Integrand size = 30, antiderivative size = 52 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=x-\frac {4 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} \sqrt {a+b} d} \]
Time = 0.25 (sec) , antiderivative size = 56, normalized size of antiderivative = 1.08 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\frac {c}{d}+x+\frac {4 b \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2} d} \]
Time = 0.45 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.00, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4530, 25, 3042, 4407, 3042, 4318, 3042, 3138, 221}
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.
\(\displaystyle \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a^2-b^2 \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4530 |
\(\displaystyle -\int -\frac {a-b \sec (c+d x)}{a+b \sec (c+d x)}dx\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \int \frac {a-b \sec (c+d x)}{a+b \sec (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {a-b \csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4407 |
\(\displaystyle x-2 b \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle x-2 b \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4318 |
\(\displaystyle x-2 \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle x-2 \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle x-\frac {4 \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle x-\frac {4 b \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{d \sqrt {a-b} \sqrt {a+b}}\) |
3.8.6.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x /Rt[-a/b, 2]], x] /; FreeQ[{a, b}, x] && NegQ[a/b]
Int[((a_) + (b_.)*sin[Pi/2 + (c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> With[{ e = FreeFactors[Tan[(c + d*x)/2], x]}, Simp[2*(e/d) Subst[Int[1/(a + b + (a - b)*e^2*x^2), x], x, Tan[(c + d*x)/2]/e], x]] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[1/b Int[1/(1 + (a/b)*Sin[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[c*(x/a), x] - Simp[(b*c - a*d)/a Int[Csc[e + f* x]/(a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. ) + (a_))^(m_.), x_Symbol] :> Simp[C/b^2 Int[(a + b*Csc[e + f*x])^(m + 1) *Simp[-a + b*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, C, m}, x] && EqQ[A*b^2 + a^2*C, 0]
Time = 0.19 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.13
method | result | size |
derivativedivides | \(\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(59\) |
default | \(\frac {2 \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\frac {4 b \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}}{d}\) | \(59\) |
risch | \(x +\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d}-\frac {2 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+\sqrt {a^{2}-b^{2}}\, b}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d}\) | \(141\) |
1/d*(2*arctan(tan(1/2*d*x+1/2*c))-4*b/((a+b)*(a-b))^(1/2)*arctanh((a-b)*ta n(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))
Time = 0.35 (sec) , antiderivative size = 226, normalized size of antiderivative = 4.35 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\left [\frac {{\left (a^{2} - b^{2}\right )} d x + \sqrt {a^{2} - b^{2}} b \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right )}{{\left (a^{2} - b^{2}\right )} d}, \frac {{\left (a^{2} - b^{2}\right )} d x - 2 \, \sqrt {-a^{2} + b^{2}} b \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right )}{{\left (a^{2} - b^{2}\right )} d}\right ] \]
[((a^2 - b^2)*d*x + sqrt(a^2 - b^2)*b*log((2*a*b*cos(d*x + c) - (a^2 - 2*b ^2)*cos(d*x + c)^2 - 2*sqrt(a^2 - b^2)*(b*cos(d*x + c) + a)*sin(d*x + c) + 2*a^2 - b^2)/(a^2*cos(d*x + c)^2 + 2*a*b*cos(d*x + c) + b^2)))/((a^2 - b^ 2)*d), ((a^2 - b^2)*d*x - 2*sqrt(-a^2 + b^2)*b*arctan(-sqrt(-a^2 + b^2)*(b *cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))))/((a^2 - b^2)*d)]
\[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\int \frac {a - b \sec {\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=\text {Exception raised: ValueError} \]
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(4*a^2-4*b^2>0)', see `assume?` f or more de
Leaf count of result is larger than twice the leaf count of optimal. 235 vs. \(2 (43) = 86\).
Time = 0.30 (sec) , antiderivative size = 235, normalized size of antiderivative = 4.52 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=-\frac {\frac {{\left (\sqrt {-a^{2} + b^{2}} {\left (a + b\right )} {\left | a \right |} {\left | -a + b \right |} - {\left (a^{2} - 3 \, a b\right )} \sqrt {-a^{2} + b^{2}} {\left | -a + b \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b + \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )}}{{\left (a^{2} - 2 \, a b + b^{2}\right )} a^{2} + {\left (a^{2} b - 2 \, a b^{2} + b^{3}\right )} {\left | a \right |}} - \frac {{\left (a^{2} - 3 \, a b + a {\left | a \right |} + b {\left | a \right |}\right )} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {b - \sqrt {{\left (a + b\right )} {\left (a - b\right )} + b^{2}}}{a - b}}}\right )\right )}}{a^{2} - b {\left | a \right |}}}{d} \]
-((sqrt(-a^2 + b^2)*(a + b)*abs(a)*abs(-a + b) - (a^2 - 3*a*b)*sqrt(-a^2 + b^2)*abs(-a + b))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(b + sqrt((a + b)*(a - b) + b^2))/(a - b))))/((a^2 - 2*a*b + b^2)*a^2 + (a^2*b - 2*a*b^2 + b^3)*abs(a)) - (a^2 - 3*a*b + a*abs(a) + b *abs(a))*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/s qrt(-(b - sqrt((a + b)*(a - b) + b^2))/(a - b))))/(a^2 - b*abs(a)))/d
Time = 16.72 (sec) , antiderivative size = 182, normalized size of antiderivative = 3.50 \[ \int \frac {a^2-b^2 \sec ^2(c+d x)}{(a+b \sec (c+d x))^2} \, dx=x-\frac {4\,b\,\mathrm {atanh}\left (\frac {8\,b^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+5\,b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-8\,a\,b^3\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,\left (a\,\left (a^2-b^2\right )+b\,\left (a^2-b^2\right )+4\,a\,b^2-2\,a^2\,b-2\,b^3\right )}\right )}{d\,\sqrt {a^2-b^2}} \]
x - (4*b*atanh((8*b^4*sin(c/2 + (d*x)/2) + a^2*sin(c/2 + (d*x)/2)*(a^2 - b ^2) + 5*b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2) - 8*a*b^3*sin(c/2 + (d*x)/2) - 2*a*b*sin(c/2 + (d*x)/2)*(a^2 - b^2))/(cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2 )*(a*(a^2 - b^2) + b*(a^2 - b^2) + 4*a*b^2 - 2*a^2*b - 2*b^3))))/(d*(a^2 - b^2)^(1/2))