\(\int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) [341]

Optimal result

Integrand size = 23, antiderivative size = 74 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\csc ^2(e+f x)}{2 (a+b) f}-\frac {b^2 \log \left (b+a \cos ^2(e+f x)\right )}{2 a (a+b)^2 f}-\frac {(a+2 b) \log (\sin (e+f x))}{(a+b)^2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {4223, 457, 90} \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {b^2 \log \left (a \cos ^2(e+f x)+b\right )}{2 a f (a+b)^2}-\frac {\csc ^2(e+f x)}{2 f (a+b)}-\frac {(a+2 b) \log (\sin (e+f x))}{f (a+b)^2} \]

[In]

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

[Out]

-1/2*Csc[e + f*x]^2/((a + b)*f) - (b^2*Log[b + a*Cos[e + f*x]^2])/(2*a*(a + b)^2*f) - ((a + 2*b)*Log[Sin[e + f
*x]])/((a + b)^2*f)

Rule 90

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

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

Mathematica [A] (verified)

Time = 0.24 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (a (a+b) \csc ^2(e+f x)+2 a (a+2 b) \log (\sin (e+f x))+b^2 \log \left (a+b-a \sin ^2(e+f x)\right )\right ) \sec ^2(e+f x)}{4 a (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )} \]

[In]

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

[Out]

-1/4*((a + 2*b + a*Cos[2*(e + f*x)])*(a*(a + b)*Csc[e + f*x]^2 + 2*a*(a + 2*b)*Log[Sin[e + f*x]] + b^2*Log[a +
b - a*Sin[e + f*x]^2])*Sec[e + f*x]^2)/(a*(a + b)^2*f*(a + b*Sec[e + f*x]^2))

Maple [A] (verified)

Time = 1.27 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.61

method	result	size
derivativedivides	\(\frac {-\frac {1}{\left (4 a +4 b \right ) \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{2}}-\frac {b^{2} \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 \left (a +b \right )^{2} a}+\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{2}}}{f}\)	\(119\)
default	\(\frac {-\frac {1}{\left (4 a +4 b \right ) \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{2}}-\frac {b^{2} \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 \left (a +b \right )^{2} a}+\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{2}}}{f}\)	\(119\)
risch	\(-\frac {i x}{a}+\frac {2 i a x}{a^{2}+2 a b +b^{2}}+\frac {2 i a e}{f \left (a^{2}+2 a b +b^{2}\right )}+\frac {4 i b x}{a^{2}+2 a b +b^{2}}+\frac {4 i b e}{f \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i b^{2} x}{a \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i b^{2} e}{a f \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 \,{\mathrm e}^{2 i \left (f x +e \right )}}{f \left (a +b \right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a}{f \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b}{f \left (a^{2}+2 a b +b^{2}\right )}-\frac {b^{2} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a f \left (a^{2}+2 a b +b^{2}\right )}\)	\(285\)

[In]

int(cot(f*x+e)^3/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.70 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {a^{2} + a b - {\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \, {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{2 \, {\left ({\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f\right )}} \]

[In]

integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\cot ^{3}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \]

[In]

integrate(cot(f*x+e)**3/(a+b*sec(f*x+e)**2),x)

[Out]

Integral(cot(e + f*x)**3/(a + b*sec(e + f*x)**2), x)

Maxima [A] (verification not implemented)

none

Time = 0.18 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {b^{2} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{3} + 2 \, a^{2} b + a b^{2}} + \frac {{\left (a + 2 \, b\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {1}{{\left (a + b\right )} \sin \left (f x + e\right )^{2}}}{2 \, f} \]

[In]

integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="maxima")

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 294 vs. \(2 (70) = 140\).

Time = 0.34 (sec) , antiderivative size = 294, normalized size of antiderivative = 3.97 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {4 \, b^{2} \log \left (a + b + \frac {2 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} - \frac {2 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {a {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {b {\left (\cos \left (f x + e\right ) - 1\right )}^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}{a^{3} + 2 \, a^{2} b + a b^{2}} + \frac {4 \, {\left (a + 2 \, b\right )} \log \left (\frac {{\left | -\cos \left (f x + e\right ) + 1 \right |}}{{\left | \cos \left (f x + e\right ) + 1 \right |}}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac {{\left (a + b + \frac {4 \, a {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1} + \frac {8 \, b {\left (\cos \left (f x + e\right ) - 1\right )}}{\cos \left (f x + e\right ) + 1}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} {\left (\cos \left (f x + e\right ) - 1\right )}} - \frac {8 \, \log \left ({\left | -\frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 1 \right |}\right )}{a} - \frac {\cos \left (f x + e\right ) - 1}{{\left (a + b\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}}{8 \, f} \]

[In]

integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="giac")

[Out]

-1/8*(4*b^2*log(a + b + 2*a*(cos(f*x + e) - 1)/(cos(f*x + e) + 1) - 2*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)/(a^3 + 2*a^2*b +
a*b^2) + 4*(a + 2*b)*log(abs(-cos(f*x + e) + 1)/abs(cos(f*x + e) + 1))/(a^2 + 2*a*b + b^2) - (a + b + 4*a*(cos
(f*x + e) - 1)/(cos(f*x + e) + 1) + 8*b*(cos(f*x + e) - 1)/(cos(f*x + e) + 1))*(cos(f*x + e) + 1)/((a^2 + 2*a*
b + b^2)*(cos(f*x + e) - 1)) - 8*log(abs(-(cos(f*x + e) - 1)/(cos(f*x + e) + 1) + 1))/a - (cos(f*x + e) - 1)/(
(a + b)*(cos(f*x + e) + 1)))/f

Mupad [B] (verification not implemented)

Time = 20.76 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.32 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a\,f}-\frac {{\mathrm {cot}\left (e+f\,x\right )}^2}{2\,f\,\left (a+b\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a+2\,b\right )}{f\,\left (a^2+2\,a\,b+b^2\right )}-\frac {b^2\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,a\,f\,{\left (a+b\right )}^2} \]

[In]

int(cot(e + f*x)^3/(a + b/cos(e + f*x)^2),x)

[Out]

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