3.65 \(\int \frac{\csc ^2(e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=48 \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{3/2} f}-\frac{\cot (e+f x)}{a f} \]

[Out]

-((Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(3/2)*f)) - Cot[e + f*x]/(a*f)

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Rubi [A]  time = 0.0616512, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {3663, 325, 205} \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{3/2} f}-\frac{\cot (e+f x)}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(3/2)*f)) - Cot[e + f*x]/(a*f)

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 325

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rubi steps

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

Mathematica [A]  time = 0.117831, size = 48, normalized size = 1. \[ -\frac{\sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{3/2} f}-\frac{\cot (e+f x)}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(a^(3/2)*f)) - Cot[e + f*x]/(a*f)

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Maple [A]  time = 0.063, size = 46, normalized size = 1. \begin{align*} -{\frac{1}{fa\tan \left ( fx+e \right ) }}-{\frac{b}{fa}\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^2/(a+b*tan(f*x+e)^2),x)

[Out]

-1/f/a/tan(f*x+e)-1/f*b/a/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2/(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.82102, size = 606, normalized size = 12.62 \begin{align*} \left [\frac{\sqrt{-\frac{b}{a}} \log \left (\frac{{\left (a^{2} + 6 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \,{\left (3 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \,{\left ({\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{3} - a b \cos \left (f x + e\right )\right )} \sqrt{-\frac{b}{a}} \sin \left (f x + e\right ) + b^{2}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{4} + 2 \,{\left (a b - b^{2}\right )} \cos \left (f x + e\right )^{2} + b^{2}}\right ) \sin \left (f x + e\right ) - 4 \, \cos \left (f x + e\right )}{4 \, a f \sin \left (f x + e\right )}, \frac{\sqrt{\frac{b}{a}} \arctan \left (\frac{{\left ({\left (a + b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt{\frac{b}{a}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) \sin \left (f x + e\right ) - 2 \, \cos \left (f x + e\right )}{2 \, a f \sin \left (f x + e\right )}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2/(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

[1/4*(sqrt(-b/a)*log(((a^2 + 6*a*b + b^2)*cos(f*x + e)^4 - 2*(3*a*b + b^2)*cos(f*x + e)^2 + 4*((a^2 + a*b)*cos
(f*x + e)^3 - a*b*cos(f*x + e))*sqrt(-b/a)*sin(f*x + e) + b^2)/((a^2 - 2*a*b + b^2)*cos(f*x + e)^4 + 2*(a*b -
b^2)*cos(f*x + e)^2 + b^2))*sin(f*x + e) - 4*cos(f*x + e))/(a*f*sin(f*x + e)), 1/2*(sqrt(b/a)*arctan(1/2*((a +
b)*cos(f*x + e)^2 - b)*sqrt(b/a)/(b*cos(f*x + e)*sin(f*x + e)))*sin(f*x + e) - 2*cos(f*x + e))/(a*f*sin(f*x +
e))]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)**2/(a+b*tan(f*x+e)**2),x)

[Out]

Integral(csc(e + f*x)**2/(a + b*tan(e + f*x)**2), x)

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Giac [A]  time = 1.4093, size = 84, normalized size = 1.75 \begin{align*} -\frac{\frac{{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor \mathrm{sgn}\left (b\right ) + \arctan \left (\frac{b \tan \left (f x + e\right )}{\sqrt{a b}}\right )\right )} b}{\sqrt{a b} a} + \frac{1}{a \tan \left (f x + e\right )}}{f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(f*x+e)^2/(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

-((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*b/(sqrt(a*b)*a) + 1/(a*tan(f*x + e)
))/f