Integrand size = 23, antiderivative size = 73 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 \sqrt {b} (a+b)^{3/2} f}+\frac {\tan (e+f x)}{2 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )} \]
1/2*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/(a+b)^(3/2)/f/b^(1/2)+1/2*tan(f *x+e)/(a+b)/f/(a+b+b*tan(f*x+e)^2)
Result contains complex when optimal does not.
Time = 2.52 (sec) , antiderivative size = 211, normalized size of antiderivative = 2.89 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^4(e+f x) \left (-\frac {\arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x))) (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {-((a+2 b) \sin (2 e))+a \sin (2 f x)}{a (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right )}{8 (a+b) f \left (a+b \sec ^2(e+f x)\right )^2} \]
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^4*(-((ArcTan[(Sec[f*x]*(Cos[2 *e] - I*Sin[2*e])*(-((a + 2*b)*Sin[f*x]) + a*Sin[2*e + f*x]))/(2*Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])]*(a + 2*b + a*Cos[2*(e + f*x)])*(Cos[2*e ] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4])) + (-((a + 2* b)*Sin[2*e]) + a*Sin[2*f*x])/(a*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(8* (a + b)*f*(a + b*Sec[e + f*x]^2)^2)
Time = 0.24 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3042, 4634, 215, 218}
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.
\(\displaystyle \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\sec (e+f x)^2}{\left (a+b \sec (e+f x)^2\right )^2}dx\) |
\(\Big \downarrow \) 4634 |
\(\displaystyle \frac {\int \frac {1}{\left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{f}\) |
\(\Big \downarrow \) 215 |
\(\displaystyle \frac {\frac {\int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{2 (a+b)}+\frac {\tan (e+f x)}{2 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
\(\Big \downarrow \) 218 |
\(\displaystyle \frac {\frac {\arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 \sqrt {b} (a+b)^{3/2}}+\frac {\tan (e+f x)}{2 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{f}\) |
(ArcTan[(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]]/(2*Sqrt[b]*(a + b)^(3/2)) + Ta n[e + f*x]/(2*(a + b)*(a + b + b*Tan[e + f*x]^2)))/f
3.3.1.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(-x)*((a + b*x^2)^(p + 1) /(2*a*(p + 1))), x] + Simp[(2*p + 3)/(2*a*(p + 1)) Int[(a + b*x^2)^(p + 1 ), x], x] /; FreeQ[{a, b}, x] && LtQ[p, -1] && (IntegerQ[4*p] || IntegerQ[6 *p])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/R t[a/b, 2]], x] /; FreeQ[{a, b}, x] && PosQ[a/b]
Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^(n_) )^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(1 + ff^2*x^2)^(m/2 - 1)*ExpandToSum[a + b*(1 + ff^2*x^2)^(n/2), x]^p, x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ [m/2] && IntegerQ[n/2]
Time = 0.47 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.88
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}}{f}\) | \(64\) |
default | \(\frac {\frac {\tan \left (f x +e \right )}{2 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {\arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{2 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}}{f}\) | \(64\) |
risch | \(\frac {i \left (a \,{\mathrm e}^{2 i \left (f x +e \right )}+2 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}{a f \left (a +b \right ) \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f}+\frac {\ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i b a +2 i b^{2}-a \sqrt {-a b -b^{2}}-2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{4 \sqrt {-a b -b^{2}}\, \left (a +b \right ) f}\) | \(262\) |
1/f*(1/2*tan(f*x+e)/(a+b)/(a+b+b*tan(f*x+e)^2)+1/2/(a+b)/((a+b)*b)^(1/2)*a rctan(b*tan(f*x+e)/((a+b)*b)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 144 vs. \(2 (61) = 122\).
Time = 0.28 (sec) , antiderivative size = 368, normalized size of antiderivative = 5.04 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\left [\frac {4 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (a \cos \left (f x + e\right )^{2} + b\right )} \sqrt {-a b - b^{2}} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 2 \, {\left (3 \, a b + 4 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{3} - b \cos \left (f x + e\right )\right )} \sqrt {-a b - b^{2}} \sin \left (f x + e\right ) + b^{2}}{a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}}\right )}{8 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f\right )}}, \frac {2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right ) \sin \left (f x + e\right ) - {\left (a \cos \left (f x + e\right )^{2} + b\right )} \sqrt {a b + b^{2}} \arctan \left (\frac {{\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b}{2 \, \sqrt {a b + b^{2}} \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right )}{4 \, {\left ({\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f\right )}}\right ] \]
[1/8*(4*(a*b + b^2)*cos(f*x + e)*sin(f*x + e) - (a*cos(f*x + e)^2 + b)*sqr t(-a*b - b^2)*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 2*(3*a*b + 4*b^2 )*cos(f*x + e)^2 + 4*((a + 2*b)*cos(f*x + e)^3 - b*cos(f*x + e))*sqrt(-a*b - b^2)*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b ^2)))/((a^3*b + 2*a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a^2*b^2 + 2*a*b^3 + b^4)*f), 1/4*(2*(a*b + b^2)*cos(f*x + e)*sin(f*x + e) - (a*cos(f*x + e)^2 + b)*sqrt(a*b + b^2)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)/(sqrt(a*b + b^2)*cos(f*x + e)*sin(f*x + e))))/((a^3*b + 2*a^2*b^2 + a*b^3)*f*cos(f*x + e)^2 + (a^2*b^2 + 2*a*b^3 + b^4)*f)]
\[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\int \frac {\sec ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{2}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 71, normalized size of antiderivative = 0.97 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {\tan \left (f x + e\right )}{{\left (a b + b^{2}\right )} \tan \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}} + \frac {\arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{\sqrt {{\left (a + b\right )} b} {\left (a + b\right )}}}{2 \, f} \]
1/2*(tan(f*x + e)/((a*b + b^2)*tan(f*x + e)^2 + a^2 + 2*a*b + b^2) + arcta n(b*tan(f*x + e)/sqrt((a + b)*b))/(sqrt((a + b)*b)*(a + b)))/f
Time = 0.30 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.14 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\frac {\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )}{\sqrt {a b + b^{2}} {\left (a + b\right )}} + \frac {\tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )} {\left (a + b\right )}}}{2 \, f} \]
1/2*((pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/(sqrt(a*b + b^2)*(a + b)) + tan(f*x + e)/((b*tan(f*x + e)^2 + a + b)*(a + b)))/f
Time = 19.35 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.95 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^2} \, dx=\frac {\mathrm {tan}\left (e+f\,x\right )}{2\,f\,\left (a+b\right )\,\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a+b\right )}+\frac {\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )\,\left (2\,a+2\,b\right )}{2\,{\left (a+b\right )}^{3/2}}\right )}{2\,\sqrt {b}\,f\,{\left (a+b\right )}^{3/2}} \]