Integrand size = 23, antiderivative size = 57 \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {(a+b)^2 \csc ^2(e+f x)}{2 f}-\frac {b^2 \log (\cos (e+f x))}{f}-\frac {\left (a^2-b^2\right ) \log (\sin (e+f x))}{f} \] Output:
-1/2*(a+b)^2*csc(f*x+e)^2/f-b^2*ln(cos(f*x+e))/f-(a^2-b^2)*ln(sin(f*x+e))/ f
Time = 0.12 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.42 \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {2 \left (b+a \cos ^2(e+f x)\right )^2 \left ((a+b)^2 \csc ^2(e+f x)+2 b^2 \log (\cos (e+f x))+2 \left (a^2-b^2\right ) \log (\sin (e+f x))\right )}{f (a+2 b+a \cos (2 (e+f x)))^2} \] Input:
Integrate[Cot[e + f*x]^3*(a + b*Sec[e + f*x]^2)^2,x]
Output:
(-2*(b + a*Cos[e + f*x]^2)^2*((a + b)^2*Csc[e + f*x]^2 + 2*b^2*Log[Cos[e + f*x]] + 2*(a^2 - b^2)*Log[Sin[e + f*x]]))/(f*(a + 2*b + a*Cos[2*(e + f*x) ])^2)
Time = 0.29 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.12, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4626, 354, 99, 2009}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a+b \sec (e+f x)^2\right )^2}{\tan (e+f x)^3}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {\left (a \cos ^2(e+f x)+b\right )^2 \sec (e+f x)}{\left (1-\cos ^2(e+f x)\right )^2}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 354 |
\(\displaystyle -\frac {\int \frac {\left (a \cos ^2(e+f x)+b\right )^2 \sec (e+f x)}{\left (1-\cos ^2(e+f x)\right )^2}d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {\int \left (\sec (e+f x) b^2+\frac {a^2-b^2}{\cos ^2(e+f x)-1}+\frac {(a+b)^2}{\left (\cos ^2(e+f x)-1\right )^2}\right )d\cos ^2(e+f x)}{2 f}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\left (a^2-b^2\right ) \log \left (1-\cos ^2(e+f x)\right )+\frac {(a+b)^2}{1-\cos ^2(e+f x)}+b^2 \log \left (\cos ^2(e+f x)\right )}{2 f}\) |
Input:
Int[Cot[e + f*x]^3*(a + b*Sec[e + f*x]^2)^2,x]
Output:
-1/2*((a + b)^2/(1 - Cos[e + f*x]^2) + b^2*Log[Cos[e + f*x]^2] + (a^2 - b^ 2)*Log[1 - Cos[e + f*x]^2])/f
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(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]))
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2)^(q_.), x_S ymbol] :> Simp[1/2 Subst[Int[x^((m - 1)/2)*(a + b*x)^p*(c + d*x)^q, x], x , x^2], x] /; FreeQ[{a, b, c, d, p, q}, x] && NeQ[b*c - a*d, 0] && IntegerQ [(m - 1)/2]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_))^(p_.)*tan[(e_.) + (f_.)*(x_ )]^(m_.), x_Symbol] :> Module[{ff = FreeFactors[Cos[e + f*x], x]}, Simp[-(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]
Time = 1.76 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.12
method | result | size |
derivativedivides | \(\frac {a^{2} \left (-\frac {\cot \left (f x +e \right )^{2}}{2}-\ln \left (\sin \left (f x +e \right )\right )\right )-\frac {a b}{\sin \left (f x +e \right )^{2}}+b^{2} \left (-\frac {1}{2 \sin \left (f x +e \right )^{2}}+\ln \left (\tan \left (f x +e \right )\right )\right )}{f}\) | \(64\) |
default | \(\frac {a^{2} \left (-\frac {\cot \left (f x +e \right )^{2}}{2}-\ln \left (\sin \left (f x +e \right )\right )\right )-\frac {a b}{\sin \left (f x +e \right )^{2}}+b^{2} \left (-\frac {1}{2 \sin \left (f x +e \right )^{2}}+\ln \left (\tan \left (f x +e \right )\right )\right )}{f}\) | \(64\) |
risch | \(i a^{2} x +\frac {2 i a^{2} e}{f}+\frac {2 \left (a^{2}+2 a b +b^{2}\right ) {\mathrm e}^{2 i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {b^{2} \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+1\right )}{f}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a^{2}}{f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b^{2}}{f}\) | \(116\) |
Input:
int(cot(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x,method=_RETURNVERBOSE)
Output:
1/f*(a^2*(-1/2*cot(f*x+e)^2-ln(sin(f*x+e)))-a*b/sin(f*x+e)^2+b^2*(-1/2/sin (f*x+e)^2+ln(tan(f*x+e))))
Time = 0.10 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.75 \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {a^{2} + 2 \, a b + b^{2} - {\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} \log \left (\cos \left (f x + e\right )^{2}\right ) - {\left ({\left (a^{2} - b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} + b^{2}\right )} \log \left (-\frac {1}{4} \, \cos \left (f x + e\right )^{2} + \frac {1}{4}\right )}{2 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}} \] Input:
integrate(cot(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="fricas")
Output:
1/2*(a^2 + 2*a*b + b^2 - (b^2*cos(f*x + e)^2 - b^2)*log(cos(f*x + e)^2) - ((a^2 - b^2)*cos(f*x + e)^2 - a^2 + b^2)*log(-1/4*cos(f*x + e)^2 + 1/4))/( f*cos(f*x + e)^2 - f)
\[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\int \left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2} \cot ^{3}{\left (e + f x \right )}\, dx \] Input:
integrate(cot(f*x+e)**3*(a+b*sec(f*x+e)**2)**2,x)
Output:
Integral((a + b*sec(e + f*x)**2)**2*cot(e + f*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.05 \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {b^{2} \log \left (\sin \left (f x + e\right )^{2} - 1\right ) + {\left (a^{2} - b^{2}\right )} \log \left (\sin \left (f x + e\right )^{2}\right ) + \frac {a^{2} + 2 \, a b + b^{2}}{\sin \left (f x + e\right )^{2}}}{2 \, f} \] Input:
integrate(cot(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="maxima")
Output:
-1/2*(b^2*log(sin(f*x + e)^2 - 1) + (a^2 - b^2)*log(sin(f*x + e)^2) + (a^2 + 2*a*b + b^2)/sin(f*x + e)^2)/f
Time = 0.12 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.25 \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=-\frac {b^{2} \log \left ({\left | \cos \left (f x + e\right ) \right |}\right )}{f} - \frac {{\left (a^{2} - b^{2}\right )} \log \left ({\left | \cos \left (f x + e\right )^{2} - 1 \right |}\right )}{2 \, f} + \frac {a^{2} + 2 \, a b + b^{2}}{2 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} f} \] Input:
integrate(cot(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x, algorithm="giac")
Output:
-b^2*log(abs(cos(f*x + e)))/f - 1/2*(a^2 - b^2)*log(abs(cos(f*x + e)^2 - 1 ))/f + 1/2*(a^2 + 2*a*b + b^2)/((cos(f*x + e)^2 - 1)*f)
Time = 15.46 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.19 \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {a^2\,\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,f}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a^2-b^2\right )}{f}-\frac {{\mathrm {cot}\left (e+f\,x\right )}^2\,\left (\frac {a^2}{2}+a\,b+\frac {b^2}{2}\right )}{f} \] Input:
int(cot(e + f*x)^3*(a + b/cos(e + f*x)^2)^2,x)
Output:
(a^2*log(tan(e + f*x)^2 + 1))/(2*f) - (log(tan(e + f*x))*(a^2 - b^2))/f - (cot(e + f*x)^2*(a*b + a^2/2 + b^2/2))/f
Time = 0.15 (sec) , antiderivative size = 187, normalized size of antiderivative = 3.28 \[ \int \cot ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^2 \, dx=\frac {4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{2} a^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \sin \left (f x +e \right )^{2} b^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \sin \left (f x +e \right )^{2} b^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} a^{2}+4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} b^{2}+\sin \left (f x +e \right )^{2} a^{2}+2 \sin \left (f x +e \right )^{2} a b +\sin \left (f x +e \right )^{2} b^{2}-2 a^{2}-4 a b -2 b^{2}}{4 \sin \left (f x +e \right )^{2} f} \] Input:
int(cot(f*x+e)^3*(a+b*sec(f*x+e)^2)^2,x)
Output:
(4*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a**2 - 4*log(tan((e + f*x) /2) - 1)*sin(e + f*x)**2*b**2 - 4*log(tan((e + f*x)/2) + 1)*sin(e + f*x)** 2*b**2 - 4*log(tan((e + f*x)/2))*sin(e + f*x)**2*a**2 + 4*log(tan((e + f*x )/2))*sin(e + f*x)**2*b**2 + sin(e + f*x)**2*a**2 + 2*sin(e + f*x)**2*a*b + sin(e + f*x)**2*b**2 - 2*a**2 - 4*a*b - 2*b**2)/(4*sin(e + f*x)**2*f)