Integrand size = 31, antiderivative size = 115 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {(c-d)^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {2 (c-d) (c+4 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {\left (2 c^2+6 c d+7 d^2\right ) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )} \]
1/5*(c-d)^2*tan(f*x+e)/f/(a+a*sec(f*x+e))^3+2/15*(c-d)*(c+4*d)*tan(f*x+e)/ a/f/(a+a*sec(f*x+e))^2+1/15*(2*c^2+6*c*d+7*d^2)*tan(f*x+e)/f/(a^3+a^3*sec( f*x+e))
Time = 0.79 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.73 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {\left (2 c^2+6 c d+7 d^2+6 \left (c^2+3 c d+d^2\right ) \cos (e+f x)+\left (7 c^2+6 c d+2 d^2\right ) \cos ^2(e+f x)\right ) \sin (e+f x)}{15 a^3 f (1+\cos (e+f x))^3} \]
((2*c^2 + 6*c*d + 7*d^2 + 6*(c^2 + 3*c*d + d^2)*Cos[e + f*x] + (7*c^2 + 6* c*d + 2*d^2)*Cos[e + f*x]^2)*Sin[e + f*x])/(15*a^3*f*(1 + Cos[e + f*x])^3)
Time = 0.36 (sec) , antiderivative size = 192, normalized size of antiderivative = 1.67, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3042, 4475, 100, 27, 87, 48}
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.
| \(\displaystyle \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a \sec (e+f x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4475 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {(c+d \sec (e+f x))^2}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{7/2}}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 100 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\int \frac {a^3 \left (2 c^2+6 d c-3 d^2+5 d^2 \sec (e+f x)\right )}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{5/2}}d\sec (e+f x)}{5 a^4}-\frac {(c-d)^2 \sqrt {a-a \sec (e+f x)}}{5 a^2 (a \sec (e+f x)+a)^{5/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\int \frac {2 c^2+6 d c-3 d^2+5 d^2 \sec (e+f x)}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{5/2}}d\sec (e+f x)}{5 a}-\frac {(c-d)^2 \sqrt {a-a \sec (e+f x)}}{5 a^2 (a \sec (e+f x)+a)^{5/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 87 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\frac {\left (2 c^2+6 c d+7 d^2\right ) \int \frac {1}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{3/2}}d\sec (e+f x)}{3 a}-\frac {2 (c-d) (c+4 d) \sqrt {a-a \sec (e+f x)}}{3 a^2 (a \sec (e+f x)+a)^{3/2}}}{5 a}-\frac {(c-d)^2 \sqrt {a-a \sec (e+f x)}}{5 a^2 (a \sec (e+f x)+a)^{5/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 48 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {-\frac {\left (2 c^2+6 c d+7 d^2\right ) \sqrt {a-a \sec (e+f x)}}{3 a^3 \sqrt {a \sec (e+f x)+a}}-\frac {2 (c-d) (c+4 d) \sqrt {a-a \sec (e+f x)}}{3 a^2 (a \sec (e+f x)+a)^{3/2}}}{5 a}-\frac {(c-d)^2 \sqrt {a-a \sec (e+f x)}}{5 a^2 (a \sec (e+f x)+a)^{5/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^2*(-1/5*((c - d)^2*Sqrt[a - a*Sec[e + f*x]])/(a^2*(a + a*Sec[e + f*x] )^(5/2)) + ((-2*(c - d)*(c + 4*d)*Sqrt[a - a*Sec[e + f*x]])/(3*a^2*(a + a* Sec[e + f*x])^(3/2)) - ((2*c^2 + 6*c*d + 7*d^2)*Sqrt[a - a*Sec[e + f*x]])/ (3*a^3*Sqrt[a + a*Sec[e + f*x]]))/(5*a))*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]]*Sqrt[a + a*Sec[e + f*x]]))
3.3.29.3.1 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_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp [(a + b*x)^(m + 1)*((c + d*x)^(n + 1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{ a, b, c, d, m, n}, x] && EqQ[m + n + 2, 0] && NeQ[m, -1]
Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p _.), x_] :> Simp[(-(b*e - a*f))*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(f*(p + 1)*(c*f - d*e))), x] - Simp[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)) Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || Intege rQ[p] || !(IntegerQ[n] || !(EqQ[e, 0] || !(EqQ[c, 0] || LtQ[p, n]))))
Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_))^( p_), x_] :> Simp[(b*c - a*d)^2*(c + d*x)^(n + 1)*((e + f*x)^(p + 1)/(d^2*(d *e - c*f)*(n + 1))), x] - Simp[1/(d^2*(d*e - c*f)*(n + 1)) Int[(c + d*x)^ (n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*( p + 1)) - 2*a*b*d*(d*e*(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x , x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] || !SumSimplerQ[p, 1])))
Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_), x_Symbol] :> Simp[a ^2*g*(Cot[e + f*x]/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]])) Subst[Int[(g*x)^(p - 1)*(a + b*x)^(m - 1/2)*((c + d*x)^n/Sqrt[a - b*x]), x ], x, Csc[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, g, m, n, p}, x] && NeQ[ b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && (EqQ[p, 1] || In tegerQ[m - 1/2])
Time = 0.64 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.58
| method | result | size |
| parallelrisch | \(\frac {\left (\left (c -d \right )^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}+\frac {10 \left (-c^{2}+d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3}+5 \left (c +d \right )^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{20 a^{3} f}\) | \(67\) |
| derivativedivides | \(\frac {\frac {\left (c -d \right )^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\frac {2 \left (-c -d \right ) \left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (-c -d \right )^{2}}{4 f \,a^{3}}\) | \(74\) |
| default | \(\frac {\frac {\left (c -d \right )^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\frac {2 \left (-c -d \right ) \left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (-c -d \right )^{2}}{4 f \,a^{3}}\) | \(74\) |
| risch | \(\frac {2 i \left (15 c^{2} {\mathrm e}^{4 i \left (f x +e \right )}+30 c^{2} {\mathrm e}^{3 i \left (f x +e \right )}+30 c d \,{\mathrm e}^{3 i \left (f x +e \right )}+40 c^{2} {\mathrm e}^{2 i \left (f x +e \right )}+30 c d \,{\mathrm e}^{2 i \left (f x +e \right )}+20 d^{2} {\mathrm e}^{2 i \left (f x +e \right )}+20 c^{2} {\mathrm e}^{i \left (f x +e \right )}+30 d \,{\mathrm e}^{i \left (f x +e \right )} c +10 d^{2} {\mathrm e}^{i \left (f x +e \right )}+7 c^{2}+6 c d +2 d^{2}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}\) | \(161\) |
| norman | \(\frac {\frac {\left (c^{2}-2 c d +d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{20 a f}+\frac {\left (c^{2}+2 c d +d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 a f}-\frac {\left (2 c^{2}+3 c d +d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}-\frac {\left (4 c^{2}-3 c d -d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{15 a f}+\frac {\left (19 c^{2}+12 c d -d^{2}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{30 a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{2} a^{2}}\) | \(179\) |
1/20*((c-d)^2*tan(1/2*f*x+1/2*e)^4+10/3*(-c^2+d^2)*tan(1/2*f*x+1/2*e)^2+5* (c+d)^2)*tan(1/2*f*x+1/2*e)/a^3/f
Time = 0.27 (sec) , antiderivative size = 113, normalized size of antiderivative = 0.98 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {{\left ({\left (7 \, c^{2} + 6 \, c d + 2 \, d^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, c^{2} + 6 \, c d + 7 \, d^{2} + 6 \, {\left (c^{2} + 3 \, c d + d^{2}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]
1/15*((7*c^2 + 6*c*d + 2*d^2)*cos(f*x + e)^2 + 2*c^2 + 6*c*d + 7*d^2 + 6*( c^2 + 3*c*d + d^2)*cos(f*x + e))*sin(f*x + e)/(a^3*f*cos(f*x + e)^3 + 3*a^ 3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a^3*f)
\[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {\int \frac {c^{2} \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {2 c d \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx}{a^{3}} \]
(Integral(c**2*sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(d**2*sec(e + f*x)**3/(sec(e + f*x)**3 + 3*sec( e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(2*c*d*sec(e + f*x)**2/(se c(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x))/a**3
Time = 0.22 (sec) , antiderivative size = 184, normalized size of antiderivative = 1.60 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {\frac {d^{2} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {c^{2} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {6 \, c d {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]
1/60*(d^2*(15*sin(f*x + e)/(cos(f*x + e) + 1) + 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 + c^2*(15*sin(f* x + e)/(cos(f*x + e) + 1) - 10*sin(f*x + e)^3/(cos(f*x + e) + 1)^3 + 3*sin (f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3 + 6*c*d*(5*sin(f*x + e)/(cos(f*x + e ) + 1) - sin(f*x + e)^5/(cos(f*x + e) + 1)^5)/a^3)/f
Time = 0.33 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.12 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {3 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 6 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 3 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 10 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, c^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 30 \, c d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{60 \, a^{3} f} \]
1/60*(3*c^2*tan(1/2*f*x + 1/2*e)^5 - 6*c*d*tan(1/2*f*x + 1/2*e)^5 + 3*d^2* tan(1/2*f*x + 1/2*e)^5 - 10*c^2*tan(1/2*f*x + 1/2*e)^3 + 10*d^2*tan(1/2*f* x + 1/2*e)^3 + 15*c^2*tan(1/2*f*x + 1/2*e) + 30*c*d*tan(1/2*f*x + 1/2*e) + 15*d^2*tan(1/2*f*x + 1/2*e))/(a^3*f)
Time = 13.75 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.69 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^2}{(a+a \sec (e+f x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,{\left (c+d\right )}^2}{4\,a^3\,f}-\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (2\,c^2-2\,d^2\right )}{12\,a^3\,f}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\left (c-d\right )}^2}{20\,a^3\,f} \]