3.4.51 \(\int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [351]

Optimal. Leaf size=51 \[ \frac {F\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*(1+b*sin(f*x+e)^2/a)^(1/2)/f/(a+b*sin(f*x+e
)^2)^(1/2)

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Rubi [A]
time = 0.02, antiderivative size = 51, 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 = {3262, 3261} \begin {gather*} \frac {\sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac {b}{a}\right .\right )}{f \sqrt {a+b \sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

(EllipticF[e + f*x, -(b/a)]*Sqrt[1 + (b*Sin[e + f*x]^2)/a])/(f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 3261

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Simp[(1/(Sqrt[a]*f))*EllipticF[e + f*x, -b/a]
, x] /; FreeQ[{a, b, e, f}, x] && GtQ[a, 0]

Rule 3262

Int[1/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[Sqrt[1 + b*(Sin[e + f*x]^2/a)]/Sqrt[a +
b*Sin[e + f*x]^2], Int[1/Sqrt[1 + (b*Sin[e + f*x]^2)/a], x], x] /; FreeQ[{a, b, e, f}, x] &&  !GtQ[a, 0]

Rubi steps

\begin {align*} \int \frac {1}{\sqrt {a+b \sin ^2(e+f x)}} \, dx &=\frac {\sqrt {1+\frac {b \sin ^2(e+f x)}{a}} \int \frac {1}{\sqrt {1+\frac {b \sin ^2(e+f x)}{a}}} \, dx}{\sqrt {a+b \sin ^2(e+f x)}}\\ &=\frac {F\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{f \sqrt {a+b \sin ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.06, size = 60, normalized size = 1.18 \begin {gather*} \frac {\sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )}{f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

(Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)])/(f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.24, size = 60, normalized size = 1.18

method	result	size
default	\(\frac {\sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )-a -b}{a}}\, \mathrm {am}^{-1}\left (f x +e \bigg | \frac {i \sqrt {b}}{\sqrt {a}}\right )}{f \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}}\)	\(60\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/f/(a+b-b*cos(f*x+e)^2)^(1/2)*(-(b*cos(f*x+e)^2-a-b)/a)^(1/2)*InverseJacobiAM(f*x+e,I/a^(1/2)*b^(1/2))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(1/sqrt(b*sin(f*x + e)^2 + a), x)

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Fricas [C] Result contains complex when optimal does not.
time = 0.12, size = 305, normalized size = 5.98 \begin {gather*} -\frac {{\left (2 i \, \sqrt {-b} b \sqrt {\frac {a^{2} + a b}{b^{2}}} + {\left (-2 i \, a - i \, b\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) + i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}}) + {\left (-2 i \, \sqrt {-b} b \sqrt {\frac {a^{2} + a b}{b^{2}}} + {\left (2 i \, a + i \, b\right )} \sqrt {-b}\right )} \sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} F(\arcsin \left (\sqrt {\frac {2 \, b \sqrt {\frac {a^{2} + a b}{b^{2}}} + 2 \, a + b}{b}} {\left (\cos \left (f x + e\right ) - i \, \sin \left (f x + e\right )\right )}\right )\,|\,\frac {8 \, a^{2} + 8 \, a b + b^{2} - 4 \, {\left (2 \, a b + b^{2}\right )} \sqrt {\frac {a^{2} + a b}{b^{2}}}}{b^{2}})}{b^{2} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

-((2*I*sqrt(-b)*b*sqrt((a^2 + a*b)/b^2) + (-2*I*a - I*b)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/
b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 +
8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (-2*I*sqrt(-b)*b*sqrt((a^2 + a*b)/b^2) + (2*I*a +
I*b)*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2)
+ 2*a + b)/b)*(cos(f*x + e) - I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))
/b^2))/(b^2*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\sqrt {a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e)**2)**(1/2),x)

[Out]

Integral(1/sqrt(a + b*sin(e + f*x)**2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(1/sqrt(b*sin(f*x + e)^2 + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {1}{\sqrt {b\,{\sin \left (e+f\,x\right )}^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

int(1/(a + b*sin(e + f*x)^2)^(1/2), x)

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