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(f*x+e)*sin(f*x+e)/(a+b)/f/(a+b*sin(f*x+e)^2)^(1/2)-(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/(a+b)/f/(1+b*sin(f*x+e)^2/a)^(1/2)+(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*Ell
ipticF(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 = 0.19, antiderivative size = 153, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, 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 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 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 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 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 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]

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]

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*}

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Mathematica [A]  time = 0.45, size = 138, normalized size = 0.90 \[ \frac {\sqrt {2} (a+b) \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )-\sqrt {2} a \sqrt {\frac {2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )-b \sin (2 (e+f x))}{\sqrt {2} b f (a+b) \sqrt {2 a-b \cos (2 (e+f x))+b}} \]

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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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} {\left (\cos \left (f x + e\right )^{2} - 1\right )}}{b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

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)

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maple [A]  time = 1.52, size = 191, normalized size = 1.25 \[ \frac {a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )+\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -a \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {\frac {a +b \left (\sin ^{2}\left (f x +e \right )\right )}{a}}\, \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )+b \left (\sin ^{3}\left (f x +e \right )\right )-b \sin \left (f x +e \right )}{b \left (a +b \right ) \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sin \left (f x + e\right )^{2}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\sin \left (e+f\,x\right )}^2}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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