\(\int \cot ^6(e+f x) (a+b \sec ^2(e+f x)) \, dx\) [323]

Optimal result

Integrand size = 21, antiderivative size = 51 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-a x-\frac {a \cot (e+f x)}{f}+\frac {a \cot ^3(e+f x)}{3 f}-\frac {(a+b) \cot ^5(e+f x)}{5 f} \]

[Out]

-a*x-a*cot(f*x+e)/f+1/3*a*cot(f*x+e)^3/f-1/5*(a+b)*cot(f*x+e)^5/f

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {4226, 1816, 209} \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {(a+b) \cot ^5(e+f x)}{5 f}+\frac {a \cot ^3(e+f x)}{3 f}-\frac {a \cot (e+f x)}{f}-a x \]

[In]

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

[Out]

-(a*x) - (a*Cot[e + f*x])/f + (a*Cot[e + f*x]^3)/(3*f) - ((a + b)*Cot[e + f*x]^5)/(5*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 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 4226

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {a+b \left (1+x^2\right )}{x^6 \left (1+x^2\right )} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\text {Subst}\left (\int \left (\frac {a+b}{x^6}-\frac {a}{x^4}+\frac {a}{x^2}-\frac {a}{1+x^2}\right ) \, dx,x,\tan (e+f x)\right )}{f} \\ & = -\frac {a \cot (e+f x)}{f}+\frac {a \cot ^3(e+f x)}{3 f}-\frac {(a+b) \cot ^5(e+f x)}{5 f}-\frac {a \text {Subst}\left (\int \frac {1}{1+x^2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = -a x-\frac {a \cot (e+f x)}{f}+\frac {a \cot ^3(e+f x)}{3 f}-\frac {(a+b) \cot ^5(e+f x)}{5 f} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.03 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.00 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {b \cot ^5(e+f x)}{5 f}-\frac {a \cot ^5(e+f x) \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},1,-\frac {3}{2},-\tan ^2(e+f x)\right )}{5 f} \]

[In]

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

[Out]

-1/5*(b*Cot[e + f*x]^5)/f - (a*Cot[e + f*x]^5*Hypergeometric2F1[-5/2, 1, -3/2, -Tan[e + f*x]^2])/(5*f)

Maple [A] (verified)

Time = 1.89 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.24

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cot \left (f x +e \right )^{5}}{5}+\frac {\cot \left (f x +e \right )^{3}}{3}-\cot \left (f x +e \right )-f x -e \right )-\frac {b \cos \left (f x +e \right )^{5}}{5 \sin \left (f x +e \right )^{5}}}{f}\)	\(63\)
default	\(\frac {a \left (-\frac {\cot \left (f x +e \right )^{5}}{5}+\frac {\cot \left (f x +e \right )^{3}}{3}-\cot \left (f x +e \right )-f x -e \right )-\frac {b \cos \left (f x +e \right )^{5}}{5 \sin \left (f x +e \right )^{5}}}{f}\)	\(63\)
risch	\(-a x -\frac {2 i \left (45 a \,{\mathrm e}^{8 i \left (f x +e \right )}+15 b \,{\mathrm e}^{8 i \left (f x +e \right )}-90 a \,{\mathrm e}^{6 i \left (f x +e \right )}+140 a \,{\mathrm e}^{4 i \left (f x +e \right )}+30 b \,{\mathrm e}^{4 i \left (f x +e \right )}-70 a \,{\mathrm e}^{2 i \left (f x +e \right )}+23 a +3 b \right )}{15 f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{5}}\)	\(104\)

[In]

int(cot(f*x+e)^6*(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

1/f*(a*(-1/5*cot(f*x+e)^5+1/3*cot(f*x+e)^3-cot(f*x+e)-f*x-e)-1/5*b/sin(f*x+e)^5*cos(f*x+e)^5)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 110 vs. \(2 (47) = 94\).

Time = 0.24 (sec) , antiderivative size = 110, normalized size of antiderivative = 2.16 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {{\left (23 \, a + 3 \, b\right )} \cos \left (f x + e\right )^{5} - 35 \, a \cos \left (f x + e\right )^{3} + 15 \, a \cos \left (f x + e\right ) + 15 \, {\left (a f x \cos \left (f x + e\right )^{4} - 2 \, a f x \cos \left (f x + e\right )^{2} + a f x\right )} \sin \left (f x + e\right )}{15 \, {\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )} \sin \left (f x + e\right )} \]

[In]

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

[Out]

-1/15*((23*a + 3*b)*cos(f*x + e)^5 - 35*a*cos(f*x + e)^3 + 15*a*cos(f*x + e) + 15*(a*f*x*cos(f*x + e)^4 - 2*a*
f*x*cos(f*x + e)^2 + a*f*x)*sin(f*x + e))/((f*cos(f*x + e)^4 - 2*f*cos(f*x + e)^2 + f)*sin(f*x + e))

Sympy [F]

\[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \cot ^{6}{\left (e + f x \right )}\, dx \]

[In]

integrate(cot(f*x+e)**6*(a+b*sec(f*x+e)**2),x)

[Out]

Integral((a + b*sec(e + f*x)**2)*cot(e + f*x)**6, x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.02 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {15 \, {\left (f x + e\right )} a + \frac {15 \, a \tan \left (f x + e\right )^{4} - 5 \, a \tan \left (f x + e\right )^{2} + 3 \, a + 3 \, b}{\tan \left (f x + e\right )^{5}}}{15 \, f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 170 vs. \(2 (47) = 94\).

Time = 0.34 (sec) , antiderivative size = 170, normalized size of antiderivative = 3.33 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {3 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 3 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 35 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 15 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 480 \, {\left (f x + e\right )} a + 330 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 30 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - \frac {330 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} + 30 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 35 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 15 \, b \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 3 \, a + 3 \, b}{\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5}}}{480 \, f} \]

[In]

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

[Out]

1/480*(3*a*tan(1/2*f*x + 1/2*e)^5 + 3*b*tan(1/2*f*x + 1/2*e)^5 - 35*a*tan(1/2*f*x + 1/2*e)^3 - 15*b*tan(1/2*f*
x + 1/2*e)^3 - 480*(f*x + e)*a + 330*a*tan(1/2*f*x + 1/2*e) + 30*b*tan(1/2*f*x + 1/2*e) - (330*a*tan(1/2*f*x +
1/2*e)^4 + 30*b*tan(1/2*f*x + 1/2*e)^4 - 35*a*tan(1/2*f*x + 1/2*e)^2 - 15*b*tan(1/2*f*x + 1/2*e)^2 + 3*a + 3*
b)/tan(1/2*f*x + 1/2*e)^5)/f

Mupad [B] (verification not implemented)

Time = 19.80 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.90 \[ \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-a\,x-\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^4-\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2}{3}+\frac {a}{5}+\frac {b}{5}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^5} \]

[In]

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

[Out]

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