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

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

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Rubi [A]  time = 0.06, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, 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 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 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 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]

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*}

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Mathematica [A]  time = 0.15, 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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fricas [F]  time = 0.46, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(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)

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maple [A]  time = 1.82, size = 103, normalized size = 1.02 \[ \frac {\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, a \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right )+\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) b}{a \left (a +b \right ) \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

((cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*a*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))+sin(f*x+e)*cos(
f*x+e)^2*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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a + b*sin(e + f*x)^2)^(3/2),x)

[Out]

int(1/(a + b*sin(e + f*x)^2)^(3/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

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

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