Integrand size = 12, antiderivative size = 66 \[ \int (a+a \sec (c+d x))^3 \, dx=a^3 x+\frac {7 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {5 a^3 \tan (c+d x)}{2 d}+\frac {\left (a^3+a^3 \sec (c+d x)\right ) \tan (c+d x)}{2 d} \]
a^3*x+7/2*a^3*arctanh(sin(d*x+c))/d+5/2*a^3*tan(d*x+c)/d+1/2*(a^3+a^3*sec( d*x+c))*tan(d*x+c)/d
Time = 0.25 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.89 \[ \int (a+a \sec (c+d x))^3 \, dx=a^3 x+\frac {7 a^3 \text {arctanh}(\sin (c+d x))}{2 d}+\frac {3 a^3 \tan (c+d x)}{d}+\frac {a^3 \sec (c+d x) \tan (c+d x)}{2 d} \]
a^3*x + (7*a^3*ArcTanh[Sin[c + d*x]])/(2*d) + (3*a^3*Tan[c + d*x])/d + (a^ 3*Sec[c + d*x]*Tan[c + d*x])/(2*d)
Time = 0.41 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.05, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.667, Rules used = {3042, 4262, 3042, 4402, 3042, 4254, 24, 4257}
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.
\(\displaystyle \int (a \sec (c+d x)+a)^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3dx\) |
\(\Big \downarrow \) 4262 |
\(\displaystyle \frac {1}{2} a \int (\sec (c+d x) a+a) (5 \sec (c+d x) a+2 a)dx+\frac {\tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} a \int \left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right ) \left (5 \csc \left (c+d x+\frac {\pi }{2}\right ) a+2 a\right )dx+\frac {\tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\) |
\(\Big \downarrow \) 4402 |
\(\displaystyle \frac {1}{2} a \left (5 a^2 \int \sec ^2(c+d x)dx+7 a^2 \int \sec (c+d x)dx+2 a^2 x\right )+\frac {\tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{2} a \left (7 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+5 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )^2dx+2 a^2 x\right )+\frac {\tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\) |
\(\Big \downarrow \) 4254 |
\(\displaystyle \frac {1}{2} a \left (-\frac {5 a^2 \int 1d(-\tan (c+d x))}{d}+7 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+2 a^2 x\right )+\frac {\tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\) |
\(\Big \downarrow \) 24 |
\(\displaystyle \frac {1}{2} a \left (7 a^2 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {5 a^2 \tan (c+d x)}{d}+2 a^2 x\right )+\frac {\tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\tan (c+d x) \left (a^3 \sec (c+d x)+a^3\right )}{2 d}+\frac {1}{2} a \left (\frac {7 a^2 \text {arctanh}(\sin (c+d x))}{d}+\frac {5 a^2 \tan (c+d x)}{d}+2 a^2 x\right )\) |
((a^3 + a^3*Sec[c + d*x])*Tan[c + d*x])/(2*d) + (a*(2*a^2*x + (7*a^2*ArcTa nh[Sin[c + d*x]])/d + (5*a^2*Tan[c + d*x])/d))/2
3.1.23.3.1 Defintions of rubi rules used
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[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Simp[(-b^2)*C ot[c + d*x]*((a + b*Csc[c + d*x])^(n - 2)/(d*(n - 1))), x] + Simp[a/(n - 1) Int[(a + b*Csc[c + d*x])^(n - 2)*(a*(n - 1) + b*(3*n - 4)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && GtQ[n, 1] && Inte gerQ[2*n]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_)), x_Symbol] :> Simp[a*c*x, x] + (Simp[b*d Int[Csc[e + f*x]^2, x], x ] + Simp[(b*c + a*d) Int[Csc[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f }, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
Time = 0.44 (sec) , antiderivative size = 80, normalized size of antiderivative = 1.21
method | result | size |
derivativedivides | \(\frac {a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \tan \left (d x +c \right )+3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} \left (d x +c \right )}{d}\) | \(80\) |
default | \(\frac {a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )+3 a^{3} \tan \left (d x +c \right )+3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )+a^{3} \left (d x +c \right )}{d}\) | \(80\) |
parts | \(a^{3} x +\frac {a^{3} \left (\frac {\sec \left (d x +c \right ) \tan \left (d x +c \right )}{2}+\frac {\ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2}\right )}{d}+\frac {3 a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}+\frac {3 a^{3} \tan \left (d x +c \right )}{d}\) | \(81\) |
risch | \(a^{3} x -\frac {i a^{3} \left ({\mathrm e}^{3 i \left (d x +c \right )}-6 \,{\mathrm e}^{2 i \left (d x +c \right )}-{\mathrm e}^{i \left (d x +c \right )}-6\right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}+\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{2 d}-\frac {7 a^{3} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{2 d}\) | \(104\) |
norman | \(\frac {a^{3} x +a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\frac {7 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{d}-\frac {5 a^{3} \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{d}-2 a^{3} x \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2}}-\frac {7 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{2 d}+\frac {7 a^{3} \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d}\) | \(133\) |
parallelrisch | \(\frac {a^{3} \left (2 d x \cos \left (2 d x +2 c \right )-7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \cos \left (2 d x +2 c \right )+7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \cos \left (2 d x +2 c \right )+2 d x +2 \sin \left (d x +c \right )+6 \sin \left (2 d x +2 c \right )-7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+7 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )\right )}{2 d \left (1+\cos \left (2 d x +2 c \right )\right )}\) | \(133\) |
1/d*(a^3*(1/2*sec(d*x+c)*tan(d*x+c)+1/2*ln(sec(d*x+c)+tan(d*x+c)))+3*a^3*t an(d*x+c)+3*a^3*ln(sec(d*x+c)+tan(d*x+c))+a^3*(d*x+c))
Time = 0.27 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.48 \[ \int (a+a \sec (c+d x))^3 \, dx=\frac {4 \, a^{3} d x \cos \left (d x + c\right )^{2} + 7 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 7 \, a^{3} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (6 \, a^{3} \cos \left (d x + c\right ) + a^{3}\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \]
1/4*(4*a^3*d*x*cos(d*x + c)^2 + 7*a^3*cos(d*x + c)^2*log(sin(d*x + c) + 1) - 7*a^3*cos(d*x + c)^2*log(-sin(d*x + c) + 1) + 2*(6*a^3*cos(d*x + c) + a ^3)*sin(d*x + c))/(d*cos(d*x + c)^2)
\[ \int (a+a \sec (c+d x))^3 \, dx=a^{3} \left (\int 1\, dx + \int 3 \sec {\left (c + d x \right )}\, dx + \int 3 \sec ^{2}{\left (c + d x \right )}\, dx + \int \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]
a**3*(Integral(1, x) + Integral(3*sec(c + d*x), x) + Integral(3*sec(c + d* x)**2, x) + Integral(sec(c + d*x)**3, x))
Time = 0.25 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.38 \[ \int (a+a \sec (c+d x))^3 \, dx=a^{3} x - \frac {a^{3} {\left (\frac {2 \, \sin \left (d x + c\right )}{\sin \left (d x + c\right )^{2} - 1} - \log \left (\sin \left (d x + c\right ) + 1\right ) + \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{4 \, d} + \frac {3 \, a^{3} \log \left (\sec \left (d x + c\right ) + \tan \left (d x + c\right )\right )}{d} + \frac {3 \, a^{3} \tan \left (d x + c\right )}{d} \]
a^3*x - 1/4*a^3*(2*sin(d*x + c)/(sin(d*x + c)^2 - 1) - log(sin(d*x + c) + 1) + log(sin(d*x + c) - 1))/d + 3*a^3*log(sec(d*x + c) + tan(d*x + c))/d + 3*a^3*tan(d*x + c)/d
Time = 0.30 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.52 \[ \int (a+a \sec (c+d x))^3 \, dx=\frac {2 \, {\left (d x + c\right )} a^{3} + 7 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 7 \, a^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (5 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 7 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}}}{2 \, d} \]
1/2*(2*(d*x + c)*a^3 + 7*a^3*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 7*a^3*lo g(abs(tan(1/2*d*x + 1/2*c) - 1)) - 2*(5*a^3*tan(1/2*d*x + 1/2*c)^3 - 7*a^3 *tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^2)/d
Time = 13.47 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.33 \[ \int (a+a \sec (c+d x))^3 \, dx=a^3\,x+\frac {7\,a^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{d}-\frac {5\,a^3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-7\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]