3.370 \(\int \frac{1}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=223 \[ \frac{2 b (2 a+b) \sin (e+f x) \cos (e+f x)}{3 a^2 f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}+\frac{2 (2 a+b) \sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 a^2 f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}+\frac{b \sin (e+f x) \cos (e+f x)}{3 a f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 a f (a+b) \sqrt{a+b \sin ^2(e+f x)}} \]

[Out]

(b*Cos[e + f*x]*Sin[e + f*x])/(3*a*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*b*(2*a + b)*Cos[e + f*x]*Sin[e
+ f*x])/(3*a^2*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) + (2*(2*a + b)*EllipticE[e + f*x, -(b/a)]*Sqrt[a + b*S
in[e + f*x]^2])/(3*a^2*(a + b)^2*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) - (EllipticF[e + f*x, -(b/a)]*Sqrt[1 + (b*S
in[e + f*x]^2)/a])/(3*a*(a + b)*f*Sqrt[a + b*Sin[e + f*x]^2])

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Rubi [A]  time = 0.265512, antiderivative size = 223, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.438, Rules used = {3184, 3173, 3172, 3178, 3177, 3183, 3182} \[ \frac{2 b (2 a+b) \sin (e+f x) \cos (e+f x)}{3 a^2 f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}}+\frac{2 (2 a+b) \sqrt{a+b \sin ^2(e+f x)} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 a^2 f (a+b)^2 \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}+\frac{b \sin (e+f x) \cos (e+f x)}{3 a f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 a f (a+b) \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[e + f*x]^2)^(-5/2),x]

[Out]

(b*Cos[e + f*x]*Sin[e + f*x])/(3*a*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) + (2*b*(2*a + b)*Cos[e + f*x]*Sin[e
+ f*x])/(3*a^2*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) + (2*(2*a + b)*EllipticE[e + f*x, -(b/a)]*Sqrt[a + b*S
in[e + f*x]^2])/(3*a^2*(a + b)^2*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) - (EllipticF[e + f*x, -(b/a)]*Sqrt[1 + (b*S
in[e + f*x]^2)/a])/(3*a*(a + b)*f*Sqrt[a + b*Sin[e + f*x]^2])

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

Mathematica [A]  time = 1.22509, size = 172, normalized size = 0.77 \[ \frac{-\sqrt{2} b \sin (2 (e+f x)) \left (-5 a^2+b (2 a+b) \cos (2 (e+f x))-5 a b-b^2\right )-a^2 (a+b) \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} F\left (e+f x\left |-\frac{b}{a}\right .\right )+2 a^2 (2 a+b) \left (\frac{2 a-b \cos (2 (e+f x))+b}{a}\right )^{3/2} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{3 a^2 f (a+b)^2 (2 a-b \cos (2 (e+f x))+b)^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a^2*(2*a + b)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticE[e + f*x, -(b/a)] - 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*(-5*a^2 - 5*a*b - b^2 + b*(2*a + b)*Cos[2
*(e + f*x)])*Sin[2*(e + f*x)])/(3*a^2*(a + b)^2*f*(2*a + b - b*Cos[2*(e + f*x)])^(3/2))

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Maple [B]  time = 1.664, size = 547, normalized size = 2.5 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(a+b*sin(f*x+e)^2)^(5/2),x)

[Out]

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

[Out]

integrate((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}}{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(1/(a+b*sin(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

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

[Out]

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