3.2.49 \(\int \frac {\sin ^2(e+f x)}{\sqrt {a+b \sin ^2(e+f x)}} \, dx\) [149]

Optimal. Leaf size=111 \[ \frac {E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {a+b \sin ^2(e+f x)}}{b f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {a F\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{b f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 111, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3251, 3257, 3256, 3262, 3261} \begin {gather*} \frac {\sqrt {a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac {b}{a}\right .\right )}{b f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {a \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac {b}{a}\right .\right )}{b f \sqrt {a+b \sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3251

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2], x_Symbol] :> Dist[
B/b, Int[Sqrt[a + b*Sin[e + f*x]^2], x], x] + Dist[(A*b - a*B)/b, Int[1/Sqrt[a + b*Sin[e + f*x]^2], x], x] /;
FreeQ[{a, b, e, f, A, B}, x]

Rule 3256

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

Rule 3257

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

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

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Mathematica [A]
time = 0.17, size = 78, normalized size = 0.70 \begin {gather*} \frac {\sqrt {2 a+b-b \cos (2 (e+f x))} \left (E\left (e+f x\left |-\frac {b}{a}\right .\right )-F\left (e+f x\left |-\frac {b}{a}\right .\right )\right )}{b f \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 4.92, size = 93, normalized size = 0.84

method	result	size
default	\(-\frac {a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \left (\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )-\EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )\right )}{b \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(93\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [F]
time = 0.10, size = 48, normalized size = 0.43 \begin {gather*} {\rm integral}\left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (\cos \left (f x + e\right )^{2} - 1\right )}}{b \cos \left (f x + e\right )^{2} - a - b}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (e+f\,x\right )}^2}{\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(sin(e + f*x)^2/(a + b*sin(e + f*x)^2)^(1/2),x)

[Out]

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

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