\(\int \frac {\sin (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [243]

Optimal result

Integrand size = 23, antiderivative size = 171 \[ \int \frac {\sin (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\sqrt [4]{a+b} \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right ) \sqrt {\frac {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}{(a+b) \left (1+\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (1+\frac {\sqrt {b}}{\sqrt {a+b}}\right )\right )}{2 \sqrt [4]{b} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \]

[Out]

-1/2*(a+b)^(1/4)*(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))*((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 = 0.08 (sec) , antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {3294, 1117} \[ \int \frac {\sin (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {\sqrt [4]{a+b} \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right ) \sqrt {\frac {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}{(a+b) \left (\frac {\sqrt {b} \cos ^2(c+d x)}{\sqrt {a+b}}+1\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{b} \cos (c+d x)}{\sqrt [4]{a+b}}\right ),\frac {1}{2} \left (\frac {\sqrt {b}}{\sqrt {a+b}}+1\right )\right )}{2 \sqrt [4]{b} d \sqrt {a+b \cos ^4(c+d x)-2 b \cos ^2(c+d x)+b}} \]

[In]

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

[Out]

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

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 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 \frac {1}{\sqrt {a+b-2 b x^2+b x^4}} \, dx,x,\cos (c+d x)\right )}{d} \\ & = -\frac {\sqrt [4]{a+b} \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 )}{2 \sqrt [4]{b} d \sqrt {a+b-2 b \cos ^2(c+d x)+b \cos ^4(c+d x)}} \\ \end{align*}

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 7.81 (sec) , antiderivative size = 294, normalized size of antiderivative = 1.72 \[ \int \frac {\sin (c+d x)}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {2 \sqrt {2} \cos ^3(c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}}}{\sqrt {2}}\right ),\frac {2 \sqrt {a}}{\sqrt {a}-i \sqrt {b}}\right ) \sqrt {\left (1-\frac {i \sqrt {a}}{\sqrt {b}}\right ) \sec ^2(c+d x)} \left (\sqrt {a}+\left (\sqrt {a}+i \sqrt {b}\right ) \tan ^2(c+d x)\right ) \sqrt {-\frac {i \left (a+i \sqrt {a} \sqrt {b}+(a+b) \tan ^2(c+d x)\right )}{\sqrt {a} \sqrt {b}}}}{\left (\sqrt {a}+i \sqrt {b}\right ) d \sqrt {8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))} \sqrt {1+\frac {i \sqrt {a}}{\sqrt {b}}+\frac {i (a+b) \tan ^2(c+d x)}{\sqrt {a} \sqrt {b}}}} \]

[In]

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

[Out]

(2*Sqrt[2]*Cos[c + d*x]^3*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)*Sqrt[((-I)*(a + I*Sqrt[a]*Sqrt[b] + (a + b)*Tan[c + d*x]^2))/(Sqrt[a]
*Sqrt[b])])/((Sqrt[a] + I*Sqrt[b])*d*Sqrt[8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)]]*Sqrt[1 + (I*S
qrt[a])/Sqrt[b] + (I*(a + b)*Tan[c + d*x]^2)/(Sqrt[a]*Sqrt[b])])

Maple [C] (verified)

Result contains complex when optimal does not.

Time = 2.05 (sec) , antiderivative size = 163, normalized size of antiderivative = 0.95

method	result	size
default	\(-\frac {\sqrt {1-\frac {\left (i \sqrt {a}\, \sqrt {b}+b \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{a +b}}\, \sqrt {1+\frac {\left (i \sqrt {a}\, \sqrt {b}-b \right ) \left (\cos ^{2}\left (d x +c \right )\right )}{a +b}}\, F\left (\cos \left (d x +c \right ) \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}, \sqrt {-1-\frac {2 \left (i \sqrt {a}\, \sqrt {b}-b \right )}{a +b}}\right )}{d \sqrt {\frac {i \sqrt {a}\, \sqrt {b}+b}{a +b}}\, \sqrt {a +b -2 b \left (\cos ^{2}\left (d x +c \right )\right )+b \left (\cos ^{4}\left (d x +c \right )\right )}}\)	\(163\)

[In]

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

[Out]

-1/d/((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))

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

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

[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)