3.3.23 \(\int \frac {\cot ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx\) [223]

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

[Out]

-x/(a-b)+b^(7/2)*arctan(b^(1/2)*tan(f*x+e)/a^(1/2))/a^(7/2)/(a-b)/f-(a^2+a*b+b^2)*cot(f*x+e)/a^3/f+1/3*(a+b)*c
ot(f*x+e)^3/a^2/f-1/5*cot(f*x+e)^5/a/f

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Rubi [A]
time = 0.16, antiderivative size = 113, normalized size of antiderivative = 1.00, 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 = {3751, 491, 597, 536, 209, 211} \begin {gather*} \frac {b^{7/2} \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )}{a^{7/2} f (a-b)}+\frac {(a+b) \cot ^3(e+f x)}{3 a^2 f}-\frac {\left (a^2+a b+b^2\right ) \cot (e+f x)}{a^3 f}-\frac {x}{a-b}-\frac {\cot ^5(e+f x)}{5 a f} \end {gather*}

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 209

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

Rule 211

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

Rule 491

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 536

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 597

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 3751

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]))

Rubi steps

\begin {align*} \int \frac {\cot ^6(e+f x)}{a+b \tan ^2(e+f x)} \, dx &=\frac {\text {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 {\text {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 {\text {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 {\text {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 {\text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{(a-b) f}+\frac {b^4 \text {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*}

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Mathematica [A]
time = 1.28, size = 121, normalized size = 1.07 \begin {gather*} \frac {15 b^{7/2} \text {ArcTan}\left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a}}\right )+\sqrt {a} \left (-15 a^3 (e+f x)-(a-b) \cot (e+f x) \left (23 a^2+20 a b+15 b^2-a (11 a+5 b) \csc ^2(e+f x)+3 a^2 \csc ^4(e+f x)\right )\right )}{15 a^{7/2} (a-b) f} \end {gather*}

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 = 0.34, size = 111, normalized size = 0.98

method	result	size
derivativedivides	\(\frac {\frac {b^{4} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{a^{3} \left (a -b \right ) \sqrt {a b}}-\frac {1}{5 a \tan \left (f x +e \right )^{5}}-\frac {-a -b}{3 a^{2} \tan \left (f x +e \right )^{3}}-\frac {a^{2}+a b +b^{2}}{a^{3} \tan \left (f x +e \right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a -b}}{f}\)	\(111\)
default	\(\frac {\frac {b^{4} \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {a b}}\right )}{a^{3} \left (a -b \right ) \sqrt {a b}}-\frac {1}{5 a \tan \left (f x +e \right )^{5}}-\frac {-a -b}{3 a^{2} \tan \left (f x +e \right )^{3}}-\frac {a^{2}+a b +b^{2}}{a^{3} \tan \left (f x +e \right )}-\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a -b}}{f}\)	\(111\)
risch	\(-\frac {x}{a -b}-\frac {2 i \left (45 a^{2} {\mathrm e}^{8 i \left (f x +e \right )}+30 a b \,{\mathrm e}^{8 i \left (f x +e \right )}+15 b^{2} {\mathrm e}^{8 i \left (f x +e \right )}-90 a^{2} {\mathrm e}^{6 i \left (f x +e \right )}-90 a b \,{\mathrm e}^{6 i \left (f x +e \right )}-60 b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+140 a^{2} {\mathrm e}^{4 i \left (f x +e \right )}+110 a b \,{\mathrm e}^{4 i \left (f x +e \right )}+90 b^{2} {\mathrm e}^{4 i \left (f x +e \right )}-70 a^{2} {\mathrm e}^{2 i \left (f x +e \right )}-70 a b \,{\mathrm e}^{2 i \left (f x +e \right )}-60 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+23 a^{2}+20 a b +15 b^{2}\right )}{15 f \,a^{3} \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5}}-\frac {\sqrt {-a b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-a b}+a +b}{a -b}\right )}{2 a^{4} \left (a -b \right ) f}+\frac {\sqrt {-a b}\, b^{3} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-a b}-a -b}{a -b}\right )}{2 a^{4} \left (a -b \right ) f}\)	\(328\)

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,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.49, size = 117, normalized size = 1.04 \begin {gather*} \frac {\frac {15 \, b^{4} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b}}\right )}{{\left (a^{4} - a^{3} b\right )} \sqrt {a b}} - \frac {15 \, {\left (f x + e\right )}}{a - b} - \frac {15 \, {\left (a^{2} + a b + b^{2}\right )} \tan \left (f x + e\right )^{4} - 5 \, {\left (a^{2} + a b\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2}}{a^{3} \tan \left (f x + e\right )^{5}}}{15 \, f} \end {gather*}

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]

1/15*(15*b^4*arctan(b*tan(f*x + e)/sqrt(a*b))/((a^4 - a^3*b)*sqrt(a*b)) - 15*(f*x + e)/(a - b) - (15*(a^2 + a*
b + b^2)*tan(f*x + e)^4 - 5*(a^2 + a*b)*tan(f*x + e)^2 + 3*a^2)/(a^3*tan(f*x + e)^5))/f

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Fricas [A]
time = 4.31, size = 370, normalized size = 3.27 \begin {gather*} \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 {gather*}

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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 [A]
time = 1.13, size = 164, normalized size = 1.45 \begin {gather*} \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^{4}}{{\left (a^{4} - a^{3} b\right )} \sqrt {a b}} - \frac {15 \, {\left (f x + e\right )}}{a - b} - \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 {gather*}

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*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b)))*b^4/((a^4 - a^3*b)*sqrt(a*b)
) - 15*(f*x + e)/(a - 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

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Mupad [B]
time = 13.74, size = 524, normalized size = 4.64 \begin {gather*} \frac {\mathrm {atan}\left (\frac {a^{14}\,b\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^7\,b^7}\,1{}\mathrm {i}-a^7\,b^8\,\mathrm {tan}\left (e+f\,x\right )\,\sqrt {-a^7\,b^7}\,1{}\mathrm {i}}{a^{11}\,b^{11}-a^{18}\,b^4}\right )\,\sqrt {-a^7\,b^7}\,15{}\mathrm {i}-15\,a^7\,\mathrm {atan}\left (\frac {\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (2\,a^{15}\,b^3+2\,a^9\,b^9\right )+\frac {\left (4\,a^{13}\,b^6-4\,a^{12}\,b^7+4\,a^{16}\,b^3-4\,a^{17}\,b^2+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^{18}\,b^2-8\,a^{17}\,b^3-8\,a^{16}\,b^4+8\,a^{15}\,b^5\right )\,1{}\mathrm {i}}{2\,a-2\,b}\right )\,1{}\mathrm {i}}{2\,a-2\,b}}{2\,a-2\,b}+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (2\,a^{15}\,b^3+2\,a^9\,b^9\right )+\frac {\left (4\,a^{12}\,b^7-4\,a^{13}\,b^6-4\,a^{16}\,b^3+4\,a^{17}\,b^2+\frac {\mathrm {tan}\left (e+f\,x\right )\,\left (8\,a^{18}\,b^2-8\,a^{17}\,b^3-8\,a^{16}\,b^4+8\,a^{15}\,b^5\right )\,1{}\mathrm {i}}{2\,a-2\,b}\right )\,1{}\mathrm {i}}{2\,a-2\,b}}{2\,a-2\,b}}{2\,a^{15}\,b^2+2\,a^{14}\,b^3+2\,a^{13}\,b^4+2\,a^{12}\,b^5+2\,a^{11}\,b^6+2\,a^{10}\,b^7+2\,a^9\,b^8}\right )}{f\,\left (15\,a^7\,b-15\,a^8\right )}+\frac {3\,a^6\,b+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (5\,a^7-5\,a^5\,b^2\right )-{\mathrm {tan}\left (e+f\,x\right )}^4\,\left (15\,a^7-15\,a^4\,b^3\right )-3\,a^7}{f\,\left (15\,a^8\,{\mathrm {tan}\left (e+f\,x\right )}^5-15\,a^7\,b\,{\mathrm {tan}\left (e+f\,x\right )}^5\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(atan((a^14*b*tan(e + f*x)*(-a^7*b^7)^(1/2)*1i - a^7*b^8*tan(e + f*x)*(-a^7*b^7)^(1/2)*1i)/(a^11*b^11 - a^18*b
^4))*(-a^7*b^7)^(1/2)*15i - 15*a^7*atan(((((4*a^13*b^6 - 4*a^12*b^7 + 4*a^16*b^3 - 4*a^17*b^2 + (tan(e + f*x)*
(8*a^15*b^5 - 8*a^16*b^4 - 8*a^17*b^3 + 8*a^18*b^2)*1i)/(2*a - 2*b))*1i)/(2*a - 2*b) + tan(e + f*x)*(2*a^9*b^9
+ 2*a^15*b^3))/(2*a - 2*b) + (((4*a^12*b^7 - 4*a^13*b^6 - 4*a^16*b^3 + 4*a^17*b^2 + (tan(e + f*x)*(8*a^15*b^5
- 8*a^16*b^4 - 8*a^17*b^3 + 8*a^18*b^2)*1i)/(2*a - 2*b))*1i)/(2*a - 2*b) + tan(e + f*x)*(2*a^9*b^9 + 2*a^15*b
^3))/(2*a - 2*b))/(2*a^9*b^8 + 2*a^10*b^7 + 2*a^11*b^6 + 2*a^12*b^5 + 2*a^13*b^4 + 2*a^14*b^3 + 2*a^15*b^2)))/
(f*(15*a^7*b - 15*a^8)) + (3*a^6*b + tan(e + f*x)^2*(5*a^7 - 5*a^5*b^2) - tan(e + f*x)^4*(15*a^7 - 15*a^4*b^3)
- 3*a^7)/(f*(15*a^8*tan(e + f*x)^5 - 15*a^7*b*tan(e + f*x)^5))

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