3.361 \(\int \frac{1}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=101 \[ \frac{b \sin (e+f x) \cos (e+f x)}{a f (a+b) \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{a f (a+b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

[Out]

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

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Rubi [A]  time = 0.0599659, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {3184, 21, 3178, 3177} \[ \frac{b \sin (e+f x) \cos (e+f x)}{a f (a+b) \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{a f (a+b) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x]^2)^(-3/2),x]

[Out]

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

Rule 3184

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, 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]

Rubi steps

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

Mathematica [A]  time = 0.14595, size = 90, normalized size = 0.89 \[ \frac{2 a \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )+\sqrt{2} b \sin (2 (e+f x))}{2 a f (a+b) \sqrt{2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 1.351, size = 103, normalized size = 1. \begin{align*}{\frac{1}{a \left ( a+b \right ) \cos \left ( fx+e \right ) f} \left ( a\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) + \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) b \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(1/(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

(a*(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))+cos(f*x+e)^2*si
n(f*x+e)*b)/a/(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{1}{{\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(1/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((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}}{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(1/(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

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

[Out]

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