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

Optimal. Leaf size=82 \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 a^{5/2} f}-\frac{3 \cot (e+f x)}{2 a^2 f}+\frac{\cot (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)\right )} \]

[Out]

(-3*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*a^(5/2)*f) - (3*Cot[e + f*x])/(2*a^2*f) + Cot[e + f*x]/
(2*a*f*(a + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.0741918, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3663, 290, 325, 205} \[ -\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 a^{5/2} f}-\frac{3 \cot (e+f x)}{2 a^2 f}+\frac{\cot (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(2*a^(5/2)*f) - (3*Cot[e + f*x])/(2*a^2*f) + Cot[e + f*x]/
(2*a*f*(a + b*Tan[e + f*x]^2))

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 290

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

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)}{\left (a+b \tan ^2(e+f x)\right )^2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cot (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)\right )}+\frac{3 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{2 a f}\\ &=-\frac{3 \cot (e+f x)}{2 a^2 f}+\frac{\cot (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)\right )}-\frac{(3 b) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{2 a^2 f}\\ &=-\frac{3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 a^{5/2} f}-\frac{3 \cot (e+f x)}{2 a^2 f}+\frac{\cot (e+f x)}{2 a f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.569976, size = 83, normalized size = 1.01 \[ \frac{\sqrt{a} \left (-\frac{b \sin (2 (e+f x))}{(a-b) \cos (2 (e+f x))+a+b}-2 \cot (e+f x)\right )-3 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{2 a^{5/2} f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] + Sqrt[a]*(-2*Cot[e + f*x] - (b*Sin[2*(e + f*x)])/(a + b +
(a - b)*Cos[2*(e + f*x)])))/(2*a^(5/2)*f)

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Maple [A]  time = 0.08, size = 75, normalized size = 0.9 \begin{align*} -{\frac{1}{f{a}^{2}\tan \left ( fx+e \right ) }}-{\frac{b\tan \left ( fx+e \right ) }{2\,f{a}^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) }}-{\frac{3\,b}{2\,f{a}^{2}}\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)^2,x)

[Out]

-1/f/a^2/tan(f*x+e)-1/2/f/a^2*b*tan(f*x+e)/(a+b*tan(f*x+e)^2)-3/2/f/a^2*b/(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)^2,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 1.97383, size = 888, normalized size = 10.83 \begin{align*} \left [-\frac{4 \,{\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + b\right )} \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 ) + 12 \, b \cos \left (f x + e\right )}{8 \,{\left (a^{2} b f +{\left (a^{3} - a^{2} b\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}, -\frac{2 \,{\left (2 \, a - 3 \, b\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left ({\left (a - b\right )} \cos \left (f x + e\right )^{2} + b\right )} \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 ) + 6 \, b \cos \left (f x + e\right )}{4 \,{\left (a^{2} b f +{\left (a^{3} - a^{2} b\right )} f \cos \left (f x + e\right )^{2}\right )} \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)^2,x, algorithm="fricas")

[Out]

[-1/8*(4*(2*a - 3*b)*cos(f*x + e)^3 - 3*((a - b)*cos(f*x + e)^2 + b)*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) + 12*b
*cos(f*x + e))/((a^2*b*f + (a^3 - a^2*b)*f*cos(f*x + e)^2)*sin(f*x + e)), -1/4*(2*(2*a - 3*b)*cos(f*x + e)^3 -
3*((a - b)*cos(f*x + e)^2 + b)*sqrt(b/a)*arctan(1/2*((a + b)*cos(f*x + e)^2 - b)*sqrt(b/a)/(b*cos(f*x + e)*si
n(f*x + e)))*sin(f*x + e) + 6*b*cos(f*x + e))/((a^2*b*f + (a^3 - a^2*b)*f*cos(f*x + e)^2)*sin(f*x + e))]

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

[Out]

Timed out

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Giac [A]  time = 1.41373, size = 126, normalized size = 1.54 \begin{align*} -\frac{\frac{3 \,{\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^{2}} + \frac{3 \, b \tan \left (f x + e\right )^{2} + 2 \, a}{{\left (b \tan \left (f x + e\right )^{3} + a \tan \left (f x + e\right )\right )} a^{2}}}{2 \, 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)^2,x, algorithm="giac")

[Out]

-1/2*(3*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*b/(sqrt(a*b)*a^2) + (3*b*tan(
f*x + e)^2 + 2*a)/((b*tan(f*x + e)^3 + a*tan(f*x + e))*a^2))/f