3.159 \(\int \frac{\sin ^2(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=153 \[ -\frac{\sin (e+f x) \cos (e+f x)}{f (a+b) \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{b f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{b f (a+b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

[Out]

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

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Rubi [A]  time = 0.191641, antiderivative size = 153, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3173, 3172, 3178, 3177, 3183, 3182} \[ -\frac{\sin (e+f x) \cos (e+f x)}{f (a+b) \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{b f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{b f (a+b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^2/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

-((Cos[e + f*x]*Sin[e + f*x])/((a + b)*f*Sqrt[a + b*Sin[e + f*x]^2])) - (EllipticE[e + f*x, -(b/a)]*Sqrt[a + b
*Sin[e + f*x]^2])/(b*(a + b)*f*Sqrt[1 + (b*Sin[e + f*x]^2)/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 3173

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> -Sim
p[((A*b - a*B)*Cos[e + f*x]*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(p + 1))/(2*a*f*(a + b)*(p + 1)), x] - Dist[1/
(2*a*(a + b)*(p + 1)), Int[(a + b*Sin[e + f*x]^2)^(p + 1)*Simp[a*B - A*(2*a*(p + 1) + b*(2*p + 3)) + 2*(A*b -
a*B)*(p + 2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && LtQ[p, -1] && NeQ[a + b, 0]

Rule 3172

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 3178

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 3177

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

Rule 3183

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]

Rule 3182

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

Rubi steps

\begin{align*} \int \frac{\sin ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx &=-\frac{\cos (e+f x) \sin (e+f x)}{(a+b) f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\int \frac{a-a \sin ^2(e+f x)}{\sqrt{a+b \sin ^2(e+f x)}} \, dx}{a (a+b)}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{(a+b) f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\int \frac{1}{\sqrt{a+b \sin ^2(e+f x)}} \, dx}{b}-\frac{\int \sqrt{a+b \sin ^2(e+f x)} \, dx}{b (a+b)}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{(a+b) f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\sqrt{a+b \sin ^2(e+f x)} \int \sqrt{1+\frac{b \sin ^2(e+f x)}{a}} \, dx}{b (a+b) \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\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}{b \sqrt{a+b \sin ^2(e+f x)}}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{(a+b) f \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 (a+b) f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{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*}

Mathematica [A]  time = 0.442664, size = 138, normalized size = 0.9 \[ \frac{\sqrt{2} (a+b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )-\sqrt{2} a \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )-b \sin (2 (e+f x))}{\sqrt{2} b f (a+b) \sqrt{2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^2/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

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

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Maple [A]  time = 1.262, size = 191, normalized size = 1.3 \begin{align*}{\frac{1}{ \left ( a+b \right ) b\cos \left ( fx+e \right ) f} \left ( a\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) +\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) b-a\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) +b \left ( \sin \left ( fx+e \right ) \right ) ^{3}-b\sin \left ( fx+e \right ) \right ){\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

(a*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))+(cos(f*x+e)^2)^(1/2)
*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b-a*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2
)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))+b*sin(f*x+e)^3-b*sin(f*x+e))/b/(a+b)/cos(f*x+e)/(a+b*sin(f*x+e
)^2)^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^2/(b*sin(f*x + e)^2 + a)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\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^{2} \cos \left (f x + e\right )^{4} - 2 \,{\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

integral(-sqrt(-b*cos(f*x + e)^2 + a + b)*(cos(f*x + e)^2 - 1)/(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e
)^2 + a^2 + 2*a*b + b^2), 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(sin(f*x+e)**2/(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sin \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^2/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^2/(b*sin(f*x + e)^2 + a)^(3/2), x)