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

Optimal. Leaf size=65 \[ \frac{\left (a^2-b^2\right ) \cot ^3(e+f x)}{3 f}-\frac{a^2 \cot (e+f x)}{f}-a^2 x-\frac{(a+b)^2 \cot ^5(e+f x)}{5 f} \]

[Out]

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

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Rubi [A]  time = 0.0933106, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.13, Rules used = {4141, 1802, 203} \[ \frac{\left (a^2-b^2\right ) \cot ^3(e+f x)}{3 f}-\frac{a^2 \cot (e+f x)}{f}-a^2 x-\frac{(a+b)^2 \cot ^5(e+f x)}{5 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Mathematica [B]  time = 1.06115, size = 256, normalized size = 3.94 \[ \frac{\csc (e) \csc ^5(e+f x) \left (180 a^2 \sin (2 e+f x)-140 a^2 \sin (2 e+3 f x)-90 a^2 \sin (4 e+3 f x)+46 a^2 \sin (4 e+5 f x)+150 a^2 f x \cos (2 e+f x)+75 a^2 f x \cos (2 e+3 f x)-75 a^2 f x \cos (4 e+3 f x)-15 a^2 f x \cos (4 e+5 f x)+15 a^2 f x \cos (6 e+5 f x)+280 a^2 \sin (f x)-150 a^2 f x \cos (f x)-60 a b \sin (4 e+3 f x)+12 a b \sin (4 e+5 f x)+120 a b \sin (f x)-60 b^2 \sin (2 e+f x)+20 b^2 \sin (2 e+3 f x)-4 b^2 \sin (4 e+5 f x)+20 b^2 \sin (f x)\right )}{480 f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Csc[e]*Csc[e + f*x]^5*(-150*a^2*f*x*Cos[f*x] + 150*a^2*f*x*Cos[2*e + f*x] + 75*a^2*f*x*Cos[2*e + 3*f*x] - 75*
a^2*f*x*Cos[4*e + 3*f*x] - 15*a^2*f*x*Cos[4*e + 5*f*x] + 15*a^2*f*x*Cos[6*e + 5*f*x] + 280*a^2*Sin[f*x] + 120*
a*b*Sin[f*x] + 20*b^2*Sin[f*x] + 180*a^2*Sin[2*e + f*x] - 60*b^2*Sin[2*e + f*x] - 140*a^2*Sin[2*e + 3*f*x] + 2
0*b^2*Sin[2*e + 3*f*x] - 90*a^2*Sin[4*e + 3*f*x] - 60*a*b*Sin[4*e + 3*f*x] + 46*a^2*Sin[4*e + 5*f*x] + 12*a*b*
Sin[4*e + 5*f*x] - 4*b^2*Sin[4*e + 5*f*x]))/(480*f)

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Maple [A]  time = 0.059, size = 107, normalized size = 1.7 \begin{align*}{\frac{1}{f} \left ({a}^{2} \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{2\,ab \left ( \cos \left ( fx+e \right ) \right ) ^{5}}{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{5}}}+{b}^{2} \left ( -{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{5\, \left ( \sin \left ( fx+e \right ) \right ) ^{5}}}-{\frac{2\, \left ( \cos \left ( fx+e \right ) \right ) ^{3}}{15\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}} \right ) \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)^2,x)

[Out]

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

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Maxima [A]  time = 1.47225, size = 97, normalized size = 1.49 \begin{align*} -\frac{15 \,{\left (f x + e\right )} a^{2} + \frac{15 \, a^{2} \tan \left (f x + e\right )^{4} - 5 \,{\left (a^{2} - b^{2}\right )} \tan \left (f x + e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{\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)^2,x, algorithm="maxima")

[Out]

-1/15*(15*(f*x + e)*a^2 + (15*a^2*tan(f*x + e)^4 - 5*(a^2 - b^2)*tan(f*x + e)^2 + 3*a^2 + 6*a*b + 3*b^2)/tan(f
*x + e)^5)/f

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Fricas [B]  time = 0.499277, size = 328, normalized size = 5.05 \begin{align*} -\frac{{\left (23 \, a^{2} + 6 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )^{5} - 5 \,{\left (7 \, a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{3} + 15 \, a^{2} \cos \left (f x + e\right ) + 15 \,{\left (a^{2} f x \cos \left (f x + e\right )^{4} - 2 \, a^{2} f x \cos \left (f x + e\right )^{2} + a^{2} 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)^2,x, algorithm="fricas")

[Out]

-1/15*((23*a^2 + 6*a*b - 2*b^2)*cos(f*x + e)^5 - 5*(7*a^2 - b^2)*cos(f*x + e)^3 + 15*a^2*cos(f*x + e) + 15*(a^
2*f*x*cos(f*x + e)^4 - 2*a^2*f*x*cos(f*x + e)^2 + a^2*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)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.50462, size = 392, normalized size = 6.03 \begin{align*} \frac{3 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 6 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} + 3 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{5} - 35 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 30 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 5 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} - 480 \,{\left (f x + e\right )} a^{2} + 330 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 60 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 30 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - \frac{330 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} + 60 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 30 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{4} - 35 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 30 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 5 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 3 \, a^{2} + 6 \, a b + 3 \, b^{2}}{\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)^2,x, algorithm="giac")

[Out]

1/480*(3*a^2*tan(1/2*f*x + 1/2*e)^5 + 6*a*b*tan(1/2*f*x + 1/2*e)^5 + 3*b^2*tan(1/2*f*x + 1/2*e)^5 - 35*a^2*tan
(1/2*f*x + 1/2*e)^3 - 30*a*b*tan(1/2*f*x + 1/2*e)^3 + 5*b^2*tan(1/2*f*x + 1/2*e)^3 - 480*(f*x + e)*a^2 + 330*a
^2*tan(1/2*f*x + 1/2*e) + 60*a*b*tan(1/2*f*x + 1/2*e) - 30*b^2*tan(1/2*f*x + 1/2*e) - (330*a^2*tan(1/2*f*x + 1
/2*e)^4 + 60*a*b*tan(1/2*f*x + 1/2*e)^4 - 30*b^2*tan(1/2*f*x + 1/2*e)^4 - 35*a^2*tan(1/2*f*x + 1/2*e)^2 - 30*a
*b*tan(1/2*f*x + 1/2*e)^2 + 5*b^2*tan(1/2*f*x + 1/2*e)^2 + 3*a^2 + 6*a*b + 3*b^2)/tan(1/2*f*x + 1/2*e)^5)/f