Integrand size = 21, antiderivative size = 66 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^2 d}-\frac {5 \tan (c+d x)}{3 a^2 d (1+\sec (c+d x))}+\frac {\tan (c+d x)}{3 d (a+a \sec (c+d x))^2} \]
arctanh(sin(d*x+c))/a^2/d-5/3*tan(d*x+c)/a^2/d/(1+sec(d*x+c))+1/3*tan(d*x+ c)/d/(a+a*sec(d*x+c))^2
Time = 0.21 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.27 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x) \left (6 \text {arctanh}(\sin (c+d x)) \cos ^3\left (\frac {1}{2} (c+d x)\right )-3 \sin \left (\frac {1}{2} (c+d x)\right )-2 \sin \left (\frac {3}{2} (c+d x)\right )\right )}{3 a^2 d (1+\sec (c+d x))^2} \]
(2*Cos[(c + d*x)/2]*Sec[c + d*x]^2*(6*ArcTanh[Sin[c + d*x]]*Cos[(c + d*x)/ 2]^3 - 3*Sin[(c + d*x)/2] - 2*Sin[(3*(c + d*x))/2]))/(3*a^2*d*(1 + Sec[c + d*x])^2)
Time = 0.51 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4286, 25, 3042, 4486, 3042, 4257, 4281}
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.
\(\displaystyle \int \frac {\sec ^3(c+d x)}{(a \sec (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 4286 |
\(\displaystyle \frac {\int -\frac {\sec (c+d x) (2 a-3 a \sec (c+d x))}{\sec (c+d x) a+a}dx}{3 a^2}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}-\frac {\int \frac {\sec (c+d x) (2 a-3 a \sec (c+d x))}{\sec (c+d x) a+a}dx}{3 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (2 a-3 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}\) |
\(\Big \downarrow \) 4486 |
\(\displaystyle \frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}-\frac {5 a \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx-3 \int \sec (c+d x)dx}{3 a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}-\frac {5 a \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-3 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{3 a^2}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}-\frac {5 a \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {3 \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}\) |
\(\Big \downarrow \) 4281 |
\(\displaystyle \frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}-\frac {\frac {5 a \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {3 \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}\) |
Tan[c + d*x]/(3*d*(a + a*Sec[c + d*x])^2) - ((-3*ArcTanh[Sin[c + d*x]])/d + (5*a*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2)
3.1.54.3.1 Defintions of rubi rules used
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} , x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] - Simp[1/(a^2*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1) *(a*m - b*(2*m + 1)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[ a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Time = 0.29 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\) | \(62\) |
default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\) | \(62\) |
parallelrisch | \(\frac {-\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{6 a^{2} d}\) | \(62\) |
risch | \(-\frac {2 i \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+4\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{2} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}\) | \(89\) |
norman | \(\frac {-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 a d}+\frac {17 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}-\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{6 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{6 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{2} a}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{2} d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}\) | \(135\) |
1/2/d/a^2*(-1/3*tan(1/2*d*x+1/2*c)^3-3*tan(1/2*d*x+1/2*c)-2*ln(tan(1/2*d*x +1/2*c)-1)+2*ln(tan(1/2*d*x+1/2*c)+1))
Time = 0.28 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.73 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {3 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
1/6*(3*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(sin(d*x + c) + 1) - 3*(co s(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(-sin(d*x + c) + 1) - 2*(4*cos(d*x + c) + 5)*sin(d*x + c))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2* d)
\[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\int \frac {\sec ^{3}{\left (c + d x \right )}}{\sec ^{2}{\left (c + d x \right )} + 2 \sec {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.48 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{6 \, d} \]
-1/6*((9*sin(d*x + c)/(cos(d*x + c) + 1) + sin(d*x + c)^3/(cos(d*x + c) + 1)^3)/a^2 - 6*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^2 + 6*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^2)/d
Time = 0.31 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=\frac {\frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
1/6*(6*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^2 - 6*log(abs(tan(1/2*d*x + 1/ 2*c) - 1))/a^2 - (a^4*tan(1/2*d*x + 1/2*c)^3 + 9*a^4*tan(1/2*d*x + 1/2*c)) /a^6)/d
Time = 13.15 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.65 \[ \int \frac {\sec ^3(c+d x)}{(a+a \sec (c+d x))^2} \, dx=-\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-12\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6\,a^2\,d} \]