Integrand size = 25, antiderivative size = 307 \[ \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\frac {\sqrt {b} \arctan \left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{3/4} d \sqrt {e}}+\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sqrt {e \sin (c+d x)}}{\sqrt [4]{-a^2+b^2} \sqrt {e}}\right )}{\left (-a^2+b^2\right )^{3/4} d \sqrt {e}}+\frac {a \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b \left (b-\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}}+\frac {a \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {-a^2+b^2}},\frac {1}{2} \left (c-\frac {\pi }{2}+d x\right ),2\right ) \sqrt {\sin (c+d x)}}{\left (a^2-b \left (b+\sqrt {-a^2+b^2}\right )\right ) d \sqrt {e \sin (c+d x)}} \]
arctan(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/4)/e^(1/2))*b^(1/2)/(-a^ 2+b^2)^(3/4)/d/e^(1/2)+arctanh(b^(1/2)*(e*sin(d*x+c))^(1/2)/(-a^2+b^2)^(1/ 4)/e^(1/2))*b^(1/2)/(-a^2+b^2)^(3/4)/d/e^(1/2)-a*(sin(1/2*c+1/4*Pi+1/2*d*x )^2)^(1/2)/sin(1/2*c+1/4*Pi+1/2*d*x)*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x), 2*b/(b-(-a^2+b^2)^(1/2)),2^(1/2))*sin(d*x+c)^(1/2)/d/(a^2-b*(b-(-a^2+b^2)^ (1/2)))/(e*sin(d*x+c))^(1/2)-a*(sin(1/2*c+1/4*Pi+1/2*d*x)^2)^(1/2)/sin(1/2 *c+1/4*Pi+1/2*d*x)*EllipticPi(cos(1/2*c+1/4*Pi+1/2*d*x),2*b/(b+(-a^2+b^2)^ (1/2)),2^(1/2))*sin(d*x+c)^(1/2)/d/(a^2-b*(b+(-a^2+b^2)^(1/2)))/(e*sin(d*x +c))^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 1.62 (sec) , antiderivative size = 261, normalized size of antiderivative = 0.85 \[ \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\frac {10 (a+b) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right ) \sqrt {e \sin (c+d x)}}{d e (a+b \cos (c+d x)) \left (5 (a+b) \operatorname {AppellF1}\left (\frac {1}{4},-\frac {1}{2},1,\frac {5}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+2 \left (-2 (a-b) \operatorname {AppellF1}\left (\frac {5}{4},-\frac {1}{2},2,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )+(a+b) \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2\left (\frac {1}{2} (c+d x)\right ),\frac {(-a+b) \tan ^2\left (\frac {1}{2} (c+d x)\right )}{a+b}\right )\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )\right )} \]
(10*(a + b)*AppellF1[1/4, -1/2, 1, 5/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan [(c + d*x)/2]^2)/(a + b)]*Sqrt[e*Sin[c + d*x]])/(d*e*(a + b*Cos[c + d*x])* (5*(a + b)*AppellF1[1/4, -1/2, 1, 5/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[ (c + d*x)/2]^2)/(a + b)] + 2*(-2*(a - b)*AppellF1[5/4, -1/2, 2, 9/4, -Tan[ (c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)] + (a + b)*AppellF1[ 5/4, 1/2, 1, 9/4, -Tan[(c + d*x)/2]^2, ((-a + b)*Tan[(c + d*x)/2]^2)/(a + b)])*Tan[(c + d*x)/2]^2))
Time = 1.17 (sec) , antiderivative size = 316, normalized size of antiderivative = 1.03, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 3181, 266, 756, 218, 221, 3042, 3286, 3042, 3284}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {1}{\sqrt {e \sin (c+d x)} (a+b \cos (c+d x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\sqrt {e \cos \left (c+d x-\frac {\pi }{2}\right )} \left (a-b \sin \left (c+d x-\frac {\pi }{2}\right )\right )}dx\) |
| \(\Big \downarrow \) 3181 |
| \(\displaystyle -\frac {b e \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b^2 \sin ^2(c+d x) e^2+\left (a^2-b^2\right ) e^2\right )}d(e \sin (c+d x))}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle -\frac {2 b e \int \frac {1}{b^2 e^4 \sin ^4(c+d x)+\left (a^2-b^2\right ) e^2}d\sqrt {e \sin (c+d x)}}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\) |
| \(\Big \downarrow \) 756 |
| \(\displaystyle -\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\int \frac {1}{b e^2 \sin ^2(c+d x)+\sqrt {b^2-a^2} e}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {2 b e \left (-\frac {\int \frac {1}{\sqrt {b^2-a^2} e-b e^2 \sin ^2(c+d x)}d\sqrt {e \sin (c+d x)}}{2 e \sqrt {b^2-a^2}}-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle -\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {a \int \frac {1}{\sqrt {e \sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\) |
| \(\Big \downarrow \) 3286 |
| \(\displaystyle -\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (\sqrt {b^2-a^2}-b \sin (c+d x)\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \int \frac {1}{\sqrt {\sin (c+d x)} \left (b \sin (c+d x)+\sqrt {b^2-a^2}\right )}dx}{2 \sqrt {b^2-a^2} \sqrt {e \sin (c+d x)}}-\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}\) |
| \(\Big \downarrow \) 3284 |
| \(\displaystyle -\frac {2 b e \left (-\frac {\arctan \left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}-\frac {\text {arctanh}\left (\frac {\sqrt {b} \sqrt {e} \sin (c+d x)}{\sqrt [4]{b^2-a^2}}\right )}{2 \sqrt {b} e^{3/2} \left (b^2-a^2\right )^{3/4}}\right )}{d}+\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b-\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (b-\sqrt {b^2-a^2}\right ) \sqrt {e \sin (c+d x)}}-\frac {a \sqrt {\sin (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{b+\sqrt {b^2-a^2}},\frac {1}{2} \left (c+d x-\frac {\pi }{2}\right ),2\right )}{d \sqrt {b^2-a^2} \left (\sqrt {b^2-a^2}+b\right ) \sqrt {e \sin (c+d x)}}\) |
(-2*b*e*(-1/2*ArcTan[(Sqrt[b]*Sqrt[e]*Sin[c + d*x])/(-a^2 + b^2)^(1/4)]/(S qrt[b]*(-a^2 + b^2)^(3/4)*e^(3/2)) - ArcTanh[(Sqrt[b]*Sqrt[e]*Sin[c + d*x] )/(-a^2 + b^2)^(1/4)]/(2*Sqrt[b]*(-a^2 + b^2)^(3/4)*e^(3/2))))/d + (a*Elli pticPi[(2*b)/(b - Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin[c + d *x]])/(Sqrt[-a^2 + b^2]*(b - Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x]]) - ( a*EllipticPi[(2*b)/(b + Sqrt[-a^2 + b^2]), (c - Pi/2 + d*x)/2, 2]*Sqrt[Sin [c + d*x]])/(Sqrt[-a^2 + b^2]*(b + Sqrt[-a^2 + b^2])*d*Sqrt[e*Sin[c + d*x] ])
3.1.64.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]*((a_) + (b_.)*sin[(e_.) + (f_.)* (x_)])), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[-a/(2*q) Int[1/( Sqrt[g*Cos[e + f*x]]*(q + b*Cos[e + f*x])), x], x] + (Simp[b*(g/f) Subst[ Int[1/(Sqrt[x]*(g^2*(a^2 - b^2) + b^2*x^2)), x], x, g*Cos[e + f*x]], x] - S imp[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]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c , d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && GtQ[c + d, 0]
Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Simp[Sqrt[(c + d*Sin[e + f*x])/(c + d)]/Sqrt [c + d*Sin[e + f*x]] Int[1/((a + b*Sin[e + f*x])*Sqrt[c/(c + d) + (d/(c + d))*Sin[e + f*x]]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a* d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && !GtQ[c + d, 0]
Time = 2.23 (sec) , antiderivative size = 519, normalized size of antiderivative = 1.69
| method | result | size |
| default | \(\frac {-\frac {b e \left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {e \sin \left (d x +c \right )+\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}{e \sin \left (d x +c \right )-\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}} \sqrt {e \sin \left (d x +c \right )}\, \sqrt {2}+\sqrt {\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {e \sin \left (d x +c \right )}}{\left (\frac {e^{2} \left (a^{2}-b^{2}\right )}{b^{2}}\right )^{\frac {1}{4}}}-1\right )\right )}{4 \left (a^{2} e^{2}-b^{2} e^{2}\right )}+\frac {a \sqrt {1-\sin \left (d x +c \right )}\, \sqrt {2 \sin \left (d x +c \right )+2}\, \left (\sqrt {\sin }\left (d x +c \right )\right ) \left (\Pi \left (\sqrt {1-\sin \left (d x +c \right )}, -\frac {b}{-b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}+\Pi \left (\sqrt {1-\sin \left (d x +c \right )}, -\frac {b}{-b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) b +\Pi \left (\sqrt {1-\sin \left (d x +c \right )}, \frac {b}{b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a^{2}+b^{2}}-\Pi \left (\sqrt {1-\sin \left (d x +c \right )}, \frac {b}{b +\sqrt {-a^{2}+b^{2}}}, \frac {\sqrt {2}}{2}\right ) b \right )}{2 \sqrt {-a^{2}+b^{2}}\, \left (-b +\sqrt {-a^{2}+b^{2}}\right ) \left (b +\sqrt {-a^{2}+b^{2}}\right ) \cos \left (d x +c \right ) \sqrt {e \sin \left (d x +c \right )}}}{d}\) | \(519\) |
(-1/4*b*e*(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)))+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^ (1/4)*(e*sin(d*x+c))^(1/2)+1)+2*arctan(2^(1/2)/(e^2*(a^2-b^2)/b^2)^(1/4)*( e*sin(d*x+c))^(1/2)-1))+1/2*a*(1-sin(d*x+c))^(1/2)*(2*sin(d*x+c)+2)^(1/2)* sin(d*x+c)^(1/2)*(EllipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^2+b^2)^(1/2)) ,1/2*2^(1/2))*(-a^2+b^2)^(1/2)+EllipticPi((1-sin(d*x+c))^(1/2),-b/(-b+(-a^ 2+b^2)^(1/2)),1/2*2^(1/2))*b+EllipticPi((1-sin(d*x+c))^(1/2),1/(b+(-a^2+b^ 2)^(1/2))*b,1/2*2^(1/2))*(-a^2+b^2)^(1/2)-EllipticPi((1-sin(d*x+c))^(1/2), 1/(b+(-a^2+b^2)^(1/2))*b,1/2*2^(1/2))*b)/(-a^2+b^2)^(1/2)/(-b+(-a^2+b^2)^( 1/2))/(b+(-a^2+b^2)^(1/2))/cos(d*x+c)/(e*sin(d*x+c))^(1/2))/d
Timed out. \[ \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {e \sin {\left (c + d x \right )}} \left (a + b \cos {\left (c + d x \right )}\right )}\, dx \]
\[ \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}} \,d x } \]
\[ \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\int { \frac {1}{{\left (b \cos \left (d x + c\right ) + a\right )} \sqrt {e \sin \left (d x + c\right )}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+b \cos (c+d x)) \sqrt {e \sin (c+d x)}} \, dx=\int \frac {1}{\sqrt {e\,\sin \left (c+d\,x\right )}\,\left (a+b\,\cos \left (c+d\,x\right )\right )} \,d x \]