3.335 \(\int \cot ^4(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=45 \[ \frac{\left (a^2-b^2\right ) \cot (e+f x)}{f}+a^2 x-\frac{(a+b)^2 \cot ^3(e+f x)}{3 f} \]

[Out]

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

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Rubi [A]  time = 0.0868653, antiderivative size = 45, 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} \[ \frac{\left (a^2-b^2\right ) \cot (e+f x)}{f}+a^2 x-\frac{(a+b)^2 \cot ^3(e+f x)}{3 f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

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

Mathematica [B]  time = 0.830663, size = 160, normalized size = 3.56 \[ \frac{\csc (e) \csc ^3(e+f x) \left (-12 a^2 \sin (2 e+f x)+8 a^2 \sin (2 e+3 f x)-9 a^2 f x \cos (2 e+f x)-3 a^2 f x \cos (2 e+3 f x)+3 a^2 f x \cos (4 e+3 f x)-12 a^2 \sin (f x)+9 a^2 f x \cos (f x)-12 a b \sin (2 e+f x)+4 a b \sin (2 e+3 f x)-4 b^2 \sin (2 e+3 f x)+12 b^2 \sin (f x)\right )}{24 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^4*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

(Csc[e]*Csc[e + f*x]^3*(9*a^2*f*x*Cos[f*x] - 9*a^2*f*x*Cos[2*e + f*x] - 3*a^2*f*x*Cos[2*e + 3*f*x] + 3*a^2*f*x
*Cos[4*e + 3*f*x] - 12*a^2*Sin[f*x] + 12*b^2*Sin[f*x] - 12*a^2*Sin[2*e + f*x] - 12*a*b*Sin[2*e + f*x] + 8*a^2*
Sin[2*e + 3*f*x] + 4*a*b*Sin[2*e + 3*f*x] - 4*b^2*Sin[2*e + 3*f*x]))/(24*f)

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Maple [A]  time = 0.053, size = 73, normalized size = 1.6 \begin{align*}{\frac{1}{f} \left ({a}^{2} \left ( -{\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 ) ^{3}}{3\, \left ( \sin \left ( fx+e \right ) \right ) ^{3}}}+{b}^{2} \left ( -{\frac{2}{3}}-{\frac{ \left ( \csc \left ( fx+e \right ) \right ) ^{2}}{3}} \right ) \cot \left ( fx+e \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^2,x)

[Out]

1/f*(a^2*(-1/3*cot(f*x+e)^3+cot(f*x+e)+f*x+e)-2/3*a*b/sin(f*x+e)^3*cos(f*x+e)^3+b^2*(-2/3-1/3*csc(f*x+e)^2)*co
t(f*x+e))

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Maxima [A]  time = 1.51048, size = 80, normalized size = 1.78 \begin{align*} \frac{3 \,{\left (f x + e\right )} a^{2} + \frac{3 \,{\left (a^{2} - b^{2}\right )} \tan \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}}{\tan \left (f x + e\right )^{3}}}{3 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 0.492148, size = 220, normalized size = 4.89 \begin{align*} \frac{2 \,{\left (2 \, a^{2} + a b - b^{2}\right )} \cos \left (f x + e\right )^{3} - 3 \,{\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right ) + 3 \,{\left (a^{2} f x \cos \left (f x + e\right )^{2} - a^{2} f x\right )} \sin \left (f x + e\right )}{3 \,{\left (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)^4*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

1/3*(2*(2*a^2 + a*b - b^2)*cos(f*x + e)^3 - 3*(a^2 - b^2)*cos(f*x + e) + 3*(a^2*f*x*cos(f*x + e)^2 - a^2*f*x)*
sin(f*x + e))/((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)**4*(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.35105, size = 252, normalized size = 5.6 \begin{align*} \frac{a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 2 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3} + 24 \,{\left (f x + e\right )} a^{2} - 15 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) - 6 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + 9 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right ) + \frac{15 \, a^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} + 6 \, a b \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - 9 \, b^{2} \tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{2} - a^{2} - 2 \, a b - b^{2}}{\tan \left (\frac{1}{2} \, f x + \frac{1}{2} \, e\right )^{3}}}{24 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^4*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

1/24*(a^2*tan(1/2*f*x + 1/2*e)^3 + 2*a*b*tan(1/2*f*x + 1/2*e)^3 + b^2*tan(1/2*f*x + 1/2*e)^3 + 24*(f*x + e)*a^
2 - 15*a^2*tan(1/2*f*x + 1/2*e) - 6*a*b*tan(1/2*f*x + 1/2*e) + 9*b^2*tan(1/2*f*x + 1/2*e) + (15*a^2*tan(1/2*f*
x + 1/2*e)^2 + 6*a*b*tan(1/2*f*x + 1/2*e)^2 - 9*b^2*tan(1/2*f*x + 1/2*e)^2 - a^2 - 2*a*b - b^2)/tan(1/2*f*x +
1/2*e)^3)/f