Integrand size = 21, antiderivative size = 68 \[ \int \frac {\sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{\sqrt {a-b} b \sqrt {a+b} d} \]
arctanh(sin(d*x+c))/b/d-2*a*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2*c)/(a+b)^( 1/2))/b/d/(a-b)^(1/2)/(a+b)^(1/2)
Time = 0.18 (sec) , antiderivative size = 102, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {\frac {2 a \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )}{\sqrt {a^2-b^2}}-\log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+\log \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )}{b d} \]
((2*a*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sqrt[a^2 - b^2]])/Sqrt[a^2 - b^2 ] - Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + Log[Cos[(c + d*x)/2] + Sin[ (c + d*x)/2]])/(b*d)
Time = 0.44 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4276, 3042, 4257, 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.
\(\displaystyle \int \frac {\sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4276 |
\(\displaystyle \frac {\int \sec (c+d x)dx}{b}-\frac {a \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{b}-\frac {a \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\) |
\(\Big \downarrow \) 4257 |
\(\displaystyle \frac {\text {arctanh}(\sin (c+d x))}{b d}-\frac {a \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{b}\) |
\(\Big \downarrow \) 4318 |
\(\displaystyle \frac {\text {arctanh}(\sin (c+d x))}{b d}-\frac {a \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{b^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\text {arctanh}(\sin (c+d x))}{b d}-\frac {a \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{b^2}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {\text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{b^2 d}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {\text {arctanh}(\sin (c+d x))}{b d}-\frac {2 a \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{b d \sqrt {a-b} \sqrt {a+b}}\) |
ArcTanh[Sin[c + d*x]]/(b*d) - (2*a*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/ Sqrt[a + b]])/(Sqrt[a - b]*b*Sqrt[a + b]*d)
3.5.90.3.1 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[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[csc[(e_.) + (f_.)*(x_)]^2/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Sym bol] :> Simp[1/b Int[Csc[e + f*x], x], x] - Simp[a/b Int[Csc[e + f*x]/( a + b*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f}, x]
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]
Time = 0.42 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.22
method | result | size |
derivativedivides | \(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}-\frac {2 a \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(83\) |
default | \(\frac {\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{b}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{b}-\frac {2 a \,\operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a -b \right ) \left (a +b \right )}}\right )}{b \sqrt {\left (a -b \right ) \left (a +b \right )}}}{d}\) | \(83\) |
risch | \(\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {-i a^{2}+i b^{2}+b \sqrt {a^{2}-b^{2}}}{a \sqrt {a^{2}-b^{2}}}\right )}{\sqrt {a^{2}-b^{2}}\, d b}-\frac {a \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i a^{2}-i b^{2}+b \sqrt {a^{2}-b^{2}}}{\sqrt {a^{2}-b^{2}}\, a}\right )}{\sqrt {a^{2}-b^{2}}\, d b}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d b}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d b}\) | \(186\) |
1/d*(1/b*ln(tan(1/2*d*x+1/2*c)+1)-1/b*ln(tan(1/2*d*x+1/2*c)-1)-2*a/b/((a-b )*(a+b))^(1/2)*arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a-b)*(a+b))^(1/2)))
Time = 0.32 (sec) , antiderivative size = 290, normalized size of antiderivative = 4.26 \[ \int \frac {\sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\left [\frac {\sqrt {a^{2} - b^{2}} a \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right ) + {\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} - b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d}, -\frac {2 \, \sqrt {-a^{2} + b^{2}} a \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right ) - {\left (a^{2} - b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a^{2} - b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} b - b^{3}\right )} d}\right ] \]
[1/2*(sqrt(a^2 - b^2)*a*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)*log(sin(d*x + c) + 1) - (a^2 - b^2)*log(-sin(d*x + c) + 1))/((a^2*b - b^3)*d), -1/2*( 2*sqrt(-a^2 + b^2)*a*arctan(-sqrt(-a^2 + b^2)*(b*cos(d*x + c) + a)/((a^2 - b^2)*sin(d*x + c))) - (a^2 - b^2)*log(sin(d*x + c) + 1) + (a^2 - b^2)*log (-sin(d*x + c) + 1))/((a^2*b - b^3)*d)]
\[ \int \frac {\sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\int \frac {\sec ^{2}{\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx \]
Exception generated. \[ \int \frac {\sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \]
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. 120 vs. \(2 (59) = 118\).
Time = 0.32 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.76 \[ \int \frac {\sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=-\frac {\frac {2 \, {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (-2 \, a + 2 \, b\right ) + \arctan \left (-\frac {a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-a^{2} + b^{2}}}\right )\right )} a}{\sqrt {-a^{2} + b^{2}} b} - \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{b} + \frac {\log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{b}}{d} \]
-(2*(pi*floor(1/2*(d*x + c)/pi + 1/2)*sgn(-2*a + 2*b) + arctan(-(a*tan(1/2 *d*x + 1/2*c) - b*tan(1/2*d*x + 1/2*c))/sqrt(-a^2 + b^2)))*a/(sqrt(-a^2 + b^2)*b) - log(abs(tan(1/2*d*x + 1/2*c) + 1))/b + log(abs(tan(1/2*d*x + 1/2 *c) - 1))/b)/d
Time = 13.88 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.74 \[ \int \frac {\sec ^2(c+d x)}{a+b \sec (c+d x)} \, dx=\frac {2\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{b\,d}+\frac {2\,a\,\mathrm {atanh}\left (\frac {2\,a^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )-2\,a^4\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )+b^2\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )+2\,a^3\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )-2\,a\,b\,\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a^2-b^2\right )}{b\,\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}\,\left (a\,b-b^2\right )}\right )}{b\,d\,\sqrt {a^2-b^2}} \]
(2*atanh(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(b*d) + (2*a*atanh((2*a^2 *sin(c/2 + (d*x)/2)*(a^2 - b^2) - 2*a^4*sin(c/2 + (d*x)/2) + b^2*sin(c/2 + (d*x)/2)*(a^2 - b^2) + 2*a^3*b*sin(c/2 + (d*x)/2) - 2*a*b*sin(c/2 + (d*x) /2)*(a^2 - b^2))/(b*cos(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2)*(a*b - b^2))))/(b *d*(a^2 - b^2)^(1/2))