Integrand size = 37, antiderivative size = 208 \[ \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {2 \sqrt {2} \sqrt {g} \operatorname {EllipticPi}\left (-\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{\sqrt {-a+b} \sqrt {a+b} f \sqrt {d \sin (e+f x)}}-\frac {2 \sqrt {2} \sqrt {g} \operatorname {EllipticPi}\left (\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{\sqrt {-a+b} \sqrt {a+b} f \sqrt {d \sin (e+f x)}} \]
2*EllipticPi((g*cos(f*x+e))^(1/2)/g^(1/2)/(1+sin(f*x+e))^(1/2),-(-a+b)^(1/ 2)/(a+b)^(1/2),I)*2^(1/2)*g^(1/2)*sin(f*x+e)^(1/2)/f/(-a+b)^(1/2)/(a+b)^(1 /2)/(d*sin(f*x+e))^(1/2)-2*EllipticPi((g*cos(f*x+e))^(1/2)/g^(1/2)/(1+sin( f*x+e))^(1/2),(-a+b)^(1/2)/(a+b)^(1/2),I)*2^(1/2)*g^(1/2)*sin(f*x+e)^(1/2) /f/(-a+b)^(1/2)/(a+b)^(1/2)/(d*sin(f*x+e))^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 13.32 (sec) , antiderivative size = 371, normalized size of antiderivative = 1.78 \[ \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx=\frac {\sqrt {g \cos (e+f x)} \sqrt {\tan (e+f x)} \left (a \sqrt {\sec ^2(e+f x)}+b \tan (e+f x)\right ) \left (\frac {3 \sqrt {2} a^{3/2} \left (-2 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}}{\sqrt {a}}\right )+2 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}}{\sqrt {a}}\right )-\log \left (-a+\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}-\sqrt {a^2-b^2} \tan (e+f x)\right )+\log \left (a+\sqrt {2} \sqrt {a} \sqrt [4]{a^2-b^2} \sqrt {\tan (e+f x)}+\sqrt {a^2-b^2} \tan (e+f x)\right )\right )}{\sqrt [4]{a^2-b^2}}-8 b \operatorname {AppellF1}\left (\frac {3}{4},\frac {1}{2},1,\frac {7}{4},-\tan ^2(e+f x),\frac {\left (-a^2+b^2\right ) \tan ^2(e+f x)}{a^2}\right ) \tan ^{\frac {3}{2}}(e+f x)\right )}{12 a^2 f \sqrt {\sec ^2(e+f x)} \sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \]
(Sqrt[g*Cos[e + f*x]]*Sqrt[Tan[e + f*x]]*(a*Sqrt[Sec[e + f*x]^2] + b*Tan[e + f*x])*((3*Sqrt[2]*a^(3/2)*(-2*ArcTan[1 - (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqr t[Tan[e + f*x]])/Sqrt[a]] + 2*ArcTan[1 + (Sqrt[2]*(a^2 - b^2)^(1/4)*Sqrt[T an[e + f*x]])/Sqrt[a]] - Log[-a + Sqrt[2]*Sqrt[a]*(a^2 - b^2)^(1/4)*Sqrt[T an[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]]))/(a^2 - b^2)^(1/4) - 8*b*AppellF1[3/4, 1/2, 1, 7/4, -Tan[e + f*x]^2, ((-a^2 + b^2 )*Tan[e + f*x]^2)/a^2]*Tan[e + f*x]^(3/2)))/(12*a^2*f*Sqrt[Sec[e + f*x]^2] *Sqrt[d*Sin[e + f*x]]*(a + b*Sin[e + f*x]))
Time = 0.65 (sec) , antiderivative size = 185, normalized size of antiderivative = 0.89, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.162, Rules used = {3042, 3385, 3042, 3384, 993, 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.
| \(\displaystyle \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3385 |
| \(\displaystyle \frac {\sqrt {\sin (e+f x)} \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)} (a+b \sin (e+f x))}dx}{\sqrt {d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\sqrt {\sin (e+f x)} \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)} (a+b \sin (e+f x))}dx}{\sqrt {d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 3384 |
| \(\displaystyle -\frac {4 \sqrt {2} g \sqrt {\sin (e+f x)} \int \frac {g \cos (e+f x)}{(\sin (e+f x)+1) \sqrt {1-\frac {\cos ^2(e+f x)}{(\sin (e+f x)+1)^2}} \left ((a+b) g^2+\frac {(a-b) \cos ^2(e+f x) g^2}{(\sin (e+f x)+1)^2}\right )}d\frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)+1}}}{f \sqrt {d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 993 |
| \(\displaystyle -\frac {4 \sqrt {2} g \sqrt {\sin (e+f x)} \left (\frac {\int \frac {1}{\sqrt {1-\frac {\cos ^2(e+f x)}{(\sin (e+f x)+1)^2}} \left (\sqrt {a+b} g-\frac {\sqrt {b-a} g \cos (e+f x)}{\sin (e+f x)+1}\right )}d\frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)+1}}}{2 \sqrt {b-a}}-\frac {\int \frac {1}{\sqrt {1-\frac {\cos ^2(e+f x)}{(\sin (e+f x)+1)^2}} \left (\sqrt {a+b} g+\frac {\sqrt {b-a} \cos (e+f x) g}{\sin (e+f x)+1}\right )}d\frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)+1}}}{2 \sqrt {b-a}}\right )}{f \sqrt {d \sin (e+f x)}}\) |
| \(\Big \downarrow \) 1542 |
| \(\displaystyle -\frac {4 \sqrt {2} g \sqrt {\sin (e+f x)} \left (\frac {\operatorname {EllipticPi}\left (\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right ),-1\right )}{2 \sqrt {g} \sqrt {b-a} \sqrt {a+b}}-\frac {\operatorname {EllipticPi}\left (-\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right ),-1\right )}{2 \sqrt {g} \sqrt {b-a} \sqrt {a+b}}\right )}{f \sqrt {d \sin (e+f x)}}\) |
(-4*Sqrt[2]*g*(-1/2*EllipticPi[-(Sqrt[-a + b]/Sqrt[a + b]), ArcSin[Sqrt[g* Cos[e + f*x]]/(Sqrt[g]*Sqrt[1 + Sin[e + f*x]])], -1]/(Sqrt[-a + b]*Sqrt[a + b]*Sqrt[g]) + EllipticPi[Sqrt[-a + b]/Sqrt[a + b], ArcSin[Sqrt[g*Cos[e + f*x]]/(Sqrt[g]*Sqrt[1 + Sin[e + f*x]])], -1]/(2*Sqrt[-a + b]*Sqrt[a + b]* Sqrt[g]))*Sqrt[Sin[e + f*x]])/(f*Sqrt[d*Sin[e + f*x]])
3.15.11.3.1 Defintions of rubi rules used
Int[(x_)^2/(((a_) + (b_.)*(x_)^4)*Sqrt[(c_) + (d_.)*(x_)^4]), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2* b) Int[1/((r + s*x^2)*Sqrt[c + d*x^4]), x], x] - Simp[s/(2*b) Int[1/((r - s*x^2)*Sqrt[c + d*x^4]), x], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]
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[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[sin[(e_.) + (f_.)*(x_)]]*((a_ ) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[-4*Sqrt[2]*(g/f) S ubst[Int[x^2/(((a + b)*g^2 + (a - b)*x^4)*Sqrt[1 - x^4/g^2]), x], x, Sqrt[g *Cos[e + f*x]]/Sqrt[1 + Sin[e + f*x]]], x] /; FreeQ[{a, b, e, f, g}, x] && NeQ[a^2 - b^2, 0]
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/(Sqrt[(d_)*sin[(e_.) + (f_.)*(x_)]] *((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])), x_Symbol] :> Simp[Sqrt[Sin[e + f* x]]/Sqrt[d*Sin[e + f*x]] Int[Sqrt[g*Cos[e + f*x]]/(Sqrt[Sin[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. \(481\) vs. \(2(164)=328\).
Time = 1.81 (sec) , antiderivative size = 482, normalized size of antiderivative = 2.32
| method | result | size |
| default | \(\frac {\left (a -b \right ) \left (2 \sqrt {-a^{2}+b^{2}}\, F\left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {\sqrt {2}}{2}\right )-\sqrt {-a^{2}+b^{2}}\, \Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right )-\sqrt {-a^{2}+b^{2}}\, \Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right )+\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right ) a +\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, \frac {a}{-b +\sqrt {-a^{2}+b^{2}}+a}, \frac {\sqrt {2}}{2}\right ) b -\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) a -\Pi \left (\sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}, -\frac {a}{b +\sqrt {-a^{2}+b^{2}}-a}, \frac {\sqrt {2}}{2}\right ) b \right ) \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )}\, \sqrt {-\csc \left (f x +e \right )+\cot \left (f x +e \right )+1}\, \sqrt {-\cot \left (f x +e \right )+\csc \left (f x +e \right )+1}\, \sqrt {g \cos \left (f x +e \right )}\, \left (\sec \left (f x +e \right )+1\right ) \sqrt {2}}{f \sqrt {d \sin \left (f x +e \right )}\, \sqrt {-a^{2}+b^{2}}\, \left (-b +\sqrt {-a^{2}+b^{2}}+a \right ) \left (b +\sqrt {-a^{2}+b^{2}}-a \right )}\) | \(482\) |
1/f*(a-b)*(2*(-a^2+b^2)^(1/2)*EllipticF((-cot(f*x+e)+csc(f*x+e)+1)^(1/2),1 /2*2^(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)*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))+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)*(-csc(f*x+e)+cot(f*x+e))^(1/2)*(-csc(f* x+e)+cot(f*x+e)+1)^(1/2)*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(g*cos(f*x+e))^( 1/2)/(d*sin(f*x+e))^(1/2)*(sec(f*x+e)+1)*2^(1/2)/(-a^2+b^2)^(1/2)/(-b+(-a^ 2+b^2)^(1/2)+a)/(b+(-a^2+b^2)^(1/2)-a)
Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
\[ \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {\sqrt {g \cos {\left (e + f x \right )}}}{\sqrt {d \sin {\left (e + f x \right )}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]
\[ \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )}}{{\left (b \sin \left (f x + e\right ) + a\right )} \sqrt {d \sin \left (f x + e\right )}} \,d x } \]
\[ \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx=\int { \frac {\sqrt {g \cos \left (f x + e\right )}}{{\left (b \sin \left (f x + e\right ) + a\right )} \sqrt {d \sin \left (f x + e\right )}} \,d x } \]
Timed out. \[ \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))} \, dx=\int \frac {\sqrt {g\,\cos \left (e+f\,x\right )}}{\sqrt {d\,\sin \left (e+f\,x\right )}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]