Integrand size = 25, antiderivative size = 237 \[ \int \frac {(a+b \sec (e+f x))^2}{(c+d \sec (e+f x))^3} \, dx=\frac {a^2 x}{c^3}-\frac {\left (3 b^2 c^4 d-2 a b c^3 \left (2 c^2+d^2\right )+a^2 \left (6 c^4 d-5 c^2 d^3+2 d^5\right )\right ) \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{c^3 (c-d)^{5/2} (c+d)^{5/2} f}-\frac {d (b c-a d)^2 \sin (e+f x)}{2 c^2 \left (c^2-d^2\right ) f (d+c \cos (e+f x))^2}-\frac {(b c-a d) \left (3 a d \left (2 c^2-d^2\right )-b c \left (2 c^2+d^2\right )\right ) \sin (e+f x)}{2 c^2 \left (c^2-d^2\right )^2 f (d+c \cos (e+f x))} \]
a^2*x/c^3-(3*b^2*c^4*d-2*a*b*c^3*(2*c^2+d^2)+a^2*(6*c^4*d-5*c^2*d^3+2*d^5) )*arctanh((c-d)^(1/2)*tan(1/2*f*x+1/2*e)/(c+d)^(1/2))/c^3/(c-d)^(5/2)/(c+d )^(5/2)/f-1/2*d*(-a*d+b*c)^2*sin(f*x+e)/c^2/(c^2-d^2)/f/(d+c*cos(f*x+e))^2 -1/2*(-a*d+b*c)*(3*a*d*(2*c^2-d^2)-b*c*(2*c^2+d^2))*sin(f*x+e)/c^2/(c^2-d^ 2)^2/f/(d+c*cos(f*x+e))
Leaf count is larger than twice the leaf count of optimal. \(493\) vs. \(2(237)=474\).
Time = 2.69 (sec) , antiderivative size = 493, normalized size of antiderivative = 2.08 \[ \int \frac {(a+b \sec (e+f x))^2}{(c+d \sec (e+f x))^3} \, dx=\frac {(d+c \cos (e+f x)) \sec (e+f x) (a+b \sec (e+f x))^2 \left (\frac {4 \left (3 b^2 c^4 d-2 a b c^3 \left (2 c^2+d^2\right )+a^2 \left (6 c^4 d-5 c^2 d^3+2 d^5\right )\right ) \text {arctanh}\left (\frac {(-c+d) \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c^2-d^2}}\right ) (d+c \cos (e+f x))^2}{\left (c^2-d^2\right )^{5/2}}+\frac {2 a^2 c^6 e-6 a^2 c^2 d^4 e+4 a^2 d^6 e+2 a^2 c^6 f x-6 a^2 c^2 d^4 f x+4 a^2 d^6 f x+8 a^2 c d \left (c^2-d^2\right )^2 (e+f x) \cos (e+f x)+2 a^2 c^2 \left (c^2-d^2\right )^2 (e+f x) \cos (2 (e+f x))+2 b^2 c^5 d \sin (e+f x)-12 a b c^4 d^2 \sin (e+f x)+10 a^2 c^3 d^3 \sin (e+f x)+4 b^2 c^3 d^3 \sin (e+f x)-4 a^2 c d^5 \sin (e+f x)+2 b^2 c^6 \sin (2 (e+f x))-8 a b c^5 d \sin (2 (e+f x))+6 a^2 c^4 d^2 \sin (2 (e+f x))+b^2 c^4 d^2 \sin (2 (e+f x))+2 a b c^3 d^3 \sin (2 (e+f x))-3 a^2 c^2 d^4 \sin (2 (e+f x))}{\left (c^2-d^2\right )^2}\right )}{4 c^3 f (b+a \cos (e+f x))^2 (c+d \sec (e+f x))^3} \]
((d + c*Cos[e + f*x])*Sec[e + f*x]*(a + b*Sec[e + f*x])^2*((4*(3*b^2*c^4*d - 2*a*b*c^3*(2*c^2 + d^2) + a^2*(6*c^4*d - 5*c^2*d^3 + 2*d^5))*ArcTanh[(( -c + d)*Tan[(e + f*x)/2])/Sqrt[c^2 - d^2]]*(d + c*Cos[e + f*x])^2)/(c^2 - d^2)^(5/2) + (2*a^2*c^6*e - 6*a^2*c^2*d^4*e + 4*a^2*d^6*e + 2*a^2*c^6*f*x - 6*a^2*c^2*d^4*f*x + 4*a^2*d^6*f*x + 8*a^2*c*d*(c^2 - d^2)^2*(e + f*x)*Co s[e + f*x] + 2*a^2*c^2*(c^2 - d^2)^2*(e + f*x)*Cos[2*(e + f*x)] + 2*b^2*c^ 5*d*Sin[e + f*x] - 12*a*b*c^4*d^2*Sin[e + f*x] + 10*a^2*c^3*d^3*Sin[e + f* x] + 4*b^2*c^3*d^3*Sin[e + f*x] - 4*a^2*c*d^5*Sin[e + f*x] + 2*b^2*c^6*Sin [2*(e + f*x)] - 8*a*b*c^5*d*Sin[2*(e + f*x)] + 6*a^2*c^4*d^2*Sin[2*(e + f* x)] + b^2*c^4*d^2*Sin[2*(e + f*x)] + 2*a*b*c^3*d^3*Sin[2*(e + f*x)] - 3*a^ 2*c^2*d^4*Sin[2*(e + f*x)])/(c^2 - d^2)^2))/(4*c^3*f*(b + a*Cos[e + f*x])^ 2*(c + d*Sec[e + f*x])^3)
Time = 1.19 (sec) , antiderivative size = 269, normalized size of antiderivative = 1.14, number of steps used = 13, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.480, Rules used = {3042, 4429, 3042, 3467, 25, 3042, 3500, 3042, 3214, 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.
| \(\displaystyle \int \frac {(a+b \sec (e+f x))^2}{(c+d \sec (e+f x))^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (a+b \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}{\left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx\) |
| \(\Big \downarrow \) 4429 |
| \(\displaystyle \int \frac {\cos (e+f x) (a \cos (e+f x)+b)^2}{(c \cos (e+f x)+d)^3}dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\sin \left (e+f x+\frac {\pi }{2}\right ) \left (a \sin \left (e+f x+\frac {\pi }{2}\right )+b\right )^2}{\left (c \sin \left (e+f x+\frac {\pi }{2}\right )+d\right )^3}dx\) |
| \(\Big \downarrow \) 3467 |
| \(\displaystyle -\frac {\int -\frac {2 c (b c-a d)^2+2 a^2 c \left (c^2-d^2\right ) \cos ^2(e+f x)-\left (\left (2 c^2 d-d^3\right ) a^2-2 b c \left (2 c^2-d^2\right ) a+b^2 c^2 d\right ) \cos (e+f x)}{(d+c \cos (e+f x))^2}dx}{2 c^2 \left (c^2-d^2\right )}-\frac {d (b c-a d)^2 \sin (e+f x)}{2 c^2 f \left (c^2-d^2\right ) (c \cos (e+f x)+d)^2}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle \frac {\int \frac {2 c (b c-a d)^2+2 a^2 c \left (c^2-d^2\right ) \cos ^2(e+f x)-\left (\left (2 c^2 d-d^3\right ) a^2-2 b c \left (2 c^2-d^2\right ) a+b^2 c^2 d\right ) \cos (e+f x)}{(d+c \cos (e+f x))^2}dx}{2 c^2 \left (c^2-d^2\right )}-\frac {d (b c-a d)^2 \sin (e+f x)}{2 c^2 f \left (c^2-d^2\right ) (c \cos (e+f x)+d)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {2 c (b c-a d)^2+2 a^2 c \left (c^2-d^2\right ) \sin \left (e+f x+\frac {\pi }{2}\right )^2+\left (-\left (\left (2 c^2 d-d^3\right ) a^2\right )+2 b c \left (2 c^2-d^2\right ) a-b^2 c^2 d\right ) \sin \left (e+f x+\frac {\pi }{2}\right )}{\left (d+c \sin \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx}{2 c^2 \left (c^2-d^2\right )}-\frac {d (b c-a d)^2 \sin (e+f x)}{2 c^2 f \left (c^2-d^2\right ) (c \cos (e+f x)+d)^2}\) |
| \(\Big \downarrow \) 3500 |
| \(\displaystyle \frac {\frac {\int \frac {(b c-a d) \left (4 a c^2-3 b d c-a d^2\right ) c^2+2 a^2 \left (c^2-d^2\right )^2 \cos (e+f x) c}{d+c \cos (e+f x)}dx}{c \left (c^2-d^2\right )}+\frac {(b c-a d) \left (-6 a c^2 d+3 a d^3+2 b c^3+b c d^2\right ) \sin (e+f x)}{f \left (c^2-d^2\right ) (c \cos (e+f x)+d)}}{2 c^2 \left (c^2-d^2\right )}-\frac {d (b c-a d)^2 \sin (e+f x)}{2 c^2 f \left (c^2-d^2\right ) (c \cos (e+f x)+d)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {\int \frac {(b c-a d) \left (4 a c^2-3 b d c-a d^2\right ) c^2+2 a^2 \left (c^2-d^2\right )^2 \sin \left (e+f x+\frac {\pi }{2}\right ) c}{d+c \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{c \left (c^2-d^2\right )}+\frac {(b c-a d) \left (-6 a c^2 d+3 a d^3+2 b c^3+b c d^2\right ) \sin (e+f x)}{f \left (c^2-d^2\right ) (c \cos (e+f x)+d)}}{2 c^2 \left (c^2-d^2\right )}-\frac {d (b c-a d)^2 \sin (e+f x)}{2 c^2 f \left (c^2-d^2\right ) (c \cos (e+f x)+d)^2}\) |
| \(\Big \downarrow \) 3214 |
| \(\displaystyle \frac {\frac {2 a^2 x \left (c^2-d^2\right )^2-\left (a^2 \left (6 c^4 d-5 c^2 d^3+2 d^5\right )-2 a b c^3 \left (2 c^2+d^2\right )+3 b^2 c^4 d\right ) \int \frac {1}{d+c \cos (e+f x)}dx}{c \left (c^2-d^2\right )}+\frac {(b c-a d) \left (-6 a c^2 d+3 a d^3+2 b c^3+b c d^2\right ) \sin (e+f x)}{f \left (c^2-d^2\right ) (c \cos (e+f x)+d)}}{2 c^2 \left (c^2-d^2\right )}-\frac {d (b c-a d)^2 \sin (e+f x)}{2 c^2 f \left (c^2-d^2\right ) (c \cos (e+f x)+d)^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\frac {2 a^2 x \left (c^2-d^2\right )^2-\left (a^2 \left (6 c^4 d-5 c^2 d^3+2 d^5\right )-2 a b c^3 \left (2 c^2+d^2\right )+3 b^2 c^4 d\right ) \int \frac {1}{d+c \sin \left (e+f x+\frac {\pi }{2}\right )}dx}{c \left (c^2-d^2\right )}+\frac {(b c-a d) \left (-6 a c^2 d+3 a d^3+2 b c^3+b c d^2\right ) \sin (e+f x)}{f \left (c^2-d^2\right ) (c \cos (e+f x)+d)}}{2 c^2 \left (c^2-d^2\right )}-\frac {d (b c-a d)^2 \sin (e+f x)}{2 c^2 f \left (c^2-d^2\right ) (c \cos (e+f x)+d)^2}\) |
| \(\Big \downarrow \) 3138 |
| \(\displaystyle \frac {\frac {2 a^2 x \left (c^2-d^2\right )^2-\frac {2 \left (a^2 \left (6 c^4 d-5 c^2 d^3+2 d^5\right )-2 a b c^3 \left (2 c^2+d^2\right )+3 b^2 c^4 d\right ) \int \frac {1}{-\left ((c-d) \tan ^2\left (\frac {1}{2} (e+f x)\right )\right )+c+d}d\tan \left (\frac {1}{2} (e+f x)\right )}{f}}{c \left (c^2-d^2\right )}+\frac {(b c-a d) \left (-6 a c^2 d+3 a d^3+2 b c^3+b c d^2\right ) \sin (e+f x)}{f \left (c^2-d^2\right ) (c \cos (e+f x)+d)}}{2 c^2 \left (c^2-d^2\right )}-\frac {d (b c-a d)^2 \sin (e+f x)}{2 c^2 f \left (c^2-d^2\right ) (c \cos (e+f x)+d)^2}\) |
| \(\Big \downarrow \) 221 |
| \(\displaystyle \frac {\frac {2 a^2 x \left (c^2-d^2\right )^2-\frac {2 \left (a^2 \left (6 c^4 d-5 c^2 d^3+2 d^5\right )-2 a b c^3 \left (2 c^2+d^2\right )+3 b^2 c^4 d\right ) \text {arctanh}\left (\frac {\sqrt {c-d} \tan \left (\frac {1}{2} (e+f x)\right )}{\sqrt {c+d}}\right )}{f \sqrt {c-d} \sqrt {c+d}}}{c \left (c^2-d^2\right )}+\frac {(b c-a d) \left (-6 a c^2 d+3 a d^3+2 b c^3+b c d^2\right ) \sin (e+f x)}{f \left (c^2-d^2\right ) (c \cos (e+f x)+d)}}{2 c^2 \left (c^2-d^2\right )}-\frac {d (b c-a d)^2 \sin (e+f x)}{2 c^2 f \left (c^2-d^2\right ) (c \cos (e+f x)+d)^2}\) |
-1/2*(d*(b*c - a*d)^2*Sin[e + f*x])/(c^2*(c^2 - d^2)*f*(d + c*Cos[e + f*x] )^2) + ((2*a^2*(c^2 - d^2)^2*x - (2*(3*b^2*c^4*d - 2*a*b*c^3*(2*c^2 + d^2) + a^2*(6*c^4*d - 5*c^2*d^3 + 2*d^5))*ArcTanh[(Sqrt[c - d]*Tan[(e + f*x)/2 ])/Sqrt[c + d]])/(Sqrt[c - d]*Sqrt[c + d]*f))/(c*(c^2 - d^2)) + ((b*c - a* d)*(2*b*c^3 - 6*a*c^2*d + b*c*d^2 + 3*a*d^3)*Sin[e + f*x])/((c^2 - d^2)*f* (d + c*Cos[e + f*x])))/(2*c^2*(c^2 - d^2))
3.2.93.3.1 Defintions of rubi rules used
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[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[b*(x/d), x] - Simp[(b*c - a*d)/d Int[1/(c + d *Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*sin[(e_.) + (f _.)*(x_)])*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[ (B*c - A*d)*(b*c - a*d)^2*Cos[e + f*x]*((c + d*Sin[e + f*x])^(n + 1)/(f*d^2 *(n + 1)*(c^2 - d^2))), x] - Simp[1/(d^2*(n + 1)*(c^2 - d^2)) Int[(c + d* Sin[e + f*x])^(n + 1)*Simp[d*(n + 1)*(B*(b*c - a*d)^2 - A*d*(a^2*c + b^2*c - 2*a*b*d)) - ((B*c - A*d)*(a^2*d^2*(n + 2) + b^2*(c^2 + d^2*(n + 1))) + 2* a*b*d*(A*c*d*(n + 2) - B*(c^2 + d^2*(n + 1))))*Sin[e + f*x] - b^2*B*d*(n + 1)*(c^2 - d^2)*Sin[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B} , x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[ n, -1]
Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[(-(A*b^2 - a*b*B + a^2*C))*Cos[e + f*x]*((a + b*Sin[e + f*x])^(m + 1)/(b*f*(m + 1)* (a^2 - b^2))), x] + Simp[1/(b*(m + 1)*(a^2 - b^2)) Int[(a + b*Sin[e + f*x ])^(m + 1)*Simp[b*(a*A - b*B + a*C)*(m + 1) - (A*b^2 - a*b*B + a^2*C + b*(A *b - a*B + b*C)*(m + 1))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B, C}, x] && LtQ[m, -1] && NeQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_))^(n_), x_Symbol] :> Int[(b + a*Sin[e + f*x])^m*((d + c*Sin[e + f *x])^n/Sin[e + f*x]^(m + n)), x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && N eQ[b*c - a*d, 0] && IntegerQ[m] && IntegerQ[n] && LeQ[-2, m + n, 0]
Time = 1.01 (sec) , antiderivative size = 386, normalized size of antiderivative = 1.63
| method | result | size |
| derivativedivides | \(\frac {\frac {2 a^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c^{3}}+\frac {\frac {2 \left (-\frac {\left (6 a^{2} c^{2} d^{2}+a^{2} d^{3} c -2 a^{2} d^{4}-8 a b \,c^{3} d -2 a b \,c^{2} d^{2}+2 b^{2} c^{4}+b^{2} c^{3} d +2 b^{2} c^{2} d^{2}\right ) c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 \left (c -d \right ) \left (c^{2}+2 c d +d^{2}\right )}+\frac {c \left (6 a^{2} c^{2} d^{2}-a^{2} d^{3} c -2 a^{2} d^{4}-8 a b \,c^{3} d +2 a b \,c^{2} d^{2}+2 b^{2} c^{4}-b^{2} c^{3} d +2 b^{2} c^{2} d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 \left (c +d \right ) \left (c -d \right )^{2}}\right )}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )^{2}}-\frac {\left (6 a^{2} c^{4} d -5 a^{2} c^{2} d^{3}+2 a^{2} d^{5}-4 a b \,c^{5}-2 a b \,c^{3} d^{2}+3 b^{2} c^{4} d \right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}}{c^{3}}}{f}\) | \(386\) |
| default | \(\frac {\frac {2 a^{2} \arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{c^{3}}+\frac {\frac {2 \left (-\frac {\left (6 a^{2} c^{2} d^{2}+a^{2} d^{3} c -2 a^{2} d^{4}-8 a b \,c^{3} d -2 a b \,c^{2} d^{2}+2 b^{2} c^{4}+b^{2} c^{3} d +2 b^{2} c^{2} d^{2}\right ) c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{2 \left (c -d \right ) \left (c^{2}+2 c d +d^{2}\right )}+\frac {c \left (6 a^{2} c^{2} d^{2}-a^{2} d^{3} c -2 a^{2} d^{4}-8 a b \,c^{3} d +2 a b \,c^{2} d^{2}+2 b^{2} c^{4}-b^{2} c^{3} d +2 b^{2} c^{2} d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{2 \left (c +d \right ) \left (c -d \right )^{2}}\right )}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c -\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} d -c -d \right )^{2}}-\frac {\left (6 a^{2} c^{4} d -5 a^{2} c^{2} d^{3}+2 a^{2} d^{5}-4 a b \,c^{5}-2 a b \,c^{3} d^{2}+3 b^{2} c^{4} d \right ) \operatorname {arctanh}\left (\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{\sqrt {\left (c +d \right ) \left (c -d \right )}}\right )}{\left (c^{4}-2 c^{2} d^{2}+d^{4}\right ) \sqrt {\left (c +d \right ) \left (c -d \right )}}}{c^{3}}}{f}\) | \(386\) |
| risch | \(\text {Expression too large to display}\) | \(1538\) |
1/f*(2*a^2/c^3*arctan(tan(1/2*f*x+1/2*e))+2/c^3*((-1/2*(6*a^2*c^2*d^2+a^2* c*d^3-2*a^2*d^4-8*a*b*c^3*d-2*a*b*c^2*d^2+2*b^2*c^4+b^2*c^3*d+2*b^2*c^2*d^ 2)*c/(c-d)/(c^2+2*c*d+d^2)*tan(1/2*f*x+1/2*e)^3+1/2*c*(6*a^2*c^2*d^2-a^2*c *d^3-2*a^2*d^4-8*a*b*c^3*d+2*a*b*c^2*d^2+2*b^2*c^4-b^2*c^3*d+2*b^2*c^2*d^2 )/(c+d)/(c-d)^2*tan(1/2*f*x+1/2*e))/(tan(1/2*f*x+1/2*e)^2*c-tan(1/2*f*x+1/ 2*e)^2*d-c-d)^2-1/2*(6*a^2*c^4*d-5*a^2*c^2*d^3+2*a^2*d^5-4*a*b*c^5-2*a*b*c ^3*d^2+3*b^2*c^4*d)/(c^4-2*c^2*d^2+d^4)/((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. 675 vs. \(2 (223) = 446\).
Time = 0.36 (sec) , antiderivative size = 1409, normalized size of antiderivative = 5.95 \[ \int \frac {(a+b \sec (e+f x))^2}{(c+d \sec (e+f x))^3} \, dx=\text {Too large to display} \]
[1/4*(4*(a^2*c^8 - 3*a^2*c^6*d^2 + 3*a^2*c^4*d^4 - a^2*c^2*d^6)*f*x*cos(f* x + e)^2 + 8*(a^2*c^7*d - 3*a^2*c^5*d^3 + 3*a^2*c^3*d^5 - a^2*c*d^7)*f*x*c os(f*x + e) + 4*(a^2*c^6*d^2 - 3*a^2*c^4*d^4 + 3*a^2*c^2*d^6 - a^2*d^8)*f* x - (4*a*b*c^5*d^2 + 2*a*b*c^3*d^4 + 5*a^2*c^2*d^5 - 2*a^2*d^7 - 3*(2*a^2 + b^2)*c^4*d^3 + (4*a*b*c^7 + 2*a*b*c^5*d^2 + 5*a^2*c^4*d^3 - 2*a^2*c^2*d^ 5 - 3*(2*a^2 + b^2)*c^6*d)*cos(f*x + e)^2 + 2*(4*a*b*c^6*d + 2*a*b*c^4*d^3 + 5*a^2*c^3*d^4 - 2*a^2*c*d^6 - 3*(2*a^2 + b^2)*c^5*d^2)*cos(f*x + e))*sq rt(c^2 - d^2)*log((2*c*d*cos(f*x + e) - (c^2 - 2*d^2)*cos(f*x + e)^2 - 2*s qrt(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*(b^2*c^7*d - 6*a*b*c^6*d^2 + 6* a*b*c^4*d^4 + 2*a^2*c*d^7 + (5*a^2 + b^2)*c^5*d^3 - (7*a^2 + 2*b^2)*c^3*d^ 5 + (2*b^2*c^8 - 8*a*b*c^7*d + 10*a*b*c^5*d^3 - 2*a*b*c^3*d^5 + 3*a^2*c^2* d^6 + (6*a^2 - b^2)*c^6*d^2 - (9*a^2 + b^2)*c^4*d^4)*cos(f*x + e))*sin(f*x + e))/((c^11 - 3*c^9*d^2 + 3*c^7*d^4 - c^5*d^6)*f*cos(f*x + e)^2 + 2*(c^1 0*d - 3*c^8*d^3 + 3*c^6*d^5 - c^4*d^7)*f*cos(f*x + e) + (c^9*d^2 - 3*c^7*d ^4 + 3*c^5*d^6 - c^3*d^8)*f), 1/2*(2*(a^2*c^8 - 3*a^2*c^6*d^2 + 3*a^2*c^4* d^4 - a^2*c^2*d^6)*f*x*cos(f*x + e)^2 + 4*(a^2*c^7*d - 3*a^2*c^5*d^3 + 3*a ^2*c^3*d^5 - a^2*c*d^7)*f*x*cos(f*x + e) + 2*(a^2*c^6*d^2 - 3*a^2*c^4*d^4 + 3*a^2*c^2*d^6 - a^2*d^8)*f*x + (4*a*b*c^5*d^2 + 2*a*b*c^3*d^4 + 5*a^2*c^ 2*d^5 - 2*a^2*d^7 - 3*(2*a^2 + b^2)*c^4*d^3 + (4*a*b*c^7 + 2*a*b*c^5*d^...
\[ \int \frac {(a+b \sec (e+f x))^2}{(c+d \sec (e+f x))^3} \, dx=\int \frac {\left (a + b \sec {\left (e + f x \right )}\right )^{2}}{\left (c + d \sec {\left (e + f x \right )}\right )^{3}}\, dx \]
Exception generated. \[ \int \frac {(a+b \sec (e+f x))^2}{(c+d \sec (e+f x))^3} \, dx=\text {Exception raised: ValueError} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 658 vs. \(2 (223) = 446\).
Time = 0.40 (sec) , antiderivative size = 658, normalized size of antiderivative = 2.78 \[ \int \frac {(a+b \sec (e+f x))^2}{(c+d \sec (e+f x))^3} \, dx=\frac {\frac {{\left (4 \, a b c^{5} - 6 \, a^{2} c^{4} d - 3 \, b^{2} c^{4} d + 2 \, a b c^{3} d^{2} + 5 \, a^{2} c^{2} d^{3} - 2 \, a^{2} d^{5}\right )} {\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 )}}{{\left (c^{7} - 2 \, c^{5} d^{2} + c^{3} d^{4}\right )} \sqrt {-c^{2} + d^{2}}} + \frac {{\left (f x + e\right )} a^{2}}{c^{3}} - \frac {2 \, b^{2} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 8 \, a b c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - b^{2} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 6 \, a^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 6 \, a b c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + b^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 5 \, a^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a b c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, b^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 3 \, a^{2} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 2 \, a^{2} d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 2 \, b^{2} c^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 8 \, a b c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b^{2} c^{4} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 6 \, a^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 6 \, a b c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - b^{2} c^{3} d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 5 \, a^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, a b c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 2 \, b^{2} c^{2} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 3 \, a^{2} c d^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 2 \, a^{2} d^{5} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (c^{6} - 2 \, c^{4} d^{2} + c^{2} d^{4}\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 )}^{2}}}{f} \]
((4*a*b*c^5 - 6*a^2*c^4*d - 3*b^2*c^4*d + 2*a*b*c^3*d^2 + 5*a^2*c^2*d^3 - 2*a^2*d^5)*(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)))/((c^7 - 2*c^5*d^2 + c^3*d^4)*sqrt(-c^2 + d^2)) + (f*x + e)*a^2/c^3 - (2*b^2*c^5*ta n(1/2*f*x + 1/2*e)^3 - 8*a*b*c^4*d*tan(1/2*f*x + 1/2*e)^3 - b^2*c^4*d*tan( 1/2*f*x + 1/2*e)^3 + 6*a^2*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 + 6*a*b*c^3*d^2* tan(1/2*f*x + 1/2*e)^3 + b^2*c^3*d^2*tan(1/2*f*x + 1/2*e)^3 - 5*a^2*c^2*d^ 3*tan(1/2*f*x + 1/2*e)^3 + 2*a*b*c^2*d^3*tan(1/2*f*x + 1/2*e)^3 - 2*b^2*c^ 2*d^3*tan(1/2*f*x + 1/2*e)^3 - 3*a^2*c*d^4*tan(1/2*f*x + 1/2*e)^3 + 2*a^2* d^5*tan(1/2*f*x + 1/2*e)^3 - 2*b^2*c^5*tan(1/2*f*x + 1/2*e) + 8*a*b*c^4*d* tan(1/2*f*x + 1/2*e) - b^2*c^4*d*tan(1/2*f*x + 1/2*e) - 6*a^2*c^3*d^2*tan( 1/2*f*x + 1/2*e) + 6*a*b*c^3*d^2*tan(1/2*f*x + 1/2*e) - b^2*c^3*d^2*tan(1/ 2*f*x + 1/2*e) - 5*a^2*c^2*d^3*tan(1/2*f*x + 1/2*e) - 2*a*b*c^2*d^3*tan(1/ 2*f*x + 1/2*e) - 2*b^2*c^2*d^3*tan(1/2*f*x + 1/2*e) + 3*a^2*c*d^4*tan(1/2* f*x + 1/2*e) + 2*a^2*d^5*tan(1/2*f*x + 1/2*e))/((c^6 - 2*c^4*d^2 + c^2*d^4 )*(c*tan(1/2*f*x + 1/2*e)^2 - d*tan(1/2*f*x + 1/2*e)^2 - c - d)^2))/f
Time = 24.42 (sec) , antiderivative size = 8682, normalized size of antiderivative = 36.63 \[ \int \frac {(a+b \sec (e+f x))^2}{(c+d \sec (e+f x))^3} \, dx=\text {Too large to display} \]
((tan(e/2 + (f*x)/2)^3*(2*b^2*c^4 - 2*a^2*d^4 + a^2*c*d^3 + b^2*c^3*d + 6* a^2*c^2*d^2 + 2*b^2*c^2*d^2 - 8*a*b*c^3*d - 2*a*b*c^2*d^2))/((c^2*d - c^3) *(c + d)^2) - (tan(e/2 + (f*x)/2)*(2*a^2*d^4 - 2*b^2*c^4 + a^2*c*d^3 + b^2 *c^3*d - 6*a^2*c^2*d^2 - 2*b^2*c^2*d^2 + 8*a*b*c^3*d - 2*a*b*c^2*d^2))/((c + d)*(c^4 - 2*c^3*d + c^2*d^2)))/(f*(2*c*d - tan(e/2 + (f*x)/2)^2*(2*c^2 - 2*d^2) + tan(e/2 + (f*x)/2)^4*(c^2 - 2*c*d + d^2) + c^2 + d^2)) - (2*a^2 *atan(((a^2*((a^2*((8*(4*a^2*c^15 - 12*a^2*c^14*d - 6*b^2*c^14*d - 4*a^2*c ^6*d^9 + 2*a^2*c^7*d^8 + 18*a^2*c^8*d^7 - 4*a^2*c^9*d^6 - 36*a^2*c^10*d^5 + 6*a^2*c^11*d^4 + 34*a^2*c^12*d^3 - 8*a^2*c^13*d^2 + 6*b^2*c^9*d^6 - 6*b^ 2*c^10*d^5 - 12*b^2*c^11*d^4 + 12*b^2*c^12*d^3 + 6*b^2*c^13*d^2 + 8*a*b*c^ 15 - 8*a*b*c^14*d - 4*a*b*c^8*d^7 + 4*a*b*c^9*d^6 + 12*a*b*c^12*d^3 - 12*a *b*c^13*d^2))/(c^12*d + c^13 - c^6*d^7 - c^7*d^6 + 3*c^8*d^5 + 3*c^9*d^4 - 3*c^10*d^3 - 3*c^11*d^2) - (a^2*tan(e/2 + (f*x)/2)*(8*c^15*d - 8*c^6*d^10 + 8*c^7*d^9 + 32*c^8*d^8 - 32*c^9*d^7 - 48*c^10*d^6 + 48*c^11*d^5 + 32*c^ 12*d^4 - 32*c^13*d^3 - 8*c^14*d^2)*8i)/(c^3*(c^10*d + c^11 - c^4*d^7 - c^5 *d^6 + 3*c^6*d^5 + 3*c^7*d^4 - 3*c^8*d^3 - 3*c^9*d^2)))*1i)/c^3 + (8*tan(e /2 + (f*x)/2)*(4*a^4*c^10 + 8*a^4*d^10 - 8*a^4*c*d^9 - 8*a^4*c^9*d + 16*a^ 2*b^2*c^10 - 32*a^4*c^2*d^8 + 32*a^4*c^3*d^7 + 57*a^4*c^4*d^6 - 48*a^4*c^5 *d^5 - 52*a^4*c^6*d^4 + 32*a^4*c^7*d^3 + 24*a^4*c^8*d^2 + 9*b^4*c^8*d^2 - 12*a*b^3*c^7*d^3 - 8*a^3*b*c^3*d^7 + 4*a^3*b*c^5*d^5 + 16*a^3*b*c^7*d^3...