Integrand size = 26, antiderivative size = 102 \[ \int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx=\frac {c^4 x}{a^2}-\frac {6 c^4 \text {arctanh}(\sin (e+f x))}{a^2 f}-\frac {16 c^4 \cot (e+f x)}{a^2 f}-\frac {32 c^4 \cot ^3(e+f x)}{3 a^2 f}+\frac {32 c^4 \csc ^3(e+f x)}{3 a^2 f}+\frac {c^4 \tan (e+f x)}{a^2 f} \]
c^4*x/a^2-6*c^4*arctanh(sin(f*x+e))/a^2/f-16*c^4*cot(f*x+e)/a^2/f-32/3*c^4 *cot(f*x+e)^3/a^2/f+32/3*c^4*csc(f*x+e)^3/a^2/f+c^4*tan(f*x+e)/a^2/f
Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.
Time = 2.44 (sec) , antiderivative size = 298, normalized size of antiderivative = 2.92 \[ \int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx=-\frac {c^{7/2} \tan (e+f x) \left (-8 \sqrt {a} \sqrt {c}-8 \sqrt {2} \sqrt {a} \sqrt {c} \operatorname {Hypergeometric2F1}\left (-\frac {5}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1+\sec (e+f x))\right ) \sqrt {1-\sec (e+f x)}-4 \sqrt {2} \sqrt {a} \sqrt {c} \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},-\frac {3}{2},-\frac {1}{2},\frac {1}{2} (1+\sec (e+f x))\right ) \sqrt {1-\sec (e+f x)}+4 \sqrt {a} \sqrt {c} \sec (e+f x)+4 \sqrt {a} \sqrt {c} \sec ^2(e+f x)+3 \text {arctanh}\left (\frac {\sqrt {-a c \tan ^2(e+f x)}}{\sqrt {a} \sqrt {c}}\right ) \sqrt {-a c \tan ^2(e+f x)}+3 \text {arctanh}\left (\frac {\sqrt {-a c \tan ^2(e+f x)}}{\sqrt {a} \sqrt {c}}\right ) \sec (e+f x) \sqrt {-a c \tan ^2(e+f x)}\right )}{3 a^{5/2} f (-1+\sec (e+f x)) (1+\sec (e+f x))^2} \]
-1/3*(c^(7/2)*Tan[e + f*x]*(-8*Sqrt[a]*Sqrt[c] - 8*Sqrt[2]*Sqrt[a]*Sqrt[c] *Hypergeometric2F1[-5/2, -3/2, -1/2, (1 + Sec[e + f*x])/2]*Sqrt[1 - Sec[e + f*x]] - 4*Sqrt[2]*Sqrt[a]*Sqrt[c]*Hypergeometric2F1[-3/2, -3/2, -1/2, (1 + Sec[e + f*x])/2]*Sqrt[1 - Sec[e + f*x]] + 4*Sqrt[a]*Sqrt[c]*Sec[e + f*x ] + 4*Sqrt[a]*Sqrt[c]*Sec[e + f*x]^2 + 3*ArcTanh[Sqrt[-(a*c*Tan[e + f*x]^2 )]/(Sqrt[a]*Sqrt[c])]*Sqrt[-(a*c*Tan[e + f*x]^2)] + 3*ArcTanh[Sqrt[-(a*c*T an[e + f*x]^2)]/(Sqrt[a]*Sqrt[c])]*Sec[e + f*x]*Sqrt[-(a*c*Tan[e + f*x]^2) ]))/(a^(5/2)*f*(-1 + Sec[e + f*x])*(1 + Sec[e + f*x])^2)
Time = 0.53 (sec) , antiderivative size = 91, normalized size of antiderivative = 0.89, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {3042, 4392, 3042, 4374, 2009}
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.
| \(\displaystyle \int \frac {(c-c \sec (e+f x))^4}{(a \sec (e+f x)+a)^2} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^4}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^2}dx\) |
| \(\Big \downarrow \) 4392 |
| \(\displaystyle \frac {\int \cot ^4(e+f x) (c-c \sec (e+f x))^6dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {\int \frac {\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^6}{\cot \left (e+f x+\frac {\pi }{2}\right )^4}dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 4374 |
| \(\displaystyle \frac {\int \left (\cot ^4(e+f x) c^6+15 \csc ^4(e+f x) c^6-20 \cot (e+f x) \csc ^3(e+f x) c^6+15 \cot ^2(e+f x) \csc ^2(e+f x) c^6+\csc ^4(e+f x) \sec ^2(e+f x) c^6-6 \cot ^3(e+f x) \csc (e+f x) c^6-6 \csc ^4(e+f x) \sec (e+f x) c^6\right )dx}{a^2 c^2}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {-\frac {6 c^6 \text {arctanh}(\sin (e+f x))}{f}+\frac {c^6 \tan (e+f x)}{f}-\frac {32 c^6 \cot ^3(e+f x)}{3 f}-\frac {16 c^6 \cot (e+f x)}{f}+\frac {32 c^6 \csc ^3(e+f x)}{3 f}+c^6 x}{a^2 c^2}\) |
(c^6*x - (6*c^6*ArcTanh[Sin[e + f*x]])/f - (16*c^6*Cot[e + f*x])/f - (32*c ^6*Cot[e + f*x]^3)/(3*f) + (32*c^6*Csc[e + f*x]^3)/(3*f) + (c^6*Tan[e + f* x])/f)/(a^2*c^2)
3.1.22.3.1 Defintions of rubi rules used
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_))^(n_), x_Symbol] :> Int[ExpandIntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[ c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.)*(csc[(e_.) + (f_.)*(x_)]*( d_.) + (c_))^(n_.), x_Symbol] :> Simp[((-a)*c)^m Int[Cot[e + f*x]^(2*m)*( c + d*Csc[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && E qQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && IntegerQ[m] && RationalQ[n] && !( IntegerQ[n] && GtQ[m - n, 0])
Time = 0.68 (sec) , antiderivative size = 105, normalized size of antiderivative = 1.03
| method | result | size |
| derivativedivides | \(\frac {8 c^{4} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}-\frac {1}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}+\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}\right )}{f \,a^{2}}\) | \(105\) |
| default | \(\frac {8 c^{4} \left (\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-\frac {1}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}-\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{4}-\frac {1}{8 \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}+\frac {3 \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{4}+\frac {\arctan \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )\right )}{4}\right )}{f \,a^{2}}\) | \(105\) |
| parallelrisch | \(\frac {6 \left (\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right ) \cos \left (f x +e \right )-\ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right ) \cos \left (f x +e \right )+\frac {19 \left (\cos \left (f x +e \right )+\frac {\cos \left (2 f x +2 e \right )}{4}+\frac {25}{76}\right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \sec \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{18}+\frac {f x \cos \left (f x +e \right )}{6}\right ) c^{4}}{f \,a^{2} \cos \left (f x +e \right )}\) | \(111\) |
| risch | \(\frac {c^{4} x}{a^{2}}+\frac {2 i c^{4} \left (51 \,{\mathrm e}^{3 i \left (f x +e \right )}+25 \,{\mathrm e}^{2 i \left (f x +e \right )}+57 \,{\mathrm e}^{i \left (f x +e \right )}+19\right )}{3 f \,a^{2} \left (1+{\mathrm e}^{2 i \left (f x +e \right )}\right ) \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{3}}+\frac {6 c^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}-i\right )}{a^{2} f}-\frac {6 c^{4} \ln \left ({\mathrm e}^{i \left (f x +e \right )}+i\right )}{a^{2} f}\) | \(131\) |
| norman | \(\frac {\frac {c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{a}-\frac {c^{4} x}{a}-\frac {10 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{a f}+\frac {76 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3 a f}-\frac {18 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{a f}+\frac {8 c^{4} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{9}}{3 a f}+\frac {3 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{a}-\frac {3 c^{4} x \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{a}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right )^{3} a}+\frac {6 c^{4} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )-1\right )}{a^{2} f}-\frac {6 c^{4} \ln \left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )+1\right )}{a^{2} f}\) | \(222\) |
8/f*c^4/a^2*(1/3*tan(1/2*f*x+1/2*e)^3+tan(1/2*f*x+1/2*e)-1/8/(tan(1/2*f*x+ 1/2*e)+1)-3/4*ln(tan(1/2*f*x+1/2*e)+1)-1/8/(tan(1/2*f*x+1/2*e)-1)+3/4*ln(t an(1/2*f*x+1/2*e)-1)+1/4*arctan(tan(1/2*f*x+1/2*e)))
Leaf count of result is larger than twice the leaf count of optimal. 220 vs. \(2 (98) = 196\).
Time = 0.30 (sec) , antiderivative size = 220, normalized size of antiderivative = 2.16 \[ \int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx=\frac {3 \, c^{4} f x \cos \left (f x + e\right )^{3} + 6 \, c^{4} f x \cos \left (f x + e\right )^{2} + 3 \, c^{4} f x \cos \left (f x + e\right ) - 9 \, {\left (c^{4} \cos \left (f x + e\right )^{3} + 2 \, c^{4} \cos \left (f x + e\right )^{2} + c^{4} \cos \left (f x + e\right )\right )} \log \left (\sin \left (f x + e\right ) + 1\right ) + 9 \, {\left (c^{4} \cos \left (f x + e\right )^{3} + 2 \, c^{4} \cos \left (f x + e\right )^{2} + c^{4} \cos \left (f x + e\right )\right )} \log \left (-\sin \left (f x + e\right ) + 1\right ) + {\left (19 \, c^{4} \cos \left (f x + e\right )^{2} + 38 \, c^{4} \cos \left (f x + e\right ) + 3 \, c^{4}\right )} \sin \left (f x + e\right )}{3 \, {\left (a^{2} f \cos \left (f x + e\right )^{3} + 2 \, a^{2} f \cos \left (f x + e\right )^{2} + a^{2} f \cos \left (f x + e\right )\right )}} \]
1/3*(3*c^4*f*x*cos(f*x + e)^3 + 6*c^4*f*x*cos(f*x + e)^2 + 3*c^4*f*x*cos(f *x + e) - 9*(c^4*cos(f*x + e)^3 + 2*c^4*cos(f*x + e)^2 + c^4*cos(f*x + e)) *log(sin(f*x + e) + 1) + 9*(c^4*cos(f*x + e)^3 + 2*c^4*cos(f*x + e)^2 + c^ 4*cos(f*x + e))*log(-sin(f*x + e) + 1) + (19*c^4*cos(f*x + e)^2 + 38*c^4*c os(f*x + e) + 3*c^4)*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))
\[ \int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx=\frac {c^{4} \left (\int \left (- \frac {4 \sec {\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {6 \sec ^{2}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \left (- \frac {4 \sec ^{3}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\right )\, dx + \int \frac {\sec ^{4}{\left (e + f x \right )}}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {1}{\sec ^{2}{\left (e + f x \right )} + 2 \sec {\left (e + f x \right )} + 1}\, dx\right )}{a^{2}} \]
c**4*(Integral(-4*sec(e + f*x)/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(6*sec(e + f*x)**2/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integral(-4*sec(e + f*x)**3/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + I ntegral(sec(e + f*x)**4/(sec(e + f*x)**2 + 2*sec(e + f*x) + 1), x) + Integ ral(1/(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. 413 vs. \(2 (98) = 196\).
Time = 0.32 (sec) , antiderivative size = 413, normalized size of antiderivative = 4.05 \[ \int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx=\frac {c^{4} {\left (\frac {\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {12 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}} + \frac {12 \, \sin \left (f x + e\right )}{{\left (a^{2} - \frac {a^{2} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (\cos \left (f x + e\right ) + 1\right )}}\right )} + 4 \, c^{4} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + 1\right )}{a^{2}} + \frac {6 \, \log \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - 1\right )}{a^{2}}\right )} - c^{4} {\left (\frac {\frac {9 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}}{a^{2}} - \frac {12 \, \arctan \left (\frac {\sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1}\right )}{a^{2}}\right )} + \frac {6 \, c^{4} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}} - \frac {4 \, c^{4} {\left (\frac {3 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}}\right )}}{a^{2}}}{6 \, f} \]
1/6*(c^4*((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*sin (f*x + e)^2/(cos(f*x + e) + 1)^2)*(cos(f*x + e) + 1))) + 4*c^4*((9*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 6*lo g(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) - c^4*((9*sin(f*x + e)/(cos(f*x + e) + 1) - sin(f*x + e )^3/(cos(f*x + e) + 1)^3)/a^2 - 12*arctan(sin(f*x + e)/(cos(f*x + e) + 1)) /a^2) + 6*c^4*(3*sin(f*x + e)/(cos(f*x + e) + 1) + sin(f*x + e)^3/(cos(f*x + e) + 1)^3)/a^2 - 4*c^4*(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 = 0.33 (sec) , antiderivative size = 134, normalized size of antiderivative = 1.31 \[ \int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx=\frac {\frac {3 \, {\left (f x + e\right )} c^{4}}{a^{2}} - \frac {18 \, c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 1 \right |}\right )}{a^{2}} + \frac {18 \, c^{4} \log \left ({\left | \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) - 1 \right |}\right )}{a^{2}} - \frac {6 \, c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{{\left (\tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} - 1\right )} a^{2}} + \frac {8 \, {\left (a^{4} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 3 \, a^{4} c^{4} \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )\right )}}{a^{6}}}{3 \, f} \]
1/3*(3*(f*x + e)*c^4/a^2 - 18*c^4*log(abs(tan(1/2*f*x + 1/2*e) + 1))/a^2 + 18*c^4*log(abs(tan(1/2*f*x + 1/2*e) - 1))/a^2 - 6*c^4*tan(1/2*f*x + 1/2*e )/((tan(1/2*f*x + 1/2*e)^2 - 1)*a^2) + 8*(a^4*c^4*tan(1/2*f*x + 1/2*e)^3 + 3*a^4*c^4*tan(1/2*f*x + 1/2*e))/a^6)/f
Time = 13.31 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.10 \[ \int \frac {(c-c \sec (e+f x))^4}{(a+a \sec (e+f x))^2} \, dx=\frac {c^4\,x}{a^2}+\frac {8\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{a^2\,f}+\frac {8\,c^4\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{3\,a^2\,f}-\frac {12\,c^4\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\right )}{a^2\,f}-\frac {2\,c^4\,\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}{f\,\left (a^2\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2-a^2\right )} \]