Integrand size = 28, antiderivative size = 61 \[ \int \frac {\frac {b B}{a}+B \sec (c+d x)}{a+b \sec (c+d x)} \, dx=\frac {b B x}{a^2}+\frac {2 \sqrt {a-b} \sqrt {a+b} B \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d} \] Output:
b*B*x/a^2+2*(a-b)^(1/2)*(a+b)^(1/2)*B*arctanh((a-b)^(1/2)*tan(1/2*d*x+1/2* c)/(a+b)^(1/2))/a^2/d
Time = 0.15 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.00 \[ \int \frac {\frac {b B}{a}+B \sec (c+d x)}{a+b \sec (c+d x)} \, dx=\frac {B \left (b (c+d x)-2 \sqrt {a^2-b^2} \text {arctanh}\left (\frac {(-a+b) \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a^2-b^2}}\right )\right )}{a^2 d} \] Input:
Integrate[((b*B)/a + B*Sec[c + d*x])/(a + b*Sec[c + d*x]),x]
Output:
(B*(b*(c + d*x) - 2*Sqrt[a^2 - b^2]*ArcTanh[((-a + b)*Tan[(c + d*x)/2])/Sq rt[a^2 - b^2]]))/(a^2*d)
Time = 0.41 (sec) , antiderivative size = 70, normalized size of antiderivative = 1.15, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {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.
\(\displaystyle \int \frac {\frac {b B}{a}+B \sec (c+d x)}{a+b \sec (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\frac {b B}{a}+B \csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx\) |
\(\Big \downarrow \) 4407 |
\(\displaystyle \frac {B \left (a^2-b^2\right ) \int \frac {\sec (c+d x)}{a+b \sec (c+d x)}dx}{a^2}+\frac {b B x}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {B \left (a^2-b^2\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{a+b \csc \left (c+d x+\frac {\pi }{2}\right )}dx}{a^2}+\frac {b B x}{a^2}\) |
\(\Big \downarrow \) 4318 |
\(\displaystyle \frac {B \left (a^2-b^2\right ) \int \frac {1}{\frac {a \cos (c+d x)}{b}+1}dx}{a^2 b}+\frac {b B x}{a^2}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {B \left (a^2-b^2\right ) \int \frac {1}{\frac {a \sin \left (c+d x+\frac {\pi }{2}\right )}{b}+1}dx}{a^2 b}+\frac {b B x}{a^2}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {2 B \left (a^2-b^2\right ) \int \frac {1}{\left (1-\frac {a}{b}\right ) \tan ^2\left (\frac {1}{2} (c+d x)\right )+\frac {a+b}{b}}d\tan \left (\frac {1}{2} (c+d x)\right )}{a^2 b d}+\frac {b B x}{a^2}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 B \left (a^2-b^2\right ) \text {arctanh}\left (\frac {\sqrt {a-b} \tan \left (\frac {1}{2} (c+d x)\right )}{\sqrt {a+b}}\right )}{a^2 d \sqrt {a-b} \sqrt {a+b}}+\frac {b B x}{a^2}\) |
Input:
Int[((b*B)/a + B*Sec[c + d*x])/(a + b*Sec[c + d*x]),x]
Output:
(b*B*x)/a^2 + (2*(a^2 - b^2)*B*ArcTanh[(Sqrt[a - b]*Tan[(c + d*x)/2])/Sqrt [a + b]])/(a^2*Sqrt[a - b]*Sqrt[a + b]*d)
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]
Time = 0.23 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.25
method | result | size |
derivativedivides | \(\frac {2 B \left (\frac {b \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {\left (a +b \right ) \left (a -b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{d a}\) | \(76\) |
default | \(\frac {2 B \left (\frac {b \arctan \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}+\frac {\left (a +b \right ) \left (a -b \right ) \operatorname {arctanh}\left (\frac {\left (a -b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{\sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{a \sqrt {\left (a +b \right ) \left (a -b \right )}}\right )}{d a}\) | \(76\) |
risch | \(\frac {b B x}{a^{2}}+\frac {\sqrt {a^{2}-b^{2}}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}+\frac {i \sqrt {a^{2}-b^{2}}+b}{a}\right )}{d \,a^{2}}-\frac {\sqrt {a^{2}-b^{2}}\, B \ln \left ({\mathrm e}^{i \left (d x +c \right )}-\frac {i \sqrt {a^{2}-b^{2}}-b}{a}\right )}{d \,a^{2}}\) | \(113\) |
Input:
int((b*B/a+B*sec(d*x+c))/(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
2/d*B/a*(b/a*arctan(tan(1/2*d*x+1/2*c))+(a+b)*(a-b)/a/((a+b)*(a-b))^(1/2)* arctanh((a-b)*tan(1/2*d*x+1/2*c)/((a+b)*(a-b))^(1/2)))
Time = 0.09 (sec) , antiderivative size = 197, normalized size of antiderivative = 3.23 \[ \int \frac {\frac {b B}{a}+B \sec (c+d x)}{a+b \sec (c+d x)} \, dx=\left [\frac {2 \, B b d x + \sqrt {a^{2} - b^{2}} B \log \left (\frac {2 \, a b \cos \left (d x + c\right ) - {\left (a^{2} - 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} + 2 \, \sqrt {a^{2} - b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, a^{2} - b^{2}}{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a b \cos \left (d x + c\right ) + b^{2}}\right )}{2 \, a^{2} d}, \frac {B b d x + \sqrt {-a^{2} + b^{2}} B \arctan \left (-\frac {\sqrt {-a^{2} + b^{2}} {\left (b \cos \left (d x + c\right ) + a\right )}}{{\left (a^{2} - b^{2}\right )} \sin \left (d x + c\right )}\right )}{a^{2} d}\right ] \] Input:
integrate((b*B/a+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="fricas")
Output:
[1/2*(2*B*b*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*d), (B* b*d*x + 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*d)]
\[ \int \frac {\frac {b B}{a}+B \sec (c+d x)}{a+b \sec (c+d x)} \, dx=\frac {B \left (\int \frac {b}{a + b \sec {\left (c + d x \right )}}\, dx + \int \frac {a \sec {\left (c + d x \right )}}{a + b \sec {\left (c + d x \right )}}\, dx\right )}{a} \] Input:
integrate((b*B/a+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)
Output:
B*(Integral(b/(a + b*sec(c + d*x)), x) + Integral(a*sec(c + d*x)/(a + b*se c(c + d*x)), x))/a
Exception generated. \[ \int \frac {\frac {b B}{a}+B \sec (c+d x)}{a+b \sec (c+d x)} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((b*B/a+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="maxima")
Output:
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. 187 vs. \(2 (52) = 104\).
Time = 0.21 (sec) , antiderivative size = 187, normalized size of antiderivative = 3.07 \[ \int \frac {\frac {b B}{a}+B \sec (c+d x)}{a+b \sec (c+d x)} \, dx=\frac {2 \, {\left (\frac {\sqrt {-a^{2} + b^{2}} {\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {a b + \sqrt {a^{2} b^{2} + {\left (a^{2} + a b\right )} {\left (a^{2} - a b\right )}}}{a^{2} - a b}}}\right )\right )} B {\left | -a + b \right |}}{a^{3} - a^{2} b} + \frac {{\left (\pi \left \lfloor \frac {d x + c}{2 \, \pi } + \frac {1}{2} \right \rfloor + \arctan \left (\frac {\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{\sqrt {-\frac {a b - \sqrt {a^{2} b^{2} + {\left (a^{2} + a b\right )} {\left (a^{2} - a b\right )}}}{a^{2} - a b}}}\right )\right )} B b}{a^{2}}\right )}}{d} \] Input:
integrate((b*B/a+B*sec(d*x+c))/(a+b*sec(d*x+c)),x, algorithm="giac")
Output:
2*(sqrt(-a^2 + b^2)*(pi*floor(1/2*(d*x + c)/pi + 1/2) + arctan(tan(1/2*d*x + 1/2*c)/sqrt(-(a*b + sqrt(a^2*b^2 + (a^2 + a*b)*(a^2 - a*b)))/(a^2 - a*b ))))*B*abs(-a + b)/(a^3 - a^2*b) + (pi*floor(1/2*(d*x + c)/pi + 1/2) + arc tan(tan(1/2*d*x + 1/2*c)/sqrt(-(a*b - sqrt(a^2*b^2 + (a^2 + a*b)*(a^2 - a* b)))/(a^2 - a*b))))*B*b/a^2)/d
Time = 10.97 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.49 \[ \int \frac {\frac {b B}{a}+B \sec (c+d x)}{a+b \sec (c+d x)} \, dx=\frac {2\,B\,b\,\mathrm {atan}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )}\right )}{a^2\,d}+\frac {2\,B\,\mathrm {atanh}\left (\frac {\sin \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\sqrt {a^2-b^2}}{\cos \left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (a+b\right )}\right )\,\sqrt {a^2-b^2}}{a^2\,d} \] Input:
int((B/cos(c + d*x) + (B*b)/a)/(a + b/cos(c + d*x)),x)
Output:
(2*B*b*atan(sin(c/2 + (d*x)/2)/cos(c/2 + (d*x)/2)))/(a^2*d) + (2*B*atanh(( sin(c/2 + (d*x)/2)*(a^2 - b^2)^(1/2))/(cos(c/2 + (d*x)/2)*(a + b)))*(a^2 - b^2)^(1/2))/(a^2*d)
Time = 0.16 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.03 \[ \int \frac {\frac {b B}{a}+B \sec (c+d x)}{a+b \sec (c+d x)} \, dx=\frac {b \left (2 \sqrt {-a^{2}+b^{2}}\, \mathit {atan} \left (\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) a -\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) b}{\sqrt {-a^{2}+b^{2}}}\right )+b d x \right )}{a^{2} d} \] Input:
int((b*B/a+B*sec(d*x+c))/(a+b*sec(d*x+c)),x)
Output:
(b*(2*sqrt( - a**2 + b**2)*atan((tan((c + d*x)/2)*a - tan((c + d*x)/2)*b)/ sqrt( - a**2 + b**2)) + b*d*x))/(a**2*d)