\(\int \cot ^5(e+f x) (a+b \sec ^2(e+f x)) \, dx\) [316]

Optimal result

Integrand size = 21, antiderivative size = 51 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {(2 a+b) \csc ^2(e+f x)}{2 f}-\frac {(a+b) \csc ^4(e+f x)}{4 f}+\frac {a \log (\sin (e+f x))}{f} \]

[Out]

1/2*(2*a+b)*csc(f*x+e)^2/f-1/4*(a+b)*csc(f*x+e)^4/f+a*ln(sin(f*x+e))/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 = {4223, 457, 78} \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {(a+b) \csc ^4(e+f x)}{4 f}+\frac {(2 a+b) \csc ^2(e+f x)}{2 f}+\frac {a \log (\sin (e+f x))}{f} \]

[In]

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

[Out]

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 457

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(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] && IntegerQ[Simplify[(m + 1)/n]]

Rule 4223

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]

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

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.43 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {a \cot ^2(e+f x)}{2 f}-\frac {a \cot ^4(e+f x)}{4 f}-\frac {b \cot ^4(e+f x)}{4 f}+\frac {a \log (\cos (e+f x))}{f}+\frac {a \log (\tan (e+f x))}{f} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.25 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.08

method	result	size
derivativedivides	\(\frac {a \left (-\frac {\cot \left (f x +e \right )^{4}}{4}+\frac {\cot \left (f x +e \right )^{2}}{2}+\ln \left (\sin \left (f x +e \right )\right )\right )-\frac {b \cos \left (f x +e \right )^{4}}{4 \sin \left (f x +e \right )^{4}}}{f}\)	\(55\)
default	\(\frac {a \left (-\frac {\cot \left (f x +e \right )^{4}}{4}+\frac {\cot \left (f x +e \right )^{2}}{2}+\ln \left (\sin \left (f x +e \right )\right )\right )-\frac {b \cos \left (f x +e \right )^{4}}{4 \sin \left (f x +e \right )^{4}}}{f}\)	\(55\)
risch	\(-i a x -\frac {2 i a e}{f}-\frac {2 \left (2 a \,{\mathrm e}^{6 i \left (f x +e \right )}+b \,{\mathrm e}^{6 i \left (f x +e \right )}-2 a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+b \,{\mathrm e}^{2 i \left (f x +e \right )}\right )}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{4}}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a}{f}\)	\(109\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.63 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=-\frac {2 \, {\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left (a \cos \left (f x + e\right )^{4} - 2 \, a \cos \left (f x + e\right )^{2} + a\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right ) - 3 \, a - b}{4 \, {\left (f \cos \left (f x + e\right )^{4} - 2 \, f \cos \left (f x + e\right )^{2} + f\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right ) \cot ^{5}{\left (e + f x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.96 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {2 \, a \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac {2 \, {\left (2 \, a + b\right )} \sin \left (f x + e\right )^{2} - a - b}{\sin \left (f x + e\right )^{4}}}{4 \, f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 238 vs. \(2 (47) = 94\).

Time = 0.35 (sec) , antiderivative size = 238, normalized size of antiderivative = 4.67 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {32 \, a \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right ) - 64 \, a \log \left ({\left | -\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1 \right |}\right ) - \frac {{\left (a + b + \frac {12 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {4 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {48 \, a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) - 1\right )}^{2}} - \frac {12 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {4 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}}{64 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 19.88 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.20 \[ \int \cot ^5(e+f x) \left (a+b \sec ^2(e+f x)\right ) \, dx=\frac {a\,\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )}{f}-\frac {a\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f}-\frac {-\frac {a\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2}+\frac {a}{4}+\frac {b}{4}}{f\,{\mathrm {tan}\left (e+f\,x\right )}^4} \]

[In]

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

[Out]

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