3.519 \(\int \frac{\cot ^2(e+f x)}{\sqrt{a+b \sin ^2(e+f x)}} \, dx\)

Optimal. Leaf size=106 \[ -\frac{\cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{a f}-\frac{\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{a f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

[Out]

-((Cot[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(a*f)) - (Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], -(b
/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(a*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a])

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Rubi [A]  time = 0.115462, antiderivative size = 106, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3196, 475, 21, 426, 424} \[ -\frac{\cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{a f}-\frac{\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{a f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^2/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

-((Cot[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(a*f)) - (Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], -(b
/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(a*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a])

Rule 3196

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[(ff^(m + 1)*Sqrt[Cos[e + f*x]^2])/(f*Cos[e + f*x]), Subst[Int[(x^m*(a + b*ff^2*
x^2)^p)/(1 - ff^2*x^2)^((m + 1)/2), x], x, Sin[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[m/2]
&&  !IntegerQ[p]

Rule 475

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[((e*x)^(m
+ 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*
x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*b*(m + 1) + n*(b*c*(p + 1) + a*d*q) + d*(b*(m + 1) + b*n*(p + q + 1))*x^n, x
], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[0, q, 1] && LtQ[m, -1] &&
IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

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 426

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[a + b*x^2]/Sqrt[1 + (b*x^2)/a]
, Int[Sqrt[1 + (b*x^2)/a]/Sqrt[c + d*x^2], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] &&  !GtQ
[a, 0]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rubi steps

\begin{align*} \int \frac{\cot ^2(e+f x)}{\sqrt{a+b \sin ^2(e+f x)}} \, dx &=\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-x^2}}{x^2 \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac{\cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{a f}+\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{-a-b x^2}{\sqrt{1-x^2} \sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{a f}\\ &=-\frac{\cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{a f}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{a f}\\ &=-\frac{\cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{a f}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{b x^2}{a}}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{a f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}\\ &=-\frac{\cot (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{a f}-\frac{\sqrt{\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right ) \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{a f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}\\ \end{align*}

Mathematica [A]  time = 0.371811, size = 101, normalized size = 0.95 \[ -\frac{\cot (e+f x) \sqrt{2 a-b \cos (2 (e+f x))+b}}{\sqrt{2} a f}-\frac{\sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{f \sqrt{2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^2/Sqrt[a + b*Sin[e + f*x]^2],x]

[Out]

-((Sqrt[2*a + b - b*Cos[2*(e + f*x)]]*Cot[e + f*x])/(Sqrt[2]*a*f)) - (Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*E
llipticE[e + f*x, -(b/a)])/(f*Sqrt[2*a + b - b*Cos[2*(e + f*x)]])

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Maple [A]  time = 1.09, size = 120, normalized size = 1.1 \begin{align*} -{\frac{1}{a\sin \left ( fx+e \right ) \cos \left ( fx+e \right ) f} \left ( \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}}}a\sin \left ( fx+e \right ){\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) -b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2} \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(cot(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2),x)

[Out]

-((cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*a*sin(f*x+e)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))-b*c
os(f*x+e)^4+(a+b)*cos(f*x+e)^2)/a/sin(f*x+e)/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{\cot \left (f x + e\right )^{2}}{\sqrt{b \sin \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(cot(f*x + e)^2/sqrt(b*sin(f*x + e)^2 + a), 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} \cot \left (f x + e\right )^{2}}{b \cos \left (f x + e\right )^{2} - a - b}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(-sqrt(-b*cos(f*x + e)^2 + a + b)*cot(f*x + e)^2/(b*cos(f*x + e)^2 - a - b), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot ^{2}{\left (e + f x \right )}}{\sqrt{a + b \sin ^{2}{\left (e + f x \right )}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(cot(e + f*x)**2/sqrt(a + b*sin(e + f*x)**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cot \left (f x + e\right )^{2}}{\sqrt{b \sin \left (f x + e\right )^{2} + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^2/(a+b*sin(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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