Integrand size = 25, antiderivative size = 499 \[ \int \frac {\sin ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}}\right ) \cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}}{2 \sqrt {b} d \sqrt {a+b \sin ^4(c+d x)}}-\frac {\sqrt [4]{a} \left (\sqrt {a}+\sqrt {a+b}\right ) \cos ^2(c+d x) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{2 b \sqrt [4]{a+b} d \sqrt {a+b \sin ^4(c+d x)}}+\frac {\left (\sqrt {a}+\sqrt {a+b}\right )^2 \cos ^2(c+d x) \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {a}-\sqrt {a+b}\right )^2}{4 \sqrt {a} \sqrt {a+b}},2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{4 \sqrt [4]{a} b \sqrt [4]{a+b} d \sqrt {a+b \sin ^4(c+d x)}} \]
-1/2*arctan(b^(1/2)*tan(d*x+c)/(a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)^(1/ 2))*cos(d*x+c)^2*(a+2*a*tan(d*x+c)^2+(a+b)*tan(d*x+c)^4)^(1/2)/d/b^(1/2)/( a+b*sin(d*x+c)^4)^(1/2)-1/2*a^(1/4)*cos(d*x+c)^2*(cos(2*arctan((a+b)^(1/4) *tan(d*x+c)/a^(1/4)))^2)^(1/2)/cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4) ))*EllipticF(sin(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4))),1/2*(2-2*a^(1/2 )/(a+b)^(1/2))^(1/2))*(a^(1/2)+(a+b)^(1/2))*((a+2*a*tan(d*x+c)^2+(a+b)*tan (d*x+c)^4)/(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)^2)^(1/2)*(a^(1/2)+(a+b)^(1/2 )*tan(d*x+c)^2)/b/(a+b)^(1/4)/d/(a+b*sin(d*x+c)^4)^(1/2)+1/4*cos(d*x+c)^2* (cos(2*arctan((a+b)^(1/4)*tan(d*x+c)/a^(1/4)))^2)^(1/2)/cos(2*arctan((a+b) ^(1/4)*tan(d*x+c)/a^(1/4)))*EllipticPi(sin(2*arctan((a+b)^(1/4)*tan(d*x+c) /a^(1/4))),-1/4*(a^(1/2)-(a+b)^(1/2))^2/a^(1/2)/(a+b)^(1/2),1/2*(2-2*a^(1/ 2)/(a+b)^(1/2))^(1/2))*(a^(1/2)+(a+b)^(1/2))^2*((a+2*a*tan(d*x+c)^2+(a+b)* tan(d*x+c)^4)/(a^(1/2)+(a+b)^(1/2)*tan(d*x+c)^2)^2)^(1/2)*(a^(1/2)+(a+b)^( 1/2)*tan(d*x+c)^2)/a^(1/4)/b/(a+b)^(1/4)/d/(a+b*sin(d*x+c)^4)^(1/2)
Result contains complex when optimal does not.
Time = 3.32 (sec) , antiderivative size = 287, normalized size of antiderivative = 0.58 \[ \int \frac {\sin ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {2 i \cos ^2(c+d x) \left (\operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right ),\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )-\operatorname {EllipticPi}\left (\frac {\sqrt {a}}{\sqrt {a}-i \sqrt {b}},i \text {arcsinh}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right ),\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right )\right ) \sqrt {1+\left (1+\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \sqrt {2+\left (2-\frac {2 i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)}}{\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} d \sqrt {8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))}} \]
((-2*I)*Cos[c + d*x]^2*(EllipticF[I*ArcSinh[Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]* Tan[c + d*x]], (Sqrt[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b])] - EllipticPi[S qrt[a]/(Sqrt[a] - I*Sqrt[b]), I*ArcSinh[Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*Tan[ c + d*x]], (Sqrt[a] + I*Sqrt[b])/(Sqrt[a] - I*Sqrt[b])])*Sqrt[1 + (1 + (I* Sqrt[b])/Sqrt[a])*Tan[c + d*x]^2]*Sqrt[2 + (2 - ((2*I)*Sqrt[b])/Sqrt[a])*T an[c + d*x]^2])/(Sqrt[1 - (I*Sqrt[b])/Sqrt[a]]*d*Sqrt[8*a + 3*b - 4*b*Cos[ 2*(c + d*x)] + b*Cos[4*(c + d*x)]])
Time = 0.80 (sec) , antiderivative size = 541, normalized size of antiderivative = 1.08, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3042, 3698, 1656, 27, 1416, 2220}
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.
| \(\displaystyle \int \frac {\sin ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin (c+d x)^2}{\sqrt {a+b \sin (c+d x)^4}}dx\) |
| \(\Big \downarrow \) 3698 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \int \frac {\tan ^2(c+d x)}{\left (\tan ^2(c+d x)+1\right ) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
| \(\Big \downarrow \) 1656 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {a \left (\frac {\sqrt {a+b}}{\sqrt {a}}+1\right ) \int \frac {\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}}{\sqrt {a} \left (\tan ^2(c+d x)+1\right ) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{b}-\frac {a \left (\frac {\sqrt {a+b}}{\sqrt {a}}+1\right ) \int \frac {1}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{b}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {\sqrt {a} \left (\frac {\sqrt {a+b}}{\sqrt {a}}+1\right ) \int \frac {\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}}{\left (\tan ^2(c+d x)+1\right ) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{b}-\frac {a \left (\frac {\sqrt {a+b}}{\sqrt {a}}+1\right ) \int \frac {1}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{b}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
| \(\Big \downarrow \) 1416 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {\sqrt {a} \left (\frac {\sqrt {a+b}}{\sqrt {a}}+1\right ) \int \frac {\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}}{\left (\tan ^2(c+d x)+1\right ) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}d\tan (c+d x)}{b}-\frac {a^{3/4} \left (\frac {\sqrt {a+b}}{\sqrt {a}}+1\right ) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 b \sqrt [4]{a+b} \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
| \(\Big \downarrow \) 2220 |
| \(\displaystyle \frac {\cos ^2(c+d x) \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a} \left (\frac {\sqrt {a} \left (\frac {\sqrt {a+b}}{\sqrt {a}}+1\right ) \left (\frac {\left (\sqrt {a}-\sqrt {a+b}\right ) \arctan \left (\frac {\sqrt {b} \tan (c+d x)}{\sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}\right )}{2 \sqrt {b}}+\frac {\left (\sqrt {a+b}+\sqrt {a}\right ) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} \operatorname {EllipticPi}\left (-\frac {\left (\sqrt {a}-\sqrt {a+b}\right )^2}{4 \sqrt {a} \sqrt {a+b}},2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{4 \sqrt [4]{a} \sqrt [4]{a+b} \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}\right )}{b}-\frac {a^{3/4} \left (\frac {\sqrt {a+b}}{\sqrt {a}}+1\right ) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 b \sqrt [4]{a+b} \sqrt {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}}\right )}{d \sqrt {a+b \sin ^4(c+d x)}}\) |
(Cos[c + d*x]^2*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4]*(-1/ 2*(a^(3/4)*(1 + Sqrt[a + b]/Sqrt[a])*EllipticF[2*ArcTan[((a + b)^(1/4)*Tan [c + d*x])/a^(1/4)], (1 - Sqrt[a]/Sqrt[a + b])/2]*(Sqrt[a] + Sqrt[a + b]*T an[c + d*x]^2)*Sqrt[(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4)/(Sqr t[a] + Sqrt[a + b]*Tan[c + d*x]^2)^2])/(b*(a + b)^(1/4)*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4]) + (Sqrt[a]*(1 + Sqrt[a + b]/Sqrt[a])* (((Sqrt[a] - Sqrt[a + b])*ArcTan[(Sqrt[b]*Tan[c + d*x])/Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4]])/(2*Sqrt[b]) + ((Sqrt[a] + Sqrt[a + b ])*EllipticPi[-1/4*(Sqrt[a] - Sqrt[a + b])^2/(Sqrt[a]*Sqrt[a + b]), 2*ArcT an[((a + b)^(1/4)*Tan[c + d*x])/a^(1/4)], (1 - Sqrt[a]/Sqrt[a + b])/2]*(Sq rt[a] + Sqrt[a + b]*Tan[c + d*x]^2)*Sqrt[(a + 2*a*Tan[c + d*x]^2 + (a + b) *Tan[c + d*x]^4)/(Sqrt[a] + Sqrt[a + b]*Tan[c + d*x]^2)^2])/(4*a^(1/4)*(a + b)^(1/4)*Sqrt[a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4])))/b))/(d *Sqrt[a + b*Sin[c + d*x]^4])
3.3.46.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[1/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c /a, 4]}, Simp[(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/ (2*q*Sqrt[a + b*x^2 + c*x^4]))*EllipticF[2*ArcTan[q*x], 1/2 - b*(q^2/(4*c)) ], x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
Int[(x_)^2/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]) , x_Symbol] :> With[{q = Rt[c/a, 2]}, Simp[(-a)*((e + d*q)/(c*d^2 - a*e^2)) Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] + Simp[a*d*((e + d*q)/(c*d^2 - a*e ^2)) Int[(1 + q*x^2)/((d + e*x^2)*Sqrt[a + b*x^2 + c*x^4]), x], x]] /; Fr eeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a] && NeQ[c*d^2 - a*e^2, 0]
Int[((A_) + (B_.)*(x_)^2)/(((d_) + (e_.)*(x_)^2)*Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4]), x_Symbol] :> With[{q = Rt[B/A, 2]}, Simp[(-(B*d - A*e))*(A rcTan[Rt[-b + c*(d/e) + a*(e/d), 2]*(x/Sqrt[a + b*x^2 + c*x^4])]/(2*d*e*Rt[ -b + c*(d/e) + a*(e/d), 2])), x] + Simp[(B*d + A*e)*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2*x^2)^2)]/(4*d*e*q*Sqrt[a + b*x^2 + c*x^4]))*El lipticPi[-(e - d*q^2)^2/(4*d*e*q^2), 2*ArcTan[q*x], 1/2 - b/(4*a*q^2)], x]] /; FreeQ[{a, b, c, d, e, A, B}, x] && NeQ[c*d^2 - a*e^2, 0] && PosQ[c/a] & & EqQ[c*A^2 - a*B^2, 0] && PosQ[B/A] && PosQ[-b + c*(d/e) + a*(e/d)]
Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^( p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff^(m + 1) *(a + b*Sin[e + f*x]^4)^p*((Sec[e + f*x]^2)^(2*p)/(f*Apart[a*(1 + Tan[e + f *x]^2)^2 + b*Tan[e + f*x]^4]^p)) Subst[Int[x^m*(ExpandToSum[a*(1 + ff^2*x ^2)^2 + b*ff^4*x^4, x]^p/(1 + ff^2*x^2)^(m/2 + 2*p + 1)), x], x, Tan[e + f* x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2] && IntegerQ[p - 1 /2]
Time = 4.70 (sec) , antiderivative size = 881, normalized size of antiderivative = 1.77
| method | result | size |
| default | \(-\frac {\sqrt {\left (\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )+4 a \right ) \left (\sin ^{2}\left (2 d x +2 c \right )\right )}\, \sqrt {-a b}\, \sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}\, \left (1+\cos \left (2 d x +2 c \right )\right )^{2} \sqrt {\frac {-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}\, \sqrt {\frac {b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}\, \left (F\left (\sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}, \sqrt {\frac {b +\sqrt {-a b}}{-b +\sqrt {-a b}}}\right )-2 \Pi \left (\sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}, \frac {\sqrt {-a b}}{-b +\sqrt {-a b}}, \sqrt {\frac {b +\sqrt {-a b}}{-b +\sqrt {-a b}}}\right )\right )}{2 \left (-b +\sqrt {-a b}\right ) \sqrt {\frac {\left (-1+\cos \left (2 d x +2 c \right )\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \left (-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b \right ) \left (b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b \right )}{b}}\, \sin \left (2 d x +2 c \right ) \sqrt {\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )+4 a}\, d}-\frac {\sqrt {\left (\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )+4 a \right ) \left (\sin ^{2}\left (2 d x +2 c \right )\right )}\, \sqrt {-a b}\, \sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}\, \left (1+\cos \left (2 d x +2 c \right )\right )^{2} \sqrt {\frac {-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}\, \sqrt {\frac {b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}\, F\left (\sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}, \sqrt {\frac {b +\sqrt {-a b}}{-b +\sqrt {-a b}}}\right )}{2 \left (-b +\sqrt {-a b}\right ) \sqrt {\frac {\left (-1+\cos \left (2 d x +2 c \right )\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \left (-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b \right ) \left (b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b \right )}{b}}\, \sin \left (2 d x +2 c \right ) \sqrt {\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )+4 a}\, d}\) | \(881\) |
-1/2*((cos(2*d*x+2*c)^2*b+b-2*b*cos(2*d*x+2*c)+4*a)*sin(2*d*x+2*c)^2)^(1/2 )*(-a*b)^(1/2)*((-b+(-a*b)^(1/2))*(-1+cos(2*d*x+2*c))/(-a*b)^(1/2)/(1+cos( 2*d*x+2*c)))^(1/2)*(1+cos(2*d*x+2*c))^2*((-b*cos(2*d*x+2*c)+2*(-a*b)^(1/2) +b)/(-a*b)^(1/2)/(1+cos(2*d*x+2*c)))^(1/2)*((b*cos(2*d*x+2*c)+2*(-a*b)^(1/ 2)-b)/(-a*b)^(1/2)/(1+cos(2*d*x+2*c)))^(1/2)*(EllipticF(((-b+(-a*b)^(1/2)) *(-1+cos(2*d*x+2*c))/(-a*b)^(1/2)/(1+cos(2*d*x+2*c)))^(1/2),((b+(-a*b)^(1/ 2))/(-b+(-a*b)^(1/2)))^(1/2))-2*EllipticPi(((-b+(-a*b)^(1/2))*(-1+cos(2*d* x+2*c))/(-a*b)^(1/2)/(1+cos(2*d*x+2*c)))^(1/2),(-a*b)^(1/2)/(-b+(-a*b)^(1/ 2)),((b+(-a*b)^(1/2))/(-b+(-a*b)^(1/2)))^(1/2)))/(-b+(-a*b)^(1/2))/(1/b*(- 1+cos(2*d*x+2*c))*(1+cos(2*d*x+2*c))*(-b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)+b)* (b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)-b))^(1/2)/sin(2*d*x+2*c)/(cos(2*d*x+2*c)^ 2*b+b-2*b*cos(2*d*x+2*c)+4*a)^(1/2)/d-1/2*((cos(2*d*x+2*c)^2*b+b-2*b*cos(2 *d*x+2*c)+4*a)*sin(2*d*x+2*c)^2)^(1/2)*(-a*b)^(1/2)*((-b+(-a*b)^(1/2))*(-1 +cos(2*d*x+2*c))/(-a*b)^(1/2)/(1+cos(2*d*x+2*c)))^(1/2)*(1+cos(2*d*x+2*c)) ^2*((-b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)+b)/(-a*b)^(1/2)/(1+cos(2*d*x+2*c)))^ (1/2)*((b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)-b)/(-a*b)^(1/2)/(1+cos(2*d*x+2*c)) )^(1/2)*EllipticF(((-b+(-a*b)^(1/2))*(-1+cos(2*d*x+2*c))/(-a*b)^(1/2)/(1+c os(2*d*x+2*c)))^(1/2),((b+(-a*b)^(1/2))/(-b+(-a*b)^(1/2)))^(1/2))/(-b+(-a* b)^(1/2))/(1/b*(-1+cos(2*d*x+2*c))*(1+cos(2*d*x+2*c))*(-b*cos(2*d*x+2*c)+2 *(-a*b)^(1/2)+b)*(b*cos(2*d*x+2*c)+2*(-a*b)^(1/2)-b))^(1/2)/sin(2*d*x+2...
\[ \int \frac {\sin ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\sin \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]
Timed out. \[ \int \frac {\sin ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\text {Timed out} \]
\[ \int \frac {\sin ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\sin \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]
\[ \int \frac {\sin ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {\sin \left (d x + c\right )^{2}}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]
Timed out. \[ \int \frac {\sin ^2(c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {{\sin \left (c+d\,x\right )}^2}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]