\(\int \cot (e+f x) (a+b \sec ^2(e+f x))^2 \, dx\) [327]

Optimal result

Integrand size = 21, antiderivative size = 53 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {b (2 a+b) \log (\cos (e+f x))}{f}+\frac {(a+b)^2 \log (\sin (e+f x))}{f}+\frac {b^2 \sec ^2(e+f x)}{2 f} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 53, 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, 90} \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {(a+b)^2 \log (\sin (e+f x))}{f}-\frac {b (2 a+b) \log (\cos (e+f x))}{f}+\frac {b^2 \sec ^2(e+f x)}{2 f} \]

[In]

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

[Out]

-((b*(2*a + b)*Log[Cos[e + f*x]])/f) + ((a + b)^2*Log[Sin[e + f*x]])/f + (b^2*Sec[e + f*x]^2)/(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 {\left (b+a x^2\right )^2}{x^3 \left (1-x^2\right )} \, dx,x,\cos (e+f x)\right )}{f} \\ & = -\frac {\text {Subst}\left (\int \frac {(b+a x)^2}{(1-x) x^2} \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {\text {Subst}\left (\int \left (-\frac {(a+b)^2}{-1+x}+\frac {b^2}{x^2}+\frac {b (2 a+b)}{x}\right ) \, dx,x,\cos ^2(e+f x)\right )}{2 f} \\ & = -\frac {b (2 a+b) \log (\cos (e+f x))}{f}+\frac {(a+b)^2 \log (\sin (e+f x))}{f}+\frac {b^2 \sec ^2(e+f x)}{2 f} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.25 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.58 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {2 \left (b^2+2 \cos ^2(e+f x) \left (-b (2 a+b) \log (\cos (e+f x))+(a+b)^2 \log (\sin (e+f x))\right )\right ) (a \cos (e+f x)+b \sec (e+f x))^2}{f (a+2 b+a \cos (2 (e+f x)))^2} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 1.85 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.94

method	result	size
derivativedivides	\(\frac {a^{2} \ln \left (\sin \left (f x +e \right )\right )+2 a b \ln \left (\tan \left (f x +e \right )\right )+b^{2} \left (\frac {1}{2 \cos \left (f x +e \right )^{2}}+\ln \left (\tan \left (f x +e \right )\right )\right )}{f}\)	\(50\)
default	\(\frac {a^{2} \ln \left (\sin \left (f x +e \right )\right )+2 a b \ln \left (\tan \left (f x +e \right )\right )+b^{2} \left (\frac {1}{2 \cos \left (f x +e \right )^{2}}+\ln \left (\tan \left (f x +e \right )\right )\right )}{f}\)	\(50\)
risch	\(-i a^{2} x -\frac {2 i a^{2} e}{f}+\frac {2 b^{2} {\mathrm e}^{2 i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )^{2}}-\frac {2 b \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right ) a}{f}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a^{2}}{f}+\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a b}{f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b^{2}}{f}\)	\(145\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.49 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {{\left (2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (\cos \left (f x + e\right )^{2}\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{2} \log \left (-\frac {1}{4} \, \cos \left (f x + e\right )^{2} + \frac {1}{4}\right ) - b^{2}}{2 \, f \cos \left (f x + e\right )^{2}} \]

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.19 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.21 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {{\left (2 \, a b + b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) - {\left (a^{2} + 2 \, a b + b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac {b^{2}}{\sin \left (f x + e\right )^{2} - 1}}{2 \, f} \]

[In]

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

[Out]

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 247 vs. \(2 (51) = 102\).

Time = 0.32 (sec) , antiderivative size = 247, normalized size of antiderivative = 4.66 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {a^{2} \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2 \right |}\right ) + {\left (2 \, a b + b^{2}\right )} \log \left ({\left | -\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} - \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} - 2 \right |}\right ) - \frac {2 \, a b {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + b^{2} {\left (\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1}\right )} + 4 \, a b - 2 \, b^{2}}{\frac {\cos \left (f x + e\right ) + 1}{\cos \left (f x + e\right ) - 1} + \frac {\cos \left (f x + e\right ) - 1}{\cos \left (f x + e\right ) + 1} + 2}}{2 \, f} \]

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 20.39 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.09 \[ \int \cot (e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2+2\,a\,b+b^2\right )}{f}-\frac {a^2\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f}+\frac {b^2\,{\mathrm {tan}\left (e+f\,x\right )}^2}{2\,f} \]

[In]

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

[Out]

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