Integrand size = 14, antiderivative size = 144 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {x}{a^3}-\frac {\sqrt {b} \left (15 a^2+20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{8 a^3 (a+b)^{5/2} f}-\frac {b \tan (e+f x)}{4 a (a+b) f \left (a+b+b \tan ^2(e+f x)\right )^2}-\frac {b (7 a+4 b) \tan (e+f x)}{8 a^2 (a+b)^2 f \left (a+b+b \tan ^2(e+f x)\right )} \]
x/a^3-1/8*(15*a^2+20*a*b+8*b^2)*arctan(b^(1/2)*tan(f*x+e)/(a+b)^(1/2))*b^( 1/2)/a^3/(a+b)^(5/2)/f-1/4*b*tan(f*x+e)/a/(a+b)/f/(a+b+b*tan(f*x+e)^2)^2-1 /8*b*(7*a+4*b)*tan(f*x+e)/a^2/(a+b)^2/f/(a+b+b*tan(f*x+e)^2)
Result contains complex when optimal does not.
Time = 6.39 (sec) , antiderivative size = 332, normalized size of antiderivative = 2.31 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\frac {(a+2 b+a \cos (2 (e+f x))) \sec ^6(e+f x) \left (8 x (a+2 b+a \cos (2 (e+f x)))^2+\frac {b \left (15 a^2+20 a b+8 b^2\right ) \arctan \left (\frac {\sec (f x) (\cos (2 e)-i \sin (2 e)) (-((a+2 b) \sin (f x))+a \sin (2 e+f x))}{2 \sqrt {a+b} \sqrt {b (\cos (e)-i \sin (e))^4}}\right ) (a+2 b+a \cos (2 (e+f x)))^2 (\cos (2 e)-i \sin (2 e))}{(a+b)^{5/2} f \sqrt {b (\cos (e)-i \sin (e))^4}}-\frac {4 b^2 ((a+2 b) \sin (2 e)-a \sin (2 f x))}{(a+b) f (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}+\frac {b (a+2 b+a \cos (2 (e+f x))) \left (\left (9 a^2+28 a b+16 b^2\right ) \sin (2 e)-3 a (3 a+2 b) \sin (2 f x)\right )}{(a+b)^2 f (\cos (e)-\sin (e)) (\cos (e)+\sin (e))}\right )}{64 a^3 \left (a+b \sec ^2(e+f x)\right )^3} \]
((a + 2*b + a*Cos[2*(e + f*x)])*Sec[e + f*x]^6*(8*x*(a + 2*b + a*Cos[2*(e + f*x)])^2 + (b*(15*a^2 + 20*a*b + 8*b^2)*ArcTan[(Sec[f*x]*(Cos[2*e] - I*S in[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*S in[2*e]))/((a + b)^(5/2)*f*Sqrt[b*(Cos[e] - I*Sin[e])^4]) - (4*b^2*((a + 2 *b)*Sin[2*e] - a*Sin[2*f*x]))/((a + b)*f*(Cos[e] - Sin[e])*(Cos[e] + Sin[e ])) + (b*(a + 2*b + a*Cos[2*(e + f*x)])*((9*a^2 + 28*a*b + 16*b^2)*Sin[2*e ] - 3*a*(3*a + 2*b)*Sin[2*f*x]))/((a + b)^2*f*(Cos[e] - Sin[e])*(Cos[e] + Sin[e]))))/(64*a^3*(a + b*Sec[e + f*x]^2)^3)
Time = 0.32 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.22, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {3042, 4616, 316, 402, 397, 216, 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.
| \(\displaystyle \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {1}{\left (a+b \sec (e+f x)^2\right )^3}dx\) |
| \(\Big \downarrow \) 4616 |
| \(\displaystyle \frac {\int \frac {1}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^3}d\tan (e+f x)}{f}\) |
| \(\Big \downarrow \) 316 |
| \(\displaystyle \frac {\frac {\int \frac {-3 b \tan ^2(e+f x)+4 a+b}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )^2}d\tan (e+f x)}{4 a (a+b)}-\frac {b \tan (e+f x)}{4 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 402 |
| \(\displaystyle \frac {\frac {\frac {\int \frac {8 a^2+9 b a+4 b^2-b (7 a+4 b) \tan ^2(e+f x)}{\left (\tan ^2(e+f x)+1\right ) \left (b \tan ^2(e+f x)+a+b\right )}d\tan (e+f x)}{2 a (a+b)}-\frac {b (7 a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a (a+b)}-\frac {b \tan (e+f x)}{4 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 397 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 (a+b)^2 \int \frac {1}{\tan ^2(e+f x)+1}d\tan (e+f x)}{a}-\frac {b \left (15 a^2+20 a b+8 b^2\right ) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a (a+b)}-\frac {b (7 a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a (a+b)}-\frac {b \tan (e+f x)}{4 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 216 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 (a+b)^2 \arctan (\tan (e+f x))}{a}-\frac {b \left (15 a^2+20 a b+8 b^2\right ) \int \frac {1}{b \tan ^2(e+f x)+a+b}d\tan (e+f x)}{a}}{2 a (a+b)}-\frac {b (7 a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a (a+b)}-\frac {b \tan (e+f x)}{4 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle \frac {\frac {\frac {\frac {8 (a+b)^2 \arctan (\tan (e+f x))}{a}-\frac {\sqrt {b} \left (15 a^2+20 a b+8 b^2\right ) \arctan \left (\frac {\sqrt {b} \tan (e+f x)}{\sqrt {a+b}}\right )}{a \sqrt {a+b}}}{2 a (a+b)}-\frac {b (7 a+4 b) \tan (e+f x)}{2 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )}}{4 a (a+b)}-\frac {b \tan (e+f x)}{4 a (a+b) \left (a+b \tan ^2(e+f x)+b\right )^2}}{f}\) |
(-1/4*(b*Tan[e + f*x])/(a*(a + b)*(a + b + b*Tan[e + f*x]^2)^2) + (((8*(a + b)^2*ArcTan[Tan[e + f*x]])/a - (Sqrt[b]*(15*a^2 + 20*a*b + 8*b^2)*ArcTan [(Sqrt[b]*Tan[e + f*x])/Sqrt[a + b]])/(a*Sqrt[a + b]))/(2*a*(a + b)) - (b* (7*a + 4*b)*Tan[e + f*x])/(2*a*(a + b)*(a + b + b*Tan[e + f*x]^2)))/(4*a*( a + b)))/f
3.1.63.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
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[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_), x_Symbol] :> Sim p[(-b)*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^(q + 1)/(2*a*(p + 1)*(b*c - a*d)) ), x] + Simp[1/(2*a*(p + 1)*(b*c - a*d)) Int[(a + b*x^2)^(p + 1)*(c + d*x ^2)^q*Simp[b*c + 2*(p + 1)*(b*c - a*d) + d*b*(2*(p + q + 2) + 1)*x^2, x], x ], x] /; FreeQ[{a, b, c, d, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] && ! ( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomialQ[a, b, c, d, 2, p, q, x]
Int[((e_) + (f_.)*(x_)^2)/(((a_) + (b_.)*(x_)^2)*((c_) + (d_.)*(x_)^2)), x_ Symbol] :> Simp[(b*e - a*f)/(b*c - a*d) Int[1/(a + b*x^2), x], x] - Simp[ (d*e - c*f)/(b*c - a*d) Int[1/(c + d*x^2), x], x] /; FreeQ[{a, b, c, d, e , f}, x]
Int[((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_.)*((e_) + (f_.)*(x _)^2), x_Symbol] :> Simp[(-(b*e - a*f))*x*(a + b*x^2)^(p + 1)*((c + d*x^2)^ (q + 1)/(a*2*(b*c - a*d)*(p + 1))), x] + Simp[1/(a*2*(b*c - a*d)*(p + 1)) Int[(a + b*x^2)^(p + 1)*(c + d*x^2)^q*Simp[c*(b*e - a*f) + e*2*(b*c - a*d) *(p + 1) + d*(b*e - a*f)*(2*(p + q + 2) + 1)*x^2, x], x], x] /; FreeQ[{a, b , c, d, e, f, q}, x] && LtQ[p, -1]
Int[((a_) + (b_.)*sec[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Simp[ff/f Subst[Int[(a + b + b*ff^2*x^2)^p /(1 + ff^2*x^2), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && NeQ[a + b, 0] && NeQ[p, -1]
Time = 1.54 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.02
| method | result | size |
| derivativedivides | \(\frac {\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}-\frac {b \left (\frac {\frac {a b \left (7 a +4 b \right ) \tan \left (f x +e \right )^{3}}{8 a^{2}+16 a b +8 b^{2}}+\frac {\left (9 a +4 b \right ) a \tan \left (f x +e \right )}{8 a +8 b}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}+20 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}}{f}\) | \(147\) |
| default | \(\frac {\frac {\arctan \left (\tan \left (f x +e \right )\right )}{a^{3}}-\frac {b \left (\frac {\frac {a b \left (7 a +4 b \right ) \tan \left (f x +e \right )^{3}}{8 a^{2}+16 a b +8 b^{2}}+\frac {\left (9 a +4 b \right ) a \tan \left (f x +e \right )}{8 a +8 b}}{\left (a +b +b \tan \left (f x +e \right )^{2}\right )^{2}}+\frac {\left (15 a^{2}+20 a b +8 b^{2}\right ) \arctan \left (\frac {b \tan \left (f x +e \right )}{\sqrt {\left (a +b \right ) b}}\right )}{8 \left (a^{2}+2 a b +b^{2}\right ) \sqrt {\left (a +b \right ) b}}\right )}{a^{3}}}{f}\) | \(147\) |
| risch | \(\frac {x}{a^{3}}-\frac {i b \left (9 a^{3} {\mathrm e}^{6 i \left (f x +e \right )}+28 a^{2} b \,{\mathrm e}^{6 i \left (f x +e \right )}+16 a \,b^{2} {\mathrm e}^{6 i \left (f x +e \right )}+27 a^{3} {\mathrm e}^{4 i \left (f x +e \right )}+90 a^{2} b \,{\mathrm e}^{4 i \left (f x +e \right )}+120 a \,b^{2} {\mathrm e}^{4 i \left (f x +e \right )}+48 b^{3} {\mathrm e}^{4 i \left (f x +e \right )}+27 a^{3} {\mathrm e}^{2 i \left (f x +e \right )}+68 a^{2} b \,{\mathrm e}^{2 i \left (f x +e \right )}+32 a \,b^{2} {\mathrm e}^{2 i \left (f x +e \right )}+9 a^{3}+6 a^{2} b \right )}{4 a^{3} \left (a +b \right )^{2} f \left (a \,{\mathrm e}^{4 i \left (f x +e \right )}+2 a \,{\mathrm e}^{2 i \left (f x +e \right )}+4 b \,{\mathrm e}^{2 i \left (f x +e \right )}+a \right )^{2}}+\frac {15 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right )}{16 \left (a +b \right )^{3} f a}+\frac {5 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right ) b}{4 \left (a +b \right )^{3} f \,a^{2}}+\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}+\frac {2 i \sqrt {-\left (a +b \right ) b}+a +2 b}{a}\right ) b^{2}}{2 \left (a +b \right )^{3} f \,a^{3}}-\frac {15 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right )}{16 \left (a +b \right )^{3} f a}-\frac {5 \sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right ) b}{4 \left (a +b \right )^{3} f \,a^{2}}-\frac {\sqrt {-\left (a +b \right ) b}\, \ln \left ({\mathrm e}^{2 i \left (f x +e \right )}-\frac {2 i \sqrt {-\left (a +b \right ) b}-a -2 b}{a}\right ) b^{2}}{2 \left (a +b \right )^{3} f \,a^{3}}\) | \(548\) |
1/f*(1/a^3*arctan(tan(f*x+e))-b/a^3*((1/8*a*b*(7*a+4*b)/(a^2+2*a*b+b^2)*ta n(f*x+e)^3+1/8*(9*a+4*b)*a/(a+b)*tan(f*x+e))/(a+b+b*tan(f*x+e)^2)^2+1/8*(1 5*a^2+20*a*b+8*b^2)/(a^2+2*a*b+b^2)/((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. 366 vs. \(2 (130) = 260\).
Time = 0.33 (sec) , antiderivative size = 819, normalized size of antiderivative = 5.69 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\left [\frac {32 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f x \cos \left (f x + e\right )^{4} + 64 \, {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f x \cos \left (f x + e\right )^{2} + 32 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f x + {\left ({\left (15 \, a^{4} + 20 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 20 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 20 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-\frac {b}{a + b}} \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} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (a b + b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {-\frac {b}{a + b}} \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 (3 \, {\left (3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (7 \, a^{2} b^{2} + 4 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{32 \, {\left ({\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 2 \, a^{4} b^{3} + a^{3} b^{4}\right )} f\right )}}, \frac {16 \, {\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} f x \cos \left (f x + e\right )^{4} + 32 \, {\left (a^{3} b + 2 \, a^{2} b^{2} + a b^{3}\right )} f x \cos \left (f x + e\right )^{2} + 16 \, {\left (a^{2} b^{2} + 2 \, a b^{3} + b^{4}\right )} f x + {\left ({\left (15 \, a^{4} + 20 \, a^{3} b + 8 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{4} + 15 \, a^{2} b^{2} + 20 \, a b^{3} + 8 \, b^{4} + 2 \, {\left (15 \, a^{3} b + 20 \, a^{2} b^{2} + 8 \, a b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {\frac {b}{a + b}} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {b}{a + b}}}{2 \, b \cos \left (f x + e\right ) \sin \left (f x + e\right )}\right ) - 2 \, {\left (3 \, {\left (3 \, a^{3} b + 2 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (7 \, a^{2} b^{2} + 4 \, a b^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{16 \, {\left ({\left (a^{7} + 2 \, a^{6} b + a^{5} b^{2}\right )} f \cos \left (f x + e\right )^{4} + 2 \, {\left (a^{6} b + 2 \, a^{5} b^{2} + a^{4} b^{3}\right )} f \cos \left (f x + e\right )^{2} + {\left (a^{5} b^{2} + 2 \, a^{4} b^{3} + a^{3} b^{4}\right )} f\right )}}\right ] \]
[1/32*(32*(a^4 + 2*a^3*b + a^2*b^2)*f*x*cos(f*x + e)^4 + 64*(a^3*b + 2*a^2 *b^2 + a*b^3)*f*x*cos(f*x + e)^2 + 32*(a^2*b^2 + 2*a*b^3 + b^4)*f*x + ((15 *a^4 + 20*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 + 15*a^2*b^2 + 20*a*b^3 + 8*b^ 4 + 2*(15*a^3*b + 20*a^2*b^2 + 8*a*b^3)*cos(f*x + e)^2)*sqrt(-b/(a + b))*l og(((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 + 3*a*b + 2*b^2)*cos(f*x + e)^3 - (a*b + b^2)*cos(f*x + e))*sq rt(-b/(a + b))*sin(f*x + e) + b^2)/(a^2*cos(f*x + e)^4 + 2*a*b*cos(f*x + e )^2 + b^2)) - 4*(3*(3*a^3*b + 2*a^2*b^2)*cos(f*x + e)^3 + (7*a^2*b^2 + 4*a *b^3)*cos(f*x + e))*sin(f*x + e))/((a^7 + 2*a^6*b + a^5*b^2)*f*cos(f*x + e )^4 + 2*(a^6*b + 2*a^5*b^2 + a^4*b^3)*f*cos(f*x + e)^2 + (a^5*b^2 + 2*a^4* b^3 + a^3*b^4)*f), 1/16*(16*(a^4 + 2*a^3*b + a^2*b^2)*f*x*cos(f*x + e)^4 + 32*(a^3*b + 2*a^2*b^2 + a*b^3)*f*x*cos(f*x + e)^2 + 16*(a^2*b^2 + 2*a*b^3 + b^4)*f*x + ((15*a^4 + 20*a^3*b + 8*a^2*b^2)*cos(f*x + e)^4 + 15*a^2*b^2 + 20*a*b^3 + 8*b^4 + 2*(15*a^3*b + 20*a^2*b^2 + 8*a*b^3)*cos(f*x + e)^2)* sqrt(b/(a + b))*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - b)*sqrt(b/(a + b))/ (b*cos(f*x + e)*sin(f*x + e))) - 2*(3*(3*a^3*b + 2*a^2*b^2)*cos(f*x + e)^3 + (7*a^2*b^2 + 4*a*b^3)*cos(f*x + e))*sin(f*x + e))/((a^7 + 2*a^6*b + a^5 *b^2)*f*cos(f*x + e)^4 + 2*(a^6*b + 2*a^5*b^2 + a^4*b^3)*f*cos(f*x + e)^2 + (a^5*b^2 + 2*a^4*b^3 + a^3*b^4)*f)]
\[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\int \frac {1}{\left (a + b \sec ^{2}{\left (e + f x \right )}\right )^{3}}\, dx \]
Time = 0.27 (sec) , antiderivative size = 231, normalized size of antiderivative = 1.60 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {{\left (a + b\right )} b}}\right )}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {{\left (a + b\right )} b}} + \frac {{\left (7 \, a b^{2} + 4 \, b^{3}\right )} \tan \left (f x + e\right )^{3} + {\left (9 \, a^{2} b + 13 \, a b^{2} + 4 \, b^{3}\right )} \tan \left (f x + e\right )}{a^{6} + 4 \, a^{5} b + 6 \, a^{4} b^{2} + 4 \, a^{3} b^{3} + a^{2} b^{4} + {\left (a^{4} b^{2} + 2 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (f x + e\right )^{4} + 2 \, {\left (a^{5} b + 3 \, a^{4} b^{2} + 3 \, a^{3} b^{3} + a^{2} b^{4}\right )} \tan \left (f x + e\right )^{2}} - \frac {8 \, {\left (f x + e\right )}}{a^{3}}}{8 \, f} \]
-1/8*((15*a^2*b + 20*a*b^2 + 8*b^3)*arctan(b*tan(f*x + e)/sqrt((a + b)*b)) /((a^5 + 2*a^4*b + a^3*b^2)*sqrt((a + b)*b)) + ((7*a*b^2 + 4*b^3)*tan(f*x + e)^3 + (9*a^2*b + 13*a*b^2 + 4*b^3)*tan(f*x + e))/(a^6 + 4*a^5*b + 6*a^4 *b^2 + 4*a^3*b^3 + a^2*b^4 + (a^4*b^2 + 2*a^3*b^3 + a^2*b^4)*tan(f*x + e)^ 4 + 2*(a^5*b + 3*a^4*b^2 + 3*a^3*b^3 + a^2*b^4)*tan(f*x + e)^2) - 8*(f*x + e)/a^3)/f
Time = 0.29 (sec) , antiderivative size = 196, normalized size of antiderivative = 1.36 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=-\frac {\frac {{\left (15 \, a^{2} b + 20 \, a b^{2} + 8 \, b^{3}\right )} {\left (\pi \left \lfloor \frac {f x + e}{\pi } + \frac {1}{2} \right \rfloor \mathrm {sgn}\left (b\right ) + \arctan \left (\frac {b \tan \left (f x + e\right )}{\sqrt {a b + b^{2}}}\right )\right )}}{{\left (a^{5} + 2 \, a^{4} b + a^{3} b^{2}\right )} \sqrt {a b + b^{2}}} + \frac {7 \, a b^{2} \tan \left (f x + e\right )^{3} + 4 \, b^{3} \tan \left (f x + e\right )^{3} + 9 \, a^{2} b \tan \left (f x + e\right ) + 13 \, a b^{2} \tan \left (f x + e\right ) + 4 \, b^{3} \tan \left (f x + e\right )}{{\left (a^{4} + 2 \, a^{3} b + a^{2} b^{2}\right )} {\left (b \tan \left (f x + e\right )^{2} + a + b\right )}^{2}} - \frac {8 \, {\left (f x + e\right )}}{a^{3}}}{8 \, f} \]
-1/8*((15*a^2*b + 20*a*b^2 + 8*b^3)*(pi*floor((f*x + e)/pi + 1/2)*sgn(b) + arctan(b*tan(f*x + e)/sqrt(a*b + b^2)))/((a^5 + 2*a^4*b + a^3*b^2)*sqrt(a *b + b^2)) + (7*a*b^2*tan(f*x + e)^3 + 4*b^3*tan(f*x + e)^3 + 9*a^2*b*tan( f*x + e) + 13*a*b^2*tan(f*x + e) + 4*b^3*tan(f*x + e))/((a^4 + 2*a^3*b + a ^2*b^2)*(b*tan(f*x + e)^2 + a + b)^2) - 8*(f*x + e)/a^3)/f
Time = 22.74 (sec) , antiderivative size = 3271, normalized size of antiderivative = 22.72 \[ \int \frac {1}{\left (a+b \sec ^2(e+f x)\right )^3} \, dx=\text {Too large to display} \]
atan((((((2*a^6*b^6 + (17*a^7*b^5)/2 + 15*a^8*b^4 + (25*a^9*b^3)/2 + 4*a^1 0*b^2)*1i)/(2*(4*a^9*b + a^10 + a^6*b^4 + 4*a^7*b^3 + 6*a^8*b^2)) - (tan(e + f*x)*(512*a^6*b^7 + 2304*a^7*b^6 + 4096*a^8*b^5 + 3584*a^9*b^4 + 1536*a ^10*b^3 + 256*a^11*b^2))/(128*a^3*(4*a^7*b + a^8 + a^4*b^4 + 4*a^5*b^3 + 6 *a^6*b^2)))/(2*a^3) + (tan(e + f*x)*(576*a*b^6 + 128*b^7 + 1024*a^2*b^5 + 856*a^3*b^4 + 289*a^4*b^3))/(64*(4*a^7*b + a^8 + a^4*b^4 + 4*a^5*b^3 + 6*a ^6*b^2)))/a^3 - ((((2*a^6*b^6 + (17*a^7*b^5)/2 + 15*a^8*b^4 + (25*a^9*b^3) /2 + 4*a^10*b^2)*1i)/(2*(4*a^9*b + a^10 + a^6*b^4 + 4*a^7*b^3 + 6*a^8*b^2) ) + (tan(e + f*x)*(512*a^6*b^7 + 2304*a^7*b^6 + 4096*a^8*b^5 + 3584*a^9*b^ 4 + 1536*a^10*b^3 + 256*a^11*b^2))/(128*a^3*(4*a^7*b + a^8 + a^4*b^4 + 4*a ^5*b^3 + 6*a^6*b^2)))/(2*a^3) - (tan(e + f*x)*(576*a*b^6 + 128*b^7 + 1024* a^2*b^5 + 856*a^3*b^4 + 289*a^4*b^3))/(64*(4*a^7*b + a^8 + a^4*b^4 + 4*a^5 *b^3 + 6*a^6*b^2)))/a^3)/(((17*a*b^5)/4 + b^6 + (25*a^2*b^4)/4 + (105*a^3* b^3)/32)/(4*a^9*b + a^10 + a^6*b^4 + 4*a^7*b^3 + 6*a^8*b^2) + (((((2*a^6*b ^6 + (17*a^7*b^5)/2 + 15*a^8*b^4 + (25*a^9*b^3)/2 + 4*a^10*b^2)*1i)/(2*(4* a^9*b + a^10 + a^6*b^4 + 4*a^7*b^3 + 6*a^8*b^2)) - (tan(e + f*x)*(512*a^6* b^7 + 2304*a^7*b^6 + 4096*a^8*b^5 + 3584*a^9*b^4 + 1536*a^10*b^3 + 256*a^1 1*b^2))/(128*a^3*(4*a^7*b + a^8 + a^4*b^4 + 4*a^5*b^3 + 6*a^6*b^2)))*1i)/( 2*a^3) + (tan(e + f*x)*(576*a*b^6 + 128*b^7 + 1024*a^2*b^5 + 856*a^3*b^4 + 289*a^4*b^3)*1i)/(64*(4*a^7*b + a^8 + a^4*b^4 + 4*a^5*b^3 + 6*a^6*b^2)...