Integrand size = 23, antiderivative size = 477 \[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=-\frac {\cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{3 d}+\frac {2 \sqrt {b} \cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{3 \sqrt {a+b} d \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )}-\frac {2 \sqrt [4]{b} (a+b)^{3/4} \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{3 d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}+\frac {(a+b)^{3/4} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{3 \sqrt [4]{b} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \]
-1/3*cos(d*x+c)*(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)/d+2/3*cos(d*x+ c)*b^(1/2)*(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)/d/(1+cos(d*x+c)^2*b ^(1/2)/(a+b)^(1/2))/(a+b)^(1/2)-2/3*b^(1/4)*(a+b)^(3/4)*(cos(2*arctan(b^(1 /4)*cos(d*x+c)/(a+b)^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*cos(d*x+c)/(a+b )^(1/4)))*EllipticE(sin(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^(1/4))),1/2*(2+2 *b^(1/2)/(a+b)^(1/2))^(1/2))*(1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))*((a+b-2* b*cos(d*x+c)^2+b*cos(d*x+c)^4)/(a+b)/(1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))^ 2)^(1/2)/d/(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)+1/3*(a+b)^(3/4)*(co s(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)* cos(d*x+c)/(a+b)^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*cos(d*x+c)/(a+b)^( 1/4))),1/2*(2+2*b^(1/2)/(a+b)^(1/2))^(1/2))*(1+cos(d*x+c)^2*b^(1/2)/(a+b)^ (1/2))*(b^(1/2)-(a+b)^(1/2))*((a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)/(a+b)/ (1+cos(d*x+c)^2*b^(1/2)/(a+b)^(1/2))^2)^(1/2)/b^(1/4)/d/(a+b-2*b*cos(d*x+c )^2+b*cos(d*x+c)^4)^(1/2)
Result contains complex when optimal does not.
Time = 26.94 (sec) , antiderivative size = 3120, normalized size of antiderivative = 6.54 \[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\text {Result too large to show} \]
-1/6*(Cos[c + d*x]*Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d* x)]])/(Sqrt[2]*d) + (Sqrt[2]*a*b^(3/2)*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b])*Se c[c + d*x]^2)/Sqrt[b]]*(4*a*Sin[c + d*x] + 3*b*Sin[c + d*x] - b*Sin[3*(c + d*x)])*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b])*(1 + Tan[c + d*x]^2))/Sqrt[b]]*Sq rt[((-I)*a + Sqrt[a]*Sqrt[b] - I*a*Tan[c + d*x]^2 - I*b*Tan[c + d*x]^2)/(S qrt[a]*Sqrt[b])]*Sqrt[(I*a + Sqrt[a]*Sqrt[b] + I*a*Tan[c + d*x]^2 + I*b*Ta n[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]*Sqrt[((a + b)*(a + 2*a*Tan[c + d*x]^2 + a *Tan[c + d*x]^4 + b*Tan[c + d*x]^4))/(a*b)]*Sqrt[Cos[c + d*x]^4*(a + 2*a*T an[c + d*x]^2 + a*Tan[c + d*x]^4 + b*Tan[c + d*x]^4)]*(a + 2*a*Tan[c + d*x ]^2 + a*Tan[c + d*x]^4 + b*Tan[c + d*x]^4 + ((-(Sqrt[b]*EllipticE[ArcSin[S qrt[1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b]) ]/Sqrt[2]], (2*Sqrt[a])/(Sqrt[a] - I*Sqrt[b])]) + ((-I)*Sqrt[a] + Sqrt[b]) *EllipticF[ArcSin[Sqrt[1 + (I*Sqrt[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2 )/(Sqrt[a]*Sqrt[b])]/Sqrt[2]], (2*Sqrt[a])/(Sqrt[a] - I*Sqrt[b])])*Sqrt[(1 - (I*Sqrt[a])/Sqrt[b])*Sec[c + d*x]^2]*(Sqrt[a] + (Sqrt[a] - I*Sqrt[b])*T an[c + d*x]^2)*(a - I*Sqrt[a]*Sqrt[b] + (a + b)*Tan[c + d*x]^2))/(Sqrt[a]* Sqrt[b]*Sqrt[((a + b)*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4))/( a*b)])))/(3*d*Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]]* Sqrt[Cos[c + d*x]^4*(a + 2*a*Tan[c + d*x]^2 + (a + b)*Tan[c + d*x]^4)]*(I* a^(7/2)*Sec[c + d*x]^4*Tan[c + d*x]^3 - a^3*Sqrt[b]*Sec[c + d*x]^4*Tan[...
Time = 0.55 (sec) , antiderivative size = 477, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {3042, 3694, 1404, 27, 1511, 1416, 1509}
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.
| \(\displaystyle \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \sin (c+d x) \sqrt {a+b \sin (c+d x)^4}dx\) |
| \(\Big \downarrow \) 3694 |
| \(\displaystyle -\frac {\int \sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}d\cos (c+d x)}{d}\) |
| \(\Big \downarrow \) 1404 |
| \(\displaystyle -\frac {\frac {1}{3} \int \frac {2 \left (-b \cos ^2(c+d x)+a+b\right )}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)+\frac {1}{3} \cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{d}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {\frac {2}{3} \int \frac {-b \cos ^2(c+d x)+a+b}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)+\frac {1}{3} \cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{d}\) |
| \(\Big \downarrow \) 1511 |
| \(\displaystyle -\frac {\frac {2}{3} \left (\sqrt {b} \sqrt {a+b} \int \frac {1-\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)-\sqrt {a+b} \left (\sqrt {b}-\sqrt {a+b}\right ) \int \frac {1}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)\right )+\frac {1}{3} \cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{d}\) |
| \(\Big \downarrow \) 1416 |
| \(\displaystyle -\frac {\frac {2}{3} \left (\sqrt {b} \sqrt {a+b} \int \frac {1-\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}}{\sqrt {b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+a+b}}d\cos (c+d x)-\frac {(a+b)^{3/4} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 \sqrt [4]{b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}\right )+\frac {1}{3} \cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{d}\) |
| \(\Big \downarrow \) 1509 |
| \(\displaystyle -\frac {\frac {2}{3} \left (\sqrt {b} \sqrt {a+b} \left (\frac {\sqrt [4]{a+b} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{\sqrt [4]{b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac {\cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )}\right )-\frac {(a+b)^{3/4} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 \sqrt [4]{b} \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}\right )+\frac {1}{3} \cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{d}\) |
-(((Cos[c + d*x]*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])/3 + (2*(Sqrt[b]*Sqrt[a + b]*(-((Cos[c + d*x]*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])/((a + b)*(1 + (Sqrt[b]*Cos[c + d*x]^2)/Sqrt[a + b]))) + ((a + b)^(1/4)*(1 + (Sqrt[b]*Cos[c + d*x]^2)/Sqrt[a + b])*Sqrt[(a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4)/((a + b)*(1 + (Sqrt[b]*Cos[c + d*x] ^2)/Sqrt[a + b])^2)]*EllipticE[2*ArcTan[(b^(1/4)*Cos[c + d*x])/(a + b)^(1/ 4)], (1 + Sqrt[b]/Sqrt[a + b])/2])/(b^(1/4)*Sqrt[a + b - 2*b*Cos[c + d*x]^ 2 + b*Cos[c + d*x]^4])) - ((a + b)^(3/4)*(Sqrt[b] - Sqrt[a + b])*(1 + (Sqr t[b]*Cos[c + d*x]^2)/Sqrt[a + b])*Sqrt[(a + b - 2*b*Cos[c + d*x]^2 + b*Cos [c + d*x]^4)/((a + b)*(1 + (Sqrt[b]*Cos[c + d*x]^2)/Sqrt[a + b])^2)]*Ellip ticF[2*ArcTan[(b^(1/4)*Cos[c + d*x])/(a + b)^(1/4)], (1 + Sqrt[b]/Sqrt[a + b])/2])/(2*b^(1/4)*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4]))) /3)/d)
3.3.39.3.1 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[((a_.) + (b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp[x*((a + b *x^2 + c*x^4)^p/(4*p + 1)), x] + Simp[2*(p/(4*p + 1)) Int[(2*a + b*x^2)*( a + b*x^2 + c*x^4)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a* c, 0] && GtQ[p, 0] && IntegerQ[2*p]
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[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo l] :> With[{q = Rt[c/a, 4]}, Simp[(-d)*x*(Sqrt[a + b*x^2 + c*x^4]/(a*(1 + q ^2*x^2))), x] + Simp[d*(1 + q^2*x^2)*(Sqrt[(a + b*x^2 + c*x^4)/(a*(1 + q^2* x^2)^2)]/(q*Sqrt[a + b*x^2 + c*x^4]))*EllipticE[2*ArcTan[q*x], 1/2 - b*(q^2 /(4*c))], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]
Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbo l] :> With[{q = Rt[c/a, 2]}, Simp[(e + d*q)/q Int[1/Sqrt[a + b*x^2 + c*x^ 4], x], x] - Simp[e/q Int[(1 - q*x^2)/Sqrt[a + b*x^2 + c*x^4], x], x] /; NeQ[e + d*q, 0]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && Pos Q[c/a]
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^ (p_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-ff/f Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b - 2*b*ff^2*x^2 + b*ff^4*x^4)^p, x], x, Cos[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Result contains complex when optimal does not.
Time = 3.26 (sec) , antiderivative size = 439, normalized size of antiderivative = 0.92
| method | result | size |
| default | \(-\frac {\frac {4 \cos \left (d x +c \right ) \sqrt {a +b -2 b \left (\cos ^{2}\left (d x +c \right )\right )+b \left (\cos ^{4}\left (d x +c \right )\right )}}{3}+\frac {4 \left (\frac {2 a}{3}+\frac {2 b}{3}\right ) \sqrt {1-\frac {\left (i \sqrt {a}\, \sqrt {b}+b \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{a +b}}\, \sqrt {1+\frac {\left (i \sqrt {a}\, \sqrt {b}-b \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{a +b}}\, F\left (\cos \left (d x +c \right ) \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}, \sqrt {-1-\frac {2 \left (i \sqrt {a}\, \sqrt {b}-b \right )}{a +b}}\right )}{\sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}\, \sqrt {a +b -2 b \left (\cos ^{2}\left (d x +c \right )\right )+b \left (\cos ^{4}\left (d x +c \right )\right )}}+\frac {16 b \left (a +b \right ) \sqrt {1-\frac {\left (i \sqrt {a}\, \sqrt {b}+b \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{a +b}}\, \sqrt {1+\frac {\left (i \sqrt {a}\, \sqrt {b}-b \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{a +b}}\, \left (F\left (\cos \left (d x +c \right ) \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}, \sqrt {-1-\frac {2 \left (i \sqrt {a}\, \sqrt {b}-b \right )}{a +b}}\right )-E\left (\cos \left (d x +c \right ) \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}, \sqrt {-1-\frac {2 \left (i \sqrt {a}\, \sqrt {b}-b \right )}{a +b}}\right )\right )}{3 \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}\, \sqrt {a +b -2 b \left (\cos ^{2}\left (d x +c \right )\right )+b \left (\cos ^{4}\left (d x +c \right )\right )}\, \left (-2 b +2 i \sqrt {a}\, \sqrt {b}\right )}}{4 d}\) | \(439\) |
-1/4/d*(4/3*cos(d*x+c)*(a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)+4*(2/3* a+2/3*b)/((I*a^(1/2)*b^(1/2)+b)/(a+b))^(1/2)*(1-(I*a^(1/2)*b^(1/2)+b)/(a+b )*cos(d*x+c)^2)^(1/2)*(1+(I*a^(1/2)*b^(1/2)-b)/(a+b)*cos(d*x+c)^2)^(1/2)/( a+b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)*EllipticF(cos(d*x+c)*((I*a^(1/2 )*b^(1/2)+b)/(a+b))^(1/2),(-1-2*(I*a^(1/2)*b^(1/2)-b)/(a+b))^(1/2))+16/3*b *(a+b)/((I*a^(1/2)*b^(1/2)+b)/(a+b))^(1/2)*(1-(I*a^(1/2)*b^(1/2)+b)/(a+b)* cos(d*x+c)^2)^(1/2)*(1+(I*a^(1/2)*b^(1/2)-b)/(a+b)*cos(d*x+c)^2)^(1/2)/(a+ b-2*b*cos(d*x+c)^2+b*cos(d*x+c)^4)^(1/2)/(-2*b+2*I*a^(1/2)*b^(1/2))*(Ellip ticF(cos(d*x+c)*((I*a^(1/2)*b^(1/2)+b)/(a+b))^(1/2),(-1-2*(I*a^(1/2)*b^(1/ 2)-b)/(a+b))^(1/2))-EllipticE(cos(d*x+c)*((I*a^(1/2)*b^(1/2)+b)/(a+b))^(1/ 2),(-1-2*(I*a^(1/2)*b^(1/2)-b)/(a+b))^(1/2))))
\[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right )^{4} + a} \sin \left (d x + c\right ) \,d x } \]
\[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int \sqrt {a + b \sin ^{4}{\left (c + d x \right )}} \sin {\left (c + d x \right )}\, dx \]
\[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right )^{4} + a} \sin \left (d x + c\right ) \,d x } \]
\[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right )^{4} + a} \sin \left (d x + c\right ) \,d x } \]
Timed out. \[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int \sin \left (c+d\,x\right )\,\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a} \,d x \]