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

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]

(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])

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Rubi [A]  time = 0.08279, antiderivative size = 162, normalized size of antiderivative = 1., 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 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]

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]

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*}

Mathematica [C]  time = 2.78717, 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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Maple [B]  time = 2.372, size = 396, normalized size = 2.4 \begin{align*} -{\frac{ \left ( \cos \left ( 2\,dx+2\,c \right ) +1 \right ) ^{2}}{\sin \left ( 2\,dx+2\,c \right ) d}\sqrt{ \left ( 4\,a+ \left ( \cos \left ( 2\,dx+2\,c \right ) \right ) ^{2}b+b-2\,b\cos \left ( 2\,dx+2\,c \right ) \right ) \left ( \sin \left ( 2\,dx+2\,c \right ) \right ) ^{2}}\sqrt{-ab}\sqrt{{\frac{-1+\cos \left ( 2\,dx+2\,c \right ) }{\cos \left ( 2\,dx+2\,c \right ) +1} \left ( -b+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{{\frac{1}{\cos \left ( 2\,dx+2\,c \right ) +1} \left ( -b\cos \left ( 2\,dx+2\,c \right ) +2\,\sqrt{-ab}+b \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{{\frac{1}{\cos \left ( 2\,dx+2\,c \right ) +1} \left ( b\cos \left ( 2\,dx+2\,c \right ) +2\,\sqrt{-ab}-b \right ){\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{\frac{-1+\cos \left ( 2\,dx+2\,c \right ) }{\cos \left ( 2\,dx+2\,c \right ) +1} \left ( -b+\sqrt{-ab} \right ){\frac{1}{\sqrt{-ab}}}}},\sqrt{{ \left ( b+\sqrt{-ab} \right ) \left ( -b+\sqrt{-ab} \right ) ^{-1}}} \right ) \left ( -b+\sqrt{-ab} \right ) ^{-1}{\frac{1}{\sqrt{{\frac{ \left ( -1+\cos \left ( 2\,dx+2\,c \right ) \right ) \left ( \cos \left ( 2\,dx+2\,c \right ) +1 \right ) }{b} \left ( -b\cos \left ( 2\,dx+2\,c \right ) +2\,\sqrt{-ab}+b \right ) \left ( b\cos \left ( 2\,dx+2\,c \right ) +2\,\sqrt{-ab}-b \right ) }}}}{\frac{1}{\sqrt{4\,a+ \left ( \cos \left ( 2\,dx+2\,c \right ) \right ) ^{2}b+b-2\,b\cos \left ( 2\,dx+2\,c \right ) }}}} \end{align*}

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., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sin \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

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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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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 ) \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \sin \left (d x + c\right )^{4} + a}}\,{d x} \end{align*}

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)