Integrand size = 39, antiderivative size = 166 \[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\frac {\sqrt {2} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {g \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d) f}-\frac {2 \sqrt {c} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {c} \sqrt {g} \cos (e+f x)}{\sqrt {c+d} \sqrt {g \sin (e+f x)} \sqrt {a+a \sin (e+f x)}}\right )}{\sqrt {a} (c-d) \sqrt {c+d} f} \]
arctan(1/2*cos(f*x+e)*a^(1/2)*g^(1/2)*2^(1/2)/(g*sin(f*x+e))^(1/2)/(a+a*si n(f*x+e))^(1/2))*2^(1/2)*g^(1/2)/(c-d)/f/a^(1/2)-2*arctan(cos(f*x+e)*a^(1/ 2)*c^(1/2)*g^(1/2)/(c+d)^(1/2)/(g*sin(f*x+e))^(1/2)/(a+a*sin(f*x+e))^(1/2) )*c^(1/2)*g^(1/2)/(c-d)/f/a^(1/2)/(c+d)^(1/2)
Time = 5.90 (sec) , antiderivative size = 280, normalized size of antiderivative = 1.69 \[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=-\frac {\left (2 \sqrt {c} \sqrt {-c^2+d^2} \arctan \left (\sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}\right )-\sqrt {d-\sqrt {-c^2+d^2}} \left (-c+d+\sqrt {-c^2+d^2}\right ) \arctan \left (\frac {\sqrt {c} \sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}}{\sqrt {d-\sqrt {-c^2+d^2}}}\right )-\left (c-d+\sqrt {-c^2+d^2}\right ) \sqrt {d+\sqrt {-c^2+d^2}} \arctan \left (\frac {\sqrt {c} \sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}}{\sqrt {d+\sqrt {-c^2+d^2}}}\right )\right ) \sqrt {g \sin (e+f x)} \left (1+\tan \left (\frac {1}{2} (e+f x)\right )\right )}{\sqrt {c} (c-d) \sqrt {-c^2+d^2} f \sqrt {a (1+\sin (e+f x))} \sqrt {\tan \left (\frac {1}{2} (e+f x)\right )}} \]
-(((2*Sqrt[c]*Sqrt[-c^2 + d^2]*ArcTan[Sqrt[Tan[(e + f*x)/2]]] - Sqrt[d - S qrt[-c^2 + d^2]]*(-c + d + Sqrt[-c^2 + d^2])*ArcTan[(Sqrt[c]*Sqrt[Tan[(e + f*x)/2]])/Sqrt[d - Sqrt[-c^2 + d^2]]] - (c - d + Sqrt[-c^2 + d^2])*Sqrt[d + Sqrt[-c^2 + d^2]]*ArcTan[(Sqrt[c]*Sqrt[Tan[(e + f*x)/2]])/Sqrt[d + Sqrt [-c^2 + d^2]]])*Sqrt[g*Sin[e + f*x]]*(1 + Tan[(e + f*x)/2]))/(Sqrt[c]*(c - d)*Sqrt[-c^2 + d^2]*f*Sqrt[a*(1 + Sin[e + f*x])]*Sqrt[Tan[(e + f*x)/2]]))
Time = 0.90 (sec) , antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {3042, 3415, 3042, 3261, 218, 3409, 218}
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.
| \(\displaystyle \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3415 |
| \(\displaystyle \frac {c g \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {g \sin (e+f x)} (c+d \sin (e+f x))}dx}{a (c-d)}-\frac {g \int \frac {1}{\sqrt {g \sin (e+f x)} \sqrt {\sin (e+f x) a+a}}dx}{c-d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {c g \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {g \sin (e+f x)} (c+d \sin (e+f x))}dx}{a (c-d)}-\frac {g \int \frac {1}{\sqrt {g \sin (e+f x)} \sqrt {\sin (e+f x) a+a}}dx}{c-d}\) |
| \(\Big \downarrow \) 3261 |
| \(\displaystyle \frac {2 a g \int \frac {1}{\frac {\cos (e+f x) \cot (e+f x) a^3}{\sin (e+f x) a+a}+2 a^2}d\frac {a \cos (e+f x)}{\sqrt {g \sin (e+f x)} \sqrt {\sin (e+f x) a+a}}}{f (c-d)}+\frac {c g \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {g \sin (e+f x)} (c+d \sin (e+f x))}dx}{a (c-d)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {c g \int \frac {\sqrt {\sin (e+f x) a+a}}{\sqrt {g \sin (e+f x)} (c+d \sin (e+f x))}dx}{a (c-d)}+\frac {\sqrt {2} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}\right )}{\sqrt {a} f (c-d)}\) |
| \(\Big \downarrow \) 3409 |
| \(\displaystyle \frac {\sqrt {2} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}\right )}{\sqrt {a} f (c-d)}-\frac {2 c g \int \frac {1}{\frac {c \cos (e+f x) \cot (e+f x) a^2}{\sin (e+f x) a+a}+(c+d) a}d\frac {a \cos (e+f x)}{\sqrt {g \sin (e+f x)} \sqrt {\sin (e+f x) a+a}}}{f (c-d)}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\sqrt {2} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {g} \cos (e+f x)}{\sqrt {2} \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}\right )}{\sqrt {a} f (c-d)}-\frac {2 \sqrt {c} \sqrt {g} \arctan \left (\frac {\sqrt {a} \sqrt {c} \sqrt {g} \cos (e+f x)}{\sqrt {c+d} \sqrt {a \sin (e+f x)+a} \sqrt {g \sin (e+f x)}}\right )}{\sqrt {a} f (c-d) \sqrt {c+d}}\) |
(Sqrt[2]*Sqrt[g]*ArcTan[(Sqrt[a]*Sqrt[g]*Cos[e + f*x])/(Sqrt[2]*Sqrt[g*Sin [e + f*x]]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*(c - d)*f) - (2*Sqrt[c]*Sq rt[g]*ArcTan[(Sqrt[a]*Sqrt[c]*Sqrt[g]*Cos[e + f*x])/(Sqrt[c + d]*Sqrt[g*Si n[e + f*x]]*Sqrt[a + a*Sin[e + f*x]])])/(Sqrt[a]*(c - d)*Sqrt[c + d]*f)
3.1.27.3.1 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[1/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*Sqrt[(c_.) + (d_.)*sin[(e _.) + (f_.)*(x_)]]), x_Symbol] :> Simp[-2*(a/f) Subst[Int[1/(2*b^2 - (a*c - b*d)*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*S in[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]
Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[(g_.)*sin[(e_.) + (f_. )*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[-2*(b/f ) Subst[Int[1/(b*c + a*d + c*g*x^2), x], x, b*(Cos[e + f*x]/(Sqrt[g*Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]))], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
Int[Sqrt[(g_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[(a_) + (b_.)*sin[(e_.) + (f_. )*(x_)]]*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[(-a)*(g /(b*c - a*d)) Int[1/(Sqrt[g*Sin[e + f*x]]*Sqrt[a + b*Sin[e + f*x]]), x], x] + Simp[c*(g/(b*c - a*d)) Int[Sqrt[a + b*Sin[e + f*x]]/(Sqrt[g*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && N eQ[b*c - a*d, 0] && (EqQ[a^2 - b^2, 0] || EqQ[c^2 - d^2, 0])
Leaf count of result is larger than twice the leaf count of optimal. \(588\) vs. \(2(131)=262\).
Time = 3.14 (sec) , antiderivative size = 589, normalized size of antiderivative = 3.55
| method | result | size |
| default | \(\frac {\sqrt {g \sin \left (f x +e \right )}\, \left (\sqrt {-\left (c -d \right ) \left (c +d \right )}\, \sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}-d \right ) c}\, \arctan \left (\frac {\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )}\, c}{\sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}+d \right ) c}}\right ) c -\sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}-d \right ) c}\, \arctan \left (\frac {\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )}\, c}{\sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}+d \right ) c}}\right ) c^{2}+\sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}-d \right ) c}\, \arctan \left (\frac {\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )}\, c}{\sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}+d \right ) c}}\right ) c d -\sqrt {-\left (c -d \right ) \left (c +d \right )}\, \sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}+d \right ) c}\, \operatorname {arctanh}\left (\frac {\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )}\, c}{\sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}-d \right ) c}}\right ) c -\sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}+d \right ) c}\, \operatorname {arctanh}\left (\frac {\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )}\, c}{\sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}-d \right ) c}}\right ) c^{2}+\sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}+d \right ) c}\, \operatorname {arctanh}\left (\frac {\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )}\, c}{\sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}-d \right ) c}}\right ) c d -2 \arctan \left (\sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )}\right ) \sqrt {-\left (c -d \right ) \left (c +d \right )}\, \sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}-d \right ) c}\, \sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}+d \right ) c}\right ) \left (\cos \left (f x +e \right )+\sin \left (f x +e \right )+1\right )}{f \left (1+\cos \left (f x +e \right )\right ) \sqrt {a \left (1+\sin \left (f x +e \right )\right )}\, \sqrt {\csc \left (f x +e \right )-\cot \left (f x +e \right )}\, \left (c -d \right ) \sqrt {-\left (c -d \right ) \left (c +d \right )}\, \sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}-d \right ) c}\, \sqrt {\left (\sqrt {-\left (c -d \right ) \left (c +d \right )}+d \right ) c}}\) | \(589\) |
1/f*(g*sin(f*x+e))^(1/2)*((-(c-d)*(c+d))^(1/2)*(((-(c-d)*(c+d))^(1/2)-d)*c )^(1/2)*arctan((csc(f*x+e)-cot(f*x+e))^(1/2)*c/(((-(c-d)*(c+d))^(1/2)+d)*c )^(1/2))*c-(((-(c-d)*(c+d))^(1/2)-d)*c)^(1/2)*arctan((csc(f*x+e)-cot(f*x+e ))^(1/2)*c/(((-(c-d)*(c+d))^(1/2)+d)*c)^(1/2))*c^2+(((-(c-d)*(c+d))^(1/2)- d)*c)^(1/2)*arctan((csc(f*x+e)-cot(f*x+e))^(1/2)*c/(((-(c-d)*(c+d))^(1/2)+ d)*c)^(1/2))*c*d-(-(c-d)*(c+d))^(1/2)*(((-(c-d)*(c+d))^(1/2)+d)*c)^(1/2)*a rctanh((csc(f*x+e)-cot(f*x+e))^(1/2)*c/(((-(c-d)*(c+d))^(1/2)-d)*c)^(1/2)) *c-(((-(c-d)*(c+d))^(1/2)+d)*c)^(1/2)*arctanh((csc(f*x+e)-cot(f*x+e))^(1/2 )*c/(((-(c-d)*(c+d))^(1/2)-d)*c)^(1/2))*c^2+(((-(c-d)*(c+d))^(1/2)+d)*c)^( 1/2)*arctanh((csc(f*x+e)-cot(f*x+e))^(1/2)*c/(((-(c-d)*(c+d))^(1/2)-d)*c)^ (1/2))*c*d-2*arctan((csc(f*x+e)-cot(f*x+e))^(1/2))*(-(c-d)*(c+d))^(1/2)*(( (-(c-d)*(c+d))^(1/2)-d)*c)^(1/2)*(((-(c-d)*(c+d))^(1/2)+d)*c)^(1/2))*(cos( f*x+e)+sin(f*x+e)+1)/(1+cos(f*x+e))/(a*(1+sin(f*x+e)))^(1/2)/(csc(f*x+e)-c ot(f*x+e))^(1/2)/(c-d)/(-(c-d)*(c+d))^(1/2)/(((-(c-d)*(c+d))^(1/2)-d)*c)^( 1/2)/(((-(c-d)*(c+d))^(1/2)+d)*c)^(1/2)
Time = 1.33 (sec) , antiderivative size = 3048, normalized size of antiderivative = 18.36 \[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\text {Too large to display} \]
[-1/4*(sqrt(2)*sqrt(-g/a)*log((17*g*cos(f*x + e)^3 - 4*sqrt(2)*(3*cos(f*x + e)^2 + (3*cos(f*x + e) + 4)*sin(f*x + e) - cos(f*x + e) - 4)*sqrt(a*sin( f*x + e) + a)*sqrt(g*sin(f*x + e))*sqrt(-g/a) + 3*g*cos(f*x + e)^2 - 18*g* cos(f*x + e) + (17*g*cos(f*x + e)^2 + 14*g*cos(f*x + e) - 4*g)*sin(f*x + e ) - 4*g)/(cos(f*x + e)^3 + 3*cos(f*x + e)^2 + (cos(f*x + e)^2 - 2*cos(f*x + e) - 4)*sin(f*x + e) - 2*cos(f*x + e) - 4)) + sqrt(-c*g/(a*c + a*d))*log (((128*c^4 + 256*c^3*d + 160*c^2*d^2 + 32*c*d^3 + d^4)*g*cos(f*x + e)^5 - (128*c^4 + 192*c^3*d + 64*c^2*d^2 - 4*c*d^3 - d^4)*g*cos(f*x + e)^4 - 2*(2 08*c^4 + 368*c^3*d + 195*c^2*d^2 + 32*c*d^3 + d^4)*g*cos(f*x + e)^3 + 2*(6 4*c^4 + 94*c^3*d + 29*c^2*d^2 - 4*c*d^3 - d^4)*g*cos(f*x + e)^2 + (289*c^4 + 480*c^3*d + 230*c^2*d^2 + 32*c*d^3 + d^4)*g*cos(f*x + e) + 8*((16*c^4 + 40*c^3*d + 34*c^2*d^2 + 11*c*d^3 + d^4)*cos(f*x + e)^4 + 51*c^4 + 110*c^3 *d + 76*c^2*d^2 + 18*c*d^3 + d^4 - (24*c^4 + 52*c^3*d + 35*c^2*d^2 + 7*c*d ^3)*cos(f*x + e)^3 - (66*c^4 + 149*c^3*d + 110*c^2*d^2 + 29*c*d^3 + 2*d^4) *cos(f*x + e)^2 + (25*c^4 + 53*c^3*d + 35*c^2*d^2 + 7*c*d^3)*cos(f*x + e) - (51*c^4 + 110*c^3*d + 76*c^2*d^2 + 18*c*d^3 + d^4 - (16*c^4 + 40*c^3*d + 34*c^2*d^2 + 11*c*d^3 + d^4)*cos(f*x + e)^3 - (40*c^4 + 92*c^3*d + 69*c^2 *d^2 + 18*c*d^3 + d^4)*cos(f*x + e)^2 + (26*c^4 + 57*c^3*d + 41*c^2*d^2 + 11*c*d^3 + d^4)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt( g*sin(f*x + e))*sqrt(-c*g/(a*c + a*d)) + (c^4 + 4*c^3*d + 6*c^2*d^2 + 4...
\[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\int \frac {\sqrt {g \sin {\left (e + f x \right )}}}{\sqrt {a \left (\sin {\left (e + f x \right )} + 1\right )} \left (c + d \sin {\left (e + f x \right )}\right )}\, dx \]
\[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\int { \frac {\sqrt {g \sin \left (f x + e\right )}}{\sqrt {a \sin \left (f x + e\right ) + a} {\left (d \sin \left (f x + e\right ) + c\right )}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\text {Timed out} \]
Timed out. \[ \int \frac {\sqrt {g \sin (e+f x)}}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))} \, dx=\int \frac {\sqrt {g\,\sin \left (e+f\,x\right )}}{\sqrt {a+a\,\sin \left (e+f\,x\right )}\,\left (c+d\,\sin \left (e+f\,x\right )\right )} \,d x \]