Integrand size = 21, antiderivative size = 65 \[ \int \cot ^3(e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=-\frac {b \text {arctanh}(\cos (e+f x))}{2 f}-\frac {b \cot (e+f x) \csc (e+f x)}{2 f}-\frac {a \csc ^2(e+f x)}{2 f}-\frac {a \log (\sin (e+f x))}{f} \] Output:
-1/2*b*arctanh(cos(f*x+e))/f-1/2*b*cot(f*x+e)*csc(f*x+e)/f-1/2*a*csc(f*x+e )^2/f-a*ln(sin(f*x+e))/f
Time = 0.05 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.66 \[ \int \cot ^3(e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=-\frac {b \csc ^2\left (\frac {1}{2} (e+f x)\right )}{8 f}-\frac {a \csc ^2(e+f x)}{2 f}-\frac {b \log \left (\cos \left (\frac {1}{2} (e+f x)\right )\right )}{2 f}+\frac {b \log \left (\sin \left (\frac {1}{2} (e+f x)\right )\right )}{2 f}-\frac {a \log (\sin (e+f x))}{f}+\frac {b \sec ^2\left (\frac {1}{2} (e+f x)\right )}{8 f} \] Input:
Integrate[Cot[e + f*x]^3*(a + b*Sec[e + f*x]^3),x]
Output:
-1/8*(b*Csc[(e + f*x)/2]^2)/f - (a*Csc[e + f*x]^2)/(2*f) - (b*Log[Cos[(e + f*x)/2]])/(2*f) + (b*Log[Sin[(e + f*x)/2]])/(2*f) - (a*Log[Sin[e + f*x]]) /f + (b*Sec[(e + f*x)/2]^2)/(8*f)
Time = 0.26 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4626, 2345, 25, 452, 219, 240}
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.
\(\displaystyle \int \cot ^3(e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a+b \sec (e+f x)^3}{\tan (e+f x)^3}dx\) |
\(\Big \downarrow \) 4626 |
\(\displaystyle -\frac {\int \frac {a \cos ^3(e+f x)+b}{\left (1-\cos ^2(e+f x)\right )^2}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 2345 |
\(\displaystyle -\frac {\frac {a+b \cos (e+f x)}{2 \left (1-\cos ^2(e+f x)\right )}-\frac {1}{2} \int -\frac {b-2 a \cos (e+f x)}{1-\cos ^2(e+f x)}d\cos (e+f x)}{f}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\frac {1}{2} \int \frac {b-2 a \cos (e+f x)}{1-\cos ^2(e+f x)}d\cos (e+f x)+\frac {a+b \cos (e+f x)}{2 \left (1-\cos ^2(e+f x)\right )}}{f}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle -\frac {\frac {1}{2} \left (b \int \frac {1}{1-\cos ^2(e+f x)}d\cos (e+f x)-2 a \int \frac {\cos (e+f x)}{1-\cos ^2(e+f x)}d\cos (e+f x)\right )+\frac {a+b \cos (e+f x)}{2 \left (1-\cos ^2(e+f x)\right )}}{f}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle -\frac {\frac {1}{2} \left (b \text {arctanh}(\cos (e+f x))-2 a \int \frac {\cos (e+f x)}{1-\cos ^2(e+f x)}d\cos (e+f x)\right )+\frac {a+b \cos (e+f x)}{2 \left (1-\cos ^2(e+f x)\right )}}{f}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle -\frac {\frac {1}{2} \left (a \log \left (1-\cos ^2(e+f x)\right )+b \text {arctanh}(\cos (e+f x))\right )+\frac {a+b \cos (e+f x)}{2 \left (1-\cos ^2(e+f x)\right )}}{f}\) |
Input:
Int[Cot[e + f*x]^3*(a + b*Sec[e + f*x]^3),x]
Output:
-(((a + b*Cos[e + f*x])/(2*(1 - Cos[e + f*x]^2)) + (b*ArcTanh[Cos[e + f*x] ] + a*Log[1 - Cos[e + f*x]^2])/2)/f)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[(x_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[Log[RemoveContent[a + b*x ^2, x]]/(2*b), x] /; FreeQ[{a, b}, x]
Int[((c_) + (d_.)*(x_))/((a_) + (b_.)*(x_)^2), x_Symbol] :> Simp[c Int[1/ (a + b*x^2), x], x] + Simp[d Int[x/(a + b*x^2), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c^2 + a*d^2, 0]
Int[(Pq_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{Q = PolynomialQuot ient[Pq, a + b*x^2, x], f = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b *f*x)*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1)) In t[(a + b*x^2)^(p + 1)*ExpandToSum[2*a*(p + 1)*Q + f*(2*p + 3), x], x], x]] /; FreeQ[{a, b}, x] && PolyQ[Pq, x] && LtQ[p, -1]
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 = 0.80 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cot \left (f x +e \right )^{2}}{2}-\ln \left (\sin \left (f x +e \right )\right )\right )+b \left (-\frac {\csc \left (f x +e \right ) \cot \left (f x +e \right )}{2}+\frac {\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2}\right )}{f}\) | \(63\) |
default | \(\frac {a \left (-\frac {\cot \left (f x +e \right )^{2}}{2}-\ln \left (\sin \left (f x +e \right )\right )\right )+b \left (-\frac {\csc \left (f x +e \right ) \cot \left (f x +e \right )}{2}+\frac {\ln \left (\csc \left (f x +e \right )-\cot \left (f x +e \right )\right )}{2}\right )}{f}\) | \(63\) |
risch | \(i a x +\frac {2 i a e}{f}+\frac {b \,{\mathrm e}^{3 i \left (f x +e \right )}+2 \,{\mathrm e}^{2 i \left (f x +e \right )} a +b \,{\mathrm e}^{i \left (f x +e \right )}}{f \left ({\mathrm e}^{2 i \left (f x +e \right )}-1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) a}{f}+\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right ) b}{2 f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) a}{f}-\frac {\ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) b}{2 f}\) | \(139\) |
Input:
int(cot(f*x+e)^3*(a+b*sec(f*x+e)^3),x,method=_RETURNVERBOSE)
Output:
1/f*(a*(-1/2*cot(f*x+e)^2-ln(sin(f*x+e)))+b*(-1/2*csc(f*x+e)*cot(f*x+e)+1/ 2*ln(csc(f*x+e)-cot(f*x+e))))
Time = 0.09 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.52 \[ \int \cot ^3(e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {2 \, b \cos \left (f x + e\right ) - {\left ({\left (2 \, a + b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - b\right )} \log \left (\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) - {\left ({\left (2 \, a - b\right )} \cos \left (f x + e\right )^{2} - 2 \, a + b\right )} \log \left (-\frac {1}{2} \, \cos \left (f x + e\right ) + \frac {1}{2}\right ) + 2 \, a}{4 \, {\left (f \cos \left (f x + e\right )^{2} - f\right )}} \] Input:
integrate(cot(f*x+e)^3*(a+b*sec(f*x+e)^3),x, algorithm="fricas")
Output:
1/4*(2*b*cos(f*x + e) - ((2*a + b)*cos(f*x + e)^2 - 2*a - b)*log(1/2*cos(f *x + e) + 1/2) - ((2*a - b)*cos(f*x + e)^2 - 2*a + b)*log(-1/2*cos(f*x + e ) + 1/2) + 2*a)/(f*cos(f*x + e)^2 - f)
\[ \int \cot ^3(e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\int \left (a + b \sec ^{3}{\left (e + f x \right )}\right ) \cot ^{3}{\left (e + f x \right )}\, dx \] Input:
integrate(cot(f*x+e)**3*(a+b*sec(f*x+e)**3),x)
Output:
Integral((a + b*sec(e + f*x)**3)*cot(e + f*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.95 \[ \int \cot ^3(e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=-\frac {{\left (2 \, a + b\right )} \log \left (\cos \left (f x + e\right ) + 1\right ) + {\left (2 \, a - b\right )} \log \left (\cos \left (f x + e\right ) - 1\right ) - \frac {2 \, {\left (b \cos \left (f x + e\right ) + a\right )}}{\cos \left (f x + e\right )^{2} - 1}}{4 \, f} \] Input:
integrate(cot(f*x+e)^3*(a+b*sec(f*x+e)^3),x, algorithm="maxima")
Output:
-1/4*((2*a + b)*log(cos(f*x + e) + 1) + (2*a - b)*log(cos(f*x + e) - 1) - 2*(b*cos(f*x + e) + a)/(cos(f*x + e)^2 - 1))/f
Time = 0.17 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.20 \[ \int \cot ^3(e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=-\frac {{\left (2 \, a + b\right )} \log \left ({\left | \cos \left (f x + e\right ) + 1 \right |}\right )}{4 \, f} - \frac {{\left (2 \, a - b\right )} \log \left ({\left | \cos \left (f x + e\right ) - 1 \right |}\right )}{4 \, f} + \frac {b \cos \left (f x + e\right ) + a}{2 \, f {\left (\cos \left (f x + e\right ) + 1\right )} {\left (\cos \left (f x + e\right ) - 1\right )}} \] Input:
integrate(cot(f*x+e)^3*(a+b*sec(f*x+e)^3),x, algorithm="giac")
Output:
-1/4*(2*a + b)*log(abs(cos(f*x + e) + 1))/f - 1/4*(2*a - b)*log(abs(cos(f* x + e) - 1))/f + 1/2*(b*cos(f*x + e) + a)/(f*(cos(f*x + e) + 1)*(cos(f*x + e) - 1))
Time = 15.17 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.32 \[ \int \cot ^3(e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+1\right )}{f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a}{8}-\frac {b}{8}\right )}{f}-\frac {{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2\,\left (\frac {a}{8}+\frac {b}{8}\right )}{f}-\frac {\ln \left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )\,\left (a-\frac {b}{2}\right )}{f} \] Input:
int(cot(e + f*x)^3*(a + b/cos(e + f*x)^3),x)
Output:
(a*log(tan(e/2 + (f*x)/2)^2 + 1))/f - (tan(e/2 + (f*x)/2)^2*(a/8 - b/8))/f - (cot(e/2 + (f*x)/2)^2*(a/8 + b/8))/f - (log(tan(e/2 + (f*x)/2))*(a - b/ 2))/f
Time = 0.18 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.58 \[ \int \cot ^3(e+f x) \left (a+b \sec ^3(e+f x)\right ) \, dx=\frac {-2 \cos \left (f x +e \right ) b +4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+1\right ) \sin \left (f x +e \right )^{2} a -4 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} a +2 \,\mathrm {log}\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right ) \sin \left (f x +e \right )^{2} b +\sin \left (f x +e \right )^{2} a -2 a}{4 \sin \left (f x +e \right )^{2} f} \] Input:
int(cot(f*x+e)^3*(a+b*sec(f*x+e)^3),x)
Output:
( - 2*cos(e + f*x)*b + 4*log(tan((e + f*x)/2)**2 + 1)*sin(e + f*x)**2*a - 4*log(tan((e + f*x)/2))*sin(e + f*x)**2*a + 2*log(tan((e + f*x)/2))*sin(e + f*x)**2*b + sin(e + f*x)**2*a - 2*a)/(4*sin(e + f*x)**2*f)