Integrand size = 19, antiderivative size = 69 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \text {arctanh}(\sin (c+d x))}{d}-\frac {a \cot (c+d x)}{d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {a \csc (c+d x)}{d}-\frac {a \csc ^3(c+d x)}{3 d} \] Output:
a*arctanh(sin(d*x+c))/d-a*cot(d*x+c)/d-1/3*a*cot(d*x+c)^3/d-a*csc(d*x+c)/d -1/3*a*csc(d*x+c)^3/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.05 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {2 a \cot (c+d x)}{3 d}-\frac {a \cot (c+d x) \csc ^2(c+d x)}{3 d}-\frac {a \csc ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\sin ^2(c+d x)\right )}{3 d} \] Input:
Integrate[Csc[c + d*x]^4*(a + a*Sec[c + d*x]),x]
Output:
(-2*a*Cot[c + d*x])/(3*d) - (a*Cot[c + d*x]*Csc[c + d*x]^2)/(3*d) - (a*Csc [c + d*x]^3*Hypergeometric2F1[-3/2, 1, -1/2, Sin[c + d*x]^2])/(3*d)
Time = 0.47 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.87, number of steps used = 16, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.789, Rules used = {3042, 4360, 25, 25, 3042, 25, 3317, 25, 3042, 3101, 25, 254, 2009, 4254, 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.
\(\displaystyle \int \csc ^4(c+d x) (a \sec (c+d x)+a) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a-a \csc \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\left (\csc ^4(c+d x) \sec (c+d x) (a (-\cos (c+d x))-a)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\left ((\cos (c+d x) a+a) \csc ^4(c+d x) \sec (c+d x)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \csc ^4(c+d x) \sec (c+d x) (a \cos (c+d x)+a)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {a-a \sin \left (c+d x-\frac {\pi }{2}\right )}{\sin \left (c+d x-\frac {\pi }{2}\right ) \cos \left (c+d x-\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int \csc ^4(c+d x)dx-a \int -\csc ^4(c+d x) \sec (c+d x)dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle a \int \csc ^4(c+d x)dx+a \int \csc ^4(c+d x) \sec (c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \csc (c+d x)^4dx+a \int \csc (c+d x)^4 \sec (c+d x)dx\) |
\(\Big \downarrow \) 3101 |
\(\displaystyle a \int \csc (c+d x)^4dx-\frac {a \int -\frac {\csc ^4(c+d x)}{1-\csc ^2(c+d x)}d\csc (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {a \int \frac {\csc ^4(c+d x)}{1-\csc ^2(c+d x)}d\csc (c+d x)}{d}+a \int \csc (c+d x)^4dx\) |
\(\Big \downarrow \) 254 |
\(\displaystyle \frac {a \int \left (-\csc ^2(c+d x)+\frac {1}{1-\csc ^2(c+d x)}-1\right )d\csc (c+d x)}{d}+a \int \csc (c+d x)^4dx\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a \int \csc (c+d x)^4dx-\frac {a \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle -\frac {a \int \left (\cot ^2(c+d x)+1\right )d\cot (c+d x)}{d}-\frac {a \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}-\frac {a \left (\frac {1}{3} \cot ^3(c+d x)+\cot (c+d x)\right )}{d}\) |
Input:
Int[Csc[c + d*x]^4*(a + a*Sec[c + d*x]),x]
Output:
-((a*(Cot[c + d*x] + Cot[c + d*x]^3/3))/d) - (a*(-ArcTanh[Csc[c + d*x]] + Csc[c + d*x] + Csc[c + d*x]^3/3))/d
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])
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n _.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[a Int[(g*Co s[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Simp[b/d Int[(g*Cos[e + f*x])^ p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Cos[e + f*x])^p*((b + a*Sin[e + f*x])^m/Si n[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]
Time = 0.18 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )}{d}\) | \(63\) |
default | \(\frac {a \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )+a \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )}{d}\) | \(63\) |
parallelrisch | \(-\frac {a \left (\cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+12 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+12 \cot \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{12 d}\) | \(69\) |
norman | \(\frac {-\frac {a}{12 d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}-\frac {a \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{4 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {a \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(90\) |
risch | \(-\frac {2 i a \left (3 \,{\mathrm e}^{3 i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}+2\right )}{3 d \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}+\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(107\) |
Input:
int(csc(d*x+c)^4*(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(a*(-1/3/sin(d*x+c)^3-1/sin(d*x+c)+ln(sec(d*x+c)+tan(d*x+c)))+a*(-2/3- 1/3*csc(d*x+c)^2)*cot(d*x+c))
Time = 0.09 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.57 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {4 \, a \cos \left (d x + c\right )^{2} - 3 \, {\left (a \cos \left (d x + c\right ) - a\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, {\left (a \cos \left (d x + c\right ) - a\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 2 \, a \cos \left (d x + c\right ) - 8 \, a}{6 \, {\left (d \cos \left (d x + c\right ) - d\right )} \sin \left (d x + c\right )} \] Input:
integrate(csc(d*x+c)^4*(a+a*sec(d*x+c)),x, algorithm="fricas")
Output:
-1/6*(4*a*cos(d*x + c)^2 - 3*(a*cos(d*x + c) - a)*log(sin(d*x + c) + 1)*si n(d*x + c) + 3*(a*cos(d*x + c) - a)*log(-sin(d*x + c) + 1)*sin(d*x + c) + 2*a*cos(d*x + c) - 8*a)/((d*cos(d*x + c) - d)*sin(d*x + c))
\[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=a \left (\int \csc ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \csc ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate(csc(d*x+c)**4*(a+a*sec(d*x+c)),x)
Output:
a*(Integral(csc(c + d*x)**4*sec(c + d*x), x) + Integral(csc(c + d*x)**4, x ))
Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=-\frac {a {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, {\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a}{\tan \left (d x + c\right )^{3}}}{6 \, d} \] Input:
integrate(csc(d*x+c)^4*(a+a*sec(d*x+c)),x, algorithm="maxima")
Output:
-1/6*(a*(2*(3*sin(d*x + c)^2 + 1)/sin(d*x + c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)) + 2*(3*tan(d*x + c)^2 + 1)*a/tan(d*x + c)^3)/d
Time = 0.19 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.14 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=\frac {12 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 12 \, a \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {12 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{12 \, d} \] Input:
integrate(csc(d*x+c)^4*(a+a*sec(d*x+c)),x, algorithm="giac")
Output:
1/12*(12*a*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 12*a*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - 3*a*tan(1/2*d*x + 1/2*c) - (12*a*tan(1/2*d*x + 1/2*c)^2 + a)/tan(1/2*d*x + 1/2*c)^3)/d
Time = 10.24 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.94 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=\frac {2\,a\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a}{12}}{d\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}-\frac {a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4\,d} \] Input:
int((a + a/cos(c + d*x))/sin(c + d*x)^4,x)
Output:
(2*a*atanh(tan(c/2 + (d*x)/2)))/d - (a/12 + a*tan(c/2 + (d*x)/2)^2)/(d*tan (c/2 + (d*x)/2)^3) - (a*tan(c/2 + (d*x)/2))/(4*d)
Time = 0.15 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.38 \[ \int \csc ^4(c+d x) (a+a \sec (c+d x)) \, dx=\frac {a \left (-12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+12 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}{12 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3} d} \] Input:
int(csc(d*x+c)^4*(a+a*sec(d*x+c)),x)
Output:
(a*( - 12*log(tan((c + d*x)/2) - 1)*tan((c + d*x)/2)**3 + 12*log(tan((c + d*x)/2) + 1)*tan((c + d*x)/2)**3 - 3*tan((c + d*x)/2)**4 - 12*tan((c + d*x )/2)**2 - 1))/(12*tan((c + d*x)/2)**3*d)