3.329 \(\int \cot ^5(e+f x) (a+b \sec ^2(e+f x))^2 \, dx\)

Optimal. Leaf size=51 \[ \frac{a^2 \log (\sin (e+f x))}{f}-\frac{(a+b)^2 \csc ^4(e+f x)}{4 f}+\frac{a (a+b) \csc ^2(e+f x)}{f} \]

[Out]

(a*(a + b)*Csc[e + f*x]^2)/f - ((a + b)^2*Csc[e + f*x]^4)/(4*f) + (a^2*Log[Sin[e + f*x]])/f

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Rubi [A]  time = 0.0865154, antiderivative size = 51, 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 = {4138, 444, 43} \[ \frac{a^2 \log (\sin (e+f x))}{f}-\frac{(a+b)^2 \csc ^4(e+f x)}{4 f}+\frac{a (a+b) \csc ^2(e+f x)}{f} \]

Antiderivative was successfully verified.

[In]

Int[Cot[e + f*x]^5*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

(a*(a + b)*Csc[e + f*x]^2)/f - ((a + b)^2*Csc[e + f*x]^4)/(4*f) + (a^2*Log[Sin[e + f*x]])/f

Rule 4138

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

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx &=-\frac{\operatorname{Subst}\left (\int \frac{x \left (b+a x^2\right )^2}{\left (1-x^2\right )^3} \, dx,x,\cos (e+f x)\right )}{f}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{(b+a x)^2}{(1-x)^3} \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-\frac{(a+b)^2}{(-1+x)^3}-\frac{2 a (a+b)}{(-1+x)^2}-\frac{a^2}{-1+x}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f}\\ &=\frac{a (a+b) \csc ^2(e+f x)}{f}-\frac{(a+b)^2 \csc ^4(e+f x)}{4 f}+\frac{a^2 \log (\sin (e+f x))}{f}\\ \end{align*}

Mathematica [A]  time = 0.260646, size = 77, normalized size = 1.51 \[ -\frac{\left (a \cos ^2(e+f x)+b\right )^2 \left (-4 a^2 \log (\sin (e+f x))+(a+b)^2 \csc ^4(e+f x)-4 a (a+b) \csc ^2(e+f x)\right )}{f (a \cos (2 (e+f x))+a+2 b)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[e + f*x]^5*(a + b*Sec[e + f*x]^2)^2,x]

[Out]

-(((b + a*Cos[e + f*x]^2)^2*(-4*a*(a + b)*Csc[e + f*x]^2 + (a + b)^2*Csc[e + f*x]^4 - 4*a^2*Log[Sin[e + f*x]])
)/(f*(a + 2*b + a*Cos[2*(e + f*x)])^2))

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Maple [A]  time = 0.06, size = 87, normalized size = 1.7 \begin{align*} -{\frac{{a}^{2} \left ( \cot \left ( fx+e \right ) \right ) ^{4}}{4\,f}}+{\frac{{a}^{2} \left ( \cot \left ( fx+e \right ) \right ) ^{2}}{2\,f}}+{\frac{{a}^{2}\ln \left ( \sin \left ( fx+e \right ) \right ) }{f}}-{\frac{ab \left ( \cos \left ( fx+e \right ) \right ) ^{4}}{2\,f \left ( \sin \left ( fx+e \right ) \right ) ^{4}}}-{\frac{{b}^{2}}{4\,f \left ( \sin \left ( fx+e \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x)

[Out]

-1/4/f*a^2*cot(f*x+e)^4+1/2/f*a^2*cot(f*x+e)^2+a^2*ln(sin(f*x+e))/f-1/2/f*a*b/sin(f*x+e)^4*cos(f*x+e)^4-1/4/f*
b^2/sin(f*x+e)^4

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Maxima [A]  time = 1.06555, size = 82, normalized size = 1.61 \begin{align*} \frac{2 \, a^{2} \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac{4 \,{\left (a^{2} + a b\right )} \sin \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}}{\sin \left (f x + e\right )^{4}}}{4 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")

[Out]

1/4*(2*a^2*log(sin(f*x + e)^2) + (4*(a^2 + a*b)*sin(f*x + e)^2 - a^2 - 2*a*b - b^2)/sin(f*x + e)^4)/f

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Fricas [A]  time = 0.520697, size = 242, normalized size = 4.75 \begin{align*} -\frac{4 \,{\left (a^{2} + a b\right )} \cos \left (f x + e\right )^{2} - 3 \, a^{2} - 2 \, a b + b^{2} - 4 \,{\left (a^{2} \cos \left (f x + e\right )^{4} - 2 \, a^{2} \cos \left (f x + e\right )^{2} + a^{2}\right )} \log \left (\frac{1}{2} \, \sin \left (f x + e\right )\right )}{4 \,{\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")

[Out]

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

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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)**5*(a+b*sec(f*x+e)**2)**2,x)

[Out]

Timed out

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Giac [B]  time = 1.39903, size = 474, normalized size = 9.29 \begin{align*} -\frac{64 \, a^{2} \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1\right ) - 32 \, a^{2} \log \left (-\frac{\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right ) + \frac{12 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{8 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{4 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{2 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac{{\left (a^{2} + 2 \, a b + b^{2} + \frac{12 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{8 \, a b{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac{4 \, b^{2}{\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac{48 \, a^{2}{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) - 1\right )}^{2}}}{64 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(f*x+e)^5*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")

[Out]

-1/64*(64*a^2*log(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1) - 32*a^2*log(-(cos(f*x + e) - 1)/(cos(f*x + e) +
1)) + 12*a^2*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 8*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b^2*(cos
(f*x + e) - 1)/(cos(f*x + e) + 1) + a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + 2*a*b*(cos(f*x + e) - 1)^2
/(cos(f*x + e) + 1)^2 + b^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2 + (a^2 + 2*a*b + b^2 + 12*a^2*(cos(f*x +
e) - 1)/(cos(f*x + e) + 1) + 8*a*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 4*b^2*(cos(f*x + e) - 1)/(cos(f*x
+ e) + 1) + 48*a^2*(cos(f*x + e) - 1)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1)^2/(cos(f*x + e) - 1)^2)/f