Integrand size = 37, antiderivative size = 379 \[ \int \frac {(d \sin (e+f x))^{3/2}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\frac {2 \sqrt {2} a^2 d^2 \operatorname {EllipticPi}\left (-\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{(-a+b)^{3/2} (a+b)^{3/2} f g^{3/2} \sqrt {d \sin (e+f x)}}-\frac {2 \sqrt {2} a^2 d^2 \operatorname {EllipticPi}\left (\frac {\sqrt {-a+b}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {1+\sin (e+f x)}}\right ),-1\right ) \sqrt {\sin (e+f x)}}{(-a+b)^{3/2} (a+b)^{3/2} f g^{3/2} \sqrt {d \sin (e+f x)}}+\frac {2 a d \sqrt {d \sin (e+f x)}}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}}-\frac {2 b (d \sin (e+f x))^{3/2}}{\left (a^2-b^2\right ) f g \sqrt {g \cos (e+f x)}}+\frac {2 b d \sqrt {g \cos (e+f x)} E\left (\left .e-\frac {\pi }{4}+f x\right |2\right ) \sqrt {d \sin (e+f x)}}{\left (a^2-b^2\right ) f g^2 \sqrt {\sin (2 e+2 f x)}} \]
-2*b*(d*sin(f*x+e))^(3/2)/(a^2-b^2)/f/g/(g*cos(f*x+e))^(1/2)+2*a^2*d^2*Ell ipticPi((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)*sin(f*x+e)^(1/2)/(-a+b)^(3/2)/(a+b)^(3/2)/f/g^(3/2)/( d*sin(f*x+e))^(1/2)-2*a^2*d^2*EllipticPi((g*cos(f*x+e))^(1/2)/g^(1/2)/(1+s in(f*x+e))^(1/2),(-a+b)^(1/2)/(a+b)^(1/2),I)*2^(1/2)*sin(f*x+e)^(1/2)/(-a+ b)^(3/2)/(a+b)^(3/2)/f/g^(3/2)/(d*sin(f*x+e))^(1/2)+2*a*d*(d*sin(f*x+e))^( 1/2)/(a^2-b^2)/f/g/(g*cos(f*x+e))^(1/2)-2*b*d*(sin(e+1/4*Pi+f*x)^2)^(1/2)/ sin(e+1/4*Pi+f*x)*EllipticE(cos(e+1/4*Pi+f*x),2^(1/2))*(g*cos(f*x+e))^(1/2 )*(d*sin(f*x+e))^(1/2)/(a^2-b^2)/f/g^2/sin(2*f*x+2*e)^(1/2)
Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.
Time = 23.74 (sec) , antiderivative size = 1648, normalized size of antiderivative = 4.35 \[ \int \frac {(d \sin (e+f x))^{3/2}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx =\text {Too large to display} \]
(2*Cot[e + f*x]*(d*Sin[e + f*x])^(3/2)*(a - b*Sin[e + f*x]))/((a^2 - b^2)* f*(g*Cos[e + f*x])^(3/2)) - (Cos[e + f*x]^(3/2)*(d*Sin[e + f*x])^(3/2)*((4 *a*b*(-(b*AppellF1[3/4, -1/4, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2) /(-a^2 + b^2)]) + a*AppellF1[3/4, 1/4, 1, 7/4, Cos[e + f*x]^2, (b^2*Cos[e + f*x]^2)/(-a^2 + b^2)])*Cos[e + f*x]^(3/2)*(a + b*Sqrt[1 - Cos[e + f*x]^2 ])*Sin[e + f*x]^(3/2))/(3*(a^2 - b^2)*(1 - Cos[e + f*x]^2)^(3/4)*(a + b*Si n[e + f*x])) + ((a^2 - b^2)*Sqrt[Tan[e + f*x]]*((3*Sqrt[2]*a^(3/2)*(-2*Arc Tan[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 + Sq rt[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]]))/(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))*(b*Tan[e + f*x] + a*Sqrt[1 + Tan[e + f*x]^2]))/(12*a^2*Cos[e + f*x] ^(3/2)*Sqrt[Sin[e + f*x]]*(a + b*Sin[e + f*x])*(1 + Tan[e + f*x]^2)^(3/2)) + (Cos[2*(e + f*x)]*Sqrt[Tan[e + f*x]]*(b*Tan[e + f*x] + a*Sqrt[1 + Tan[e + f*x]^2])*(56*b*(-3*a^2 + b^2)*AppellF1[3/4, 1/2, 1, 7/4, -Tan[e + f*x]^ 2, (-1 + b^2/a^2)*Tan[e + f*x]^2]*Tan[e + f*x]^(3/2) + 24*b*(-a^2 + b^2)*A ppellF1[7/4, 1/2, 1, 11/4, -Tan[e + f*x]^2, (-1 + b^2/a^2)*Tan[e + f*x]^2] *Tan[e + f*x]^(7/2) + 21*a^(3/2)*(4*Sqrt[2]*a^(3/2)*ArcTan[1 - Sqrt[2]*...
Time = 1.94 (sec) , antiderivative size = 358, normalized size of antiderivative = 0.94, number of steps used = 15, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.378, Rules used = {3042, 3381, 3042, 3043, 3051, 3042, 3052, 3042, 3119, 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.
| \(\displaystyle \int \frac {(d \sin (e+f x))^{3/2}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {(d \sin (e+f x))^{3/2}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))}dx\) |
| \(\Big \downarrow \) 3381 |
| \(\displaystyle -\frac {a^2 d^2 \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}+\frac {a d^2 \int \frac {1}{(g \cos (e+f x))^{3/2} \sqrt {d \sin (e+f x)}}dx}{a^2-b^2}-\frac {b d \int \frac {\sqrt {d \sin (e+f x)}}{(g \cos (e+f x))^{3/2}}dx}{a^2-b^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a^2 d^2 \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}+\frac {a d^2 \int \frac {1}{(g \cos (e+f x))^{3/2} \sqrt {d \sin (e+f x)}}dx}{a^2-b^2}-\frac {b d \int \frac {\sqrt {d \sin (e+f x)}}{(g \cos (e+f x))^{3/2}}dx}{a^2-b^2}\) |
| \(\Big \downarrow \) 3043 |
| \(\displaystyle -\frac {a^2 d^2 \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b d \int \frac {\sqrt {d \sin (e+f x)}}{(g \cos (e+f x))^{3/2}}dx}{a^2-b^2}+\frac {2 a d \sqrt {d \sin (e+f x)}}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\) |
| \(\Big \downarrow \) 3051 |
| \(\displaystyle -\frac {a^2 d^2 \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b d \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 \int \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}dx}{g^2}\right )}{a^2-b^2}+\frac {2 a d \sqrt {d \sin (e+f x)}}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a^2 d^2 \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b d \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 \int \sqrt {g \cos (e+f x)} \sqrt {d \sin (e+f x)}dx}{g^2}\right )}{a^2-b^2}+\frac {2 a d \sqrt {d \sin (e+f x)}}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\) |
| \(\Big \downarrow \) 3052 |
| \(\displaystyle -\frac {a^2 d^2 \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b d \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\sin (2 e+2 f x)}dx}{g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{a^2-b^2}+\frac {2 a d \sqrt {d \sin (e+f x)}}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a^2 d^2 \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b d \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)} \int \sqrt {\sin (2 e+2 f x)}dx}{g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{a^2-b^2}+\frac {2 a d \sqrt {d \sin (e+f x)}}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\) |
| \(\Big \downarrow \) 3119 |
| \(\displaystyle -\frac {a^2 d^2 \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {d \sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right )}-\frac {b d \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{a^2-b^2}+\frac {2 a d \sqrt {d \sin (e+f x)}}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\) |
| \(\Big \downarrow \) 3385 |
| \(\displaystyle -\frac {a^2 d^2 \sqrt {\sin (e+f x)} \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}-\frac {b d \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{a^2-b^2}+\frac {2 a d \sqrt {d \sin (e+f x)}}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {a^2 d^2 \sqrt {\sin (e+f x)} \int \frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)} (a+b \sin (e+f x))}dx}{g^2 \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}-\frac {b d \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{a^2-b^2}+\frac {2 a d \sqrt {d \sin (e+f x)}}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\) |
| \(\Big \downarrow \) 3384 |
| \(\displaystyle \frac {4 \sqrt {2} a^2 d^2 \sqrt {\sin (e+f x)} \int \frac {g \cos (e+f x)}{(\sin (e+f x)+1) \sqrt {1-\frac {\cos ^2(e+f x)}{(\sin (e+f x)+1)^2}} \left ((a+b) g^2+\frac {(a-b) \cos ^2(e+f x) g^2}{(\sin (e+f x)+1)^2}\right )}d\frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)+1}}}{f g \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}-\frac {b d \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{a^2-b^2}+\frac {2 a d \sqrt {d \sin (e+f x)}}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\) |
| \(\Big \downarrow \) 993 |
| \(\displaystyle \frac {4 \sqrt {2} a^2 d^2 \sqrt {\sin (e+f x)} \left (\frac {\int \frac {1}{\sqrt {1-\frac {\cos ^2(e+f x)}{(\sin (e+f x)+1)^2}} \left (\sqrt {a+b} g-\frac {\sqrt {b-a} g \cos (e+f x)}{\sin (e+f x)+1}\right )}d\frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)+1}}}{2 \sqrt {b-a}}-\frac {\int \frac {1}{\sqrt {1-\frac {\cos ^2(e+f x)}{(\sin (e+f x)+1)^2}} \left (\sqrt {a+b} g+\frac {\sqrt {b-a} \cos (e+f x) g}{\sin (e+f x)+1}\right )}d\frac {\sqrt {g \cos (e+f x)}}{\sqrt {\sin (e+f x)+1}}}{2 \sqrt {b-a}}\right )}{f g \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}-\frac {b d \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{a^2-b^2}+\frac {2 a d \sqrt {d \sin (e+f x)}}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\) |
| \(\Big \downarrow \) 1542 |
| \(\displaystyle \frac {4 \sqrt {2} a^2 d^2 \sqrt {\sin (e+f x)} \left (\frac {\operatorname {EllipticPi}\left (\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right ),-1\right )}{2 \sqrt {g} \sqrt {b-a} \sqrt {a+b}}-\frac {\operatorname {EllipticPi}\left (-\frac {\sqrt {b-a}}{\sqrt {a+b}},\arcsin \left (\frac {\sqrt {g \cos (e+f x)}}{\sqrt {g} \sqrt {\sin (e+f x)+1}}\right ),-1\right )}{2 \sqrt {g} \sqrt {b-a} \sqrt {a+b}}\right )}{f g \left (a^2-b^2\right ) \sqrt {d \sin (e+f x)}}-\frac {b d \left (\frac {2 (d \sin (e+f x))^{3/2}}{d f g \sqrt {g \cos (e+f x)}}-\frac {2 E\left (\left .e+f x-\frac {\pi }{4}\right |2\right ) \sqrt {d \sin (e+f x)} \sqrt {g \cos (e+f x)}}{f g^2 \sqrt {\sin (2 e+2 f x)}}\right )}{a^2-b^2}+\frac {2 a d \sqrt {d \sin (e+f x)}}{f g \left (a^2-b^2\right ) \sqrt {g \cos (e+f x)}}\) |
(4*Sqrt[2]*a^2*d^2*(-1/2*EllipticPi[-(Sqrt[-a + b]/Sqrt[a + b]), ArcSin[Sq rt[g*Cos[e + f*x]]/(Sqrt[g]*Sqrt[1 + Sin[e + f*x]])], -1]/(Sqrt[-a + b]*Sq rt[a + b]*Sqrt[g]) + EllipticPi[Sqrt[-a + b]/Sqrt[a + b], ArcSin[Sqrt[g*Co s[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]])/((a^2 - b^2)*f*g*Sqrt[d*Sin[e + f*x]]) + (2*a*d*Sqrt[d*Sin[e + f*x]])/((a^2 - b^2)*f*g*Sqrt[g*Cos[e + f*x]]) - (b *d*((2*(d*Sin[e + f*x])^(3/2))/(d*f*g*Sqrt[g*Cos[e + f*x]]) - (2*Sqrt[g*Co s[e + f*x]]*EllipticE[e - Pi/4 + f*x, 2]*Sqrt[d*Sin[e + f*x]])/(f*g^2*Sqrt [Sin[2*e + 2*f*x]])))/(a^2 - b^2)
3.15.37.3.1 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[(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[(cos[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sin[(e_.) + (f_.)*(x_)])^(n _), x_Symbol] :> Simp[(-(b*Sin[e + f*x])^(n + 1))*((a*Cos[e + f*x])^(m + 1) /(a*b*f*(m + 1))), x] + Simp[(m + n + 2)/(a^2*(m + 1)) Int[(b*Sin[e + f*x ])^n*(a*Cos[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m , -1] && IntegersQ[2*m, 2*n]
Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]] , x_Symbol] :> Simp[Sqrt[a*Sin[e + f*x]]*(Sqrt[b*Cos[e + f*x]]/Sqrt[Sin[2*e + 2*f*x]]) Int[Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b, e, f}, x]
Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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[a*(d^2/(a^2 - b^2)) Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 2), x], x] + (-Simp[ b*(d/(a^2 - b^2)) Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n - 1), x], x] - Simp[a^2*(d^2/(g^2*(a^2 - b^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] && NeQ[a^2 - b^2, 0] && IntegersQ[2*n, 2*p] && LtQ[p, -1] && GtQ[n, 1 ]
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. \(1184\) vs. \(2(346)=692\).
Time = 2.28 (sec) , antiderivative size = 1185, normalized size of antiderivative = 3.13
1/f*(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))^2*sin(f*x+e)^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*(-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)+Ellipt icPi((-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*(-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*(-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^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*(-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)+c ot(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*(-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)+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)*EllipticF((-cot(f*x+e)+csc(f*x+e )+1)^(1/2),1/2*2^(1/2))*a+2*(-a^2+b^2)^(1/2)*(-cot(f*x+e)+csc(f*x+e)+1)...
Timed out. \[ \int \frac {(d \sin (e+f x))^{3/2}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\text {Timed out} \]
\[ \int \frac {(d \sin (e+f x))^{3/2}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {\left (d \sin {\left (e + f x \right )}\right )^{\frac {3}{2}}}{\left (g \cos {\left (e + f x \right )}\right )^{\frac {3}{2}} \left (a + b \sin {\left (e + f x \right )}\right )}\, dx \]
\[ \int \frac {(d \sin (e+f x))^{3/2}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
\[ \int \frac {(d \sin (e+f x))^{3/2}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int { \frac {\left (d \sin \left (f x + e\right )\right )^{\frac {3}{2}}}{\left (g \cos \left (f x + e\right )\right )^{\frac {3}{2}} {\left (b \sin \left (f x + e\right ) + a\right )}} \,d x } \]
Timed out. \[ \int \frac {(d \sin (e+f x))^{3/2}}{(g \cos (e+f x))^{3/2} (a+b \sin (e+f x))} \, dx=\int \frac {{\left (d\,\sin \left (e+f\,x\right )\right )}^{3/2}}{{\left (g\,\cos \left (e+f\,x\right )\right )}^{3/2}\,\left (a+b\,\sin \left (e+f\,x\right )\right )} \,d x \]