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

Optimal. Leaf size=51 \[ -\frac{(a+b) \cot ^5(e+f x)}{5 f}+\frac{a \cot ^3(e+f x)}{3 f}-\frac{a \cot (e+f x)}{f}-a x \]

[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)

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Rubi [A]  time = 0.0628051, antiderivative size = 51, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {4141, 1802, 203} \[ -\frac{(a+b) \cot ^5(e+f x)}{5 f}+\frac{a \cot ^3(e+f x)}{3 f}-\frac{a \cot (e+f x)}{f}-a x \]

Antiderivative was successfully verified.

[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 4141

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])

Rule 1802

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 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])

Rubi steps

\begin{align*} \int \cot ^6(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx &=\frac{\operatorname{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{\operatorname{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 \operatorname{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]  time = 0.029903, size = 51, normalized size = 1. \[ -\frac{a \cot ^5(e+f x) \text{Hypergeometric2F1}\left (-\frac{5}{2},1,-\frac{3}{2},-\tan ^2(e+f x)\right )}{5 f}-\frac{b \cot ^5(e+f x)}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.051, size = 63, normalized size = 1.2 \begin{align*}{\frac{1}{f} \left ( a \left ( -{\frac{ \left ( \cot \left ( fx+e \right ) \right ) ^{5}}{5}}+{\frac{ \left ( \cot \left ( fx+e \right ) \right ) ^{3}}{3}}-\cot \left ( fx+e \right ) -fx-e \right ) -{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{5}}{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{5}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[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)

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Maxima [A]  time = 1.56723, size = 70, normalized size = 1.37 \begin{align*} -\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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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

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Fricas [B]  time = 0.491892, size = 286, normalized size = 5.61 \begin{align*} -\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 )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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))

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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*sec(f*x+e)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.46628, size = 246, normalized size = 4.82 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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