Integrand size = 37, antiderivative size = 209 \[ \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=-\frac {2 \sqrt {2} \sqrt {d} \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{\sqrt {-a^2+b^2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} \sqrt {d} \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{\sqrt {-a^2+b^2} f \sqrt {g \cos (e+f x)}} \]
-2*EllipticPi((d*sin(f*x+e))^(1/2)/d^(1/2)/(1+cos(f*x+e))^(1/2),-a/(b-(-a^ 2+b^2)^(1/2)),I)*2^(1/2)*d^(1/2)*cos(f*x+e)^(1/2)/f/(-a^2+b^2)^(1/2)/(g*co s(f*x+e))^(1/2)+2*EllipticPi((d*sin(f*x+e))^(1/2)/d^(1/2)/(1+cos(f*x+e))^( 1/2),-a/(b+(-a^2+b^2)^(1/2)),I)*2^(1/2)*d^(1/2)*cos(f*x+e)^(1/2)/f/(-a^2+b ^2)^(1/2)/(g*cos(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 12.92 (sec) , antiderivative size = 373, normalized size of antiderivative = 1.78 \[ \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {2 \sqrt {d \sin (e+f x)} \left (a \sqrt {\sec ^2(e+f x)}+b \tan (e+f x)\right ) \left (\frac {\sqrt {a} \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}}{\sqrt {a}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}}{\sqrt {a}}\right )+\log \left (-a+\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}-\sqrt {a^2-b^2} \tan (e+f x)\right )-\log \left (a+\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}+\sqrt {a^2-b^2} \tan (e+f x)\right )\right )}{4 \sqrt {2} \left (a^2-b^2\right )^{3/4}}-\frac {b \operatorname {AppellF1}\left (\frac {5}{4},\frac {1}{2},1,\frac {9}{4},-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right ) \tan ^{\frac {5}{2}}(e+f x)}{5 a^2}\right )}{f \sqrt {g \cos (e+f x)} \sqrt {\sec ^2(e+f x)} (a+b \sin (e+f x)) \sqrt {\tan (e+f x)}} \]
(2*Sqrt[d*Sin[e + f*x]]*(a*Sqrt[Sec[e + f*x]^2] + b*Tan[e + f*x])*((Sqrt[a ]*(-2*ArcTan[1 - (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]] + 2*ArcTan[1 + (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]])/Sqrt[a]] + Lo g[-a + Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]] - Sqrt[a^2 - b ^2]*Tan[e + f*x]] - Log[a + Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[Tan[e + f*x]] + Sqrt[a^2 - b^2]*Tan[e + f*x]]))/(4*Sqrt[2]*(a^2 - b^2)^(3/4)) - ( b*AppellF1[5/4, 1/2, 1, 9/4, -Tan[e + f*x]^2, ((-a^2 + b^2)*Tan[e + f*x]^2 )/a^2]*Tan[e + f*x]^(5/2))/(5*a^2)))/(f*Sqrt[g*Cos[e + f*x]]*Sqrt[Sec[e + f*x]^2]*(a + b*Sin[e + f*x])*Sqrt[Tan[e + f*x]])
Time = 0.72 (sec) , antiderivative size = 233, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.135, Rules used = {3042, 3387, 3042, 3386, 1542}
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.
\(\displaystyle \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx\) |
\(\Big \downarrow \) 3387 |
\(\displaystyle \frac {\sqrt {\cos (e+f x)} \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))}dx}{\sqrt {g \cos (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\sqrt {\cos (e+f x)} \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))}dx}{\sqrt {g \cos (e+f x)}}\) |
\(\Big \downarrow \) 3386 |
\(\displaystyle \frac {\sqrt {\cos (e+f x)} \left (\frac {2 \sqrt {2} d \left (1-\frac {b}{\sqrt {b^2-a^2}}\right ) \int \frac {1}{\left (\left (b-\sqrt {b^2-a^2}\right ) d+\frac {a \sin (e+f x) d}{\cos (e+f x)+1}\right ) \sqrt {1-\frac {\sin ^2(e+f x)}{(\cos (e+f x)+1)^2}}}d\frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)+1}}}{f}+\frac {2 \sqrt {2} d \left (\frac {b}{\sqrt {b^2-a^2}}+1\right ) \int \frac {1}{\left (\left (b+\sqrt {b^2-a^2}\right ) d+\frac {a \sin (e+f x) d}{\cos (e+f x)+1}\right ) \sqrt {1-\frac {\sin ^2(e+f x)}{(\cos (e+f x)+1)^2}}}d\frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)+1}}}{f}\right )}{\sqrt {g \cos (e+f x)}}\) |
\(\Big \downarrow \) 1542 |
\(\displaystyle \frac {\sqrt {\cos (e+f x)} \left (\frac {2 \sqrt {2} \sqrt {d} \left (1-\frac {b}{\sqrt {b^2-a^2}}\right ) \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right ),-1\right )}{f \left (b-\sqrt {b^2-a^2}\right )}+\frac {2 \sqrt {2} \sqrt {d} \left (\frac {b}{\sqrt {b^2-a^2}}+1\right ) \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right ),-1\right )}{f \left (\sqrt {b^2-a^2}+b\right )}\right )}{\sqrt {g \cos (e+f x)}}\) |
(Sqrt[Cos[e + f*x]]*((2*Sqrt[2]*(1 - b/Sqrt[-a^2 + b^2])*Sqrt[d]*EllipticP i[-(a/(b - Sqrt[-a^2 + b^2])), ArcSin[Sqrt[d*Sin[e + f*x]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x]])], -1])/((b - Sqrt[-a^2 + b^2])*f) + (2*Sqrt[2]*(1 + b/Sq rt[-a^2 + b^2])*Sqrt[d]*EllipticPi[-(a/(b + Sqrt[-a^2 + b^2])), ArcSin[Sqr t[d*Sin[e + f*x]]/(Sqrt[d]*Sqrt[1 + Cos[e + f*x]])], -1])/((b + Sqrt[-a^2 + b^2])*f)))/Sqrt[g*Cos[e + f*x]]
3.15.32.3.1 Defintions of rubi rules used
Int[1/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (c_.)*(x_)^4]), x_Symbol] :> With[ {q = Rt[-c/a, 4]}, Simp[(1/(d*Sqrt[a]*q))*EllipticPi[-e/(d*q^2), ArcSin[q*x ], -1], x]] /; FreeQ[{a, c, d, e}, x] && NegQ[c/a] && GtQ[a, 0]
Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]]*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> With[{q = Rt[-a^2 + b^2, 2]}, Simp[2*Sqrt[2]*d*((b + q)/(f*q)) Subst[Int[1/((d*(b + q) + a*x^2)*Sq rt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x] - Simp[2*Sqrt[2]*d*((b - q)/(f*q)) Subst[Int[1/((d*(b - q) + a*x^2)*Sqrt[1 - x^4/d^2]), x], x, Sqrt[d*Sin[e + f*x]]/Sqrt[1 + Cos[e + f*x]]], x]] /; F reeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[Sqrt[(d_.)*sin[(e_.) + (f_.)*(x_)]]/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.) ]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[Sqrt[Cos[e + f *x]]/Sqrt[g*Cos[e + f*x]] Int[Sqrt[d*Sin[e + f*x]]/(Sqrt[Cos[e + f*x]]*(a + b*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, f, g}, x] && NeQ[a^2 - b^ 2, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(443\) vs. \(2(173)=346\).
Time = 1.76 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.12
method | result | size |
default | \(\frac {\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )+1}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \left (\sqrt {-a^{2}+b^{2}}\, \Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right )+\sqrt {-a^{2}+b^{2}}\, \Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right )+\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right ) a -\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right ) b -\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) a +\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) b \right ) \sqrt {d \sin \left (f x +e \right )}\, \left (\cot \left (f x +e \right )+\csc \left (f x +e \right )\right ) \sqrt {2}\, a}{f \sqrt {g \cos \left (f x +e \right )}\, \sqrt {-a^{2}+b^{2}}\, \left (-b +\sqrt {-a^{2}+b^{2}}+a \right ) \left (b +\sqrt {-a^{2}+b^{2}}-a \right )}\) | \(444\) |
1/f*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(-csc(f*x+e)+cot(f*x+e)+1)^(1/2)*(-cs c(f*x+e)+cot(f*x+e))^(1/2)*((-a^2+b^2)^(1/2)*EllipticPi((-cot(f*x+e)+csc(f *x+e)+1)^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a),1/2*2^(1/2))+(-a^2+b^2)^(1/2)*Ell ipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^( 1/2))+EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),a/(-b+(-a^2+b^2)^(1/2)+a ),1/2*2^(1/2))*a-EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),a/(-b+(-a^2+b ^2)^(1/2)+a),1/2*2^(1/2))*b-EllipticPi((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),-a /(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*a+EllipticPi((-cot(f*x+e)+csc(f*x+e)+ 1)^(1/2),-a/(b+(-a^2+b^2)^(1/2)-a),1/2*2^(1/2))*b)*(d*sin(f*x+e))^(1/2)/(g *cos(f*x+e))^(1/2)*(cot(f*x+e)+csc(f*x+e))*2^(1/2)*a/(-a^2+b^2)^(1/2)/(-b+ (-a^2+b^2)^(1/2)+a)/(b+(-a^2+b^2)^(1/2)-a)
Timed out. \[ \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {\sqrt {d \sin {\left (e + f x \right )}}}{\sqrt {g \cos {\left (e + f x \right )}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]
\[ \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\sqrt {d \sin \left (f x + e\right )}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
\[ \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\sqrt {d \sin \left (f x + e\right )}}{\sqrt {g \cos \left (f x + e\right )} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {\sqrt {d\,\sin \left (e+f\,x\right )}}{\sqrt {g\,\cos \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]