Integrand size = 32, antiderivative size = 119 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=\frac {5 c^3 \text {arctanh}(\sin (e+f x))}{a^2 f}-\frac {5 c^3 \tan (e+f x)}{a^2 f}+\frac {2 c (c-c \sec (e+f x))^2 \tan (e+f x)}{3 f (a+a \sec (e+f x))^2}-\frac {10 \left (c^3-c^3 \sec (e+f x)\right ) \tan (e+f x)}{3 f \left (a^2+a^2 \sec (e+f x)\right )} \]
5*c^3*arctanh(sin(f*x+e))/a^2/f-5*c^3*tan(f*x+e)/a^2/f+2/3*c*(c-c*sec(f*x+ e))^2*tan(f*x+e)/f/(a+a*sec(f*x+e))^2-10/3*(c^3-c^3*sec(f*x+e))*tan(f*x+e) /f/(a^2+a^2*sec(f*x+e))
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 0.45 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.63 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=-\frac {8 c^3 \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1+\sec (e+f x))\right ) \sqrt {2-2 \sec (e+f x)} \tan (e+f x)}{3 a^2 f (-1+\sec (e+f x)) (1+\sec (e+f x))^2} \]
(-8*c^3*Hypergeometric2F1[-5/2, -3/2, -1/2, (1 + Sec[e + f*x])/2]*Sqrt[2 - 2*Sec[e + f*x]]*Tan[e + f*x])/(3*a^2*f*(-1 + Sec[e + f*x])*(1 + Sec[e + f *x])^2)
Time = 0.72 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3042, 4445, 3042, 4445, 3042, 4274, 3042, 4254, 24, 4257}
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.
| \(\displaystyle \int \frac {\sec (e+f x) (c-c \sec (e+f x))^3}{(a \sec (e+f x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4445 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2}-\frac {5 c \int \frac {\sec (e+f x) (c-c \sec (e+f x))^2}{\sec (e+f x) a+a}dx}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2}-\frac {5 c \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}{\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}dx}{3 a}\) |
| \(\Big \downarrow \) 4445 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2}-\frac {5 c \left (\frac {2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{f (a \sec (e+f x)+a)}-\frac {3 c \int \sec (e+f x) (c-c \sec (e+f x))dx}{a}\right )}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2}-\frac {5 c \left (\frac {2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{f (a \sec (e+f x)+a)}-\frac {3 c \int \csc \left (e+f x+\frac {\pi }{2}\right ) \left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )dx}{a}\right )}{3 a}\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2}-\frac {5 c \left (\frac {2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{f (a \sec (e+f x)+a)}-\frac {3 c \left (c \int \sec (e+f x)dx-c \int \sec ^2(e+f x)dx\right )}{a}\right )}{3 a}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2}-\frac {5 c \left (\frac {2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{f (a \sec (e+f x)+a)}-\frac {3 c \left (c \int \csc \left (e+f x+\frac {\pi }{2}\right )dx-c \int \csc \left (e+f x+\frac {\pi }{2}\right )^2dx\right )}{a}\right )}{3 a}\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2}-\frac {5 c \left (\frac {2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{f (a \sec (e+f x)+a)}-\frac {3 c \left (\frac {c \int 1d(-\tan (e+f x))}{f}+c \int \csc \left (e+f x+\frac {\pi }{2}\right )dx\right )}{a}\right )}{3 a}\) |
| \(\Big \downarrow \) 24 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2}-\frac {5 c \left (\frac {2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{f (a \sec (e+f x)+a)}-\frac {3 c \left (c \int \csc \left (e+f x+\frac {\pi }{2}\right )dx-\frac {c \tan (e+f x)}{f}\right )}{a}\right )}{3 a}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle \frac {2 c \tan (e+f x) (c-c \sec (e+f x))^2}{3 f (a \sec (e+f x)+a)^2}-\frac {5 c \left (\frac {2 \tan (e+f x) \left (c^2-c^2 \sec (e+f x)\right )}{f (a \sec (e+f x)+a)}-\frac {3 c \left (\frac {c \text {arctanh}(\sin (e+f x))}{f}-\frac {c \tan (e+f x)}{f}\right )}{a}\right )}{3 a}\) |
(2*c*(c - c*Sec[e + f*x])^2*Tan[e + f*x])/(3*f*(a + a*Sec[e + f*x])^2) - ( 5*c*((2*(c^2 - c^2*Sec[e + f*x])*Tan[e + f*x])/(f*(a + a*Sec[e + f*x])) - (3*c*((c*ArcTanh[Sin[e + f*x]])/f - (c*Tan[e + f*x])/f))/a))/(3*a)
3.1.44.3.1 Defintions of rubi rules used
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[2*a*c*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^(n - 1)/(b*f*(2*m + 1))), x] - Simp[d*((2*n - 1)/(b*(2*m + 1))) Int[Csc[e + f*x]*(a + b*Csc[e + f* x])^(m + 1)*(c + d*Csc[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d, e, f }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IGtQ[n, 0] && LtQ[m, -2^ (-1)] && IntegerQ[2*m]
Time = 1.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.80
| method | result | size |
| derivativedivides | \(\frac {4 c^{3} \left (-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {1}{4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4}+\frac {5 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}+\frac {1}{4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-4}-\frac {5 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}\right )}{f \,a^{2}}\) | \(95\) |
| default | \(\frac {4 c^{3} \left (-\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}-2 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+\frac {1}{4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+4}+\frac {5 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}+\frac {1}{4 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )-4}-\frac {5 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}\right )}{f \,a^{2}}\) | \(95\) |
| parallelrisch | \(-\frac {5 \left (\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \cos \left (f x +e \right )-\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \cos \left (f x +e \right )+\frac {17 \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (\cos \left (f x +e \right )+\frac {23 \cos \left (2 f x +2 e \right )}{68}+\frac {29}{68}\right ) \sec \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{15}\right ) c^{3}}{f \,a^{2} \cos \left (f x +e \right )}\) | \(101\) |
| risch | \(-\frac {2 i c^{3} \left (12 \,{\mathrm e}^{4 i \left (f x +e \right )}+51 \,{\mathrm e}^{3 i \left (f x +e \right )}+41 \,{\mathrm e}^{2 i \left (f x +e \right )}+57 \,{\mathrm e}^{i \left (f x +e \right )}+23\right )}{3 f \,a^{2} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3} \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right )}+\frac {5 c^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{a^{2} f}-\frac {5 c^{3} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a^{2} f}\) | \(134\) |
| norman | \(\frac {\frac {10 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}-\frac {80 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}+\frac {22 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}-\frac {4 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{a f}-\frac {4 c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{3 a f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3} a}-\frac {5 c^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{a^{2} f}+\frac {5 c^{3} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a^{2} f}\) | \(176\) |
4/f/a^2*c^3*(-1/3*tan(1/2*f*x+1/2*e)^3-2*tan(1/2*f*x+1/2*e)+1/4/(tan(1/2*f *x+1/2*e)+1)+5/4*ln(tan(1/2*f*x+1/2*e)+1)+1/4/(tan(1/2*f*x+1/2*e)-1)-5/4*l n(tan(1/2*f*x+1/2*e)-1))
Time = 0.27 (sec) , antiderivative size = 178, normalized size of antiderivative = 1.50 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=\frac {15 \, {\left (c^{3} \cos \left (f x + e\right )^{3} + 2 \, c^{3} \cos \left (f x + e\right )^{2} + c^{3} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) - 15 \, {\left (c^{3} \cos \left (f x + e\right )^{3} + 2 \, c^{3} \cos \left (f x + e\right )^{2} + c^{3} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) - 2 \, {\left (23 \, c^{3} \cos \left (f x + e\right )^{2} + 34 \, c^{3} \cos \left (f x + e\right ) + 3 \, c^{3}\right )} \sin \left (f x + e\right )}{6 \, {\left (a^{2} f \cos \left (f x + e\right )^{3} + 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f \cos \left (f x + e\right )\right )}} \]
1/6*(15*(c^3*cos(f*x + e)^3 + 2*c^3*cos(f*x + e)^2 + c^3*cos(f*x + e))*log (sin(f*x + e) + 1) - 15*(c^3*cos(f*x + e)^3 + 2*c^3*cos(f*x + e)^2 + c^3*c os(f*x + e))*log(-sin(f*x + e) + 1) - 2*(23*c^3*cos(f*x + e)^2 + 34*c^3*co s(f*x + e) + 3*c^3)*sin(f*x + e))/(a^2*f*cos(f*x + e)^3 + 2*a^2*f*cos(f*x + e)^2 + a^2*f*cos(f*x + e))
\[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=- \frac {c^{3} \left (\int \left (- \frac {\sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {3 \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {3 \sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{2}} \]
-c**3*(Integral(-sec(e + f*x)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(3*sec(e + f*x)**2/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + I ntegral(-3*sec(e + f*x)**3/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + In tegral(sec(e + f*x)**4/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x))/a**2
Leaf count of result is larger than twice the leaf count of optimal. 341 vs. \(2 (117) = 234\).
Time = 0.21 (sec) , antiderivative size = 341, normalized size of antiderivative = 2.87 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=-\frac {c^{3} {\left (\frac {\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (f x + e\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + 3 \, c^{3} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} + \frac {3 \, c^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac {c^{3} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \]
-1/6*(c^3*((15*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 12*log(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 12*log (sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2 + 12*sin(f*x + e)/((a^2 - a^2*si n(f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1))) + 3*c^3*((9*sin(f* x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 6*l og(sin(f*x + e)/(cos(f*x + e) + 1) + 1)/a^2 + 6*log(sin(f*x + e)/(cos(f*x + e) + 1) - 1)/a^2) + 3*c^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - c^3*(3*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2)/f
Time = 0.35 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.02 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=\frac {\frac {15 \, c^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} - \frac {15 \, c^{3} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} + \frac {6 \, c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{2}} - \frac {4 \, {\left (a^{4} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 6 \, a^{4} c^{3} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{6}}}{3 \, f} \]
1/3*(15*c^3*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 - 15*c^3*log(abs(tan(1/ 2*f*x + 1/2*e) - 1))/a^2 + 6*c^3*tan(1/2*f*x + 1/2*e)/((tan(1/2*f*x + 1/2* e)^2 - 1)*a^2) - 4*(a^4*c^3*tan(1/2*f*x + 1/2*e)^3 + 6*a^4*c^3*tan(1/2*f*x + 1/2*e))/a^6)/f
Time = 13.04 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.87 \[ \int \frac {\sec (e+f x) (c-c \sec (e+f x))^3}{(a+a \sec (e+f x))^2} \, dx=\frac {10\,c^3\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a^2\,f}-\frac {4\,c^3\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3\,a^2\,f}-\frac {8\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a^2\,f}+\frac {2\,c^3\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-a^2\right )} \]