3.89 \(\int \frac{\csc ^2(e+f x)}{(a+b \tan ^2(e+f x))^3} \, dx\)

Optimal. Leaf size=112 \[ -\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{7/2} f}+\frac{5 \cot (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac{15 \cot (e+f x)}{8 a^3 f}+\frac{\cot (e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^2} \]

[Out]

(-15*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(8*a^(7/2)*f) - (15*Cot[e + f*x])/(8*a^3*f) + Cot[e + f*x
]/(4*a*f*(a + b*Tan[e + f*x]^2)^2) + (5*Cot[e + f*x])/(8*a^2*f*(a + b*Tan[e + f*x]^2))

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Rubi [A]  time = 0.0877426, antiderivative size = 112, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3663, 290, 325, 205} \[ -\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{7/2} f}+\frac{5 \cot (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac{15 \cot (e+f x)}{8 a^3 f}+\frac{\cot (e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^2} \]

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^2/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

(-15*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]])/(8*a^(7/2)*f) - (15*Cot[e + f*x])/(8*a^3*f) + Cot[e + f*x
]/(4*a*f*(a + b*Tan[e + f*x]^2)^2) + (5*Cot[e + f*x])/(8*a^2*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 )^3} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^3} \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac{\cot (e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{5 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )^2} \, dx,x,\tan (e+f x)\right )}{4 a f}\\ &=\frac{\cot (e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{5 \cot (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )}+\frac{15 \operatorname{Subst}\left (\int \frac{1}{x^2 \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{8 a^2 f}\\ &=-\frac{15 \cot (e+f x)}{8 a^3 f}+\frac{\cot (e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{5 \cot (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )}-\frac{(15 b) \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{8 a^3 f}\\ &=-\frac{15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{8 a^{7/2} f}-\frac{15 \cot (e+f x)}{8 a^3 f}+\frac{\cot (e+f x)}{4 a f \left (a+b \tan ^2(e+f x)\right )^2}+\frac{5 \cot (e+f x)}{8 a^2 f \left (a+b \tan ^2(e+f x)\right )}\\ \end{align*}

Mathematica [A]  time = 0.826774, size = 144, normalized size = 1.29 \[ \frac{\frac{4 a^{3/2} b^2 \sin (2 (e+f x))}{(a-b) ((a-b) \cos (2 (e+f x))+a+b)^2}-15 \sqrt{b} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )-\frac{\sqrt{a} b (9 a-7 b) \sin (2 (e+f x))}{(a-b) ((a-b) \cos (2 (e+f x))+a+b)}-8 \sqrt{a} \cot (e+f x)}{8 a^{7/2} f} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[e + f*x]^2/(a + b*Tan[e + f*x]^2)^3,x]

[Out]

(-15*Sqrt[b]*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] - 8*Sqrt[a]*Cot[e + f*x] + (4*a^(3/2)*b^2*Sin[2*(e + f*x)]
)/((a - b)*(a + b + (a - b)*Cos[2*(e + f*x)])^2) - (Sqrt[a]*(9*a - 7*b)*b*Sin[2*(e + f*x)])/((a - b)*(a + b +
(a - b)*Cos[2*(e + f*x)])))/(8*a^(7/2)*f)

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Maple [A]  time = 0.086, size = 108, normalized size = 1. \begin{align*} -{\frac{1}{f{a}^{3}\tan \left ( fx+e \right ) }}-{\frac{7\,{b}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}{8\,f{a}^{3} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{9\,b\tan \left ( fx+e \right ) }{8\,f{a}^{2} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{2}}}-{\frac{15\,b}{8\,f{a}^{3}}\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)^3,x)

[Out]

-1/f/a^3/tan(f*x+e)-7/8/f/a^3*b^2/(a+b*tan(f*x+e)^2)^2*tan(f*x+e)^3-9/8/f/a^2*b/(a+b*tan(f*x+e)^2)^2*tan(f*x+e
)-15/8/f/a^3*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)^3,x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.34278, size = 1300, normalized size = 11.61 \begin{align*} \left [-\frac{4 \,{\left (8 \, a^{2} - 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 20 \,{\left (5 \, a b - 6 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left ({\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 )} \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 ) + 60 \, b^{2} \cos \left (f x + e\right )}{32 \,{\left (a^{3} b^{2} f +{\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b - a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}, -\frac{2 \,{\left (8 \, a^{2} - 25 \, a b + 15 \, b^{2}\right )} \cos \left (f x + e\right )^{5} + 10 \,{\left (5 \, a b - 6 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - 15 \,{\left ({\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 )} \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 ) + 30 \, b^{2} \cos \left (f x + e\right )}{16 \,{\left (a^{3} b^{2} f +{\left (a^{5} - 2 \, a^{4} b + a^{3} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \,{\left (a^{4} b - a^{3} b^{2}\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)^3,x, algorithm="fricas")

[Out]

[-1/32*(4*(8*a^2 - 25*a*b + 15*b^2)*cos(f*x + e)^5 + 20*(5*a*b - 6*b^2)*cos(f*x + e)^3 - 15*((a^2 - 2*a*b + b^
2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^2)*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) + 60*b^2*cos(f*x +
e))/((a^3*b^2*f + (a^5 - 2*a^4*b + a^3*b^2)*f*cos(f*x + e)^4 + 2*(a^4*b - a^3*b^2)*f*cos(f*x + e)^2)*sin(f*x +
e)), -1/16*(2*(8*a^2 - 25*a*b + 15*b^2)*cos(f*x + e)^5 + 10*(5*a*b - 6*b^2)*cos(f*x + e)^3 - 15*((a^2 - 2*a*b
+ b^2)*cos(f*x + e)^4 + 2*(a*b - b^2)*cos(f*x + e)^2 + b^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) + 30*b^2*cos(f*x + e))/((a^3*b^2*f + (a^5 - 2*a^4*b + a
^3*b^2)*f*cos(f*x + e)^4 + 2*(a^4*b - a^3*b^2)*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)**3,x)

[Out]

Timed out

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Giac [A]  time = 1.62092, size = 147, normalized size = 1.31 \begin{align*} -\frac{\frac{15 \,{\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^{3}} + \frac{7 \, b^{2} \tan \left (f x + e\right )^{3} + 9 \, a b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a\right )}^{2} a^{3}} + \frac{8}{a^{3} \tan \left (f x + e\right )}}{8 \, 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)^3,x, algorithm="giac")

[Out]

-1/8*(15*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*b/(sqrt(a*b)*a^3) + (7*b^2*t
an(f*x + e)^3 + 9*a*b*tan(f*x + e))/((b*tan(f*x + e)^2 + a)^2*a^3) + 8/(a^3*tan(f*x + e)))/f