Integrand size = 31, antiderivative size = 133 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^3} \, dx=\frac {d^3 \text {arctanh}(\sin (e+f x))}{a^3 f}+\frac {(c-d) (c+d \sec (e+f x))^2 \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {(c-d) \left (2 \left (2 c^2+8 c d+11 d^2\right )+\left (2 c^2+11 c d+29 d^2\right ) \sec (e+f x)\right ) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2} \]
d^3*arctanh(sin(f*x+e))/a^3/f+1/5*(c-d)*(c+d*sec(f*x+e))^2*tan(f*x+e)/f/(a +a*sec(f*x+e))^3+1/15*(c-d)*(4*c^2+16*c*d+22*d^2+(2*c^2+11*c*d+29*d^2)*sec (f*x+e))*tan(f*x+e)/a/f/(a+a*sec(f*x+e))^2
Leaf count is larger than twice the leaf count of optimal. \(295\) vs. \(2(133)=266\).
Time = 2.59 (sec) , antiderivative size = 295, normalized size of antiderivative = 2.22 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^3} \, dx=\frac {-240 d^3 \cos ^6\left (\frac {1}{2} (e+f x)\right ) \left (\log \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right )-\log \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right )\right )+(c-d) \cos \left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {e}{2}\right ) \left (5 \left (8 c^2+17 c d+29 d^2\right ) \sin \left (\frac {f x}{2}\right )-15 \left (2 c^2+5 c d+5 d^2\right ) \sin \left (e+\frac {f x}{2}\right )+20 c^2 \sin \left (e+\frac {3 f x}{2}\right )+65 c d \sin \left (e+\frac {3 f x}{2}\right )+95 d^2 \sin \left (e+\frac {3 f x}{2}\right )-15 c^2 \sin \left (2 e+\frac {3 f x}{2}\right )-15 c d \sin \left (2 e+\frac {3 f x}{2}\right )-15 d^2 \sin \left (2 e+\frac {3 f x}{2}\right )+7 c^2 \sin \left (2 e+\frac {5 f x}{2}\right )+16 c d \sin \left (2 e+\frac {5 f x}{2}\right )+22 d^2 \sin \left (2 e+\frac {5 f x}{2}\right )\right )}{30 a^3 f (1+\cos (e+f x))^3} \]
(-240*d^3*Cos[(e + f*x)/2]^6*(Log[Cos[(e + f*x)/2] - Sin[(e + f*x)/2]] - L og[Cos[(e + f*x)/2] + Sin[(e + f*x)/2]]) + (c - d)*Cos[(e + f*x)/2]*Sec[e/ 2]*(5*(8*c^2 + 17*c*d + 29*d^2)*Sin[(f*x)/2] - 15*(2*c^2 + 5*c*d + 5*d^2)* Sin[e + (f*x)/2] + 20*c^2*Sin[e + (3*f*x)/2] + 65*c*d*Sin[e + (3*f*x)/2] + 95*d^2*Sin[e + (3*f*x)/2] - 15*c^2*Sin[2*e + (3*f*x)/2] - 15*c*d*Sin[2*e + (3*f*x)/2] - 15*d^2*Sin[2*e + (3*f*x)/2] + 7*c^2*Sin[2*e + (5*f*x)/2] + 16*c*d*Sin[2*e + (5*f*x)/2] + 22*d^2*Sin[2*e + (5*f*x)/2]))/(30*a^3*f*(1 + Cos[e + f*x])^3)
Time = 0.41 (sec) , antiderivative size = 225, normalized size of antiderivative = 1.69, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.258, Rules used = {3042, 4475, 109, 25, 27, 162, 45, 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.
| \(\displaystyle \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a \sec (e+f x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4475 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \int \frac {(c+d \sec (e+f x))^3}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{7/2}}d\sec (e+f x)}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 109 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (-\frac {\int -\frac {a^2 (c+d \sec (e+f x)) \left (2 c^2+5 d c-2 d^2+5 d^2 \sec (e+f x)\right )}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{5/2}}d\sec (e+f x)}{5 a^3}-\frac {(c-d) \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}{5 a^2 (a \sec (e+f x)+a)^{5/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\int \frac {a^2 (c+d \sec (e+f x)) \left (2 c^2+5 d c-2 d^2+5 d^2 \sec (e+f x)\right )}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{5/2}}d\sec (e+f x)}{5 a^3}-\frac {(c-d) \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}{5 a^2 (a \sec (e+f x)+a)^{5/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\int \frac {(c+d \sec (e+f x)) \left (2 c^2+5 d c-2 d^2+5 d^2 \sec (e+f x)\right )}{\sqrt {a-a \sec (e+f x)} (\sec (e+f x) a+a)^{5/2}}d\sec (e+f x)}{5 a}-\frac {(c-d) \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}{5 a^2 (a \sec (e+f x)+a)^{5/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 162 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\frac {5 d^3 \int \frac {1}{\sqrt {a-a \sec (e+f x)} \sqrt {\sec (e+f x) a+a}}d\sec (e+f x)}{a^2}-\frac {(c-d) \sqrt {a-a \sec (e+f x)} \left (\left (2 c^2+11 c d+29 d^2\right ) \sec (e+f x)+2 \left (2 c^2+8 c d+11 d^2\right )\right )}{3 a^2 (a \sec (e+f x)+a)^{3/2}}}{5 a}-\frac {(c-d) \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}{5 a^2 (a \sec (e+f x)+a)^{5/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 45 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {\frac {10 d^3 \int \frac {1}{-\frac {(a-a \sec (e+f x)) a}{\sec (e+f x) a+a}-a}d\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {\sec (e+f x) a+a}}}{a^2}-\frac {(c-d) \sqrt {a-a \sec (e+f x)} \left (\left (2 c^2+11 c d+29 d^2\right ) \sec (e+f x)+2 \left (2 c^2+8 c d+11 d^2\right )\right )}{3 a^2 (a \sec (e+f x)+a)^{3/2}}}{5 a}-\frac {(c-d) \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}{5 a^2 (a \sec (e+f x)+a)^{5/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
| \(\Big \downarrow \) 218 |
| \(\displaystyle -\frac {a^2 \tan (e+f x) \left (\frac {-\frac {10 d^3 \arctan \left (\frac {\sqrt {a-a \sec (e+f x)}}{\sqrt {a \sec (e+f x)+a}}\right )}{a^3}-\frac {(c-d) \sqrt {a-a \sec (e+f x)} \left (\left (2 c^2+11 c d+29 d^2\right ) \sec (e+f x)+2 \left (2 c^2+8 c d+11 d^2\right )\right )}{3 a^2 (a \sec (e+f x)+a)^{3/2}}}{5 a}-\frac {(c-d) \sqrt {a-a \sec (e+f x)} (c+d \sec (e+f x))^2}{5 a^2 (a \sec (e+f x)+a)^{5/2}}\right )}{f \sqrt {a-a \sec (e+f x)} \sqrt {a \sec (e+f x)+a}}\) |
-((a^2*(-1/5*((c - d)*Sqrt[a - a*Sec[e + f*x]]*(c + d*Sec[e + f*x])^2)/(a^ 2*(a + a*Sec[e + f*x])^(5/2)) + ((-10*d^3*ArcTan[Sqrt[a - a*Sec[e + f*x]]/ Sqrt[a + a*Sec[e + f*x]]])/a^3 - ((c - d)*Sqrt[a - a*Sec[e + f*x]]*(2*(2*c ^2 + 8*c*d + 11*d^2) + (2*c^2 + 11*c*d + 29*d^2)*Sec[e + f*x]))/(3*a^2*(a + a*Sec[e + f*x])^(3/2)))/(5*a))*Tan[e + f*x])/(f*Sqrt[a - a*Sec[e + f*x]] *Sqrt[a + a*Sec[e + f*x]]))
3.3.28.3.1 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[1/(Sqrt[(a_) + (b_.)*(x_)]*Sqrt[(c_) + (d_.)*(x_)]), x_Symbol] :> Simp[ 2 Subst[Int[1/(b - d*x^2), x], x, Sqrt[a + b*x]/Sqrt[c + d*x]], x] /; Fre eQ[{a, b, c, d}, x] && EqQ[b*c + a*d, 0] && !GtQ[c, 0]
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) )^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f *x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) + c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || IntegersQ[m, n + p] || IntegersQ[p, m + n])
Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) )*((g_.) + (h_.)*(x_)), x_] :> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g + e*h) + d*e *g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)/(b^2*(b *c - a*d)^2*(m + 1)*(m + 2)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] + Sim p[(f*(h/b^2) - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d *(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/( b^2*(b*c - a*d)^2*(m + 1)*(m + 2))) Int[(a + b*x)^(m + 2)*(c + d*x)^n, x] , x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + n + 3, 0] && !LtQ[n, -2]))
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[(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 = 0.69 (sec) , antiderivative size = 117, normalized size of antiderivative = 0.88
| method | result | size |
| parallelrisch | \(\frac {-60 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) d^{3}+60 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) d^{3}+3 \left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (\left (c -d \right )^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-\frac {10 \left (c +2 d \right ) \left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3}+5 c^{2}+20 c d +35 d^{2}\right )}{60 a^{3} f}\) | \(117\) |
| derivativedivides | \(\frac {-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} c^{3}}{3}+2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} c \,d^{2}+3 c^{2} d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c \,d^{2}-\frac {3 c^{2} d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\frac {3 c \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-7 d^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) d^{3}+\frac {c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {d^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} d^{3}}{3}-4 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) d^{3}}{4 f \,a^{3}}\) | \(216\) |
| default | \(\frac {-\frac {2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} c^{3}}{3}+2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} c \,d^{2}+3 c^{2} d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) c \,d^{2}-\frac {3 c^{2} d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+\frac {3 c \,d^{2} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}+c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-7 d^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) d^{3}+\frac {c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {d^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} d^{3}}{3}-4 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) d^{3}}{4 f \,a^{3}}\) | \(216\) |
| risch | \(\frac {2 i \left (15 c^{3} {\mathrm e}^{4 i \left (f x +e \right )}-15 d^{3} {\mathrm e}^{4 i \left (f x +e \right )}+30 c^{3} {\mathrm e}^{3 i \left (f x +e \right )}+45 c^{2} d \,{\mathrm e}^{3 i \left (f x +e \right )}-75 d^{3} {\mathrm e}^{3 i \left (f x +e \right )}+40 c^{3} {\mathrm e}^{2 i \left (f x +e \right )}+45 c^{2} d \,{\mathrm e}^{2 i \left (f x +e \right )}+60 c \,d^{2} {\mathrm e}^{2 i \left (f x +e \right )}-145 d^{3} {\mathrm e}^{2 i \left (f x +e \right )}+20 c^{3} {\mathrm e}^{i \left (f x +e \right )}+45 c^{2} d \,{\mathrm e}^{i \left (f x +e \right )}+30 c \,d^{2} {\mathrm e}^{i \left (f x +e \right )}-95 d^{3} {\mathrm e}^{i \left (f x +e \right )}+7 c^{3}+9 c^{2} d +6 c \,d^{2}-22 d^{3}\right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}-\frac {d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a^{3} f}+\frac {d^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{a^{3} f}\) | \(281\) |
| norman | \(\frac {\frac {\left (c^{3}-3 c^{2} d +3 c \,d^{2}-d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{11}}{20 a f}-\frac {\left (c^{3}+3 c^{2} d +3 c \,d^{2}-7 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 a f}+\frac {3 \left (3 c^{3}+c^{2} d -c \,d^{2}-3 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{10 a f}+\frac {\left (11 c^{3}+27 c^{2} d +21 c \,d^{2}-59 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{12 a f}-\frac {\left (13 c^{3}+21 c^{2} d +9 c \,d^{2}-43 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{10 a f}-\frac {\left (19 c^{3}-27 c^{2} d -3 c \,d^{2}+11 d^{3}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{60 a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3} a^{2}}+\frac {d^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a^{3} f}-\frac {d^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{a^{3} f}\) | \(312\) |
1/60*(-60*ln(tan(1/2*f*x+1/2*e)-1)*d^3+60*ln(tan(1/2*f*x+1/2*e)+1)*d^3+3*( c-d)*tan(1/2*f*x+1/2*e)*((c-d)^2*tan(1/2*f*x+1/2*e)^4-10/3*(c+2*d)*(c-d)*t an(1/2*f*x+1/2*e)^2+5*c^2+20*c*d+35*d^2))/a^3/f
Time = 0.27 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.86 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^3} \, dx=\frac {15 \, {\left (d^{3} \cos \left (f x + e\right )^{3} + 3 \, d^{3} \cos \left (f x + e\right )^{2} + 3 \, d^{3} \cos \left (f x + e\right ) + d^{3}\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, {\left (d^{3} \cos \left (f x + e\right )^{3} + 3 \, d^{3} \cos \left (f x + e\right )^{2} + 3 \, d^{3} \cos \left (f x + e\right ) + d^{3}\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + 2 \, {\left (2 \, c^{3} + 9 \, c^{2} d + 21 \, c d^{2} - 32 \, d^{3} + {\left (7 \, c^{3} + 9 \, c^{2} d + 6 \, c d^{2} - 22 \, d^{3}\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (2 \, c^{3} + 9 \, c^{2} d + 6 \, c d^{2} - 17 \, d^{3}\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )}{30 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \]
1/30*(15*(d^3*cos(f*x + e)^3 + 3*d^3*cos(f*x + e)^2 + 3*d^3*cos(f*x + e) + d^3)*log(sin(f*x + e) + 1) - 15*(d^3*cos(f*x + e)^3 + 3*d^3*cos(f*x + e)^ 2 + 3*d^3*cos(f*x + e) + d^3)*log(-sin(f*x + e) + 1) + 2*(2*c^3 + 9*c^2*d + 21*c*d^2 - 32*d^3 + (7*c^3 + 9*c^2*d + 6*c*d^2 - 22*d^3)*cos(f*x + e)^2 + 3*(2*c^3 + 9*c^2*d + 6*c*d^2 - 17*d^3)*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)
\[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^3} \, dx=\frac {\int \frac {c^{3} \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d^{3} \sec ^{4}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c d^{2} \sec ^{3}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {3 c^{2} d \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx}{a^{3}} \]
(Integral(c**3*sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(d**3*sec(e + f*x)**4/(sec(e + f*x)**3 + 3*sec( e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(3*c*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( 3*c**2*d*sec(e + f*x)**2/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x))/a**3
Leaf count of result is larger than twice the leaf count of optimal. 307 vs. \(2 (126) = 252\).
Time = 0.23 (sec) , antiderivative size = 307, normalized size of antiderivative = 2.31 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^3} \, dx=-\frac {d^{3} {\left (\frac {\frac {105 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {20 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{3}}\right )} - \frac {3 \, c d^{2} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {c^{3} {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} - \frac {9 \, c^{2} d {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \]
-1/60*(d^3*((105*sin(f*x + e)/(cos(f*x + e) + 1) + 20*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 - 60*log(sin( f*x + e)/(cos(f*x + e) + 1) + 1)/a^3 + 60*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^3) - 3*c*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^3*(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 - 9*c^2*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
Leaf count of result is larger than twice the leaf count of optimal. 259 vs. \(2 (126) = 252\).
Time = 0.37 (sec) , antiderivative size = 259, normalized size of antiderivative = 1.95 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^3} \, dx=\frac {\frac {60 \, d^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, d^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{3}} + \frac {3 \, a^{12} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 9 \, a^{12} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} + 9 \, a^{12} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 3 \, a^{12} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, a^{12} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 30 \, a^{12} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} - 20 \, a^{12} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, a^{12} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 45 \, a^{12} c^{2} d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 45 \, a^{12} c d^{2} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 105 \, a^{12} d^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{a^{15}}}{60 \, f} \]
1/60*(60*d^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^3 - 60*d^3*log(abs(tan(1 /2*f*x + 1/2*e) - 1))/a^3 + (3*a^12*c^3*tan(1/2*f*x + 1/2*e)^5 - 9*a^12*c^ 2*d*tan(1/2*f*x + 1/2*e)^5 + 9*a^12*c*d^2*tan(1/2*f*x + 1/2*e)^5 - 3*a^12* d^3*tan(1/2*f*x + 1/2*e)^5 - 10*a^12*c^3*tan(1/2*f*x + 1/2*e)^3 + 30*a^12* c*d^2*tan(1/2*f*x + 1/2*e)^3 - 20*a^12*d^3*tan(1/2*f*x + 1/2*e)^3 + 15*a^1 2*c^3*tan(1/2*f*x + 1/2*e) + 45*a^12*c^2*d*tan(1/2*f*x + 1/2*e) + 45*a^12* c*d^2*tan(1/2*f*x + 1/2*e) - 105*a^12*d^3*tan(1/2*f*x + 1/2*e))/a^15)/f
Time = 13.52 (sec) , antiderivative size = 147, normalized size of antiderivative = 1.11 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))^3}{(a+a \sec (e+f x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (\frac {{\left (c-d\right )}^3}{4\,a^3}-\frac {3\,\left (c+d\right )\,{\left (c-d\right )}^2}{4\,a^3}+\frac {3\,{\left (c+d\right )}^2\,\left (c-d\right )}{4\,a^3}\right )}{f}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3\,\left (\frac {{\left (c-d\right )}^3}{12\,a^3}-\frac {\left (c+d\right )\,{\left (c-d\right )}^2}{4\,a^3}\right )}{f}+\frac {2\,d^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a^3\,f}+\frac {{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5\,{\left (c-d\right )}^3}{20\,a^3\,f} \]