Integrand size = 19, antiderivative size = 66 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^2 d}-\frac {4 \sin (c+d x)}{3 a^2 d (1+\cos (c+d x))}-\frac {\sin (c+d x)}{3 d (a+a \cos (c+d x))^2} \]
arctanh(sin(d*x+c))/a^2/d-4/3*sin(d*x+c)/a^2/d/(1+cos(d*x+c))-1/3*sin(d*x+ c)/d/(a+a*cos(d*x+c))^2
Leaf count is larger than twice the leaf count of optimal. \(152\) vs. \(2(66)=132\).
Time = 0.44 (sec) , antiderivative size = 152, normalized size of antiderivative = 2.30 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {2 \cos \left (\frac {1}{2} (c+d x)\right ) \left (6 \cos ^3\left (\frac {1}{2} (c+d x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )\right )+\sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+8 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec \left (\frac {c}{2}\right ) \sin \left (\frac {d x}{2}\right )+\cos \left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {c}{2}\right )\right )}{3 a^2 d (1+\cos (c+d x))^2} \]
(-2*Cos[(c + d*x)/2]*(6*Cos[(c + d*x)/2]^3*(Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] - Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]) + Sec[c/2]*Sin[(d*x )/2] + 8*Cos[(c + d*x)/2]^2*Sec[c/2]*Sin[(d*x)/2] + Cos[(c + d*x)/2]*Tan[c /2]))/(3*a^2*d*(1 + Cos[c + d*x])^2)
Time = 0.43 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {3042, 3245, 3042, 3457, 27, 3042, 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.
\(\displaystyle \int \frac {\sec (c+d x)}{(a \cos (c+d x)+a)^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (a \sin \left (c+d x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
\(\Big \downarrow \) 3245 |
\(\displaystyle \frac {\int \frac {(3 a-a \cos (c+d x)) \sec (c+d x)}{\cos (c+d x) a+a}dx}{3 a^2}-\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {3 a-a \sin \left (c+d x+\frac {\pi }{2}\right )}{\sin \left (c+d x+\frac {\pi }{2}\right ) \left (\sin \left (c+d x+\frac {\pi }{2}\right ) a+a\right )}dx}{3 a^2}-\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3457 |
\(\displaystyle \frac {\frac {\int 3 a^2 \sec (c+d x)dx}{a^2}-\frac {4 \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {3 \int \sec (c+d x)dx-\frac {4 \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \csc \left (c+d x+\frac {\pi }{2}\right )dx-\frac {4 \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\frac {3 \text {arctanh}(\sin (c+d x))}{d}-\frac {4 \sin (c+d x)}{d (\cos (c+d x)+1)}}{3 a^2}-\frac {\sin (c+d x)}{3 d (a \cos (c+d x)+a)^2}\) |
-1/3*Sin[c + d*x]/(d*(a + a*Cos[c + d*x])^2) + ((3*ArcTanh[Sin[c + d*x]])/ d - (4*Sin[c + d*x])/(d*(1 + Cos[c + d*x])))/(3*a^2)
3.1.59.3.1 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_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b^2*Cos[e + f*x]*(a + b*Sin[e + f*x])^ m*((c + d*Sin[e + f*x])^(n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/( a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[b*c*(m + 1) - a*d*(2*m + n + 2) + b*d*(m + n + 2)*Sin[e + f*x] , x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && EqQ [a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && !GtQ[n, 0] && (Intege rsQ[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))
Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim p[b*(A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*((c + d*Sin[e + f*x])^( n + 1)/(a*f*(2*m + 1)*(b*c - a*d))), x] + Simp[1/(a*(2*m + 1)*(b*c - a*d)) Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[B*(a*c*m + b *d*(n + 1)) + A*(b*c*(m + 1) - a*d*(2*m + n + 2)) + d*(A*b - a*B)*(m + n + 2)*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ [b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -2^(-1)] && !GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Time = 0.90 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.94
method | result | size |
derivativedivides | \(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\) | \(62\) |
default | \(\frac {-\frac {\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{3}-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+2 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{2 d \,a^{2}}\) | \(62\) |
parallelrisch | \(\frac {-\left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+6 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{6 a^{2} d}\) | \(62\) |
norman | \(\frac {-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{2 d a}-\frac {\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )}{6 d a}}{a}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \,a^{2}}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{2} d}\) | \(82\) |
risch | \(-\frac {2 i \left (3 \,{\mathrm e}^{2 i \left (d x +c \right )}+9 \,{\mathrm e}^{i \left (d x +c \right )}+4\right )}{3 d \,a^{2} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{3}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \,a^{2}}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{2} d}\) | \(89\) |
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 = 0.32 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.73 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {3 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (4 \, \cos \left (d x + c\right ) + 5\right )} \sin \left (d x + c\right )}{6 \, {\left (a^{2} d \cos \left (d x + c\right )^{2} + 2 \, a^{2} d \cos \left (d x + c\right ) + a^{2} d\right )}} \]
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)
\[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\int \frac {\sec {\left (c + d x \right )}}{\cos ^{2}{\left (c + d x \right )} + 2 \cos {\left (c + d x \right )} + 1}\, dx}{a^{2}} \]
Time = 0.23 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.48 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {\frac {\frac {9 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{2}}}{6 \, d} \]
-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 = 0.33 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.17 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=\frac {\frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{2}} - \frac {6 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{2}} - \frac {a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 9 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{6}}}{6 \, d} \]
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 = 14.08 (sec) , antiderivative size = 43, normalized size of antiderivative = 0.65 \[ \int \frac {\sec (c+d x)}{(a+a \cos (c+d x))^2} \, dx=-\frac {9\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-12\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{6\,a^2\,d} \]