Integrand size = 29, antiderivative size = 79 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx=\frac {2 a \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{\sqrt {c-d} (c+d)^{3/2} f}+\frac {a \tan (e+f x)}{(c+d) f (c+d \sec (e+f x))} \]
2*a*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/(c+d)^(3/2)/f/(c-d )^(1/2)+a*tan(f*x+e)/(c+d)/f/(c+d*sec(f*x+e))
Time = 0.77 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.95 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx=\frac {a \left (-\frac {2 \text {arctanh}\left (\frac {(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right )}{\sqrt {c^2-d^2}}+\frac {\sin (e+f x)}{d+c \cos (e+f x)}\right )}{(c+d) f} \]
(a*((-2*ArcTanh[((-c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]])/Sqrt[c^2 - d ^2] + Sin[e + f*x]/(d + c*Cos[e + f*x])))/((c + d)*f)
Time = 0.47 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.14, number of steps used = 10, number of rules used = 9, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.310, Rules used = {3042, 4491, 25, 27, 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.
\(\displaystyle \int \frac {\sec (e+f x) (a \sec (e+f x)+a)}{(c+d \sec (e+f x))^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )}{\left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx\) |
\(\Big \downarrow \) 4491 |
\(\displaystyle \frac {a \tan (e+f x)}{f (c+d) (c+d \sec (e+f x))}-\frac {\int -\frac {a (c-d) \sec (e+f x)}{c+d \sec (e+f x)}dx}{c^2-d^2}\) |
\(\Big \downarrow \) 25 |
\(\displaystyle \frac {\int \frac {a (c-d) \sec (e+f x)}{c+d \sec (e+f x)}dx}{c^2-d^2}+\frac {a \tan (e+f x)}{f (c+d) (c+d \sec (e+f x))}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {a (c-d) \int \frac {\sec (e+f x)}{c+d \sec (e+f x)}dx}{c^2-d^2}+\frac {a \tan (e+f x)}{f (c+d) (c+d \sec (e+f x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (c-d) \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{c+d \csc \left (e+f x+\frac {\pi }{2}\right )}dx}{c^2-d^2}+\frac {a \tan (e+f x)}{f (c+d) (c+d \sec (e+f x))}\) |
\(\Big \downarrow \) 4318 |
\(\displaystyle \frac {a (c-d) \int \frac {1}{\frac {c \cos (e+f x)}{d}+1}dx}{d \left (c^2-d^2\right )}+\frac {a \tan (e+f x)}{f (c+d) (c+d \sec (e+f x))}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {a (c-d) \int \frac {1}{\frac {c \sin \left (e+f x+\frac {\pi }{2}\right )}{d}+1}dx}{d \left (c^2-d^2\right )}+\frac {a \tan (e+f x)}{f (c+d) (c+d \sec (e+f x))}\) |
\(\Big \downarrow \) 3138 |
\(\displaystyle \frac {2 a (c-d) \int \frac {1}{\left (1-\frac {c}{d}\right ) \tan ^2\left (\frac {1}{2} (e+f x)\right )+\frac {c+d}{d}}d\tan \left (\frac {1}{2} (e+f x)\right )}{d f \left (c^2-d^2\right )}+\frac {a \tan (e+f x)}{f (c+d) (c+d \sec (e+f x))}\) |
\(\Big \downarrow \) 221 |
\(\displaystyle \frac {2 a \sqrt {c-d} \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f \sqrt {c+d} \left (c^2-d^2\right )}+\frac {a \tan (e+f x)}{f (c+d) (c+d \sec (e+f x))}\) |
(2*a*Sqrt[c - d]*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2])/Sqrt[c + d]])/(Sqr t[c + d]*(c^2 - d^2)*f) + (a*Tan[e + f*x])/((c + d)*f*(c + d*Sec[e + f*x]) )
3.2.90.3.1 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_.)*(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_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(f*(m + 1)*(a^2 - b^2))), x] + Simp[1 /((m + 1)*(a^2 - b^2)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp [(a*A - b*B)*(m + 1) - (A*b - a*B)*(m + 2)*Csc[e + f*x], x], x], x] /; Free Q[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m , -1]
Time = 0.66 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.33
method | result | size |
derivativedivides | \(\frac {4 a \left (-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 \left (c +d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )}+\frac {\operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{2 \left (c +d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{f}\) | \(105\) |
default | \(\frac {4 a \left (-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 \left (c +d \right ) \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )}+\frac {\operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{2 \left (c +d \right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{f}\) | \(105\) |
risch | \(\frac {2 i a \left (d \,{\mathrm e}^{i \left (f x +e \right )}+c \right )}{c f \left (c +d \right ) \left ({\mathrm e}^{2 i \left (f x +e \right )} c +2 d \,{\mathrm e}^{i \left (f x +e \right )}+c \right )}+\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {i c^{2}-i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{\sqrt {c^{2}-d^{2}}\, c}\right )}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) f}-\frac {a \ln \left ({\mathrm e}^{i \left (f x +e \right )}+\frac {-i c^{2}+i d^{2}+\sqrt {c^{2}-d^{2}}\, d}{c \sqrt {c^{2}-d^{2}}}\right )}{\sqrt {c^{2}-d^{2}}\, \left (c +d \right ) f}\) | \(204\) |
4/f*a*(-1/2*tan(1/2*f*x+1/2*e)/(c+d)/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1 /2*e)^2*d-c-d)+1/2/(c+d)/((c+d)*(c-d))^(1/2)*arctanh((c-d)*tan(1/2*f*x+1/2 *e)/((c+d)*(c-d))^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (70) = 140\).
Time = 0.29 (sec) , antiderivative size = 357, normalized size of antiderivative = 4.52 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx=\left [\frac {{\left (a c \cos \left (f x + e\right ) + a d\right )} \sqrt {c^{2} - d^{2}} \log \left (\frac {2 \, c d \cos \left (f x + e\right ) - {\left (c^{2} - 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {c^{2} - d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )} \sin \left (f x + e\right ) + 2 \, c^{2} - d^{2}}{c^{2} \cos \left (f x + e\right )^{2} + 2 \, c d \cos \left (f x + e\right ) + d^{2}}\right ) + 2 \, {\left (a c^{2} - a d^{2}\right )} \sin \left (f x + e\right )}{2 \, {\left ({\left (c^{4} + c^{3} d - c^{2} d^{2} - c d^{3}\right )} f \cos \left (f x + e\right ) + {\left (c^{3} d + c^{2} d^{2} - c d^{3} - d^{4}\right )} f\right )}}, \frac {{\left (a c \cos \left (f x + e\right ) + a d\right )} \sqrt {-c^{2} + d^{2}} \arctan \left (-\frac {\sqrt {-c^{2} + d^{2}} {\left (d \cos \left (f x + e\right ) + c\right )}}{{\left (c^{2} - d^{2}\right )} \sin \left (f x + e\right )}\right ) + {\left (a c^{2} - a d^{2}\right )} \sin \left (f x + e\right )}{{\left (c^{4} + c^{3} d - c^{2} d^{2} - c d^{3}\right )} f \cos \left (f x + e\right ) + {\left (c^{3} d + c^{2} d^{2} - c d^{3} - d^{4}\right )} f}\right ] \]
[1/2*((a*c*cos(f*x + e) + a*d)*sqrt(c^2 - d^2)*log((2*c*d*cos(f*x + e) - ( c^2 - 2*d^2)*cos(f*x + e)^2 + 2*sqrt(c^2 - d^2)*(d*cos(f*x + e) + c)*sin(f *x + e) + 2*c^2 - d^2)/(c^2*cos(f*x + e)^2 + 2*c*d*cos(f*x + e) + d^2)) + 2*(a*c^2 - a*d^2)*sin(f*x + e))/((c^4 + c^3*d - c^2*d^2 - c*d^3)*f*cos(f*x + e) + (c^3*d + c^2*d^2 - c*d^3 - d^4)*f), ((a*c*cos(f*x + e) + a*d)*sqrt (-c^2 + d^2)*arctan(-sqrt(-c^2 + d^2)*(d*cos(f*x + e) + c)/((c^2 - d^2)*si n(f*x + e))) + (a*c^2 - a*d^2)*sin(f*x + e))/((c^4 + c^3*d - c^2*d^2 - c*d ^3)*f*cos(f*x + e) + (c^3*d + c^2*d^2 - c*d^3 - d^4)*f)]
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx=a \left (\int \frac {\sec {\left (e + f x \right )}}{c^{2} + 2 c d \sec {\left (e + f x \right )} + d^{2} \sec ^{2}{\left (e + f x \right )}}\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{c^{2} + 2 c d \sec {\left (e + f x \right )} + d^{2} \sec ^{2}{\left (e + f x \right )}}\, dx\right ) \]
a*(Integral(sec(e + f*x)/(c**2 + 2*c*d*sec(e + f*x) + d**2*sec(e + f*x)**2 ), x) + Integral(sec(e + f*x)**2/(c**2 + 2*c*d*sec(e + f*x) + d**2*sec(e + f*x)**2), x))
Exception generated. \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx=\text {Exception raised: ValueError} \]
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*c^2-4*d^2>0)', see `assume?` f or more de
Time = 0.32 (sec) , antiderivative size = 137, normalized size of antiderivative = 1.73 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx=-\frac {2 \, {\left (\frac {{\left (\pi \left \lfloor \frac {f x + e}{2 \, \pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (2 \, c - 2 \, d\right ) + \arctan \left (\frac {c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{\sqrt {-c^{2} + d^{2}}}\right )\right )} a}{\sqrt {-c^{2} + d^{2}} {\left (c + d\right )}} + \frac {a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - c - d\right )} {\left (c + d\right )}}\right )}}{f} \]
-2*((pi*floor(1/2*(f*x + e)/pi + 1/2)*sgn(2*c - 2*d) + arctan((c*tan(1/2*f *x + 1/2*e) - d*tan(1/2*f*x + 1/2*e))/sqrt(-c^2 + d^2)))*a/(sqrt(-c^2 + d^ 2)*(c + d)) + a*tan(1/2*f*x + 1/2*e)/((c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/ 2*f*x + 1/2*e)^2 - c - d)*(c + d)))/f
Time = 13.47 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.08 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c+d \sec (e+f x))^2} \, dx=\frac {2\,a\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (c+d\right )\,\left (\left (d-c\right )\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+c+d\right )}+\frac {2\,a\,\mathrm {atanh}\left (\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\sqrt {c-d}}{\sqrt {c+d}}\right )}{f\,{\left (c+d\right )}^{3/2}\,\sqrt {c-d}} \]