3.223 \(\int \frac{\cot ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx\)

Optimal. Leaf size=113 \[ \frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{7/2} f (a-b)}-\frac{\left (a^2+a b+b^2\right ) \cot (e+f x)}{a^3 f}+\frac{(a+b) \cot ^3(e+f x)}{3 a^2 f}-\frac{x}{a-b}-\frac{\cot ^5(e+f x)}{5 a f} \]

[Out]

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

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Rubi [A]  time = 0.241209, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.261, Rules used = {3670, 480, 583, 522, 203, 205} \[ \frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{7/2} f (a-b)}-\frac{\left (a^2+a b+b^2\right ) \cot (e+f x)}{a^3 f}+\frac{(a+b) \cot ^3(e+f x)}{3 a^2 f}-\frac{x}{a-b}-\frac{\cot ^5(e+f x)}{5 a f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^6/(a + b*Tan[e + f*x]^2),x]

[Out]

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

Rule 3670

Int[((d_.)*tan[(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)/f, Subst[Int[(((d*ff*x)/c)^m*(a + b*(ff*x)^n)^p)/(c^
2 + ff^2*x^2), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && (IGtQ[p, 0] || EqQ
[n, 2] || EqQ[n, 4] || (IntegerQ[p] && RationalQ[n]))

Rule 480

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 + 1))/(a*c*e*(m + 1)), x] - Dist[1/(a*c*e^n*(m + 1)), Int[(e*x)^(m +
n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[(b*c + a*d)*(m + n + 1) + n*(b*c*p + a*d*q) + b*d*(m + n*(p + q + 2) + 1)*
x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && IntBino
mialQ[a, b, c, d, e, m, n, p, q, x]

Rule 583

Int[((g_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)),
x_Symbol] :> Simp[(e*(g*x)^(m + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(a*c*g*(m + 1)), x] + Dist[1/(a*c*
g^n*(m + 1)), Int[(g*x)^(m + n)*(a + b*x^n)^p*(c + d*x^n)^q*Simp[a*f*c*(m + 1) - e*(b*c + a*d)*(m + n + 1) - e
*n*(b*c*p + a*d*q) - b*e*d*(m + n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, g, p, q}, x] &&
IGtQ[n, 0] && LtQ[m, -1]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

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{\cot ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{1}{x^6 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{f}\\ &=-\frac{\cot ^5(e+f x)}{5 a f}+\frac{\operatorname{Subst}\left (\int \frac{-5 (a+b)-5 b x^2}{x^4 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{5 a f}\\ &=\frac{(a+b) \cot ^3(e+f x)}{3 a^2 f}-\frac{\cot ^5(e+f x)}{5 a f}-\frac{\operatorname{Subst}\left (\int \frac{-15 \left (a^2+a b+b^2\right )-15 b (a+b) x^2}{x^2 \left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{15 a^2 f}\\ &=-\frac{\left (a^2+a b+b^2\right ) \cot (e+f x)}{a^3 f}+\frac{(a+b) \cot ^3(e+f x)}{3 a^2 f}-\frac{\cot ^5(e+f x)}{5 a f}+\frac{\operatorname{Subst}\left (\int \frac{-15 (a+b) \left (a^2+b^2\right )-15 b \left (a^2+a b+b^2\right ) x^2}{\left (1+x^2\right ) \left (a+b x^2\right )} \, dx,x,\tan (e+f x)\right )}{15 a^3 f}\\ &=-\frac{\left (a^2+a b+b^2\right ) \cot (e+f x)}{a^3 f}+\frac{(a+b) \cot ^3(e+f x)}{3 a^2 f}-\frac{\cot ^5(e+f x)}{5 a f}-\frac{\operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b) f}+\frac{b^4 \operatorname{Subst}\left (\int \frac{1}{a+b x^2} \, dx,x,\tan (e+f x)\right )}{a^3 (a-b) f}\\ &=-\frac{x}{a-b}+\frac{b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{a^{7/2} (a-b) f}-\frac{\left (a^2+a b+b^2\right ) \cot (e+f x)}{a^3 f}+\frac{(a+b) \cot ^3(e+f x)}{3 a^2 f}-\frac{\cot ^5(e+f x)}{5 a f}\\ \end{align*}

Mathematica [A]  time = 1.86771, size = 121, normalized size = 1.07 \[ \frac{\sqrt{a} \left (-(a-b) \cot (e+f x) \left (3 a^2 \csc ^4(e+f x)+23 a^2-a (11 a+5 b) \csc ^2(e+f x)+20 a b+15 b^2\right )-15 a^3 (e+f x)\right )+15 b^{7/2} \tan ^{-1}\left (\frac{\sqrt{b} \tan (e+f x)}{\sqrt{a}}\right )}{15 a^{7/2} f (a-b)} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^6/(a + b*Tan[e + f*x]^2),x]

[Out]

(15*b^(7/2)*ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a]] + Sqrt[a]*(-15*a^3*(e + f*x) - (a - b)*Cot[e + f*x]*(23*a^2
+ 20*a*b + 15*b^2 - a*(11*a + 5*b)*Csc[e + f*x]^2 + 3*a^2*Csc[e + f*x]^4)))/(15*a^(7/2)*(a - b)*f)

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Maple [A]  time = 0.082, size = 158, normalized size = 1.4 \begin{align*} -{\frac{1}{5\,fa \left ( \tan \left ( fx+e \right ) \right ) ^{5}}}+{\frac{1}{3\,fa \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}+{\frac{b}{3\,f{a}^{2} \left ( \tan \left ( fx+e \right ) \right ) ^{3}}}-{\frac{1}{fa\tan \left ( fx+e \right ) }}-{\frac{b}{f{a}^{2}\tan \left ( fx+e \right ) }}-{\frac{{b}^{2}}{f{a}^{3}\tan \left ( fx+e \right ) }}+{\frac{{b}^{4}}{f{a}^{3} \left ( a-b \right ) }\arctan \left ({b\tan \left ( fx+e \right ){\frac{1}{\sqrt{ab}}}} \right ){\frac{1}{\sqrt{ab}}}}-{\frac{\arctan \left ( \tan \left ( fx+e \right ) \right ) }{f \left ( a-b \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^6/(a+b*tan(f*x+e)^2),x)

[Out]

-1/5/f/a/tan(f*x+e)^5+1/3/f/a/tan(f*x+e)^3+1/3/f/a^2/tan(f*x+e)^3*b-1/f/a/tan(f*x+e)-1/f/a^2/tan(f*x+e)*b-1/f/
a^3/tan(f*x+e)*b^2+1/f/a^3*b^4/(a-b)/(a*b)^(1/2)*arctan(b*tan(f*x+e)/(a*b)^(1/2))-1/f/(a-b)*arctan(tan(f*x+e))

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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(cot(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 1.23079, size = 811, normalized size = 7.18 \begin{align*} \left [-\frac{60 \, a^{3} f x \tan \left (f x + e\right )^{5} + 15 \, b^{3} \sqrt{-\frac{b}{a}} \log \left (\frac{b^{2} \tan \left (f x + e\right )^{4} - 6 \, a b \tan \left (f x + e\right )^{2} + a^{2} - 4 \,{\left (a b \tan \left (f x + e\right )^{3} - a^{2} \tan \left (f x + e\right )\right )} \sqrt{-\frac{b}{a}}}{b^{2} \tan \left (f x + e\right )^{4} + 2 \, a b \tan \left (f x + e\right )^{2} + a^{2}}\right ) \tan \left (f x + e\right )^{5} + 60 \,{\left (a^{3} - b^{3}\right )} \tan \left (f x + e\right )^{4} + 12 \, a^{3} - 12 \, a^{2} b - 20 \,{\left (a^{3} - a b^{2}\right )} \tan \left (f x + e\right )^{2}}{60 \,{\left (a^{4} - a^{3} b\right )} f \tan \left (f x + e\right )^{5}}, -\frac{30 \, a^{3} f x \tan \left (f x + e\right )^{5} - 15 \, b^{3} \sqrt{\frac{b}{a}} \arctan \left (\frac{{\left (b \tan \left (f x + e\right )^{2} - a\right )} \sqrt{\frac{b}{a}}}{2 \, b \tan \left (f x + e\right )}\right ) \tan \left (f x + e\right )^{5} + 30 \,{\left (a^{3} - b^{3}\right )} \tan \left (f x + e\right )^{4} + 6 \, a^{3} - 6 \, a^{2} b - 10 \,{\left (a^{3} - a b^{2}\right )} \tan \left (f x + e\right )^{2}}{30 \,{\left (a^{4} - a^{3} b\right )} f \tan \left (f x + e\right )^{5}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="fricas")

[Out]

[-1/60*(60*a^3*f*x*tan(f*x + e)^5 + 15*b^3*sqrt(-b/a)*log((b^2*tan(f*x + e)^4 - 6*a*b*tan(f*x + e)^2 + a^2 - 4
*(a*b*tan(f*x + e)^3 - a^2*tan(f*x + e))*sqrt(-b/a))/(b^2*tan(f*x + e)^4 + 2*a*b*tan(f*x + e)^2 + a^2))*tan(f*
x + e)^5 + 60*(a^3 - b^3)*tan(f*x + e)^4 + 12*a^3 - 12*a^2*b - 20*(a^3 - a*b^2)*tan(f*x + e)^2)/((a^4 - a^3*b)
*f*tan(f*x + e)^5), -1/30*(30*a^3*f*x*tan(f*x + e)^5 - 15*b^3*sqrt(b/a)*arctan(1/2*(b*tan(f*x + e)^2 - a)*sqrt
(b/a)/(b*tan(f*x + e)))*tan(f*x + e)^5 + 30*(a^3 - b^3)*tan(f*x + e)^4 + 6*a^3 - 6*a^2*b - 10*(a^3 - a*b^2)*ta
n(f*x + e)^2)/((a^4 - a^3*b)*f*tan(f*x + e)^5)]

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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(cot(f*x+e)**6/(a+b*tan(f*x+e)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.7312, size = 602, normalized size = 5.33 \begin{align*} \frac{\frac{15 \,{\left (a^{6} b + a^{3} b^{4} - a^{2} b{\left | -a^{4} + a^{3} b \right |} - a b^{2}{\left | -a^{4} + a^{3} b \right |} - b^{3}{\left | -a^{4} + a^{3} b \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{2 \, \tan \left (f x + e\right )}{\sqrt{\frac{2 \, a^{4} + 2 \, a^{3} b + \sqrt{-16 \, a^{7} b + 4 \,{\left (a^{4} + a^{3} b\right )}^{2}}}{a^{3} b}}}\right )\right )}}{a^{4}{\left | -a^{4} + a^{3} b \right |} + a^{3} b{\left | -a^{4} + a^{3} b \right |} +{\left (a^{4} - a^{3} b\right )}^{2}} + \frac{15 \,{\left ({\left (a^{2} + a b + b^{2}\right )} \sqrt{a b}{\left | -a^{4} + a^{3} b \right |}{\left | b \right |} +{\left (a^{6} + a^{3} b^{3}\right )} \sqrt{a b}{\left | b \right |}\right )}{\left (\pi \left \lfloor \frac{f x + e}{\pi } + \frac{1}{2} \right \rfloor + \arctan \left (\frac{2 \, \tan \left (f x + e\right )}{\sqrt{\frac{2 \, a^{4} + 2 \, a^{3} b - \sqrt{-16 \, a^{7} b + 4 \,{\left (a^{4} + a^{3} b\right )}^{2}}}{a^{3} b}}}\right )\right )}}{{\left (a^{4} - a^{3} b\right )}^{2} b -{\left (a^{4} b + a^{3} b^{2}\right )}{\left | -a^{4} + a^{3} b \right |}} - \frac{15 \, a^{2} \tan \left (f x + e\right )^{4} + 15 \, a b \tan \left (f x + e\right )^{4} + 15 \, b^{2} \tan \left (f x + e\right )^{4} - 5 \, a^{2} \tan \left (f x + e\right )^{2} - 5 \, a b \tan \left (f x + e\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (f x + e\right )^{5}}}{15 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^6/(a+b*tan(f*x+e)^2),x, algorithm="giac")

[Out]

1/15*(15*(a^6*b + a^3*b^4 - a^2*b*abs(-a^4 + a^3*b) - a*b^2*abs(-a^4 + a^3*b) - b^3*abs(-a^4 + a^3*b))*(pi*flo
or((f*x + e)/pi + 1/2) + arctan(2*tan(f*x + e)/sqrt((2*a^4 + 2*a^3*b + sqrt(-16*a^7*b + 4*(a^4 + a^3*b)^2))/(a
^3*b))))/(a^4*abs(-a^4 + a^3*b) + a^3*b*abs(-a^4 + a^3*b) + (a^4 - a^3*b)^2) + 15*((a^2 + a*b + b^2)*sqrt(a*b)
*abs(-a^4 + a^3*b)*abs(b) + (a^6 + a^3*b^3)*sqrt(a*b)*abs(b))*(pi*floor((f*x + e)/pi + 1/2) + arctan(2*tan(f*x
+ e)/sqrt((2*a^4 + 2*a^3*b - sqrt(-16*a^7*b + 4*(a^4 + a^3*b)^2))/(a^3*b))))/((a^4 - a^3*b)^2*b - (a^4*b + a^
3*b^2)*abs(-a^4 + a^3*b)) - (15*a^2*tan(f*x + e)^4 + 15*a*b*tan(f*x + e)^4 + 15*b^2*tan(f*x + e)^4 - 5*a^2*tan
(f*x + e)^2 - 5*a*b*tan(f*x + e)^2 + 3*a^2)/(a^3*tan(f*x + e)^5))/f