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

Optimal. Leaf size=162 \[ \frac {\cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 \sqrt [4]{a} d \sqrt [4]{a+b} \sqrt {a+b \sin ^4(c+d x)}} \]

[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)

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Rubi [A]  time = 0.08, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3210, 1103} \[ \frac {\cos ^2(c+d x) \left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right ) \sqrt {\frac {(a+b) \tan ^4(c+d x)+2 a \tan ^2(c+d x)+a}{\left (\sqrt {a+b} \tan ^2(c+d x)+\sqrt {a}\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{a+b} \tan (c+d x)}{\sqrt [4]{a}}\right )|\frac {1}{2} \left (1-\frac {\sqrt {a}}{\sqrt {a+b}}\right )\right )}{2 \sqrt [4]{a} d \sqrt [4]{a+b} \sqrt {a+b \sin ^4(c+d x)}} \]

Antiderivative was successfully verified.

[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 1103

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)]*EllipticF[2*ArcTan[q*x], 1/2 - (b*q^2)/(4*c)])/(2*q*Sqrt[a + b*x^2 + c
*x^4]), x]] /; FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0] && PosQ[c/a]

Rule 3210

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*} \int \frac {1}{\sqrt {a+b \sin ^4(c+d x)}} \, dx &=\frac {\left (\cos ^2(c+d x) \sqrt {a+2 a \tan ^2(c+d x)+(a+b) \tan ^4(c+d x)}\right ) \operatorname {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) F\left (2 \tan ^{-1}\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*}

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Mathematica [C]  time = 8.70, size = 304, normalized size = 1.88 \[ \frac {2 \sqrt {2} \left (\sqrt {b}+i \sqrt {a}\right ) \sin ^2(c+d x) \tan (c+d x) \left (2 \sqrt {a}+i \sqrt {b} \cos (2 (c+d x))-i \sqrt {b}\right ) \left (2 i \sqrt {a}+\sqrt {b} \cos (2 (c+d x))-\sqrt {b}\right ) \sqrt {\csc ^2(c+d x) \left (-\frac {2 i \sqrt {a}}{\sqrt {b}}-\cos (2 (c+d x))+1\right )} \sqrt {\frac {\cot ^2(c+d x) \left (-a \csc ^2(c+d x)+i \sqrt {a} \sqrt {b}\right )}{\left (\sqrt {a}-i \sqrt {b}\right )^2}} F\left (\sin ^{-1}\left (\sqrt {\frac {\sqrt {a} \csc ^2(c+d x)-i \sqrt {b}}{\sqrt {a}-i \sqrt {b}}}\right )|\frac {i \sqrt {a}}{2 \sqrt {b}}+\frac {1}{2}\right )}{\sqrt {a} d (8 a-4 b \cos (2 (c+d x))+b \cos (4 (c+d x))+3 b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[2]*(I*Sqrt[a] + Sqrt[b])*(2*Sqrt[a] - I*Sqrt[b] + I*Sqrt[b]*Cos[2*(c + d*x)])*((2*I)*Sqrt[a] - Sqrt[b]
+ Sqrt[b]*Cos[2*(c + d*x)])*Sqrt[(1 - ((2*I)*Sqrt[a])/Sqrt[b] - Cos[2*(c + d*x)])*Csc[c + d*x]^2]*Sqrt[(Cot[c
+ d*x]^2*(I*Sqrt[a]*Sqrt[b] - a*Csc[c + d*x]^2))/(Sqrt[a] - I*Sqrt[b])^2]*EllipticF[ArcSin[Sqrt[((-I)*Sqrt[b]
+ Sqrt[a]*Csc[c + d*x]^2)/(Sqrt[a] - I*Sqrt[b])]], 1/2 + ((I/2)*Sqrt[a])/Sqrt[b]]*Sin[c + d*x]^2*Tan[c + d*x]
)/(Sqrt[a]*d*(8*a + 3*b - 4*b*Cos[2*(c + d*x)] + b*Cos[4*(c + d*x)])^(3/2))

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fricas [F]  time = 0.49, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {1}{\sqrt {b \cos \left (d x + c\right )^{4} - 2 \, b \cos \left (d x + c\right )^{2} + a + b}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sin \left (d x + c\right )^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 2.94, size = 396, normalized size = 2.44 \[ -\frac {\sqrt {\left (4 a +\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )\right ) \left (\sin ^{2}\left (2 d x +2 c \right )\right )}\, \sqrt {-a b}\, \sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \left (\cos \left (2 d x +2 c \right )+1\right )^{2} \sqrt {\frac {-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \sqrt {\frac {b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}\, \EllipticF \left (\sqrt {\frac {\left (-b +\sqrt {-a b}\right ) \left (-1+\cos \left (2 d x +2 c \right )\right )}{\sqrt {-a b}\, \left (\cos \left (2 d x +2 c \right )+1\right )}}, \sqrt {\frac {b +\sqrt {-a b}}{-b +\sqrt {-a b}}}\right )}{\left (-b +\sqrt {-a b}\right ) \sqrt {\frac {\left (-1+\cos \left (2 d x +2 c \right )\right ) \left (\cos \left (2 d x +2 c \right )+1\right ) \left (-b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}+b \right ) \left (b \cos \left (2 d x +2 c \right )+2 \sqrt {-a b}-b \right )}{b}}\, \sin \left (2 d x +2 c \right ) \sqrt {4 a +\left (\cos ^{2}\left (2 d x +2 c \right )\right ) b +b -2 b \cos \left (2 d x +2 c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {b \sin \left (d x + c\right )^{4} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{\sqrt {b\,{\sin \left (c+d\,x\right )}^4+a}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\sqrt {a + b \sin ^{4}{\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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