^2-b^2\right )^2 \sqrt{e \sin (c+d x)}}+\frac{3 a \left (5 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 d \left (a^2-b^2\right )^2 \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{3 a \left (5 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 d \left (a^2-b^2\right )^2 \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \sin (c+d x)}} \]

[Out]

(3*Sqrt[b]*(5*a^2 + 2*b^2)*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(8*(-a^2 + b^2
)^(11/4)*d*Sqrt[e]) + (3*Sqrt[b]*(5*a^2 + 2*b^2)*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sq
rt[e])])/(8*(-a^2 + b^2)^(11/4)*d*Sqrt[e]) - (7*a*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(4*(a^2
- b^2)^2*d*Sqrt[e*Sin[c + d*x]]) + (3*a*(5*a^2 + 2*b^2)*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 +
d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*(a^2 - b^2)^2*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*d*Sqrt[e*Sin[c + d*x]]) + (3*
a*(5*a^2 + 2*b^2)*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*(a^2
- b^2)^2*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*d*Sqrt[e*Sin[c + d*x]]) - (b*Sqrt[e*Sin[c + d*x]])/(2*(a^2 - b^2)*d*
e*(a + b*Cos[c + d*x])^2) - (7*a*b*Sqrt[e*Sin[c + d*x]])/(4*(a^2 - b^2)^2*d*e*(a + b*Cos[c + d*x]))

________________________________________________________________________________________

Rubi [A]  time = 1.22894, antiderivative size = 535, 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, 2642, 2641, 2702, 2807, 2805, 329, 212, 208, 205} \[ \frac{3 \sqrt{b} \left (5 a^2+2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 d \sqrt{e} \left (b^2-a^2\right )^{11/4}}+\frac{3 \sqrt{b} \left (5 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt{e} \sqrt [4]{b^2-a^2}}\right )}{8 d \sqrt{e} \left (b^2-a^2\right )^{11/4}}-\frac{7 a b \sqrt{e \sin (c+d x)}}{4 d e \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{b \sqrt{e \sin (c+d x)}}{2 d e \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}-\frac{7 a \sqrt{\sin (c+d x)} F\left (\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{4 d \left (a^2-b^2\right )^2 \sqrt{e \sin (c+d x)}}+\frac{3 a \left (5 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b-\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 d \left (a^2-b^2\right )^2 \left (a^2-b \left (b-\sqrt{b^2-a^2}\right )\right ) \sqrt{e \sin (c+d x)}}+\frac{3 a \left (5 a^2+2 b^2\right ) \sqrt{\sin (c+d x)} \Pi \left (\frac{2 b}{b+\sqrt{b^2-a^2}};\left .\frac{1}{2} \left (c+d x-\frac{\pi }{2}\right )\right |2\right )}{8 d \left (a^2-b^2\right )^2 \left (a^2-b \left (\sqrt{b^2-a^2}+b\right )\right ) \sqrt{e \sin (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[1/((a + b*Cos[c + d*x])^3*Sqrt[e*Sin[c + d*x]]),x]

[Out]

(3*Sqrt[b]*(5*a^2 + 2*b^2)*ArcTan[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sqrt[e])])/(8*(-a^2 + b^2
)^(11/4)*d*Sqrt[e]) + (3*Sqrt[b]*(5*a^2 + 2*b^2)*ArcTanh[(Sqrt[b]*Sqrt[e*Sin[c + d*x]])/((-a^2 + b^2)^(1/4)*Sq
rt[e])])/(8*(-a^2 + b^2)^(11/4)*d*Sqrt[e]) - (7*a*EllipticF[(c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(4*(a^2
- b^2)^2*d*Sqrt[e*Sin[c + d*x]]) + (3*a*(5*a^2 + 2*b^2)*EllipticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 +
d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*(a^2 - b^2)^2*(a^2 - b*(b - Sqrt[-a^2 + b^2]))*d*Sqrt[e*Sin[c + d*x]]) + (3*
a*(5*a^2 + 2*b^2)*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d*x]])/(8*(a^2
- b^2)^2*(a^2 - b*(b + Sqrt[-a^2 + b^2]))*d*Sqrt[e*Sin[c + d*x]]) - (b*Sqrt[e*Sin[c + d*x]])/(2*(a^2 - b^2)*d*
e*(a + b*Cos[c + d*x])^2) - (7*a*b*Sqrt[e*Sin[c + d*x]])/(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 2642

Int[1/Sqrt[(b_)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[c + d*x]]/Sqrt[b*Sin[c + d*x]], Int[1/Sqr
t[Sin[c + d*x]], x], x] /; FreeQ[{b, c, d}, x]

Rule 2641

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

Rule 2702

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[
-a^2 + b^2, 2]}, -Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Dist[(b*g)/f, Sub
st[Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - Dist[a/(2*q), Int[1/(Sqrt[g*Cos[e
+ f*x]]*(q - b*Cos[e + f*x])), 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 212

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-(a/b), 2]], s = Denominator[Rt[-(a/b), 2]
]}, Dist[r/(2*a), Int[1/(r - s*x^2), x], x] + Dist[r/(2*a), Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&
!GtQ[a/b, 0]

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]

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]

Rubi steps

\begin{align*} \int \frac{1}{(a+b \cos (c+d x))^3 \sqrt{e \sin (c+d x)}} \, dx &=-\frac{b \sqrt{e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{\int \frac{-2 a+\frac{3}{2} b \cos (c+d x)}{(a+b \cos (c+d x))^2 \sqrt{e \sin (c+d x)}} \, dx}{2 \left (a^2-b^2\right )}\\ &=-\frac{b \sqrt{e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{7 a b \sqrt{e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}+\frac{\int \frac{\frac{1}{2} \left (4 a^2+3 b^2\right )-\frac{7}{4} a b \cos (c+d x)}{(a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}} \, dx}{2 \left (a^2-b^2\right )^2}\\ &=-\frac{b \sqrt{e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{7 a b \sqrt{e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}-\frac{(7 a) \int \frac{1}{\sqrt{e \sin (c+d x)}} \, dx}{8 \left (a^2-b^2\right )^2}+\frac{\left (3 \left (5 a^2+2 b^2\right )\right ) \int \frac{1}{(a+b \cos (c+d x)) \sqrt{e \sin (c+d x)}} \, dx}{8 \left (a^2-b^2\right )^2}\\ &=-\frac{b \sqrt{e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{7 a b \sqrt{e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}-\frac{\left (3 a \left (5 a^2+2 b^2\right )\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}-b \sin (c+d x)\right )} \, dx}{16 \left (-a^2+b^2\right )^{5/2}}-\frac{\left (3 a \left (5 a^2+2 b^2\right )\right ) \int \frac{1}{\sqrt{e \sin (c+d x)} \left (\sqrt{-a^2+b^2}+b \sin (c+d x)\right )} \, dx}{16 \left (-a^2+b^2\right )^{5/2}}-\frac{\left (3 b \left (5 a^2+2 b^2\right ) e\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (\left (a^2-b^2\right ) e^2+b^2 x^2\right )} \, dx,x,e \sin (c+d x)\right )}{8 \left (a^2-b^2\right )^2 d}-\frac{\left (7 a \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\sin (c+d x)}} \, dx}{8 \left (a^2-b^2\right )^2 \sqrt{e \sin (c+d x)}}\\ &=-\frac{7 a F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt{e \sin (c+d x)}}-\frac{b \sqrt{e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{7 a b \sqrt{e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}-\frac{\left (3 b \left (5 a^2+2 b^2\right ) e\right ) \operatorname{Subst}\left (\int \frac{1}{\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 (3 a \left (5 a^2+2 b^2\right ) \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 \left (-a^2+b^2\right )^{5/2} \sqrt{e \sin (c+d x)}}-\frac{\left (3 a \left (5 a^2+2 b^2\right ) \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 \left (-a^2+b^2\right )^{5/2} \sqrt{e \sin (c+d x)}}\\ &=-\frac{7 a F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt{e \sin (c+d x)}}+\frac{3 a \left (5 a^2+2 b^2\right ) \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 \left (-a^2+b^2\right )^{5/2} \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{3 a \left (5 a^2+2 b^2\right ) \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 \left (-a^2+b^2\right )^{5/2} \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{b \sqrt{e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{7 a b \sqrt{e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}+\frac{\left (3 b \left (5 a^2+2 b^2\right )\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 )^{5/2} d}+\frac{\left (3 b \left (5 a^2+2 b^2\right )\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 )^{5/2} d}\\ &=\frac{3 \sqrt{b} \left (5 a^2+2 b^2\right ) \tan ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 \left (-a^2+b^2\right )^{11/4} d \sqrt{e}}+\frac{3 \sqrt{b} \left (5 a^2+2 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{b} \sqrt{e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt{e}}\right )}{8 \left (-a^2+b^2\right )^{11/4} d \sqrt{e}}-\frac{7 a F\left (\left .\frac{1}{2} \left (c-\frac{\pi }{2}+d x\right )\right |2\right ) \sqrt{\sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d \sqrt{e \sin (c+d x)}}+\frac{3 a \left (5 a^2+2 b^2\right ) \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 \left (-a^2+b^2\right )^{5/2} \left (b-\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{3 a \left (5 a^2+2 b^2\right ) \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 \left (-a^2+b^2\right )^{5/2} \left (b+\sqrt{-a^2+b^2}\right ) d \sqrt{e \sin (c+d x)}}-\frac{b \sqrt{e \sin (c+d x)}}{2 \left (a^2-b^2\right ) d e (a+b \cos (c+d x))^2}-\frac{7 a b \sqrt{e \sin (c+d x)}}{4 \left (a^2-b^2\right )^2 d e (a+b \cos (c+d x))}\\ \end{align*}

Mathematica [C]  time = 12.8381, size = 1226, normalized size = 2.29 \[ \frac{\left (-\frac{7 a b}{4 \left (a^2-b^2\right )^2 (a+b \cos (c+d x))}-\frac{b}{2 \left (a^2-b^2\right ) (a+b \cos (c+d x))^2}\right ) \sin (c+d x)}{d \sqrt{e \sin (c+d x)}}+\frac{\left (\frac{2 \left (8 a^2+6 b^2\right ) \cos (c+d x) \left (a+b \sqrt{1-\sin ^2(c+d x)}\right ) \left (\frac{5 a \left (a^2-b^2\right ) F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) \sqrt{\sin (c+d x)}}{\sqrt{1-\sin ^2(c+d x)} \left (5 \left (a^2-b^2\right ) F_1\left (\frac{1}{4};\frac{1}{2},1;\frac{5}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right )-2 \left (2 F_1\left (\frac{5}{4};\frac{1}{2},2;\frac{9}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) b^2+\left (b^2-a^2\right ) F_1\left (\frac{5}{4};\frac{3}{2},1;\frac{9}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right )\right ) \sin ^2(c+d x)\right ) \left (a^2+b^2 \left (\sin ^2(c+d x)-1\right )\right )}-\frac{\left (\frac{1}{8}-\frac{i}{8}\right ) \sqrt{b} \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 )}{\left (b^2-a^2\right )^{3/4}}\right )}{(a+b \cos (c+d x)) \sqrt{1-\sin ^2(c+d x)}}-\frac{14 a b \cos ^2(c+d x) \left (a+b \sqrt{1-\sin ^2(c+d x)}\right ) \left (\frac{5 b \left (a^2-b^2\right ) \sqrt{\sin (c+d x)} \sqrt{1-\sin ^2(c+d x)} F_1\left (\frac{1}{4};-\frac{1}{2},1;\frac{5}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right )}{\left (2 \left (2 F_1\left (\frac{5}{4};-\frac{1}{2},2;\frac{9}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right ) b^2+\left (a^2-b^2\right ) F_1\left (\frac{5}{4};\frac{1}{2},1;\frac{9}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right )\right ) \sin ^2(c+d x)-5 \left (a^2-b^2\right ) F_1\left (\frac{1}{4};-\frac{1}{2},1;\frac{5}{4};\sin ^2(c+d x),\frac{b^2 \sin ^2(c+d x)}{b^2-a^2}\right )\right ) \left (a^2+b^2 \left (\sin ^2(c+d x)-1\right )\right )}+\frac{a \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 )}{4 \sqrt{2} \sqrt{b} \left (a^2-b^2\right )^{3/4}}\right )}{(a+b \cos (c+d x)) \left (1-\sin ^2(c+d x)\right )}\right ) \sqrt{\sin (c+d x)}}{8 (a-b)^2 (a+b)^2 d \sqrt{e \sin (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[1/((a + b*Cos[c + d*x])^3*Sqrt[e*Sin[c + d*x]]),x]

[Out]

((-b/(2*(a^2 - b^2)*(a + b*Cos[c + d*x])^2) - (7*a*b)/(4*(a^2 - b^2)^2*(a + b*Cos[c + d*x])))*Sin[c + d*x])/(d
*Sqrt[e*Sin[c + d*x]]) + (Sqrt[Sin[c + d*x]]*((-14*a*b*Cos[c + d*x]^2*(a + b*Sqrt[1 - Sin[c + d*x]^2])*((a*(-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[Sin[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]]))
/(4*Sqrt[2]*Sqrt[b]*(a^2 - b^2)^(3/4)) + (5*b*(a^2 - b^2)*AppellF1[1/4, -1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin
[c + d*x]^2)/(-a^2 + b^2)]*Sqrt[Sin[c + d*x]]*Sqrt[1 - Sin[c + d*x]^2])/((-5*(a^2 - b^2)*AppellF1[1/4, -1/2, 1
, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + 2*(2*b^2*AppellF1[5/4, -1/2, 2, 9/4, Sin[c + d*x]^
2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] + (a^2 - b^2)*AppellF1[5/4, 1/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*
x]^2)/(-a^2 + b^2)])*Sin[c + d*x]^2)*(a^2 + b^2*(-1 + Sin[c + d*x]^2)))))/((a + b*Cos[c + d*x])*(1 - Sin[c + d
*x]^2)) + (2*(8*a^2 + 6*b^2)*Cos[c + d*x]*(a + b*Sqrt[1 - Sin[c + d*x]^2])*(((-1/8 + I/8)*Sqrt[b]*(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]]))
/(-a^2 + b^2)^(3/4) + (5*a*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 +
b^2)]*Sqrt[Sin[c + d*x]])/(Sqrt[1 - Sin[c + d*x]^2]*(5*(a^2 - b^2)*AppellF1[1/4, 1/2, 1, 5/4, Sin[c + d*x]^2,
(b^2*Sin[c + d*x]^2)/(-a^2 + b^2)] - 2*(2*b^2*AppellF1[5/4, 1/2, 2, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)
/(-a^2 + b^2)] + (-a^2 + b^2)*AppellF1[5/4, 3/2, 1, 9/4, Sin[c + d*x]^2, (b^2*Sin[c + d*x]^2)/(-a^2 + b^2)])*S
in[c + d*x]^2)*(a^2 + b^2*(-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[e*Sin[c + d*x]])

________________________________________________________________________________________

Maple [B]  time = 12.183, size = 2918, normalized size = 5.5 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2/d*b^5*e/(-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))^(5/2)-1/2/d*b^3*e^3/(-b^2*co
s(d*x+c)^2*e^2+e^2*a^2)^2/(a^2-b^2)*(e*sin(d*x+c))^(1/2)-15/8/d*a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)/b/
(-a^2+b^2)^(1/2)*(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)*EllipticP
i((1-sin(d*x+c))^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))+3/4/d/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)
*b/(-a^2+b^2)^(1/2)*(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)*Ellipt
icPi((1-sin(d*x+c))^(1/2),1/(1+(-a^2+b^2)^(1/2)/b),1/2*2^(1/2))-3/4/d/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b
^2)^2*b^3/(-a^2+b^2)^(1/2)*(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))+45/16/d*a^3/cos(d*x+c)/(e*sin(d*x+c))^(
1/2)/(a^2-b^2)^2/b/(-a^2+b^2)^(1/2)*(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))-45/16/d*a^3/cos(d*x+c)/(e*sin(
d*x+c))^(1/2)/(a^2-b^2)^2/b/(-a^2+b^2)^(1/2)*(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))+9/4/d*a/cos(d*x+c)/(e
*sin(d*x+c))^(1/2)/(a^2-b^2)^2*b/(-a^2+b^2)^(1/2)*(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))+15/8/d*a/cos(d*x
+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)/b/(-a^2+b^2)^(1/2)*(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/a/cos(
d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)*b/(-a^2+b^2)^(1/2)*(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/a/c
os(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)*Ellipt
icF((1-sin(d*x+c))^(1/2),1/2*2^(1/2))*b^2-9/4/d*a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*b/(-a^2+b^2)^(1/
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))-3/2/d*sin(d*x+c)*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)-3/8/d*b^3*e/(a^4-2*a^2*b^2+b^4)*(e^2*(a^2-b^2)/b^2)^(1/4)/(a^2*e^2-b^2*e^2)*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)-3/16/d*b^3*e/(a^4-2*a^2*b^2+b^4)*(e^2*(a^2-
b^2)/b^2)^(1/4)/(a^2*e^2-b^2*e^2)*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)))-3/8/d*b^3*e/(a^4-2*a^2*b^2+b^4)*(e^2*(a^2-b^2)/b^2)^(1/4)/(a^2*e^2-b^2*e^2)*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)-3/2/d*sin(d*x+c)*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*sin(d*x+c)*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+13/4/d*sin(d*x+c)*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)+3/4
/d/a/cos(d*x+c)/(e*sin(d*x+c))^(1/2)/(a^2-b^2)^2*b^3/(-a^2+b^2)^(1/2)*(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))-5/4/d*b^3*e/(-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))^(5/2)*a^2-9/4/d*b*e^3/(-b
^2*cos(d*x+c)^2*e^2+e^2*a^2)^2/(a^2-b^2)*(e*sin(d*x+c))^(1/2)*a^2-15/32/d*b*e/(a^4-2*a^2*b^2+b^4)*(e^2*(a^2-b^
2)/b^2)^(1/4)/(a^2*e^2-b^2*e^2)*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^2-15/16/d*b*e/(a^4-2*a^2*b^2+b^4)*(e^2*(a^2-b^2)/b^2)^(1/4)/(a^2*e^2-b^2*e^2)*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^2-15/16/d*b*e/(a^4-2*a^2*b^2+b^4)*(e^2*(a^2-b^2)/
b^2)^(1/4)/(a^2*e^2-b^2*e^2)*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+13/8
/d*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))-3/4/d/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))

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

[Out]

Timed out

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{{\left (b \cos \left (d x + c\right ) + a\right )}^{3} \sqrt{e \sin \left (d x + c\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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