\(\int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx\) [239]

Optimal result

Integrand size = 23, antiderivative size = 477 \[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=-\frac {\cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{3 d}+\frac {2 \sqrt {b} \cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{3 \sqrt {a+b} d \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )}-\frac {2 \sqrt [4]{b} (a+b)^{3/4} \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{3 d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}+\frac {(a+b)^{3/4} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{3 \sqrt [4]{b} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \]

[Out]

-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*c
os(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)*(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)))*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)

Rubi [A] (verified)

Time = 0.27 (sec) , antiderivative size = 477, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3294, 1105, 1211, 1117, 1209} \[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\frac {(a+b)^{3/4} \left (\sqrt {b}-\sqrt {a+b}\right ) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{3 \sqrt [4]{b} d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac {2 \sqrt [4]{b} (a+b)^{3/4} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right )|\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{3 d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}-\frac {\cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{3 d}+\frac {2 \sqrt {b} \cos (c+d x) \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}}{3 d \sqrt {a+b} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )} \]

[In]

Int[Sin[c + d*x]*Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-1/3*(Cos[c + d*x]*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])/d + (2*Sqrt[b]*Cos[c + d*x]*Sqrt[a + b
- 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x]^4])/(3*Sqrt[a + b]*d*(1 + (Sqrt[b]*Cos[c + d*x]^2)/Sqrt[a + b])) - (2*b
^(1/4)*(a + b)^(3/4)*(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])/(3*d*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 + (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)]*EllipticF[2*ArcTan[(b^(1/4)*Cos[c + d*
x])/(a + b)^(1/4)], (1 + Sqrt[b]/Sqrt[a + b])/2])/(3*b^(1/4)*d*Sqrt[a + b - 2*b*Cos[c + d*x]^2 + b*Cos[c + d*x
]^4])

Rule 1105

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] + Dis
t[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]

Rule 1117

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]

Rule 1209

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> 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]

Rule 1211

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 2]}, Dist[(
e + d*q)/q, Int[1/Sqrt[a + b*x^2 + c*x^4], x], x] - Dist[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] && PosQ[c/a]

Rule 3294

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Cos[e + f*x], x]}, Dist[-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]

Rubi steps \begin{align*} \text {integral}& = -\frac {\text {Subst}\left (\int \sqrt {a+b-2 b x^2+b x^4} \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {\cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{3 d}-\frac {\text {Subst}\left (\int \frac {2 (a+b)-2 b x^2}{\sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{3 d} \\ & = -\frac {\cos (c+d x) \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}}{3 d}-\frac {\left (2 \sqrt {b} \sqrt {a+b}\right ) \text {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a+b}}}{\sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{3 d}-\frac {\left (\sqrt {a+b} \left (-2 b+2 \sqrt {b} \sqrt {a+b}\right )\right ) \text {Subst}\left (\int \frac {1}{\sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{3 \sqrt {b} d} \\ & = -\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)}} \\ \end{align*}

Mathematica [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 26.94 (sec) , antiderivative size = 3120, normalized size of antiderivative = 6.54 \[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\text {Result too large to show} \]

[In]

Integrate[Sin[c + d*x]*Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

-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])*Sec[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]]*Sqrt[((-I)*a + Sqrt[a]*Sqrt[b] - I*a
*Tan[c + d*x]^2 - I*b*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]*Sqrt[(I*a + Sqrt[a]*Sqrt[b] + I*a*Tan[c + d*x]^2 + I*
b*Tan[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*Tan[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[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])]) + ((-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])*Tan[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*T
an[c + d*x]^3 - a^3*Sqrt[b]*Sec[c + d*x]^4*Tan[c + d*x]^3 + (2*I)*a^(5/2)*b*Sec[c + d*x]^4*Tan[c + d*x]^3 - 2*
a^2*b^(3/2)*Sec[c + d*x]^4*Tan[c + d*x]^3 + I*a^(3/2)*b^2*Sec[c + d*x]^4*Tan[c + d*x]^3 - a*b^(5/2)*Sec[c + d*
x]^4*Tan[c + d*x]^3 + (2*I)*a^(7/2)*Sec[c + d*x]^4*Tan[c + d*x]^5 - 2*a^3*Sqrt[b]*Sec[c + d*x]^4*Tan[c + d*x]^
5 + (4*I)*a^(5/2)*b*Sec[c + d*x]^4*Tan[c + d*x]^5 - 4*a^2*b^(3/2)*Sec[c + d*x]^4*Tan[c + d*x]^5 + (2*I)*a^(3/2
)*b^2*Sec[c + d*x]^4*Tan[c + d*x]^5 - 2*a*b^(5/2)*Sec[c + d*x]^4*Tan[c + d*x]^5 + I*a^(7/2)*Sec[c + d*x]^4*Tan
[c + d*x]^7 - a^3*Sqrt[b]*Sec[c + d*x]^4*Tan[c + d*x]^7 + (3*I)*a^(5/2)*b*Sec[c + d*x]^4*Tan[c + d*x]^7 - 3*a^
2*b^(3/2)*Sec[c + d*x]^4*Tan[c + d*x]^7 + (3*I)*a^(3/2)*b^2*Sec[c + d*x]^4*Tan[c + d*x]^7 - 3*a*b^(5/2)*Sec[c
+ d*x]^4*Tan[c + d*x]^7 + I*Sqrt[a]*b^3*Sec[c + d*x]^4*Tan[c + d*x]^7 - b^(7/2)*Sec[c + d*x]^4*Tan[c + d*x]^7
- a^2*b^(3/2)*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b])*Sec[c + d*x]^2)/Sqrt[b]]*Tan[c + d*x]*Sqrt[((-I)*(Sqrt[a] + I*S
qrt[b])*(1 + Tan[c + d*x]^2))/Sqrt[b]]*Sqrt[((-I)*a + Sqrt[a]*Sqrt[b] - I*a*Tan[c + d*x]^2 - I*b*Tan[c + d*x]^
2)/(Sqrt[a]*Sqrt[b])]*Sqrt[(I*a + Sqrt[a]*Sqrt[b] + I*a*Tan[c + d*x]^2 + I*b*Tan[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)] + 2*a^2*b^(3/2)*Sec[c +
d*x]^2*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b])*Sec[c + d*x]^2)/Sqrt[b]]*Tan[c + d*x]*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b]
)*(1 + Tan[c + d*x]^2))/Sqrt[b]]*Sqrt[((-I)*a + Sqrt[a]*Sqrt[b] - I*a*Tan[c + d*x]^2 - I*b*Tan[c + d*x]^2)/(Sq
rt[a]*Sqrt[b])]*Sqrt[(I*a + Sqrt[a]*Sqrt[b] + I*a*Tan[c + d*x]^2 + I*b*Tan[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)] - 2*a^2*b^(3/2)*Sqrt[((-I)*(S
qrt[a] + I*Sqrt[b])*Sec[c + d*x]^2)/Sqrt[b]]*Tan[c + d*x]^3*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b])*(1 + Tan[c + d*x]
^2))/Sqrt[b]]*Sqrt[((-I)*a + Sqrt[a]*Sqrt[b] - I*a*Tan[c + d*x]^2 - I*b*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]*Sqr
t[(I*a + Sqrt[a]*Sqrt[b] + I*a*Tan[c + d*x]^2 + I*b*Tan[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)] + 2*a^2*b^(3/2)*Sec[c + d*x]^2*Sqrt[((-I)*(Sqrt[
a] + I*Sqrt[b])*Sec[c + d*x]^2)/Sqrt[b]]*Tan[c + d*x]^3*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b])*(1 + Tan[c + d*x]^2))
/Sqrt[b]]*Sqrt[((-I)*a + Sqrt[a]*Sqrt[b] - I*a*Tan[c + d*x]^2 - I*b*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]*Sqrt[(I
*a + Sqrt[a]*Sqrt[b] + I*a*Tan[c + d*x]^2 + I*b*Tan[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)] + 2*a*b^(5/2)*Sec[c + d*x]^2*Sqrt[((-I)*(Sqrt[a] + I
*Sqrt[b])*Sec[c + d*x]^2)/Sqrt[b]]*Tan[c + d*x]^3*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b])*(1 + Tan[c + d*x]^2))/Sqrt[
b]]*Sqrt[((-I)*a + Sqrt[a]*Sqrt[b] - I*a*Tan[c + d*x]^2 - I*b*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]*Sqrt[(I*a + S
qrt[a]*Sqrt[b] + I*a*Tan[c + d*x]^2 + I*b*Tan[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)] - a^2*b^(3/2)*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b])*Sec[c + d*x
]^2)/Sqrt[b]]*Tan[c + d*x]^5*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b])*(1 + Tan[c + d*x]^2))/Sqrt[b]]*Sqrt[((-I)*a + Sq
rt[a]*Sqrt[b] - I*a*Tan[c + d*x]^2 - I*b*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])]*Sqrt[(I*a + Sqrt[a]*Sqrt[b] + I*a*
Tan[c + d*x]^2 + I*b*Tan[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)] - a*b^(5/2)*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b])*Sec[c + d*x]^2)/Sqrt[b]]*Tan[c + d
*x]^5*Sqrt[((-I)*(Sqrt[a] + I*Sqrt[b])*(1 + Tan[c + d*x]^2))/Sqrt[b]]*Sqrt[((-I)*a + Sqrt[a]*Sqrt[b] - I*a*Tan
[c + d*x]^2 - I*b*Tan[c + d*x]^2)/(Sqrt[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)]))

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 3.26 (sec) , antiderivative size = 439, normalized size of antiderivative = 0.92

method	result	size
default	\(-\frac {\frac {4 \cos \left (d x +c \right ) \sqrt {a +b -2 b \left (\cos ^{2}\left (d x +c \right )\right )+b \left (\cos ^{4}\left (d x +c \right )\right )}}{3}+\frac {4 \left (\frac {2 a}{3}+\frac {2 b}{3}\right ) \sqrt {1-\frac {\left (i \sqrt {a}\, \sqrt {b}+b \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{a +b}}\, \sqrt {1+\frac {\left (i \sqrt {a}\, \sqrt {b}-b \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{a +b}}\, F\left (\cos \left (d x +c \right ) \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}, \sqrt {-1-\frac {2 \left (i \sqrt {a}\, \sqrt {b}-b \right )}{a +b}}\right )}{\sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}\, \sqrt {a +b -2 b \left (\cos ^{2}\left (d x +c \right )\right )+b \left (\cos ^{4}\left (d x +c \right )\right )}}+\frac {16 b \left (a +b \right ) \sqrt {1-\frac {\left (i \sqrt {a}\, \sqrt {b}+b \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{a +b}}\, \sqrt {1+\frac {\left (i \sqrt {a}\, \sqrt {b}-b \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{a +b}}\, \left (F\left (\cos \left (d x +c \right ) \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}, \sqrt {-1-\frac {2 \left (i \sqrt {a}\, \sqrt {b}-b \right )}{a +b}}\right )-E\left (\cos \left (d x +c \right ) \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}, \sqrt {-1-\frac {2 \left (i \sqrt {a}\, \sqrt {b}-b \right )}{a +b}}\right )\right )}{3 \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}\, \sqrt {a +b -2 b \left (\cos ^{2}\left (d x +c \right )\right )+b \left (\cos ^{4}\left (d x +c \right )\right )}\, \left (-2 b +2 i \sqrt {a}\, \sqrt {b}\right )}}{4 d}\)	\(439\)

[In]

int(sin(d*x+c)*(a+b*sin(d*x+c)^4)^(1/2),x,method=_RETURNVERBOSE)

[Out]

-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))*(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))-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))))

Fricas [F]

\[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right )^{4} + a} \sin \left (d x + c\right ) \,d x } \]

[In]

integrate(sin(d*x+c)*(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b)*sin(d*x + c), x)

Sympy [F]

\[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int \sqrt {a + b \sin ^{4}{\left (c + d x \right )}} \sin {\left (c + d x \right )}\, dx \]

[In]

integrate(sin(d*x+c)*(a+b*sin(d*x+c)**4)**(1/2),x)

[Out]

Integral(sqrt(a + b*sin(c + d*x)**4)*sin(c + d*x), x)

Maxima [F]

\[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right )^{4} + a} \sin \left (d x + c\right ) \,d x } \]

[In]

integrate(sin(d*x+c)*(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(b*sin(d*x + c)^4 + a)*sin(d*x + c), x)

Giac [F]

\[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int { \sqrt {b \sin \left (d x + c\right )^{4} + a} \sin \left (d x + c\right ) \,d x } \]

[In]

integrate(sin(d*x+c)*(a+b*sin(d*x+c)^4)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(b*sin(d*x + c)^4 + a)*sin(d*x + c), x)

Mupad [F(-1)]

Timed out. \[ \int \sin (c+d x) \sqrt {a+b \sin ^4(c+d x)} \, dx=\int \sin \left (c+d\,x\right )\,\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a} \,d x \]

[In]

int(sin(c + d*x)*(a + b*sin(c + d*x)^4)^(1/2),x)

[Out]

int(sin(c + d*x)*(a + b*sin(c + d*x)^4)^(1/2), x)