Integrand size = 17, antiderivative size = 89 \[ \int \csc ^5(a+b x) \sec ^4(a+b x) \, dx=-\frac {35 \text {arctanh}(\cos (a+b x))}{8 b}+\frac {35 \sec (a+b x)}{8 b}+\frac {35 \sec ^3(a+b x)}{24 b}-\frac {7 \csc ^2(a+b x) \sec ^3(a+b x)}{8 b}-\frac {\csc ^4(a+b x) \sec ^3(a+b x)}{4 b} \]
-35/8*arctanh(cos(b*x+a))/b+35/8*sec(b*x+a)/b+35/24*sec(b*x+a)^3/b-7/8*csc (b*x+a)^2*sec(b*x+a)^3/b-1/4*csc(b*x+a)^4*sec(b*x+a)^3/b
Leaf count is larger than twice the leaf count of optimal. \(268\) vs. \(2(89)=178\).
Time = 0.45 (sec) , antiderivative size = 268, normalized size of antiderivative = 3.01 \[ \int \csc ^5(a+b x) \sec ^4(a+b x) \, dx=-\frac {\csc ^{10}(a+b x) \left (-204+658 \cos (2 (a+b x))-228 \cos (3 (a+b x))+140 \cos (4 (a+b x))-76 \cos (5 (a+b x))-210 \cos (6 (a+b x))+76 \cos (7 (a+b x))-315 \cos (3 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-105 \cos (5 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+105 \cos (7 (a+b x)) \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )+3 \cos (a+b x) \left (76+105 \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )-105 \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )+315 \cos (3 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )+105 \cos (5 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )-105 \cos (7 (a+b x)) \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )\right )}{24 b \left (\csc ^2\left (\frac {1}{2} (a+b x)\right )-\sec ^2\left (\frac {1}{2} (a+b x)\right )\right )^3} \]
-1/24*(Csc[a + b*x]^10*(-204 + 658*Cos[2*(a + b*x)] - 228*Cos[3*(a + b*x)] + 140*Cos[4*(a + b*x)] - 76*Cos[5*(a + b*x)] - 210*Cos[6*(a + b*x)] + 76* Cos[7*(a + b*x)] - 315*Cos[3*(a + b*x)]*Log[Cos[(a + b*x)/2]] - 105*Cos[5* (a + b*x)]*Log[Cos[(a + b*x)/2]] + 105*Cos[7*(a + b*x)]*Log[Cos[(a + b*x)/ 2]] + 3*Cos[a + b*x]*(76 + 105*Log[Cos[(a + b*x)/2]] - 105*Log[Sin[(a + b* x)/2]]) + 315*Cos[3*(a + b*x)]*Log[Sin[(a + b*x)/2]] + 105*Cos[5*(a + b*x) ]*Log[Sin[(a + b*x)/2]] - 105*Cos[7*(a + b*x)]*Log[Sin[(a + b*x)/2]]))/(b* (Csc[(a + b*x)/2]^2 - Sec[(a + b*x)/2]^2)^3)
Time = 0.25 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 3102, 25, 252, 252, 254, 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.
\(\displaystyle \int \csc ^5(a+b x) \sec ^4(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (a+b x)^5 \sec (a+b x)^4dx\) |
\(\Big \downarrow \) 3102 |
\(\displaystyle \frac {\int -\frac {\sec ^8(a+b x)}{\left (1-\sec ^2(a+b x)\right )^3}d\sec (a+b x)}{b}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {\int \frac {\sec ^8(a+b x)}{\left (1-\sec ^2(a+b x)\right )^3}d\sec (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {7}{4} \int \frac {\sec ^6(a+b x)}{\left (1-\sec ^2(a+b x)\right )^2}d\sec (a+b x)-\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle \frac {\frac {7}{4} \left (\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\sec ^4(a+b x)}{1-\sec ^2(a+b x)}d\sec (a+b x)\right )-\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{b}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {\frac {7}{4} \left (\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\sec ^2(a+b x)+\frac {1}{1-\sec ^2(a+b x)}-1\right )d\sec (a+b x)\right )-\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {7}{4} \left (\frac {\sec ^5(a+b x)}{2 \left (1-\sec ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\sec (a+b x))-\frac {1}{3} \sec ^3(a+b x)-\sec (a+b x)\right )\right )-\frac {\sec ^7(a+b x)}{4 \left (1-\sec ^2(a+b x)\right )^2}}{b}\) |
(-1/4*Sec[a + b*x]^7/(1 - Sec[a + b*x]^2)^2 + (7*(Sec[a + b*x]^5/(2*(1 - S ec[a + b*x]^2)) - (5*(ArcTanh[Sec[a + b*x]] - Sec[a + b*x] - Sec[a + b*x]^ 3/3))/2))/4)/b
3.2.83.3.1 Defintions of rubi rules used
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x )^(m - 1)*((a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] - Simp[c^2*((m - 1)/(2*b* (p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^(p + 1), x], x] /; FreeQ[{a, b, c }, x] && LtQ[p, -1] && GtQ[m, 1] && !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomi alQ[a, b, c, 2, m, p, x]
Int[(x_)^(m_)/((a_) + (b_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^2, x], x] /; FreeQ[{a, b}, x] && IGtQ[m, 3]
Int[csc[(e_.) + (f_.)*(x_)]^(n_.)*((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_), x_S ymbol] :> Simp[1/(f*a^n) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/ 2), x], x, a*Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1 )/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Time = 0.33 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.99
method | result | size |
derivativedivides | \(\frac {-\frac {1}{4 \cos \left (b x +a \right )^{3} \sin \left (b x +a \right )^{4}}+\frac {7}{12 \cos \left (b x +a \right )^{3} \sin \left (b x +a \right )^{2}}-\frac {35}{24 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}}+\frac {35}{8 \cos \left (b x +a \right )}+\frac {35 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{8}}{b}\) | \(88\) |
default | \(\frac {-\frac {1}{4 \cos \left (b x +a \right )^{3} \sin \left (b x +a \right )^{4}}+\frac {7}{12 \cos \left (b x +a \right )^{3} \sin \left (b x +a \right )^{2}}-\frac {35}{24 \cos \left (b x +a \right ) \sin \left (b x +a \right )^{2}}+\frac {35}{8 \cos \left (b x +a \right )}+\frac {35 \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{8}}{b}\) | \(88\) |
risch | \(\frac {105 \,{\mathrm e}^{13 i \left (b x +a \right )}-70 \,{\mathrm e}^{11 i \left (b x +a \right )}-329 \,{\mathrm e}^{9 i \left (b x +a \right )}+204 \,{\mathrm e}^{7 i \left (b x +a \right )}-329 \,{\mathrm e}^{5 i \left (b x +a \right )}-70 \,{\mathrm e}^{3 i \left (b x +a \right )}+105 \,{\mathrm e}^{i \left (b x +a \right )}}{12 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{3}}-\frac {35 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{8 b}+\frac {35 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{8 b}\) | \(145\) |
norman | \(\frac {\frac {1}{64 b}+\frac {21 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{64 b}+\frac {21 \left (\tan ^{12}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{64 b}+\frac {\tan ^{14}\left (\frac {b x}{2}+\frac {a}{2}\right )}{64 b}-\frac {21 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{2 b}+\frac {511 \left (\tan ^{6}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{32 b}-\frac {847 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{96 b}}{\left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3} \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}+\frac {35 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )}{8 b}\) | \(146\) |
parallelrisch | \(\frac {840 \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{3} \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )\right )+3 \left (\tan ^{10}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+63 \left (\tan ^{8}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+3 \left (\cot ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-2016 \left (\tan ^{4}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+63 \left (\cot ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )+3066 \left (\tan ^{2}\left (\frac {b x}{2}+\frac {a}{2}\right )\right )-1694}{192 b \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )^{3} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )^{3}}\) | \(150\) |
1/b*(-1/4/cos(b*x+a)^3/sin(b*x+a)^4+7/12/cos(b*x+a)^3/sin(b*x+a)^2-35/24/c os(b*x+a)/sin(b*x+a)^2+35/8/cos(b*x+a)+35/8*ln(csc(b*x+a)-cot(b*x+a)))
Time = 0.35 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.66 \[ \int \csc ^5(a+b x) \sec ^4(a+b x) \, dx=\frac {210 \, \cos \left (b x + a\right )^{6} - 350 \, \cos \left (b x + a\right )^{4} + 112 \, \cos \left (b x + a\right )^{2} - 105 \, {\left (\cos \left (b x + a\right )^{7} - 2 \, \cos \left (b x + a\right )^{5} + \cos \left (b x + a\right )^{3}\right )} \log \left (\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 105 \, {\left (\cos \left (b x + a\right )^{7} - 2 \, \cos \left (b x + a\right )^{5} + \cos \left (b x + a\right )^{3}\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2}\right ) + 16}{48 \, {\left (b \cos \left (b x + a\right )^{7} - 2 \, b \cos \left (b x + a\right )^{5} + b \cos \left (b x + a\right )^{3}\right )}} \]
1/48*(210*cos(b*x + a)^6 - 350*cos(b*x + a)^4 + 112*cos(b*x + a)^2 - 105*( cos(b*x + a)^7 - 2*cos(b*x + a)^5 + cos(b*x + a)^3)*log(1/2*cos(b*x + a) + 1/2) + 105*(cos(b*x + a)^7 - 2*cos(b*x + a)^5 + cos(b*x + a)^3)*log(-1/2* cos(b*x + a) + 1/2) + 16)/(b*cos(b*x + a)^7 - 2*b*cos(b*x + a)^5 + b*cos(b *x + a)^3)
\[ \int \csc ^5(a+b x) \sec ^4(a+b x) \, dx=\int \frac {\sec ^{4}{\left (a + b x \right )}}{\sin ^{5}{\left (a + b x \right )}}\, dx \]
Time = 0.19 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.02 \[ \int \csc ^5(a+b x) \sec ^4(a+b x) \, dx=\frac {\frac {2 \, {\left (105 \, \cos \left (b x + a\right )^{6} - 175 \, \cos \left (b x + a\right )^{4} + 56 \, \cos \left (b x + a\right )^{2} + 8\right )}}{\cos \left (b x + a\right )^{7} - 2 \, \cos \left (b x + a\right )^{5} + \cos \left (b x + a\right )^{3}} - 105 \, \log \left (\cos \left (b x + a\right ) + 1\right ) + 105 \, \log \left (\cos \left (b x + a\right ) - 1\right )}{48 \, b} \]
1/48*(2*(105*cos(b*x + a)^6 - 175*cos(b*x + a)^4 + 56*cos(b*x + a)^2 + 8)/ (cos(b*x + a)^7 - 2*cos(b*x + a)^5 + cos(b*x + a)^3) - 105*log(cos(b*x + a ) + 1) + 105*log(cos(b*x + a) - 1))/b
Leaf count of result is larger than twice the leaf count of optimal. 209 vs. \(2 (79) = 158\).
Time = 0.32 (sec) , antiderivative size = 209, normalized size of antiderivative = 2.35 \[ \int \csc ^5(a+b x) \sec ^4(a+b x) \, dx=\frac {\frac {3 \, {\left (\frac {24 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} - \frac {210 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} - 1\right )} {\left (\cos \left (b x + a\right ) + 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) - 1\right )}^{2}} - \frac {72 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {3 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + \frac {256 \, {\left (\frac {9 \, {\left (\cos \left (b x + a\right ) - 1\right )}}{\cos \left (b x + a\right ) + 1} + \frac {6 \, {\left (\cos \left (b x + a\right ) - 1\right )}^{2}}{{\left (\cos \left (b x + a\right ) + 1\right )}^{2}} + 5\right )}}{{\left (\frac {\cos \left (b x + a\right ) - 1}{\cos \left (b x + a\right ) + 1} + 1\right )}^{3}} + 420 \, \log \left (\frac {{\left | -\cos \left (b x + a\right ) + 1 \right |}}{{\left | \cos \left (b x + a\right ) + 1 \right |}}\right )}{192 \, b} \]
1/192*(3*(24*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) - 210*(cos(b*x + a) - 1 )^2/(cos(b*x + a) + 1)^2 - 1)*(cos(b*x + a) + 1)^2/(cos(b*x + a) - 1)^2 - 72*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 3*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 256*(9*(cos(b*x + a) - 1)/(cos(b*x + a) + 1) + 6*(cos(b*x + a) - 1)^2/(cos(b*x + a) + 1)^2 + 5)/((cos(b*x + a) - 1)/(cos(b*x + a) + 1 ) + 1)^3 + 420*log(abs(-cos(b*x + a) + 1)/abs(cos(b*x + a) + 1)))/b
Time = 0.15 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.88 \[ \int \csc ^5(a+b x) \sec ^4(a+b x) \, dx=\frac {\frac {35\,{\cos \left (a+b\,x\right )}^6}{8}-\frac {175\,{\cos \left (a+b\,x\right )}^4}{24}+\frac {7\,{\cos \left (a+b\,x\right )}^2}{3}+\frac {1}{3}}{b\,\left ({\cos \left (a+b\,x\right )}^7-2\,{\cos \left (a+b\,x\right )}^5+{\cos \left (a+b\,x\right )}^3\right )}-\frac {35\,\mathrm {atanh}\left (\cos \left (a+b\,x\right )\right )}{8\,b} \]