Integrand size = 19, antiderivative size = 69 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {b \text {arctanh}(\sin (c+d x))}{d}-\frac {a \cot (c+d x)}{d}-\frac {a \cot ^3(c+d x)}{3 d}-\frac {b \csc (c+d x)}{d}-\frac {b \csc ^3(c+d x)}{3 d} \]
b*arctanh(sin(d*x+c))/d-a*cot(d*x+c)/d-1/3*a*cot(d*x+c)^3/d-b*csc(d*x+c)/d -1/3*b*csc(d*x+c)^3/d
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.04 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {2 a \cot (c+d x)}{3 d}-\frac {a \cot (c+d x) \csc ^2(c+d x)}{3 d}-\frac {b \csc ^3(c+d x) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\sin ^2(c+d x)\right )}{3 d} \]
(-2*a*Cot[c + d*x])/(3*d) - (a*Cot[c + d*x]*Csc[c + d*x]^2)/(3*d) - (b*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.
\(\displaystyle \int \csc ^4(c+d x) (a+b \sec (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a-b \csc \left (c+d x-\frac {\pi }{2}\right )}{\cos \left (c+d x-\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int -\left (\csc ^4(c+d x) \sec (c+d x) (-a \cos (c+d x)-b)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -\left ((b+a \cos (c+d x)) \csc ^4(c+d x) \sec (c+d x)\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \csc ^4(c+d x) \sec (c+d x) (a \cos (c+d x)+b)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {b-a \sin \left (c+d x-\frac {\pi }{2}\right )}{\sin \left (c+d x-\frac {\pi }{2}\right ) \cos \left (c+d x-\frac {\pi }{2}\right )^4}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {b-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right )^4 \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )}dx\) |
\(\Big \downarrow \) 3317 |
\(\displaystyle a \int \csc ^4(c+d x)dx-b \int -\csc ^4(c+d x) \sec (c+d x)dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle a \int \csc ^4(c+d x)dx+b \int \csc ^4(c+d x) \sec (c+d x)dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \csc (c+d x)^4dx+b \int \csc (c+d x)^4 \sec (c+d x)dx\) |
\(\Big \downarrow \) 3101 |
\(\displaystyle a \int \csc (c+d x)^4dx-\frac {b \int -\frac {\csc ^4(c+d x)}{1-\csc ^2(c+d x)}d\csc (c+d x)}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle a \int \csc (c+d x)^4dx+\frac {b \int \frac {\csc ^4(c+d x)}{1-\csc ^2(c+d x)}d\csc (c+d x)}{d}\) |
\(\Big \downarrow \) 254 |
\(\displaystyle a \int \csc (c+d x)^4dx+\frac {b \int \left (-\csc ^2(c+d x)+\frac {1}{1-\csc ^2(c+d x)}-1\right )d\csc (c+d x)}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle a \int \csc (c+d x)^4dx-\frac {b \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle -\frac {a \int \left (\cot ^2(c+d x)+1\right )d\cot (c+d x)}{d}-\frac {b \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a \left (\frac {1}{3} \cot ^3(c+d x)+\cot (c+d x)\right )}{d}-\frac {b \left (-\text {arctanh}(\csc (c+d x))+\frac {1}{3} \csc ^3(c+d x)+\csc (c+d x)\right )}{d}\) |
-((a*(Cot[c + d*x] + Cot[c + d*x]^3/3))/d) - (b*(-ArcTanh[Csc[c + d*x]] + Csc[c + d*x] + Csc[c + d*x]^3/3))/d
3.2.72.3.1 Defintions of rubi rules used
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 = 0.97 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91
method | result | size |
derivativedivides | \(\frac {a \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )+b \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(63\) |
default | \(\frac {a \left (-\frac {2}{3}-\frac {\csc \left (d x +c \right )^{2}}{3}\right ) \cot \left (d x +c \right )+b \left (-\frac {1}{3 \sin \left (d x +c \right )^{3}}-\frac {1}{\sin \left (d x +c \right )}+\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )\right )}{d}\) | \(63\) |
parallelrisch | \(\frac {-24 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+24 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+\left (-a -b \right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+\left (-9 a -15 b \right ) \cot \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (a -b \right )+9 a -15 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{24 d}\) | \(107\) |
risch | \(-\frac {2 i \left (3 b \,{\mathrm e}^{5 i \left (d x +c \right )}-10 b \,{\mathrm e}^{3 i \left (d x +c \right )}-6 a \,{\mathrm e}^{2 i \left (d x +c \right )}+3 b \,{\mathrm e}^{i \left (d x +c \right )}+2 a \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}-1\right )^{3}}+\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d}-\frac {b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d}\) | \(110\) |
norman | \(\frac {-\frac {a +b}{24 d}+\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{6}}{24 d}+\frac {\left (3 a -5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}}{8 d}-\frac {\left (3 a +5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{8 d}}{\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}+\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}-\frac {b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}\) | \(125\) |
1/d*(a*(-2/3-1/3*csc(d*x+c)^2)*cot(d*x+c)+b*(-1/3/sin(d*x+c)^3-1/sin(d*x+c )+ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.29 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.81 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {4 \, a \cos \left (d x + c\right )^{3} + 6 \, b \cos \left (d x + c\right )^{2} - 3 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) + 3 \, {\left (b \cos \left (d x + c\right )^{2} - b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) \sin \left (d x + c\right ) - 6 \, a \cos \left (d x + c\right ) - 8 \, b}{6 \, {\left (d \cos \left (d x + c\right )^{2} - d\right )} \sin \left (d x + c\right )} \]
-1/6*(4*a*cos(d*x + c)^3 + 6*b*cos(d*x + c)^2 - 3*(b*cos(d*x + c)^2 - b)*l og(sin(d*x + c) + 1)*sin(d*x + c) + 3*(b*cos(d*x + c)^2 - b)*log(-sin(d*x + c) + 1)*sin(d*x + c) - 6*a*cos(d*x + c) - 8*b)/((d*cos(d*x + c)^2 - d)*s in(d*x + c))
\[ \int \csc ^4(c+d x) (a+b \sec (c+d x)) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \csc ^{4}{\left (c + d x \right )}\, dx \]
Time = 0.21 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.10 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x)) \, dx=-\frac {b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{2} + 1\right )}}{\sin \left (d x + c\right )^{3}} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )} + \frac {2 \, {\left (3 \, \tan \left (d x + c\right )^{2} + 1\right )} a}{\tan \left (d x + c\right )^{3}}}{6 \, d} \]
-1/6*(b*(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
Leaf count of result is larger than twice the leaf count of optimal. 133 vs. \(2 (65) = 130\).
Time = 0.31 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.93 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 24 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 24 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + 9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \frac {9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 15 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a + b}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3}}}{24 \, d} \]
1/24*(a*tan(1/2*d*x + 1/2*c)^3 - b*tan(1/2*d*x + 1/2*c)^3 + 24*b*log(abs(t an(1/2*d*x + 1/2*c) + 1)) - 24*b*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 9*a* tan(1/2*d*x + 1/2*c) - 15*b*tan(1/2*d*x + 1/2*c) - (9*a*tan(1/2*d*x + 1/2* c)^2 + 15*b*tan(1/2*d*x + 1/2*c)^2 + a + b)/tan(1/2*d*x + 1/2*c)^3)/d
Time = 14.85 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.46 \[ \int \csc ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a}{24}-\frac {b}{24}\right )}{d}-\frac {{\mathrm {cot}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\left (3\,a+5\,b\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\frac {a}{3}+\frac {b}{3}\right )}{8\,d}+\frac {2\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {3\,a}{8}-\frac {5\,b}{8}\right )}{d} \]