Integrand size = 37, antiderivative size = 321 \[ \int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=-\frac {2 \sqrt {2} \sqrt {-a^2+b^2} g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 d^{3/2} f \sqrt {g \cos (e+f x)}}+\frac {2 \sqrt {2} \sqrt {-a^2+b^2} g^2 \sqrt {\cos (e+f x)} \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {-a^2+b^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {1+\cos (e+f x)}}\right ),-1\right )}{a^2 d^{3/2} f \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}-\frac {b g^2 \operatorname {EllipticF}\left (e-\frac {\pi }{4}+f x,2\right ) \sqrt {\sin (2 e+2 f x)}}{a^2 d f \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}} \]
-2*g^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)*(-a^2+b^2)^(1/2)*cos(f*x+e)^(1/2)/a^2/d^(3/2) /f/(g*cos(f*x+e))^(1/2)+2*g^2*EllipticPi((d*sin(f*x+e))^(1/2)/d^(1/2)/(1+c os(f*x+e))^(1/2),-a/(b+(-a^2+b^2)^(1/2)),I)*2^(1/2)*(-a^2+b^2)^(1/2)*cos(f *x+e)^(1/2)/a^2/d^(3/2)/f/(g*cos(f*x+e))^(1/2)-2*g*(g*cos(f*x+e))^(1/2)/a/ d/f/(d*sin(f*x+e))^(1/2)+b*g^2*(sin(e+1/4*Pi+f*x)^2)^(1/2)/sin(e+1/4*Pi+f* x)*EllipticF(cos(e+1/4*Pi+f*x),2^(1/2))*sin(2*f*x+2*e)^(1/2)/a^2/d/f/(g*co s(f*x+e))^(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 = 18.55 (sec) , antiderivative size = 1095, normalized size of antiderivative = 3.41 \[ \int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=-\frac {2 (g \cos (e+f x))^{3/2} \tan (e+f x)}{a f (d \sin (e+f x))^{3/2}}-\frac {(g \cos (e+f x))^{3/2} \sin ^{\frac {3}{2}}(e+f x) \left (-\frac {2 b \left (a+b \sqrt {1-\cos ^2(e+f x)}\right ) \left (\frac {5 a \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right ) \sqrt {\cos (e+f x)}}{\left (1-\cos ^2(e+f x)\right )^{3/4} \left (5 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {1}{4},\frac {3}{4},1,\frac {5}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+\left (-4 b^2 \operatorname {AppellF1}\left (\frac {5}{4},\frac {3}{4},2,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )+3 \left (a^2-b^2\right ) \operatorname {AppellF1}\left (\frac {5}{4},\frac {7}{4},1,\frac {9}{4},\cos ^2(e+f x),\frac {b^2 \cos ^2(e+f x)}{-a^2+b^2}\right )\right ) \cos ^2(e+f x)\right ) \left (a^2+b^2 \left (-1+\cos ^2(e+f x)\right )\right )}-\frac {\left (\frac {1}{8}-\frac {i}{8}\right ) b \left (2 \arctan \left (1-\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )-2 \arctan \left (1+\frac {(1+i) \sqrt {a} \sqrt {\cos (e+f x)}}{\sqrt [4]{-a^2+b^2} \sqrt [4]{-1+\cos ^2(e+f x)}}\right )+\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}-\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )-\log \left (\sqrt {-a^2+b^2}+\frac {i a \cos (e+f x)}{\sqrt {-1+\cos ^2(e+f x)}}+\frac {(1+i) \sqrt {a} \sqrt [4]{-a^2+b^2} \sqrt {\cos (e+f x)}}{\sqrt [4]{-1+\cos ^2(e+f x)}}\right )\right )}{\sqrt {a} \left (-a^2+b^2\right )^{3/4}}\right ) \sqrt {\sin (e+f x)}}{\sqrt [4]{1-\cos ^2(e+f x)} (a+b \sin (e+f x))}+\frac {2 a \sqrt {\sin (e+f x)} \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 ) \left (b \tan (e+f x)+a \sqrt {1+\tan ^2(e+f x)}\right )}{\cos ^{\frac {5}{2}}(e+f x) (a+b \sin (e+f x)) \sqrt {\tan (e+f x)} \left (1+\tan ^2(e+f x)\right )^{3/2}}\right )}{a f \cos ^{\frac {3}{2}}(e+f x) (d \sin (e+f x))^{3/2}} \]
(-2*(g*Cos[e + f*x])^(3/2)*Tan[e + f*x])/(a*f*(d*Sin[e + f*x])^(3/2)) - (( g*Cos[e + f*x])^(3/2)*Sin[e + f*x]^(3/2)*((-2*b*(a + b*Sqrt[1 - Cos[e + f* x]^2])*((5*a*(a^2 - b^2)*AppellF1[1/4, 3/4, 1, 5/4, Cos[e + f*x]^2, (b^2*C os[e + f*x]^2)/(-a^2 + b^2)]*Sqrt[Cos[e + f*x]])/((1 - Cos[e + f*x]^2)^(3/ 4)*(5*(a^2 - b^2)*AppellF1[1/4, 3/4, 1, 5/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + (-4*b^2*AppellF1[5/4, 3/4, 2, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)] + 3*(a^2 - b^2)*AppellF1[5/4, 7/4, 1, 9/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^2)*( a^2 + b^2*(-1 + Cos[e + f*x]^2))) - ((1/8 - I/8)*b*(2*ArcTan[1 - ((1 + I)* Sqrt[a]*Sqrt[Cos[e + f*x]])/((-a^2 + b^2)^(1/4)*(-1 + Cos[e + f*x]^2)^(1/4 ))] - 2*ArcTan[1 + ((1 + I)*Sqrt[a]*Sqrt[Cos[e + f*x]])/((-a^2 + b^2)^(1/4 )*(-1 + Cos[e + f*x]^2)^(1/4))] + Log[Sqrt[-a^2 + b^2] + (I*a*Cos[e + f*x] )/Sqrt[-1 + Cos[e + f*x]^2] - ((1 + I)*Sqrt[a]*(-a^2 + b^2)^(1/4)*Sqrt[Cos [e + f*x]])/(-1 + Cos[e + f*x]^2)^(1/4)] - Log[Sqrt[-a^2 + b^2] + (I*a*Cos [e + f*x])/Sqrt[-1 + Cos[e + f*x]^2] + ((1 + I)*Sqrt[a]*(-a^2 + b^2)^(1/4) *Sqrt[Cos[e + f*x]])/(-1 + Cos[e + f*x]^2)^(1/4)]))/(Sqrt[a]*(-a^2 + b^2)^ (3/4)))*Sqrt[Sin[e + f*x]])/((1 - Cos[e + f*x]^2)^(1/4)*(a + b*Sin[e + f*x ])) + (2*a*Sqrt[Sin[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]] + Log[-a + Sqrt[2]*Sqrt[a]*(a^2 - b^2...
Time = 1.60 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.10, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.297, Rules used = {3042, 3378, 3042, 3043, 3053, 3042, 3120, 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.
\(\displaystyle \int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))}dx\) |
\(\Big \downarrow \) 3378 |
\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}dx}{a^2 d}+\frac {g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2}}dx}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}dx}{a^2 d}+\frac {g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} (d \sin (e+f x))^{3/2}}dx}{a}\) |
\(\Big \downarrow \) 3043 |
\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \int \frac {1}{\sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}}dx}{a^2 d}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3053 |
\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{a^2 d \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \int \frac {1}{\sqrt {\sin (2 e+2 f x)}}dx}{a^2 d \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3120 |
\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {g \cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a^2 d f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3387 |
\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2 \sqrt {g \cos (e+f x)}}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a^2 d f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \int \frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)} (a+b \sin (e+f x))}dx}{a^2 d^2 \sqrt {g \cos (e+f x)}}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a^2 d f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\) |
\(\Big \downarrow \) 3386 |
\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \left (\frac {2 \sqrt {2} d \left (1-\frac {b}{\sqrt {b^2-a^2}}\right ) \int \frac {1}{\left (\left (b-\sqrt {b^2-a^2}\right ) d+\frac {a \sin (e+f x) d}{\cos (e+f x)+1}\right ) \sqrt {1-\frac {\sin ^2(e+f x)}{(\cos (e+f x)+1)^2}}}d\frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)+1}}}{f}+\frac {2 \sqrt {2} d \left (\frac {b}{\sqrt {b^2-a^2}}+1\right ) \int \frac {1}{\left (\left (b+\sqrt {b^2-a^2}\right ) d+\frac {a \sin (e+f x) d}{\cos (e+f x)+1}\right ) \sqrt {1-\frac {\sin ^2(e+f x)}{(\cos (e+f x)+1)^2}}}d\frac {\sqrt {d \sin (e+f x)}}{\sqrt {\cos (e+f x)+1}}}{f}\right )}{a^2 d^2 \sqrt {g \cos (e+f x)}}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a^2 d f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\) |
\(\Big \downarrow \) 1542 |
\(\displaystyle -\frac {g^2 \left (a^2-b^2\right ) \sqrt {\cos (e+f x)} \left (\frac {2 \sqrt {2} \sqrt {d} \left (1-\frac {b}{\sqrt {b^2-a^2}}\right ) \operatorname {EllipticPi}\left (-\frac {a}{b-\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right ),-1\right )}{f \left (b-\sqrt {b^2-a^2}\right )}+\frac {2 \sqrt {2} \sqrt {d} \left (\frac {b}{\sqrt {b^2-a^2}}+1\right ) \operatorname {EllipticPi}\left (-\frac {a}{b+\sqrt {b^2-a^2}},\arcsin \left (\frac {\sqrt {d \sin (e+f x)}}{\sqrt {d} \sqrt {\cos (e+f x)+1}}\right ),-1\right )}{f \left (\sqrt {b^2-a^2}+b\right )}\right )}{a^2 d^2 \sqrt {g \cos (e+f x)}}-\frac {b g^2 \sqrt {\sin (2 e+2 f x)} \operatorname {EllipticF}\left (e+f x-\frac {\pi }{4},2\right )}{a^2 d f \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}-\frac {2 g \sqrt {g \cos (e+f x)}}{a d f \sqrt {d \sin (e+f x)}}\) |
-(((a^2 - b^2)*g^2*Sqrt[Cos[e + f*x]]*((2*Sqrt[2]*(1 - b/Sqrt[-a^2 + b^2]) *Sqrt[d]*EllipticPi[-(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/Sqrt[-a^2 + b^2])*Sqrt[d]*EllipticPi[-(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)))/(a^2*d^2*Sqrt[g*Cos[e + f*x]])) - (2*g*Sqrt [g*Cos[e + f*x]])/(a*d*f*Sqrt[d*Sin[e + f*x]]) - (b*g^2*EllipticF[e - Pi/4 + f*x, 2]*Sqrt[Sin[2*e + 2*f*x]])/(a^2*d*f*Sqrt[g*Cos[e + f*x]]*Sqrt[d*Si n[e + f*x]])
3.15.19.3.1 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[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_.)*((a_.)*sin[(e_.) + (f_.)*(x_)])^( m_.), x_Symbol] :> Simp[(a*Sin[e + f*x])^(m + 1)*((b*Cos[e + f*x])^(n + 1)/ (a*b*f*(m + 1))), x] /; FreeQ[{a, b, e, f, m, n}, x] && EqQ[m + n + 2, 0] & & NeQ[m, -1]
Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_ )]]), x_Symbol] :> Simp[Sqrt[Sin[2*e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b *Cos[e + f*x]]) Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f }, x]
Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticF[(1/2 )*(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^( n_))/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[g^2/a Int[ (g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] + (-Simp[b*(g^2/(a^2*d) ) Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] - Simp[g^ 2*((a^2 - b^2)/(a^2*d^2)) Int[(g*Cos[e + f*x])^(p - 2)*((d*Sin[e + f*x])^ (n + 2)/(a + b*Sin[e + f*x])), x], x]) /; FreeQ[{a, b, d, e, f, g}, x] && N eQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && GtQ[p, 1] && (LeQ[n, -2] || (EqQ [n, -3/2] && EqQ[p, 3/2]))
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. \(1783\) vs. \(2(299)=598\).
Time = 1.60 (sec) , antiderivative size = 1784, normalized size of antiderivative = 5.56
1/f*csc(f*x+e)/(d/((1-cos(f*x+e))^2*csc(f*x+e)^2+1)*(csc(f*x+e)-cot(f*x+e) ))^(3/2)*(1-cos(f*x+e))*(-g*((1-cos(f*x+e))^2*csc(f*x+e)^2-1)/((1-cos(f*x+ e))^2*csc(f*x+e)^2+1))^(3/2)*(2*EllipticF((-cot(f*x+e)+csc(f*x+e)+1)^(1/2) ,1/2*2^(1/2))*a*b*(-a^2+b^2)^(1/2)*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(2+2*c ot(f*x+e)-2*csc(f*x+e))^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(1/2)-2*EllipticF(( -cot(f*x+e)+csc(f*x+e)+1)^(1/2),1/2*2^(1/2))*b^2*(-a^2+b^2)^(1/2)*(-cot(f* x+e)+csc(f*x+e)+1)^(1/2)*(2+2*cot(f*x+e)-2*csc(f*x+e))^(1/2)*(-csc(f*x+e)+ cot(f*x+e))^(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^3*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(2+2*cot(f *x+e)-2*csc(f*x+e))^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(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*(- cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(2+2*cot(f*x+e)-2*csc(f*x+e))^(1/2)*(-csc(f *x+e)+cot(f*x+e))^(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*(-a^2+b^2)^(1/2)*(-cot(f*x+e)+csc(f*x +e)+1)^(1/2)*(2+2*cot(f*x+e)-2*csc(f*x+e))^(1/2)*(-csc(f*x+e)+cot(f*x+e))^ (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*b^2*(-cot(f*x+e)+csc(f*x+e)+1)^(1/2)*(2+2*cot(f*x+e)-2*cs c(f*x+e))^(1/2)*(-csc(f*x+e)+cot(f*x+e))^(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))*b^3*(-cot(f*x+e)+c sc(f*x+e)+1)^(1/2)*(2+2*cot(f*x+e)-2*csc(f*x+e))^(1/2)*(-csc(f*x+e)+cot...
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
\[ \int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {\left (g \cos {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]
\[ \int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}}}{{\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
\[ \int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}}}{{\left (b \sin \left (f x + e\right ) + a\right )} \left (d \sin \left (f x + e\right )\right )^{\frac {3}{2}}} \,d x } \]
Timed out. \[ \int \frac {(g \cos (e+f x))^{3/2}}{(d \sin (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}}{{\left (d\,\sin \left (e+f\,x\right )\right )}^{3/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]