Integrand size = 19, antiderivative size = 67 \[ \int \csc (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {4 a^3 \log (1-\cos (c+d x))}{d}-\frac {4 a^3 \log (\cos (c+d x))}{d}+\frac {3 a^3 \sec (c+d x)}{d}+\frac {a^3 \sec ^2(c+d x)}{2 d} \]
Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 1.21 \[ \int \csc (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {a^3 \left (1+6 \cos (c+d x)-4 \log (\cos (c+d x))-4 \cos (2 (c+d x)) \left (\log (\cos (c+d x))-2 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+8 \log \left (\sin \left (\frac {1}{2} (c+d x)\right )\right )\right ) \sec ^2(c+d x)}{2 d} \]
(a^3*(1 + 6*Cos[c + d*x] - 4*Log[Cos[c + d*x]] - 4*Cos[2*(c + d*x)]*(Log[C os[c + d*x]] - 2*Log[Sin[(c + d*x)/2]]) + 8*Log[Sin[(c + d*x)/2]])*Sec[c + d*x]^2)/(2*d)
Time = 0.39 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.99, number of steps used = 12, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.579, Rules used = {3042, 4360, 25, 25, 3042, 25, 3315, 25, 27, 99, 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.
\(\displaystyle \int \csc (c+d x) (a \sec (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\left (a-a \csc \left (c+d x-\frac {\pi }{2}\right )\right )^3}{\cos \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4360 |
\(\displaystyle \int \csc (c+d x) \sec ^3(c+d x) \left (-(a (-\cos (c+d x))-a)^3\right )dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int -(\cos (c+d x) a+a)^3 \csc (c+d x) \sec ^3(c+d x)dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \csc (c+d x) \sec ^3(c+d x) (a \cos (c+d x)+a)^3dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int -\frac {\left (a-a \sin \left (c+d x-\frac {\pi }{2}\right )\right )^3}{\sin \left (c+d x-\frac {\pi }{2}\right )^3 \cos \left (c+d x-\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\int \frac {\left (a-a \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )\right )^3}{\cos \left (\frac {1}{2} (2 c-\pi )+d x\right ) \sin \left (\frac {1}{2} (2 c-\pi )+d x\right )^3}dx\) |
\(\Big \downarrow \) 3315 |
\(\displaystyle \frac {a \int -\frac {(\cos (c+d x) a+a)^2 \sec ^3(c+d x)}{a-a \cos (c+d x)}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle -\frac {a \int \frac {(\cos (c+d x) a+a)^2 \sec ^3(c+d x)}{a-a \cos (c+d x)}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle -\frac {a^4 \int \frac {(\cos (c+d x) a+a)^2 \sec ^3(c+d x)}{a^3 (a-a \cos (c+d x))}d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 99 |
\(\displaystyle -\frac {a^4 \int \left (\frac {\sec ^3(c+d x)}{a^2}+\frac {3 \sec ^2(c+d x)}{a^2}+\frac {4 \sec (c+d x)}{a^2}+\frac {4}{a (a-a \cos (c+d x))}\right )d(a \cos (c+d x))}{d}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle -\frac {a^4 \left (-\frac {\sec ^2(c+d x)}{2 a}-\frac {3 \sec (c+d x)}{a}+\frac {4 \log (a \cos (c+d x))}{a}-\frac {4 \log (a-a \cos (c+d x))}{a}\right )}{d}\) |
-((a^4*((4*Log[a*Cos[c + d*x]])/a - (4*Log[a - a*Cos[c + d*x]])/a - (3*Sec [c + d*x])/a - Sec[c + d*x]^2/(2*a)))/d)
3.1.43.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] && (IntegerQ[p] | | (GtQ[m, 0] && GeQ[n, -1]))
Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_ .)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[1/(b^p* f) Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && Intege rQ[(p - 1)/2] && EqQ[a^2 - b^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 = 0.49 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.34
method | result | size |
derivativedivides | \(\frac {a^{3} \left (\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )+3 a^{3} \ln \left (\tan \left (d x +c \right )\right )+a^{3} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{d}\) | \(90\) |
default | \(\frac {a^{3} \left (\frac {1}{2 \cos \left (d x +c \right )^{2}}+\ln \left (\tan \left (d x +c \right )\right )\right )+3 a^{3} \left (\frac {1}{\cos \left (d x +c \right )}+\ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )+3 a^{3} \ln \left (\tan \left (d x +c \right )\right )+a^{3} \ln \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )}{d}\) | \(90\) |
risch | \(\frac {2 a^{3} \left (3 \,{\mathrm e}^{3 i \left (d x +c \right )}+{\mathrm e}^{2 i \left (d x +c \right )}+3 \,{\mathrm e}^{i \left (d x +c \right )}\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {8 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-1\right )}{d}-\frac {4 a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(95\) |
norman | \(\frac {\frac {6 a^{3}}{d}-\frac {4 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}+\frac {8 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d}-\frac {4 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d}-\frac {4 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d}\) | \(104\) |
parallelrisch | \(\frac {-8 a^{3} \left (1+\cos \left (2 d x +2 c \right )\right ) \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-a^{3} \left (-16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (2 d x +2 c \right )+8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (2 d x +2 c \right )-12 \cos \left (d x +c \right )-5 \cos \left (2 d x +2 c \right )-16 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )+8 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )-7\right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(144\) |
1/d*(a^3*(1/2/cos(d*x+c)^2+ln(tan(d*x+c)))+3*a^3*(1/cos(d*x+c)+ln(-cot(d*x +c)+csc(d*x+c)))+3*a^3*ln(tan(d*x+c))+a^3*ln(-cot(d*x+c)+csc(d*x+c)))
Time = 0.27 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.13 \[ \int \csc (c+d x) (a+a \sec (c+d x))^3 \, dx=-\frac {8 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\cos \left (d x + c\right )\right ) - 8 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 6 \, a^{3} \cos \left (d x + c\right ) - a^{3}}{2 \, d \cos \left (d x + c\right )^{2}} \]
-1/2*(8*a^3*cos(d*x + c)^2*log(-cos(d*x + c)) - 8*a^3*cos(d*x + c)^2*log(- 1/2*cos(d*x + c) + 1/2) - 6*a^3*cos(d*x + c) - a^3)/(d*cos(d*x + c)^2)
\[ \int \csc (c+d x) (a+a \sec (c+d x))^3 \, dx=a^{3} \left (\int 3 \csc {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 3 \csc {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \csc {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int \csc {\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(3*csc(c + d*x)*sec(c + d*x), x) + Integral(3*csc(c + d*x)*s ec(c + d*x)**2, x) + Integral(csc(c + d*x)*sec(c + d*x)**3, x) + Integral( csc(c + d*x), x))
Time = 0.21 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.84 \[ \int \csc (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {8 \, a^{3} \log \left (\cos \left (d x + c\right ) - 1\right ) - 8 \, a^{3} \log \left (\cos \left (d x + c\right )\right ) + \frac {6 \, a^{3} \cos \left (d x + c\right ) + a^{3}}{\cos \left (d x + c\right )^{2}}}{2 \, d} \]
1/2*(8*a^3*log(cos(d*x + c) - 1) - 8*a^3*log(cos(d*x + c)) + (6*a^3*cos(d* x + c) + a^3)/cos(d*x + c)^2)/d
Leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (65) = 130\).
Time = 0.33 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.12 \[ \int \csc (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {2 \, {\left (2 \, a^{3} \log \left (\frac {{\left | -\cos \left (d x + c\right ) + 1 \right |}}{{\left | \cos \left (d x + c\right ) + 1 \right |}}\right ) - 2 \, a^{3} \log \left ({\left | -\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1 \right |}\right ) + \frac {6 \, a^{3} + \frac {8 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} + \frac {3 \, a^{3} {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}}}{{\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}^{2}}\right )}}{d} \]
2*(2*a^3*log(abs(-cos(d*x + c) + 1)/abs(cos(d*x + c) + 1)) - 2*a^3*log(abs (-(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)) + (6*a^3 + 8*a^3*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 3*a^3*(cos(d*x + c) - 1)^2/(cos(d*x + c) + 1) ^2)/((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)^2)/d
Time = 13.30 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.73 \[ \int \csc (c+d x) (a+a \sec (c+d x))^3 \, dx=\frac {3\,a^3\,\cos \left (c+d\,x\right )+\frac {a^3}{2}}{d\,{\cos \left (c+d\,x\right )}^2}-\frac {8\,a^3\,\mathrm {atanh}\left (2\,\cos \left (c+d\,x\right )-1\right )}{d} \]