\(\int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx\) [562]

Optimal result

Integrand size = 16, antiderivative size = 162 \[ \int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\cos ^2(c+d x) \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right ) \left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right ) \sqrt {\frac {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}{\left (\sqrt {a}+\sqrt {a+b} \tan ^2(c+d x)\right )^2}}}{2 \sqrt [4]{a} \sqrt [4]{a+b} d \sqrt {a+b \sin ^4(c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.10 (sec) , antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3289, 1117} \[ \int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\frac {\cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right ),\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 \sqrt [4]{a} d \sqrt [4]{a+b} \sqrt {a+b \sin ^4(c+d x)}} \]

[In]

Int[1/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

(Cos[c + d*x]^2*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]*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])/(2*a^(1/4)*(a + b)^(1/4)*d*Sqrt[a + b*Sin[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 3289

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^4)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Dist
[ff*(a + b*Sin[e + f*x]^4)^p*((Sec[e + f*x]^2)^(2*p)/(f*(a + 2*a*Tan[e + f*x]^2 + (a + b)*Tan[e + f*x]^4)^p)),
Subst[Int[(a + 2*a*ff^2*x^2 + (a + b)*ff^4*x^4)^p/(1 + ff^2*x^2)^(2*p + 1), x], x, Tan[e + f*x]/ff], x]] /; F
reeQ[{a, b, e, f, p}, x] && IntegerQ[p - 1/2]

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.46 (sec) , antiderivative size = 195, normalized size of antiderivative = 1.20 \[ \int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=-\frac {2 i \cos ^2(c+d x) \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} \tan (c+d x)\right ),\frac {\sqrt {a}+i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}\right ) \sqrt {1+\left (1+\frac {i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)} \sqrt {2+\left (2-\frac {2 i \sqrt {b}}{\sqrt {a}}\right ) \tan ^2(c+d x)}}{\sqrt {1-\frac {i \sqrt {b}}{\sqrt {a}}} d \sqrt {8 a+3 b-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))}} \]

[In]

Integrate[1/Sqrt[a + b*Sin[c + d*x]^4],x]

[Out]

((-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])]*Sqrt[1 + (1 + (I*Sqrt[b])/Sqrt[a])*Tan[c + d*x]^2]*Sqrt[2 + (2 - ((2*I)*Sqrt[b])/Sqrt[a
])*Tan[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)
]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(395\) vs. \(2(181)=362\).

Time = 3.05 (sec) , antiderivative size = 396, normalized size of antiderivative = 2.44

method	result	size
default	\(-\frac {\sqrt {\left (\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )+4 a \right ) \left (\sin ^{2}\left (2 d x +2 c \right )\right )}\, \sqrt {-a b}\, \sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}\, \left (1+\cos \left (2 d x +2 c \right )\right )^{2} \sqrt {\frac {-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}\, \sqrt {\frac {b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}\, F\left (\sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (1+\cos \left (2 d x +2 c \right )\right )}}, \sqrt {\frac {b +\sqrt {-a b}}{-b +\sqrt {-a b}}}\right )}{\left (-b +\sqrt {-a b}\right ) \sqrt {\frac {\left (-1+\cos \left (2 d x +2 c \right )\right ) \left (1+\cos \left (2 d x +2 c \right )\right ) \left (-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b \right ) \left (b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b \right )}{b}}\, \sin \left (2 d x +2 c \right ) \sqrt {\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )+4 a}\, d}\)	\(396\)

[In]

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

[Out]

-((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+co
s(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))/(-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

Fricas [F]

\[ \int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]

[In]

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

[Out]

integral(1/sqrt(b*cos(d*x + c)^4 - 2*b*cos(d*x + c)^2 + a + b), x)

Sympy [F]

\[ \int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {1}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \]

[In]

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

[Out]

Integral(1/sqrt(a + b*sin(c + d*x)**4), x)

Maxima [F]

\[ \int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]

[In]

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

[Out]

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

Giac [F]

\[ \int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int { \frac {1}{\sqrt {b \sin \left (d x + c\right )^{4} + a}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx=\int \frac {1}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]

[In]

int(1/(a + b*sin(c + d*x)^4)^(1/2),x)

[Out]

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