3.168 \(\int \frac{\sin ^2(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=221 \[ -\frac{(a-b) \sin (e+f x) \cos (e+f x)}{3 a f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}-\frac{\sin (e+f x) \cos (e+f x)}{3 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 b f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{(a-b) \sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 a b f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

[Out]

-(Cos[e + f*x]*Sin[e + f*x])/(3*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) - ((a - b)*Cos[e + f*x]*Sin[e + f*x])/
(3*a*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) - ((a - b)*EllipticE[e + f*x, -(b/a)]*Sqrt[a + b*Sin[e + f*x]^2])
/(3*a*b*(a + b)^2*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])/(3*b*(a + b)*f*Sqrt[a + b*Sin[e + f*x]^2])

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Rubi [A]  time = 0.289666, antiderivative size = 221, normalized size of antiderivative = 1., number of steps used = 7, 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} \[ -\frac{(a-b) \sin (e+f x) \cos (e+f x)}{3 a f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}-\frac{\sin (e+f x) \cos (e+f x)}{3 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 b f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{(a-b) \sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 a b f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^2/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

-(Cos[e + f*x]*Sin[e + f*x])/(3*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) - ((a - b)*Cos[e + f*x]*Sin[e + f*x])/
(3*a*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) - ((a - b)*EllipticE[e + f*x, -(b/a)]*Sqrt[a + b*Sin[e + f*x]^2])
/(3*a*b*(a + b)^2*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])/(3*b*(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 )^{5/2}} \, dx &=-\frac{\cos (e+f x) \sin (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\int \frac{a+a \sin ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{3/2}} \, dx}{3 a (a+b)}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(a-b) \cos (e+f x) \sin (e+f x)}{3 a (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\int \frac{2 a^2-a (a-b) \sin ^2(e+f x)}{\sqrt{a+b \sin ^2(e+f x)}} \, dx}{3 a^2 (a+b)^2}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(a-b) \cos (e+f x) \sin (e+f x)}{3 a (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(a-b) \int \sqrt{a+b \sin ^2(e+f x)} \, dx}{3 a b (a+b)^2}+\frac{\int \frac{1}{\sqrt{a+b \sin ^2(e+f x)}} \, dx}{3 b (a+b)}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(a-b) \cos (e+f x) \sin (e+f x)}{3 a (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\left ((a-b) \sqrt{a+b \sin ^2(e+f x)}\right ) \int \sqrt{1+\frac{b \sin ^2(e+f x)}{a}} \, dx}{3 a b (a+b)^2 \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}{3 b (a+b) \sqrt{a+b \sin ^2(e+f x)}}\\ &=-\frac{\cos (e+f x) \sin (e+f x)}{3 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(a-b) \cos (e+f x) \sin (e+f x)}{3 a (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(a-b) E\left (e+f x\left |-\frac{b}{a}\right .\right ) \sqrt{a+b \sin ^2(e+f x)}}{3 a b (a+b)^2 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}}}{3 b (a+b) f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.46077, size = 174, normalized size = 0.79 \[ \frac{-\sqrt{2} b \sin (2 (e+f x)) \left (4 a^2+b (b-a) \cos (2 (e+f x))+a b-b^2\right )+2 a^2 (a+b) \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} F\left (e+f x\left |-\frac{b}{a}\right .\right )-2 a^2 (a-b) \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{6 a b f (a+b)^2 (2 a-b \cos (2 (e+f x))+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^2/(a + b*Sin[e + f*x]^2)^(5/2),x]

[Out]

(-2*a^2*(a - b)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticE[e + f*x, -(b/a)] + 2*a^2*(a + b)*((2*a + b
- b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticF[e + f*x, -(b/a)] - Sqrt[2]*b*(4*a^2 + a*b - b^2 + b*(-a + b)*Cos[2*(e
+ f*x)])*Sin[2*(e + f*x)])/(6*a*b*(a + b)^2*f*(2*a + b - b*Cos[2*(e + f*x)])^(3/2))

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Maple [A]  time = 1.424, size = 483, normalized size = 2.2 \begin{align*} \text{result too large to display} \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)^(5/2),x)

[Out]

1/3*((a*b^2-b^3)*sin(f*x+e)*cos(f*x+e)^4+(-2*a^2*b-a*b^2+b^3)*cos(f*x+e)^2*sin(f*x+e)-(-b/a*cos(f*x+e)^2+(a+b)
/a)^(1/2)*(cos(f*x+e)^2)^(1/2)*a*b*(EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a+EllipticF(sin(f*x+e),(-1/a*b)^(1/2)
)*b-EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a+EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*b)*cos(f*x+e)^2+(cos(f*x+e)^2)
^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^3+2*(cos(f*x+e)^2)^(1/2)*(-b/a
*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b+(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^
2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2-(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(
1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^3+(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE
(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2)/(a+b*sin(f*x+e)^2)^(3/2)/(a+b)^2/a/b/cos(f*x+e)/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{5}{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)^(5/2),x, algorithm="maxima")

[Out]

integrate(sin(f*x + e)^2/(b*sin(f*x + e)^2 + a)^(5/2), 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^{3} \cos \left (f x + e\right )^{6} - 3 \,{\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 3 \, a^{2} b - 3 \, a b^{2} - b^{3} + 3 \,{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \cos \left (f x + e\right )^{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)^(5/2),x, algorithm="fricas")

[Out]

integral(sqrt(-b*cos(f*x + e)^2 + a + b)*(cos(f*x + e)^2 - 1)/(b^3*cos(f*x + e)^6 - 3*(a*b^2 + b^3)*cos(f*x +
e)^4 - a^3 - 3*a^2*b - 3*a*b^2 - b^3 + 3*(a^2*b + 2*a*b^2 + b^3)*cos(f*x + e)^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)**(5/2),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{5}{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)^(5/2),x, algorithm="giac")

[Out]

integrate(sin(f*x + e)^2/(b*sin(f*x + e)^2 + a)^(5/2), x)