Integrand size = 17, antiderivative size = 81 \[ \int \csc ^4(a+b x) \sec ^5(a+b x) \, dx=\frac {35 \text {arctanh}(\sin (a+b x))}{8 b}-\frac {3 \csc (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b}+\frac {13 \sec (a+b x) \tan (a+b x)}{8 b}+\frac {\sec (a+b x) \tan ^3(a+b x)}{4 b} \] Output:
35/8*arctanh(sin(b*x+a))/b-3*csc(b*x+a)/b-1/3*csc(b*x+a)^3/b+13/8*sec(b*x+ a)*tan(b*x+a)/b+1/4*sec(b*x+a)*tan(b*x+a)^3/b
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.02 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.38 \[ \int \csc ^4(a+b x) \sec ^5(a+b x) \, dx=-\frac {\csc ^3(a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},3,-\frac {1}{2},\sin ^2(a+b x)\right )}{3 b} \] Input:
Integrate[Csc[a + b*x]^4*Sec[a + b*x]^5,x]
Output:
-1/3*(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 3, -1/2, Sin[a + b*x]^2])/b
Time = 0.42 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.17, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 3101, 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.
| \(\displaystyle \int \csc ^4(a+b x) \sec ^5(a+b x) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc (a+b x)^4 \sec (a+b x)^5dx\) |
| \(\Big \downarrow \) 3101 |
| \(\displaystyle -\frac {\int -\frac {\csc ^8(a+b x)}{\left (1-\csc ^2(a+b x)\right )^3}d\csc (a+b x)}{b}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {\csc ^8(a+b x)}{\left (1-\csc ^2(a+b x)\right )^3}d\csc (a+b x)}{b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\frac {7}{4} \int \frac {\csc ^6(a+b x)}{\left (1-\csc ^2(a+b x)\right )^2}d\csc (a+b x)-\frac {\csc ^7(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{b}\) |
| \(\Big \downarrow \) 252 |
| \(\displaystyle -\frac {\frac {7}{4} \left (\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\csc ^4(a+b x)}{1-\csc ^2(a+b x)}d\csc (a+b x)\right )-\frac {\csc ^7(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{b}\) |
| \(\Big \downarrow \) 254 |
| \(\displaystyle -\frac {\frac {7}{4} \left (\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\csc ^2(a+b x)+\frac {1}{1-\csc ^2(a+b x)}-1\right )d\csc (a+b x)\right )-\frac {\csc ^7(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{b}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {\frac {7}{4} \left (\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\csc (a+b x))-\frac {1}{3} \csc ^3(a+b x)-\csc (a+b x)\right )\right )-\frac {\csc ^7(a+b x)}{4 \left (1-\csc ^2(a+b x)\right )^2}}{b}\) |
Input:
Int[Csc[a + b*x]^4*Sec[a + b*x]^5,x]
Output:
-((-1/4*Csc[a + b*x]^7/(1 - Csc[a + b*x]^2)^2 + (7*(Csc[a + b*x]^5/(2*(1 - Csc[a + b*x]^2)) - (5*(ArcTanh[Csc[a + b*x]] - Csc[a + b*x] - Csc[a + b*x ]^3/3))/2))/4)/b)
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_)]*(a_.))^(m_)*sec[(e_.) + (f_.)*(x_)]^(n_.), x_S ymbol] :> Simp[-(f*a^n)^(-1) Subst[Int[x^(m + n - 1)/(-1 + x^2/a^2)^((n + 1)/2), x], x, a*Csc[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2] && !(IntegerQ[(m + 1)/2] && LtQ[0, m, n])
Time = 2.49 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.06
| method | result | size |
| derivativedivides | \(\frac {\frac {1}{4 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{4}}-\frac {7}{12 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{2}}+\frac {35}{24 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {35}{8 \sin \left (b x +a \right )}+\frac {35 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\) | \(86\) |
| default | \(\frac {\frac {1}{4 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{4}}-\frac {7}{12 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{2}}+\frac {35}{24 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {35}{8 \sin \left (b x +a \right )}+\frac {35 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{8}}{b}\) | \(86\) |
| risch | \(-\frac {i \left (105 \,{\mathrm e}^{13 i \left (b x +a \right )}+70 \,{\mathrm e}^{11 i \left (b x +a \right )}-329 \,{\mathrm e}^{9 i \left (b x +a \right )}-204 \,{\mathrm e}^{7 i \left (b x +a \right )}-329 \,{\mathrm e}^{5 i \left (b x +a \right )}+70 \,{\mathrm e}^{3 i \left (b x +a \right )}+105 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{12 b \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{4} \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3}}-\frac {35 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{8 b}+\frac {35 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{8 b}\) | \(148\) |
| norman | \(\frac {-\frac {1}{24 b}-\frac {35 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{24 b}+\frac {63 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}{8 b}-\frac {35 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}}{8 b}-\frac {35 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{8}}{8 b}+\frac {63 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{10}}{8 b}-\frac {35 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{12}}{24 b}-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{14}}{24 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )^{4}}-\frac {35 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{8 b}+\frac {35 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{8 b}\) | \(181\) |
| parallelrisch | \(\frac {35 \left (\left (\sin \left (7 b x +7 a \right )-3 \sin \left (b x +a \right )-3 \sin \left (3 b x +3 a \right )+\sin \left (5 b x +5 a \right )\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+\left (3 \sin \left (b x +a \right )+3 \sin \left (3 b x +3 a \right )-\sin \left (5 b x +5 a \right )-\sin \left (7 b x +7 a \right )\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )-\frac {94 \cos \left (2 b x +2 a \right )}{15}+\frac {4 \cos \left (4 b x +4 a \right )}{3}+2 \cos \left (6 b x +6 a \right )-\frac {68}{35}\right ) \csc \left (\frac {b x}{2}+\frac {a}{2}\right )^{3} \sec \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}}{512 b \left (\cos \left (4 b x +4 a \right )+4 \cos \left (2 b x +2 a \right )+3\right )}\) | \(193\) |
Input:
int(csc(b*x+a)^4*sec(b*x+a)^5,x,method=_RETURNVERBOSE)
Output:
1/b*(1/4/sin(b*x+a)^3/cos(b*x+a)^4-7/12/sin(b*x+a)^3/cos(b*x+a)^2+35/24/si n(b*x+a)/cos(b*x+a)^2-35/8/sin(b*x+a)+35/8*ln(sec(b*x+a)+tan(b*x+a)))
Time = 0.10 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.73 \[ \int \csc ^4(a+b x) \sec ^5(a+b x) \, dx=-\frac {210 \, \cos \left (b x + a\right )^{6} - 280 \, \cos \left (b x + a\right )^{4} - 105 \, {\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 105 \, {\left (\cos \left (b x + a\right )^{6} - \cos \left (b x + a\right )^{4}\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 42 \, \cos \left (b x + a\right )^{2} + 12}{48 \, {\left (b \cos \left (b x + a\right )^{6} - b \cos \left (b x + a\right )^{4}\right )} \sin \left (b x + a\right )} \] Input:
integrate(csc(b*x+a)^4*sec(b*x+a)^5,x, algorithm="fricas")
Output:
-1/48*(210*cos(b*x + a)^6 - 280*cos(b*x + a)^4 - 105*(cos(b*x + a)^6 - cos (b*x + a)^4)*log(sin(b*x + a) + 1)*sin(b*x + a) + 105*(cos(b*x + a)^6 - co s(b*x + a)^4)*log(-sin(b*x + a) + 1)*sin(b*x + a) + 42*cos(b*x + a)^2 + 12 )/((b*cos(b*x + a)^6 - b*cos(b*x + a)^4)*sin(b*x + a))
Timed out. \[ \int \csc ^4(a+b x) \sec ^5(a+b x) \, dx=\text {Timed out} \] Input:
integrate(csc(b*x+a)**4*sec(b*x+a)**5,x)
Output:
Timed out
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.12 \[ \int \csc ^4(a+b x) \sec ^5(a+b x) \, dx=-\frac {\frac {2 \, {\left (105 \, \sin \left (b x + a\right )^{6} - 175 \, \sin \left (b x + a\right )^{4} + 56 \, \sin \left (b x + a\right )^{2} + 8\right )}}{\sin \left (b x + a\right )^{7} - 2 \, \sin \left (b x + a\right )^{5} + \sin \left (b x + a\right )^{3}} - 105 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 105 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{48 \, b} \] Input:
integrate(csc(b*x+a)^4*sec(b*x+a)^5,x, algorithm="maxima")
Output:
-1/48*(2*(105*sin(b*x + a)^6 - 175*sin(b*x + a)^4 + 56*sin(b*x + a)^2 + 8) /(sin(b*x + a)^7 - 2*sin(b*x + a)^5 + sin(b*x + a)^3) - 105*log(sin(b*x + a) + 1) + 105*log(sin(b*x + a) - 1))/b
Time = 0.13 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.14 \[ \int \csc ^4(a+b x) \sec ^5(a+b x) \, dx=\frac {35 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right )}{16 \, b} - \frac {35 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{16 \, b} - \frac {11 \, \sin \left (b x + a\right )^{3} - 13 \, \sin \left (b x + a\right )}{8 \, {\left (\sin \left (b x + a\right )^{2} - 1\right )}^{2} b} - \frac {9 \, \sin \left (b x + a\right )^{2} + 1}{3 \, b \sin \left (b x + a\right )^{3}} \] Input:
integrate(csc(b*x+a)^4*sec(b*x+a)^5,x, algorithm="giac")
Output:
35/16*log(abs(sin(b*x + a) + 1))/b - 35/16*log(abs(sin(b*x + a) - 1))/b - 1/8*(11*sin(b*x + a)^3 - 13*sin(b*x + a))/((sin(b*x + a)^2 - 1)^2*b) - 1/3 *(9*sin(b*x + a)^2 + 1)/(b*sin(b*x + a)^3)
Time = 25.51 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.98 \[ \int \csc ^4(a+b x) \sec ^5(a+b x) \, dx=\frac {35\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{8\,b}-\frac {\frac {35\,{\sin \left (a+b\,x\right )}^6}{8}-\frac {175\,{\sin \left (a+b\,x\right )}^4}{24}+\frac {7\,{\sin \left (a+b\,x\right )}^2}{3}+\frac {1}{3}}{b\,\left ({\sin \left (a+b\,x\right )}^7-2\,{\sin \left (a+b\,x\right )}^5+{\sin \left (a+b\,x\right )}^3\right )} \] Input:
int(1/(cos(a + b*x)^5*sin(a + b*x)^4),x)
Output:
(35*atanh(sin(a + b*x)))/(8*b) - ((7*sin(a + b*x)^2)/3 - (175*sin(a + b*x) ^4)/24 + (35*sin(a + b*x)^6)/8 + 1/3)/(b*(sin(a + b*x)^3 - 2*sin(a + b*x)^ 5 + sin(a + b*x)^7))
Time = 0.17 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.46 \[ \int \csc ^4(a+b x) \sec ^5(a+b x) \, dx=\frac {-105 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{7}+210 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{5}-105 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{3}+105 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{7}-210 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{5}+105 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{3}-105 \sin \left (b x +a \right )^{6}+175 \sin \left (b x +a \right )^{4}-56 \sin \left (b x +a \right )^{2}-8}{24 \sin \left (b x +a \right )^{3} b \left (\sin \left (b x +a \right )^{4}-2 \sin \left (b x +a \right )^{2}+1\right )} \] Input:
int(csc(b*x+a)^4*sec(b*x+a)^5,x)
Output:
( - 105*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**7 + 210*log(tan((a + b*x)/ 2) - 1)*sin(a + b*x)**5 - 105*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**3 + 105*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**7 - 210*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**5 + 105*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**3 - 105* sin(a + b*x)**6 + 175*sin(a + b*x)**4 - 56*sin(a + b*x)**2 - 8)/(24*sin(a + b*x)**3*b*(sin(a + b*x)**4 - 2*sin(a + b*x)**2 + 1))