Integrand size = 17, antiderivative size = 60 \[ \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx=\frac {5 \text {arctanh}(\sin (a+b x))}{2 b}-\frac {2 \csc (a+b x)}{b}-\frac {\csc ^3(a+b x)}{3 b}+\frac {\sec (a+b x) \tan (a+b x)}{2 b} \] Output:
5/2*arctanh(sin(b*x+a))/b-2*csc(b*x+a)/b-1/3*csc(b*x+a)^3/b+1/2*sec(b*x+a) *tan(b*x+a)/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 = 0.52 \[ \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx=-\frac {\csc ^3(a+b x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},2,-\frac {1}{2},\sin ^2(a+b x)\right )}{3 b} \] Input:
Integrate[Csc[a + b*x]^4*Sec[a + b*x]^3,x]
Output:
-1/3*(Csc[a + b*x]^3*Hypergeometric2F1[-3/2, 2, -1/2, Sin[a + b*x]^2])/b
Time = 0.40 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {3042, 3101, 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.
\(\displaystyle \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \csc (a+b x)^4 \sec (a+b x)^3dx\) |
\(\Big \downarrow \) 3101 |
\(\displaystyle -\frac {\int \frac {\csc ^6(a+b x)}{\left (1-\csc ^2(a+b x)\right )^2}d\csc (a+b x)}{b}\) |
\(\Big \downarrow \) 252 |
\(\displaystyle -\frac {\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \int \frac {\csc ^4(a+b x)}{1-\csc ^2(a+b x)}d\csc (a+b x)}{b}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle -\frac {\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \int \left (-\csc ^2(a+b x)+\frac {1}{1-\csc ^2(a+b x)}-1\right )d\csc (a+b x)}{b}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {\frac {\csc ^5(a+b x)}{2 \left (1-\csc ^2(a+b x)\right )}-\frac {5}{2} \left (\text {arctanh}(\csc (a+b x))-\frac {1}{3} \csc ^3(a+b x)-\csc (a+b x)\right )}{b}\) |
Input:
Int[Csc[a + b*x]^4*Sec[a + b*x]^3,x]
Output:
-((Csc[a + b*x]^5/(2*(1 - Csc[a + b*x]^2)) - (5*(ArcTanh[Csc[a + b*x]] - C sc[a + b*x] - Csc[a + b*x]^3/3))/2)/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 = 2.17 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.13
method | result | size |
derivativedivides | \(\frac {-\frac {1}{3 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{2}}+\frac {5}{6 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {5}{2 \sin \left (b x +a \right )}+\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\) | \(68\) |
default | \(\frac {-\frac {1}{3 \sin \left (b x +a \right )^{3} \cos \left (b x +a \right )^{2}}+\frac {5}{6 \sin \left (b x +a \right ) \cos \left (b x +a \right )^{2}}-\frac {5}{2 \sin \left (b x +a \right )}+\frac {5 \ln \left (\sec \left (b x +a \right )+\tan \left (b x +a \right )\right )}{2}}{b}\) | \(68\) |
risch | \(-\frac {i \left (15 \,{\mathrm e}^{9 i \left (b x +a \right )}-20 \,{\mathrm e}^{7 i \left (b x +a \right )}-22 \,{\mathrm e}^{5 i \left (b x +a \right )}-20 \,{\mathrm e}^{3 i \left (b x +a \right )}+15 \,{\mathrm e}^{i \left (b x +a \right )}\right )}{3 b \left ({\mathrm e}^{2 i \left (b x +a \right )}-1\right )^{3} \left ({\mathrm e}^{2 i \left (b x +a \right )}+1\right )^{2}}-\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}-i\right )}{2 b}+\frac {5 \ln \left ({\mathrm e}^{i \left (b x +a \right )}+i\right )}{2 b}\) | \(126\) |
norman | \(\frac {-\frac {1}{24 b}-\frac {25 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}}{24 b}+\frac {25 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{4}}{12 b}+\frac {25 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{6}}{12 b}-\frac {25 \tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{8}}{24 b}-\frac {\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{10}}{24 b}}{\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{3} \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-1\right )^{2}}-\frac {5 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )}{2 b}+\frac {5 \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )}{2 b}\) | \(149\) |
parallelrisch | \(\frac {30 \left (-\cos \left (2 b x +2 a \right )-1\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right )+30 \left (1+\cos \left (2 b x +2 a \right )\right ) \ln \left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right )+60 \left (-3+\cos \left (b x +a \right )\right ) \cot \left (\frac {b x}{2}+\frac {a}{2}\right )^{3}+140 \cot \left (\frac {b x}{2}+\frac {a}{2}\right ) \csc \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}-\sec \left (\frac {b x}{2}+\frac {a}{2}\right ) \csc \left (\frac {b x}{2}+\frac {a}{2}\right )^{3} \left (\sec \left (\frac {b x}{2}+\frac {a}{2}\right )^{2}+20\right )}{12 b \left (1+\cos \left (2 b x +2 a \right )\right )}\) | \(150\) |
Input:
int(csc(b*x+a)^4*sec(b*x+a)^3,x,method=_RETURNVERBOSE)
Output:
1/b*(-1/3/sin(b*x+a)^3/cos(b*x+a)^2+5/6/sin(b*x+a)/cos(b*x+a)^2-5/2/sin(b* x+a)+5/2*ln(sec(b*x+a)+tan(b*x+a)))
Leaf count of result is larger than twice the leaf count of optimal. 130 vs. \(2 (54) = 108\).
Time = 0.11 (sec) , antiderivative size = 130, normalized size of antiderivative = 2.17 \[ \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx=-\frac {30 \, \cos \left (b x + a\right )^{4} - 15 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) + 15 \, {\left (\cos \left (b x + a\right )^{4} - \cos \left (b x + a\right )^{2}\right )} \log \left (-\sin \left (b x + a\right ) + 1\right ) \sin \left (b x + a\right ) - 40 \, \cos \left (b x + a\right )^{2} + 6}{12 \, {\left (b \cos \left (b x + a\right )^{4} - b \cos \left (b x + a\right )^{2}\right )} \sin \left (b x + a\right )} \] Input:
integrate(csc(b*x+a)^4*sec(b*x+a)^3,x, algorithm="fricas")
Output:
-1/12*(30*cos(b*x + a)^4 - 15*(cos(b*x + a)^4 - cos(b*x + a)^2)*log(sin(b* x + a) + 1)*sin(b*x + a) + 15*(cos(b*x + a)^4 - cos(b*x + a)^2)*log(-sin(b *x + a) + 1)*sin(b*x + a) - 40*cos(b*x + a)^2 + 6)/((b*cos(b*x + a)^4 - b* cos(b*x + a)^2)*sin(b*x + a))
\[ \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx=\int \csc ^{4}{\left (a + b x \right )} \sec ^{3}{\left (a + b x \right )}\, dx \] Input:
integrate(csc(b*x+a)**4*sec(b*x+a)**3,x)
Output:
Integral(csc(a + b*x)**4*sec(a + b*x)**3, x)
Time = 0.03 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.22 \[ \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx=-\frac {\frac {2 \, {\left (15 \, \sin \left (b x + a\right )^{4} - 10 \, \sin \left (b x + a\right )^{2} - 2\right )}}{\sin \left (b x + a\right )^{5} - \sin \left (b x + a\right )^{3}} - 15 \, \log \left (\sin \left (b x + a\right ) + 1\right ) + 15 \, \log \left (\sin \left (b x + a\right ) - 1\right )}{12 \, b} \] Input:
integrate(csc(b*x+a)^4*sec(b*x+a)^3,x, algorithm="maxima")
Output:
-1/12*(2*(15*sin(b*x + a)^4 - 10*sin(b*x + a)^2 - 2)/(sin(b*x + a)^5 - sin (b*x + a)^3) - 15*log(sin(b*x + a) + 1) + 15*log(sin(b*x + a) - 1))/b
Time = 0.12 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.32 \[ \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx=\frac {5 \, \log \left ({\left | \sin \left (b x + a\right ) + 1 \right |}\right )}{4 \, b} - \frac {5 \, \log \left ({\left | \sin \left (b x + a\right ) - 1 \right |}\right )}{4 \, b} - \frac {\sin \left (b x + a\right )}{2 \, {\left (\sin \left (b x + a\right )^{2} - 1\right )} b} - \frac {6 \, \sin \left (b x + a\right )^{2} + 1}{3 \, b \sin \left (b x + a\right )^{3}} \] Input:
integrate(csc(b*x+a)^4*sec(b*x+a)^3,x, algorithm="giac")
Output:
5/4*log(abs(sin(b*x + a) + 1))/b - 5/4*log(abs(sin(b*x + a) - 1))/b - 1/2* sin(b*x + a)/((sin(b*x + a)^2 - 1)*b) - 1/3*(6*sin(b*x + a)^2 + 1)/(b*sin( b*x + a)^3)
Time = 0.05 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.02 \[ \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx=\frac {5\,\mathrm {atanh}\left (\sin \left (a+b\,x\right )\right )}{2\,b}-\frac {-\frac {5\,{\sin \left (a+b\,x\right )}^4}{2}+\frac {5\,{\sin \left (a+b\,x\right )}^2}{3}+\frac {1}{3}}{b\,\left ({\sin \left (a+b\,x\right )}^3-{\sin \left (a+b\,x\right )}^5\right )} \] Input:
int(1/(cos(a + b*x)^3*sin(a + b*x)^4),x)
Output:
(5*atanh(sin(a + b*x)))/(2*b) - ((5*sin(a + b*x)^2)/3 - (5*sin(a + b*x)^4) /2 + 1/3)/(b*(sin(a + b*x)^3 - sin(a + b*x)^5))
Time = 0.17 (sec) , antiderivative size = 135, normalized size of antiderivative = 2.25 \[ \int \csc ^4(a+b x) \sec ^3(a+b x) \, dx=\frac {-15 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{5}+15 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )-1\right ) \sin \left (b x +a \right )^{3}+15 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{5}-15 \,\mathrm {log}\left (\tan \left (\frac {b x}{2}+\frac {a}{2}\right )+1\right ) \sin \left (b x +a \right )^{3}-15 \sin \left (b x +a \right )^{4}+10 \sin \left (b x +a \right )^{2}+2}{6 \sin \left (b x +a \right )^{3} b \left (\sin \left (b x +a \right )^{2}-1\right )} \] Input:
int(csc(b*x+a)^4*sec(b*x+a)^3,x)
Output:
( - 15*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**5 + 15*log(tan((a + b*x)/2) - 1)*sin(a + b*x)**3 + 15*log(tan((a + b*x)/2) + 1)*sin(a + b*x)**5 - 15* log(tan((a + b*x)/2) + 1)*sin(a + b*x)**3 - 15*sin(a + b*x)**4 + 10*sin(a + b*x)**2 + 2)/(6*sin(a + b*x)**3*b*(sin(a + b*x)**2 - 1))