Integrand size = 19, antiderivative size = 72 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {(2 a+b) \log (1-\cos (c+d x))}{4 d}-\frac {(2 a-b) \log (1+\cos (c+d x))}{4 d}-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d} \] Output:
-1/4*(2*a+b)*ln(1-cos(d*x+c))/d-1/4*(2*a-b)*ln(1+cos(d*x+c))/d-1/2*cot(d*x +c)^2*(a+b*sec(d*x+c))/d
Time = 0.07 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.50 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{8 d}-\frac {a \csc ^2(c+d x)}{2 d}+\frac {b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{2 d}-\frac {a \log (\sin (c+d x))}{d}+\frac {b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{8 d} \] Input:
Integrate[Cot[c + d*x]^3*(a + b*Sec[c + d*x]),x]
Output:
-1/8*(b*Csc[(c + d*x)/2]^2)/d - (a*Csc[c + d*x]^2)/(2*d) + (b*Log[Cos[(c + d*x)/2]])/(2*d) - (b*Log[Sin[(c + d*x)/2]])/(2*d) - (a*Log[Sin[c + d*x]]) /d + (b*Sec[(c + d*x)/2]^2)/(8*d)
Time = 0.41 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.99, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {3042, 25, 4370, 3042, 25, 4371, 3042, 25, 3147, 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.
\(\displaystyle \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}{\cot \left (c+d x+\frac {\pi }{2}\right )^3}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3}dx\) |
\(\Big \downarrow \) 4370 |
\(\displaystyle -\frac {1}{2} \int \cot (c+d x) (2 a+b \sec (c+d x))dx-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle -\frac {1}{2} \int -\frac {2 a+b \csc \left (c+d x+\frac {\pi }{2}\right )}{\cot \left (c+d x+\frac {\pi }{2}\right )}dx-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {1}{2} \int \frac {2 a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}\) |
\(\Big \downarrow \) 4371 |
\(\displaystyle \frac {1}{2} \int \sec \left (\frac {1}{2} (2 c+\pi )+d x\right ) \left (b+2 a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )dx-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} \int -\frac {b-2 a \sin \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )}dx-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {1}{2} \int \frac {b-2 a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )}dx-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}\) |
\(\Big \downarrow \) 3147 |
\(\displaystyle \frac {a \int \frac {b+2 a \cos (c+d x)}{4 a^2-4 a^2 \cos ^2(c+d x)}d(2 a \cos (c+d x))}{d}-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}\) |
\(\Big \downarrow \) 452 |
\(\displaystyle \frac {a \left (b \int \frac {1}{4 a^2-4 a^2 \cos ^2(c+d x)}d(2 a \cos (c+d x))+\int \frac {2 a \cos (c+d x)}{4 a^2-4 a^2 \cos ^2(c+d x)}d(2 a \cos (c+d x))\right )}{d}-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {a \left (\int \frac {2 a \cos (c+d x)}{4 a^2-4 a^2 \cos ^2(c+d x)}d(2 a \cos (c+d x))+\frac {b \text {arctanh}(\cos (c+d x))}{2 a}\right )}{d}-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}\) |
\(\Big \downarrow \) 240 |
\(\displaystyle \frac {a \left (\frac {b \text {arctanh}(\cos (c+d x))}{2 a}-\frac {1}{2} \log \left (4 a^2-4 a^2 \cos ^2(c+d x)\right )\right )}{d}-\frac {\cot ^2(c+d x) (a+b \sec (c+d x))}{2 d}\) |
Input:
Int[Cot[c + d*x]^3*(a + b*Sec[c + d*x]),x]
Output:
(a*((b*ArcTanh[Cos[c + d*x]])/(2*a) - Log[4*a^2 - 4*a^2*Cos[c + d*x]^2]/2) )/d - (Cot[c + d*x]^2*(a + b*Sec[c + d*x]))/(2*d)
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[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m _.), x_Symbol] :> Simp[1/(b^p*f) Subst[Int[(a + x)^m*(b^2 - x^2)^((p - 1) /2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_)), x_Symbol] :> Simp[(-(e*Cot[c + d*x])^(m + 1))*((a + b*Csc[c + d*x])/( d*e*(m + 1))), x] - Simp[1/(e^2*(m + 1)) Int[(e*Cot[c + d*x])^(m + 2)*(a* (m + 1) + b*(m + 2)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && L tQ[m, -1]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))/cot[(c_.) + (d_.)*(x_)], x_Symbo l] :> Int[(b + a*Sin[c + d*x])/Cos[c + d*x], x] /; FreeQ[{a, b, c, d}, x]
Time = 0.22 (sec) , antiderivative size = 75, normalized size of antiderivative = 1.04
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\) | \(75\) |
default | \(\frac {a \left (-\frac {\cot \left (d x +c \right )^{2}}{2}-\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos \left (d x +c \right )^{3}}{2 \sin \left (d x +c \right )^{2}}-\frac {\cos \left (d x +c \right )}{2}-\frac {\ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{2}\right )}{d}\) | \(75\) |
risch | \(i a x +\frac {2 i a c}{d}+\frac {b \,{\mathrm e}^{3 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+b \,{\mathrm e}^{i \left (d x +c \right )}}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{2}}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b}{2 d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b}{2 d}\) | \(139\) |
Input:
int(cot(d*x+c)^3*(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(-1/2*cot(d*x+c)^2-ln(sin(d*x+c)))+b*(-1/2/sin(d*x+c)^2*cos(d*x+c)^ 3-1/2*cos(d*x+c)-1/2*ln(csc(d*x+c)-cot(d*x+c))))
Time = 0.12 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.38 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=\frac {2 \, b \cos \left (d x + c\right ) - {\left ({\left (2 \, a - b\right )} \cos \left (d x + c\right )^{2} - 2 \, a + b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (2 \, a + b\right )} \cos \left (d x + c\right )^{2} - 2 \, a - b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) + 2 \, a}{4 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )}} \] Input:
integrate(cot(d*x+c)^3*(a+b*sec(d*x+c)),x, algorithm="fricas")
Output:
1/4*(2*b*cos(d*x + c) - ((2*a - b)*cos(d*x + c)^2 - 2*a + b)*log(1/2*cos(d *x + c) + 1/2) - ((2*a + b)*cos(d*x + c)^2 - 2*a - b)*log(-1/2*cos(d*x + c ) + 1/2) + 2*a)/(d*cos(d*x + c)^2 - d)
\[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \cot ^{3}{\left (c + d x \right )}\, dx \] Input:
integrate(cot(d*x+c)**3*(a+b*sec(d*x+c)),x)
Output:
Integral((a + b*sec(c + d*x))*cot(c + d*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.86 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {{\left (2 \, a - b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) + {\left (2 \, a + b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (b \cos \left (d x + c\right ) + a\right )}}{\cos \left (d x + c\right )^{2} - 1}}{4 \, d} \] Input:
integrate(cot(d*x+c)^3*(a+b*sec(d*x+c)),x, algorithm="maxima")
Output:
-1/4*((2*a - b)*log(cos(d*x + c) + 1) + (2*a + b)*log(cos(d*x + c) - 1) - 2*(b*cos(d*x + c) + a)/(cos(d*x + c)^2 - 1))/d
Time = 0.14 (sec) , antiderivative size = 78, normalized size of antiderivative = 1.08 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {{\left (2 \, a - b\right )} \log \left ({\left | \cos \left (d x + c\right ) + 1 \right |}\right )}{4 \, d} - \frac {{\left (2 \, a + b\right )} \log \left ({\left | \cos \left (d x + c\right ) - 1 \right |}\right )}{4 \, d} + \frac {b \cos \left (d x + c\right ) + a}{2 \, d {\left (\cos \left (d x + c\right ) + 1\right )} {\left (\cos \left (d x + c\right ) - 1\right )}} \] Input:
integrate(cot(d*x+c)^3*(a+b*sec(d*x+c)),x, algorithm="giac")
Output:
-1/4*(2*a - b)*log(abs(cos(d*x + c) + 1))/d - 1/4*(2*a + b)*log(abs(cos(d* x + c) - 1))/d + 1/2*(b*cos(d*x + c) + a)/(d*(cos(d*x + c) + 1)*(cos(d*x + c) - 1))
Time = 10.76 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.19 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a}{8}-\frac {b}{8}\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a}{8}+\frac {b}{8}\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a+\frac {b}{2}\right )}{d} \] Input:
int(cot(c + d*x)^3*(a + b/cos(c + d*x)),x)
Output:
(a*log(tan(c/2 + (d*x)/2)^2 + 1))/d - (tan(c/2 + (d*x)/2)^2*(a/8 - b/8))/d - (cot(c/2 + (d*x)/2)^2*(a/8 + b/8))/d - (log(tan(c/2 + (d*x)/2))*(a + b/ 2))/d
Time = 0.16 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.43 \[ \int \cot ^3(c+d x) (a+b \sec (c+d x)) \, dx=\frac {-2 \cos \left (d x +c \right ) b +4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (d x +c \right )^{2} a -4 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} a -2 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \sin \left (d x +c \right )^{2} b +\sin \left (d x +c \right )^{2} a -2 a}{4 \sin \left (d x +c \right )^{2} d} \] Input:
int(cot(d*x+c)^3*(a+b*sec(d*x+c)),x)
Output:
( - 2*cos(c + d*x)*b + 4*log(tan((c + d*x)/2)**2 + 1)*sin(c + d*x)**2*a - 4*log(tan((c + d*x)/2))*sin(c + d*x)**2*a - 2*log(tan((c + d*x)/2))*sin(c + d*x)**2*b + sin(c + d*x)**2*a - 2*a)/(4*sin(c + d*x)**2*d)