Integrand size = 23, antiderivative size = 74 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\csc ^2(e+f x)}{2 (a+b) f}-\frac {b^2 \log \left (b+a \cos ^2(e+f x)\right )}{2 a (a+b)^2 f}-\frac {(a+2 b) \log (\sin (e+f x))}{(a+b)^2 f} \] Output:
-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
Time = 0.16 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.35 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {(a+2 b+a \cos (2 (e+f x))) \left (a (a+b) \csc ^2(e+f x)+2 a (a+2 b) \log (\sin (e+f x))+b^2 \log \left (a+b-a \sin ^2(e+f x)\right )\right ) \sec ^2(e+f x)}{4 a (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )} \] Input:
Integrate[Cot[e + f*x]^3/(a + b*Sec[e + f*x]^2),x]
Output:
-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))
Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.04, 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.
| \(\displaystyle \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\tan (e+f x)^3 \left (a+b \sec (e+f x)^2\right )}dx\) |
| \(\Big \downarrow \) 4626 |
| \(\displaystyle -\frac {\int \frac {\cos ^5(e+f x)}{\left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^2(e+f x)+b\right )}d\cos (e+f x)}{f}\) |
| \(\Big \downarrow \) 354 |
| \(\displaystyle -\frac {\int \frac {\cos ^4(e+f x)}{\left (1-\cos ^2(e+f x)\right )^2 \left (a \cos ^2(e+f x)+b\right )}d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 99 |
| \(\displaystyle -\frac {\int \left (\frac {b^2}{(a+b)^2 \left (a \cos ^2(e+f x)+b\right )}+\frac {a+2 b}{(a+b)^2 \left (\cos ^2(e+f x)-1\right )}+\frac {1}{(a+b) \left (\cos ^2(e+f x)-1\right )^2}\right )d\cos ^2(e+f x)}{2 f}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {b^2 \log \left (a \cos ^2(e+f x)+b\right )}{a (a+b)^2}+\frac {1}{(a+b) \left (1-\cos ^2(e+f x)\right )}+\frac {(a+2 b) \log \left (1-\cos ^2(e+f x)\right )}{(a+b)^2}}{2 f}\) |
Input:
Int[Cot[e + f*x]^3/(a + b*Sec[e + f*x]^2),x]
Output:
-1/2*(1/((a + b)*(1 - Cos[e + f*x]^2)) + ((a + 2*b)*Log[1 - Cos[e + f*x]^2 ])/(a + b)^2 + (b^2*Log[b + a*Cos[e + f*x]^2])/(a*(a + b)^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 = 0.99 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.61
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{2}}-\frac {b^{2} \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 \left (a +b \right )^{2} a}}{f}\) | \(119\) |
| default | \(\frac {\frac {1}{\left (4 a +4 b \right ) \left (-1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (-1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{2}}-\frac {1}{\left (4 a +4 b \right ) \left (1+\cos \left (f x +e \right )\right )}+\frac {\left (-a -2 b \right ) \ln \left (1+\cos \left (f x +e \right )\right )}{2 \left (a +b \right )^{2}}-\frac {b^{2} \ln \left (b +a \cos \left (f x +e \right )^{2}\right )}{2 \left (a +b \right )^{2} a}}{f}\) | \(119\) |
| risch | \(-\frac {i x}{a}+\frac {2 i a x}{a^{2}+2 a b +b^{2}}+\frac {2 i a e}{f \left (a^{2}+2 a b +b^{2}\right )}+\frac {4 i b x}{a^{2}+2 a b +b^{2}}+\frac {4 i b e}{f \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i b^{2} x}{a \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 i b^{2} e}{a f \left (a^{2}+2 a b +b^{2}\right )}+\frac {2 \,{\mathrm e}^{2 i \left (f x +e \right )}}{f \left (a +b \right ) \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) a}{f \left (a^{2}+2 a b +b^{2}\right )}-\frac {2 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right ) b}{f \left (a^{2}+2 a b +b^{2}\right )}-\frac {b^{2} \ln \left ({\mathrm e}^{4 i \left (f x +e \right )}+\frac {2 \left (a +2 b \right ) {\mathrm e}^{2 i \left (f x +e \right )}}{a}+1\right )}{2 a f \left (a^{2}+2 a b +b^{2}\right )}\) | \(285\) |
Input:
int(cot(f*x+e)^3/(a+b*sec(f*x+e)^2),x,method=_RETURNVERBOSE)
Output:
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/ (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))
Time = 0.17 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.70 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {a^{2} + a b - {\left (b^{2} \cos \left (f x + e\right )^{2} - b^{2}\right )} \log \left (a \cos \left (f x + e\right )^{2} + b\right ) - 2 \, {\left ({\left (a^{2} + 2 \, a b\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b\right )} \log \left (\frac {1}{2} \, \sin \left (f x + e\right )\right )}{2 \, {\left ({\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f \cos \left (f x + e\right )^{2} - {\left (a^{3} + 2 \, a^{2} b + a b^{2}\right )} f\right )}} \] Input:
integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="fricas")
Output:
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)
\[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\int \frac {\cot ^{3}{\left (e + f x \right )}}{a + b \sec ^{2}{\left (e + f x \right )}}\, dx \] Input:
integrate(cot(f*x+e)**3/(a+b*sec(f*x+e)**2),x)
Output:
Integral(cot(e + f*x)**3/(a + b*sec(e + f*x)**2), x)
Time = 0.03 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.18 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {\frac {b^{2} \log \left (a \sin \left (f x + e\right )^{2} - a - b\right )}{a^{3} + 2 \, a^{2} b + a b^{2}} + \frac {{\left (a + 2 \, b\right )} \log \left (\sin \left (f x + e\right )^{2}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac {1}{{\left (a + b\right )} \sin \left (f x + e\right )^{2}}}{2 \, f} \] Input:
integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="maxima")
Output:
-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
Time = 0.19 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.38 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=-\frac {b^{2} \log \left ({\left | a \cos \left (f x + e\right )^{2} + b \right |}\right )}{2 \, {\left (a^{3} f + 2 \, a^{2} b f + a b^{2} f\right )}} - \frac {{\left (a + 2 \, b\right )} \log \left ({\left | -\cos \left (f x + e\right )^{2} + 1 \right |}\right )}{2 \, {\left (a^{2} f + 2 \, a b f + b^{2} f\right )}} + \frac {1}{2 \, {\left (\cos \left (f x + e\right )^{2} - 1\right )} {\left (a + b\right )} f} \] Input:
integrate(cot(f*x+e)^3/(a+b*sec(f*x+e)^2),x, algorithm="giac")
Output:
-1/2*b^2*log(abs(a*cos(f*x + e)^2 + b))/(a^3*f + 2*a^2*b*f + a*b^2*f) - 1/ 2*(a + 2*b)*log(abs(-cos(f*x + e)^2 + 1))/(a^2*f + 2*a*b*f + b^2*f) + 1/2/ ((cos(f*x + e)^2 - 1)*(a + b)*f)
Time = 16.07 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.32 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {\ln \left ({\mathrm {tan}\left (e+f\,x\right )}^2+1\right )}{2\,a\,f}-\frac {{\mathrm {cot}\left (e+f\,x\right )}^2}{2\,f\,\left (a+b\right )}-\frac {\ln \left (\mathrm {tan}\left (e+f\,x\right )\right )\,\left (a+2\,b\right )}{f\,\left (a^2+2\,a\,b+b^2\right )}-\frac {b^2\,\ln \left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}{2\,a\,f\,{\left (a+b\right )}^2} \] Input:
int(cot(e + f*x)^3/(a + b/cos(e + f*x)^2),x)
Output:
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)
Time = 0.17 (sec) , antiderivative size = 283, normalized size of antiderivative = 3.82 \[ \int \frac {\cot ^3(e+f x)}{a+b \sec ^2(e+f x)} \, dx=\frac {4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{2} a^{2}+8 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{2} a b +4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{2} b^{2}-2 \,\mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}-2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} b^{2}-2 \,\mathrm {log}\left (\sqrt {a +b}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+\sqrt {a +b}+2 \sqrt {a}\, \tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} b^{2}-4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} a^{2}-8 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} a b +\sin \left (f x +e \right )^{2} a^{2}+\sin \left (f x +e \right )^{2} a b -2 a^{2}-2 a b}{4 \sin \left (f x +e \right )^{2} a f \left (a^{2}+2 a b +b^{2}\right )} \] Input:
int(cot(f*x+e)^3/(a+b*sec(f*x+e)^2),x)
Output:
(4*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a**2 + 8*log(tan((e + f*x) /2)**2 + 1)*sin(e + f*x)**2*a*b + 4*log(tan((e + f*x)/2)**2 + 1)*sin(e + f *x)**2*b**2 - 2*log(sqrt(a + b)*tan((e + f*x)/2)**2 + sqrt(a + b) - 2*sqrt (a)*tan((e + f*x)/2))*sin(e + f*x)**2*b**2 - 2*log(sqrt(a + b)*tan((e + f* x)/2)**2 + sqrt(a + b) + 2*sqrt(a)*tan((e + f*x)/2))*sin(e + f*x)**2*b**2 - 4*log(tan((e + f*x)/2))*sin(e + f*x)**2*a**2 - 8*log(tan((e + f*x)/2))*s in(e + f*x)**2*a*b + sin(e + f*x)**2*a**2 + sin(e + f*x)**2*a*b - 2*a**2 - 2*a*b)/(4*sin(e + f*x)**2*a*f*(a**2 + 2*a*b + b**2))