Integrand size = 19, antiderivative size = 102 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {(8 a+3 b) \log (1-\cos (c+d x))}{16 d}+\frac {(8 a-3 b) \log (1+\cos (c+d x))}{16 d}-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}+\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{8 d} \]
1/16*(8*a+3*b)*ln(1-cos(d*x+c))/d+1/16*(8*a-3*b)*ln(1+cos(d*x+c))/d-1/4*co t(d*x+c)^4*(a+b*sec(d*x+c))/d+1/8*cot(d*x+c)^2*(4*a+3*b*sec(d*x+c))/d
Time = 0.10 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.72 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {a \cot ^2(c+d x)}{2 d}-\frac {a \cot ^4(c+d x)}{4 d}+\frac {5 b \csc ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}-\frac {b \csc ^4\left (\frac {1}{2} (c+d x)\right )}{64 d}-\frac {3 b \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {a \log (\cos (c+d x))}{d}+\frac {3 b \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )}{8 d}+\frac {a \log (\tan (c+d x))}{d}-\frac {5 b \sec ^2\left (\frac {1}{2} (c+d x)\right )}{32 d}+\frac {b \sec ^4\left (\frac {1}{2} (c+d x)\right )}{64 d} \]
(a*Cot[c + d*x]^2)/(2*d) - (a*Cot[c + d*x]^4)/(4*d) + (5*b*Csc[(c + d*x)/2 ]^2)/(32*d) - (b*Csc[(c + d*x)/2]^4)/(64*d) - (3*b*Log[Cos[(c + d*x)/2]])/ (8*d) + (a*Log[Cos[c + d*x]])/d + (3*b*Log[Sin[(c + d*x)/2]])/(8*d) + (a*L og[Tan[c + d*x]])/d - (5*b*Sec[(c + d*x)/2]^2)/(32*d) + (b*Sec[(c + d*x)/2 ]^4)/(64*d)
Time = 0.56 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.789, Rules used = {3042, 25, 4370, 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.
| \(\displaystyle \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int -\frac {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}{\cot \left (c+d x+\frac {\pi }{2}\right )^5}dx\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\int \frac {a+b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^5}dx\) |
| \(\Big \downarrow \) 4370 |
| \(\displaystyle -\frac {1}{4} \int \cot ^3(c+d x) (4 a+3 b \sec (c+d x))dx-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {1}{4} \int -\frac {4 a+3 b \csc \left (c+d x+\frac {\pi }{2}\right )}{\cot \left (c+d x+\frac {\pi }{2}\right )^3}dx-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \int \frac {4 a+3 b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )^3}dx-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 4370 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \cot (c+d x) (8 a+3 b \sec (c+d x))dx+\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{2 d}\right )-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int -\frac {8 a+3 b \csc \left (c+d x+\frac {\pi }{2}\right )}{\cot \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{2 d}\right )-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \left (\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{2 d}-\frac {1}{2} \int \frac {8 a+3 b \csc \left (\frac {1}{2} (2 c+\pi )+d x\right )}{\cot \left (\frac {1}{2} (2 c+\pi )+d x\right )}dx\right )-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 4371 |
| \(\displaystyle \frac {1}{4} \left (\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{2 d}-\frac {1}{2} \int \sec \left (\frac {1}{2} (2 c+\pi )+d x\right ) \left (3 b+8 a \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )dx\right )-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {1}{4} \left (\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{2 d}-\frac {1}{2} \int -\frac {3 b-8 a \sin \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )}dx\right )-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {1}{4} \left (\frac {1}{2} \int \frac {3 b-8 a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )}dx+\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{2 d}\right )-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 3147 |
| \(\displaystyle \frac {1}{4} \left (\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{2 d}-\frac {4 a \int \frac {3 b+8 a \cos (c+d x)}{64 a^2-64 a^2 \cos ^2(c+d x)}d(8 a \cos (c+d x))}{d}\right )-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 452 |
| \(\displaystyle \frac {1}{4} \left (\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{2 d}-\frac {4 a \left (3 b \int \frac {1}{64 a^2-64 a^2 \cos ^2(c+d x)}d(8 a \cos (c+d x))+\int \frac {8 a \cos (c+d x)}{64 a^2-64 a^2 \cos ^2(c+d x)}d(8 a \cos (c+d x))\right )}{d}\right )-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 219 |
| \(\displaystyle \frac {1}{4} \left (\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{2 d}-\frac {4 a \left (\int \frac {8 a \cos (c+d x)}{64 a^2-64 a^2 \cos ^2(c+d x)}d(8 a \cos (c+d x))+\frac {3 b \text {arctanh}(\cos (c+d x))}{8 a}\right )}{d}\right )-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
| \(\Big \downarrow \) 240 |
| \(\displaystyle \frac {1}{4} \left (\frac {\cot ^2(c+d x) (4 a+3 b \sec (c+d x))}{2 d}-\frac {4 a \left (\frac {3 b \text {arctanh}(\cos (c+d x))}{8 a}-\frac {1}{2} \log \left (64 a^2-64 a^2 \cos ^2(c+d x)\right )\right )}{d}\right )-\frac {\cot ^4(c+d x) (a+b \sec (c+d x))}{4 d}\) |
-1/4*(Cot[c + d*x]^4*(a + b*Sec[c + d*x]))/d + ((-4*a*((3*b*ArcTanh[Cos[c + d*x]])/(8*a) - Log[64*a^2 - 64*a^2*Cos[c + d*x]^2]/2))/d + (Cot[c + d*x] ^2*(4*a + 3*b*Sec[c + d*x]))/(2*d))/4
3.3.62.3.1 Defintions of rubi rules used
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 = 0.72 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {a \left (-\frac {\cot \left (d x +c \right )^{4}}{4}+\frac {\cot \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{8}\right )}{d}\) | \(111\) |
| default | \(\frac {a \left (-\frac {\cot \left (d x +c \right )^{4}}{4}+\frac {\cot \left (d x +c \right )^{2}}{2}+\ln \left (\sin \left (d x +c \right )\right )\right )+b \left (-\frac {\cos \left (d x +c \right )^{5}}{4 \sin \left (d x +c \right )^{4}}+\frac {\cos \left (d x +c \right )^{5}}{8 \sin \left (d x +c \right )^{2}}+\frac {\cos \left (d x +c \right )^{3}}{8}+\frac {3 \cos \left (d x +c \right )}{8}+\frac {3 \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{8}\right )}{d}\) | \(111\) |
| risch | \(-i a x -\frac {2 i a c}{d}-\frac {5 b \,{\mathrm e}^{7 i \left (d x +c \right )}+16 a \,{\mathrm e}^{6 i \left (d x +c \right )}+3 b \,{\mathrm e}^{5 i \left (d x +c \right )}-16 a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 b \,{\mathrm e}^{3 i \left (d x +c \right )}+16 a \,{\mathrm e}^{2 i \left (d x +c \right )}+5 b \,{\mathrm e}^{i \left (d x +c \right )}}{4 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{4}}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}+\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right ) b}{8 d}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{d}-\frac {3 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right ) b}{8 d}\) | \(188\) |
1/d*(a*(-1/4*cot(d*x+c)^4+1/2*cot(d*x+c)^2+ln(sin(d*x+c)))+b*(-1/4/sin(d*x +c)^4*cos(d*x+c)^5+1/8/sin(d*x+c)^2*cos(d*x+c)^5+1/8*cos(d*x+c)^3+3/8*cos( d*x+c)+3/8*ln(-cot(d*x+c)+csc(d*x+c))))
Time = 0.27 (sec) , antiderivative size = 168, normalized size of antiderivative = 1.65 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {10 \, b \cos \left (d x + c\right )^{3} + 16 \, a \cos \left (d x + c\right )^{2} - 6 \, b \cos \left (d x + c\right ) - {\left ({\left (8 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (8 \, a - 3 \, b\right )} \cos \left (d x + c\right )^{2} + 8 \, a - 3 \, b\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - {\left ({\left (8 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{4} - 2 \, {\left (8 \, a + 3 \, b\right )} \cos \left (d x + c\right )^{2} + 8 \, a + 3 \, b\right )} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 12 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{4} - 2 \, d \cos \left (d x + c\right )^{2} + d\right )}} \]
-1/16*(10*b*cos(d*x + c)^3 + 16*a*cos(d*x + c)^2 - 6*b*cos(d*x + c) - ((8* a - 3*b)*cos(d*x + c)^4 - 2*(8*a - 3*b)*cos(d*x + c)^2 + 8*a - 3*b)*log(1/ 2*cos(d*x + c) + 1/2) - ((8*a + 3*b)*cos(d*x + c)^4 - 2*(8*a + 3*b)*cos(d* x + c)^2 + 8*a + 3*b)*log(-1/2*cos(d*x + c) + 1/2) - 12*a)/(d*cos(d*x + c) ^4 - 2*d*cos(d*x + c)^2 + d)
\[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \cot ^{5}{\left (c + d x \right )}\, dx \]
Time = 0.24 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.97 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {{\left (8 \, a - 3 \, b\right )} \log \left (\cos \left (d x + c\right ) + 1\right ) + {\left (8 \, a + 3 \, b\right )} \log \left (\cos \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (5 \, b \cos \left (d x + c\right )^{3} + 8 \, a \cos \left (d x + c\right )^{2} - 3 \, b \cos \left (d x + c\right ) - 6 \, a\right )}}{\cos \left (d x + c\right )^{4} - 2 \, \cos \left (d x + c\right )^{2} + 1}}{16 \, d} \]
1/16*((8*a - 3*b)*log(cos(d*x + c) + 1) + (8*a + 3*b)*log(cos(d*x + c) - 1 ) - 2*(5*b*cos(d*x + c)^3 + 8*a*cos(d*x + c)^2 - 3*b*cos(d*x + c) - 6*a)/( cos(d*x + c)^4 - 2*cos(d*x + c)^2 + 1))/d
Leaf count of result is larger than twice the leaf count of optimal. 266 vs. \(2 (94) = 188\).
Time = 0.34 (sec) , antiderivative size = 266, normalized size of antiderivative = 2.61 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {4 \, {\left (8 \, a + 3 \, b\right )} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 64 \, a \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1 \right |}\right ) - \frac {{\left (a + b + \frac {12 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {8 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {48 \, a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {18 \, b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (d x + c\right ) + 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) - 1\right )}^{2}} - \frac {12 \, a {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {8 \, b {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {a {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {b {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{64 \, d} \]
1/64*(4*(8*a + 3*b)*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 64 *a*log(abs(-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) - (a + b + 12*a*(c os(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*b*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 48*a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 18*b*(cos(d*x + c ) - 1)^2/(cos(d*x + c) + 1)^2)*(cos(d*x + c) + 1)^2/(cos(d*x + c) - 1)^2 - 12*a*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 8*b*(cos(d*x + c) - 1)/(cos( d*x + c) + 1) - a*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + b*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1)^2)/d
Time = 14.49 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.25 \[ \int \cot ^5(c+d x) (a+b \sec (c+d x)) \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {3\,a}{16}-\frac {b}{8}\right )}{d}-\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a}{64}-\frac {b}{64}\right )}{d}-\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\left (-3\,a-2\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a}{4}+\frac {b}{4}\right )}{16\,d}+\frac {\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )\,\left (a+\frac {3\,b}{8}\right )}{d} \]