Integrand size = 23, antiderivative size = 106 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {3 \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 \sqrt {b} (a+b)^{5/2} f}+\frac {\tan (e+f x)}{4 (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}+\frac {3 \tan (e+f x)}{8 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )} \]
3/8*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))/(a+b)^(5/2)/f/b^(1/2)+1/4*tan(f *x+e)/(a+b)/f/(a+b+b*tan(f*x+e)^2)^2+3/8*tan(f*x+e)/(a+b)^2/f/(a+b+b*tan(f *x+e)^2)
Result contains complex when optimal does not.
Time = 4.59 (sec) , antiderivative size = 265, normalized size of antiderivative = 2.50 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^6(e+f x) \left (-\frac {3 \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x)))^2 (\cos (2 e)-i \sin (2 e))}{\sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}+\frac {4 b (a+b) \sec (2 e) ((a+2 b) \sin (2 e)-a \sin (2 f x))}{a^2}+\frac {(a+2 b+a \cos (2 (e+f x))) \sec (2 e) \left (-\left (\left (5 a^2+16 a b+8 b^2\right ) \sin (2 e)\right )+a (5 a+2 b) \sin (2 f x)\right )}{a^2}\right )}{64 (a+b)^2 f \left (a+b \sec ^2(e+f x)\right )^3} \]
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*((-3*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)])^2*(Cos[ 2*e] - I*Sin[2*e]))/(Sqrt[a + b]*Sqrt[b*(Cos[e] - I*Sin[e])^4]) + (4*b*(a + b)*Sec[2*e]*((a + 2*b)*Sin[2*e] - a*Sin[2*f*x]))/a^2 + ((a + 2*b + a*Cos [2*(e + f*x)])*Sec[2*e]*(-((5*a^2 + 16*a*b + 8*b^2)*Sin[2*e]) + a*(5*a + 2 *b)*Sin[2*f*x]))/a^2))/(64*(a + b)^2*f*(a + b*Sec[e + f*x]^2)^3)
Time = 0.26 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3042, 4634, 215, 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.
| \(\displaystyle \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sec (e+f x)^2}{\left (a+b \sec (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4634 |
| \(\displaystyle \frac {\int \frac {1}{\left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\frac {3 \int \frac {1}{\left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{4 (a+b)}+\frac {\tan (e+f x)}{4 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 215 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{2 (a+b)}+\frac {\tan (e+f x)}{2 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}\right )}{4 (a+b)}+\frac {\tan (e+f x)}{4 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {3 \left (\frac {\arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{2 \sqrt {b} (a+b)^{3/2}}+\frac {\tan (e+f x)}{2 (a+b) \left (a+b \tan ^2(e+f x)+b\right )}\right )}{4 (a+b)}+\frac {\tan (e+f x)}{4 (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
(Tan[e + f*x]/(4*(a + b)*(a + b + b*Tan[e + f*x]^2)^2) + (3*(ArcTan[(Sqrt[ b]*Tan[e + f*x])/Sqrt[a + b]]/(2*Sqrt[b]*(a + b)^(3/2)) + Tan[e + f*x]/(2* (a + b)*(a + b + b*Tan[e + f*x]^2))))/(4*(a + b)))/f
3.3.14.3.1 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 = 1.58 (sec) , antiderivative size = 100, normalized size of antiderivative = 0.94
| method | result | size |
| derivativedivides | \(\frac {\frac {\tan \left (f x +e \right )}{4 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\frac {3 \tan \left (f x +e \right )}{8 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {3 \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}}{a +b}}{f}\) | \(100\) |
| default | \(\frac {\frac {\tan \left (f x +e \right )}{4 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\frac {3 \tan \left (f x +e \right )}{8 \left (a +b \right ) \left (a +b +b \tan \left (f x +e \right )^{2}\right )}+\frac {3 \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a +b \right ) \sqrt {\left (a +b \right ) b}}}{a +b}}{f}\) | \(100\) |
| risch | \(\frac {i \left (5 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}+16 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}+8 a \,b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+15 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+46 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+56 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+16 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+15 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+32 a^{2} b \,{\mathrm e}^{2 i \left (f x +e \right )}+8 a \,b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+5 a^{3}+2 a^{2} b \right )}{4 a^{2} \left (a +b \right )^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}-\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i b a +2 i b^{2}+a \sqrt {-a b -b^{2}}+2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{16 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f}+\frac {3 \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i b a +2 i b^{2}-a \sqrt {-a b -b^{2}}-2 b \sqrt {-a b -b^{2}}}{a \sqrt {-a b -b^{2}}}\right )}{16 \sqrt {-a b -b^{2}}\, \left (a +b \right )^{2} f}\) | \(395\) |
1/f*(1/4*tan(f*x+e)/(a+b)/(a+b+b*tan(f*x+e)^2)^2+3/4/(a+b)*(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)*arctan(b*tan(f*x+e)/ ((a+b)*b)^(1/2))))
Leaf count of result is larger than twice the leaf count of optimal. 250 vs. \(2 (92) = 184\).
Time = 0.30 (sec) , antiderivative size = 580, normalized size of antiderivative = 5.47 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [-\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\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 ) - 4 \, {\left ({\left (5 \, a^{2} b + 7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\left ({\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} f\right )}}, -\frac {3 \, {\left (a^{2} \cos \left (f x + e\right )^{4} + 2 \, a b \cos \left (f x + e\right )^{2} + b^{2}\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 ) - 2 \, {\left ({\left (5 \, a^{2} b + 7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{3} + 3 \, {\left (a b^{2} + b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{4} b^{2} + 3 \, a^{3} b^{3} + 3 \, a^{2} b^{4} + a b^{5}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{3} b^{3} + 3 \, a^{2} b^{4} + 3 \, a b^{5} + b^{6}\right )} f\right )}}\right ] \]
[-1/32*(3*(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e)^2 + b^2)*sqrt(-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)) - 4*(( 5*a^2*b + 7*a*b^2 + 2*b^3)*cos(f*x + e)^3 + 3*(a*b^2 + b^3)*cos(f*x + e))* sin(f*x + e))/((a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f*cos(f*x + e)^4 + 2*(a^4*b^2 + 3*a^3*b^3 + 3*a^2*b^4 + a*b^5)*f*cos(f*x + e)^2 + (a^3*b^3 + 3*a^2*b^4 + 3*a*b^5 + b^6)*f), -1/16*(3*(a^2*cos(f*x + e)^4 + 2*a*b*cos( f*x + e)^2 + b^2)*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))) - 2*((5*a^2*b + 7*a*b^2 + 2 *b^3)*cos(f*x + e)^3 + 3*(a*b^2 + b^3)*cos(f*x + e))*sin(f*x + e))/((a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*f*cos(f*x + e)^4 + 2*(a^4*b^2 + 3*a^3* b^3 + 3*a^2*b^4 + a*b^5)*f*cos(f*x + e)^2 + (a^3*b^3 + 3*a^2*b^4 + 3*a*b^5 + b^6)*f)]
\[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\int \frac {\sec ^{2}{\left (e + f x \right )}}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \]
Time = 0.28 (sec) , antiderivative size = 156, normalized size of antiderivative = 1.47 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {3 \, b \tan \left (f x + e\right )^{3} + 5 \, {\left (a + b\right )} \tan \left (f x + e\right )}{{\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{4} + a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4} + 2 \, {\left (a^{3} b + 3 \, a^{2} b^{2} + 3 \, a b^{3} + b^{4}\right )} \tan \left (f x + e\right )^{2}} + \frac {3 \, \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {{\left (a + b\right )} b}}}{8 \, f} \]
1/8*((3*b*tan(f*x + e)^3 + 5*(a + b)*tan(f*x + e))/((a^2*b^2 + 2*a*b^3 + b ^4)*tan(f*x + e)^4 + a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4 + 2*(a^3*b + 3*a^2*b^2 + 3*a*b^3 + b^4)*tan(f*x + e)^2) + 3*arctan(b*tan(f*x + e)/sqr t((a + b)*b))/((a^2 + 2*a*b + b^2)*sqrt((a + b)*b)))/f
Time = 0.33 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.17 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {3 \, {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{2} + 2 \, a b + b^{2}\right )} \sqrt {a b + b^{2}}} + \frac {3 \, b \tan \left (f x + e\right )^{3} + 5 \, a \tan \left (f x + e\right ) + 5 \, b \tan \left (f x + e\right )}{{\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2} {\left (a^{2} + 2 \, a b + b^{2}\right )}}}{8 \, f} \]
1/8*(3*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a *b + b^2)))/((a^2 + 2*a*b + b^2)*sqrt(a*b + b^2)) + (3*b*tan(f*x + e)^3 + 5*a*tan(f*x + e) + 5*b*tan(f*x + e))/((b*tan(f*x + e)^2 + a + b)^2*(a^2 + 2*a*b + b^2)))/f
Time = 19.07 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.06 \[ \int \frac {\sec ^2(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {\frac {5\,\mathrm {tan}\left (e+f\,x\right )}{8\,\left (a+b\right )}+\frac {3\,b\,{\mathrm {tan}\left (e+f\,x\right )}^3}{8\,{\left (a+b\right )}^2}}{f\,\left (2\,a\,b+a^2+b^2+{\mathrm {tan}\left (e+f\,x\right )}^2\,\left (2\,b^2+2\,a\,b\right )+b^2\,{\mathrm {tan}\left (e+f\,x\right )}^4\right )}+\frac {3\,\mathrm {atan}\left (\frac {\sqrt {b}\,\mathrm {tan}\left (e+f\,x\right )}{\sqrt {a+b}}\right )}{8\,\sqrt {b}\,f\,{\left (a+b\right )}^{5/2}} \]