Integrand size = 30, antiderivative size = 116 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^4} \, dx=-\frac {(a+a \sec (e+f x)) \tan (e+f x)}{7 f (c-c \sec (e+f x))^4}-\frac {2 (a+a \sec (e+f x)) \tan (e+f x)}{35 c f (c-c \sec (e+f x))^3}-\frac {2 (a+a \sec (e+f x)) \tan (e+f x)}{105 f \left (c^2-c^2 \sec (e+f x)\right )^2} \]
-1/7*(a+a*sec(f*x+e))*tan(f*x+e)/f/(c-c*sec(f*x+e))^4-2/35*(a+a*sec(f*x+e) )*tan(f*x+e)/c/f/(c-c*sec(f*x+e))^3-2/105*(a+a*sec(f*x+e))*tan(f*x+e)/f/(c ^2-c^2*sec(f*x+e))^2
Time = 3.95 (sec) , antiderivative size = 55, normalized size of antiderivative = 0.47 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^4} \, dx=-\frac {a (1+\sec (e+f x)) \left (23-10 \sec (e+f x)+2 \sec ^2(e+f x)\right ) \tan (e+f x)}{105 c^4 f (-1+\sec (e+f x))^4} \]
-1/105*(a*(1 + Sec[e + f*x])*(23 - 10*Sec[e + f*x] + 2*Sec[e + f*x]^2)*Tan [e + f*x])/(c^4*f*(-1 + Sec[e + f*x])^4)
Time = 0.54 (sec) , antiderivative size = 120, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 4439, 3042, 4439, 3042, 4438}
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.
\(\displaystyle \int \frac {\sec (e+f x) (a \sec (e+f x)+a)}{(c-c \sec (e+f x))^4} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^4}dx\) |
\(\Big \downarrow \) 4439 |
\(\displaystyle \frac {2 \int \frac {\sec (e+f x) (\sec (e+f x) a+a)}{(c-c \sec (e+f x))^3}dx}{7 c}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{7 f (c-c \sec (e+f x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^3}dx}{7 c}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{7 f (c-c \sec (e+f x))^4}\) |
\(\Big \downarrow \) 4439 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {\sec (e+f x) (\sec (e+f x) a+a)}{(c-c \sec (e+f x))^2}dx}{5 c}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{5 f (c-c \sec (e+f x))^3}\right )}{7 c}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{7 f (c-c \sec (e+f x))^4}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )}{\left (c-c \csc \left (e+f x+\frac {\pi }{2}\right )\right )^2}dx}{5 c}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{5 f (c-c \sec (e+f x))^3}\right )}{7 c}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{7 f (c-c \sec (e+f x))^4}\) |
\(\Big \downarrow \) 4438 |
\(\displaystyle \frac {2 \left (-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{15 c f (c-c \sec (e+f x))^2}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{5 f (c-c \sec (e+f x))^3}\right )}{7 c}-\frac {\tan (e+f x) (a \sec (e+f x)+a)}{7 f (c-c \sec (e+f x))^4}\) |
-1/7*((a + a*Sec[e + f*x])*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^4) + (2*( -1/5*((a + a*Sec[e + f*x])*Tan[e + f*x])/(f*(c - c*Sec[e + f*x])^3) - ((a + a*Sec[e + f*x])*Tan[e + f*x])/(15*c*f*(c - c*Sec[e + f*x])^2)))/(7*c)
3.1.8.3.1 Defintions of rubi rules used
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[b*Cot[e + f*x] *(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] /; Fre eQ[{a, b, c, d, e, f, m, n}, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 - b^2, 0] & & EqQ[m + n + 1, 0] && NeQ[2*m + 1, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(d_.) + (c_))^(n_.), x_Symbol] :> Simp[b*Cot[e + f*x] *(a + b*Csc[e + f*x])^m*((c + d*Csc[e + f*x])^n/(a*f*(2*m + 1))), x] + Simp [(m + n + 1)/(a*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)* (c + d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && EqQ [b*c + a*d, 0] && EqQ[a^2 - b^2, 0] && ILtQ[m + n + 1, 0] && NeQ[2*m + 1, 0 ] && !LtQ[n, 0] && !(IGtQ[n + 1/2, 0] && LtQ[n + 1/2, -(m + n)])
Time = 0.92 (sec) , antiderivative size = 49, normalized size of antiderivative = 0.42
method | result | size |
parallelrisch | \(-\frac {a \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{3} \left (15 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-42 \cot \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}+35\right )}{420 c^{4} f}\) | \(49\) |
derivativedivides | \(\frac {a \left (-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {2}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}\right )}{4 f \,c^{4}}\) | \(50\) |
default | \(\frac {a \left (-\frac {1}{7 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}-\frac {1}{3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}+\frac {2}{5 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}\right )}{4 f \,c^{4}}\) | \(50\) |
risch | \(\frac {2 i a \left (105 \,{\mathrm e}^{6 i \left (f x +e \right )}-210 \,{\mathrm e}^{5 i \left (f x +e \right )}+455 \,{\mathrm e}^{4 i \left (f x +e \right )}-350 \,{\mathrm e}^{3 i \left (f x +e \right )}+273 \,{\mathrm e}^{2 i \left (f x +e \right )}-56 \,{\mathrm e}^{i \left (f x +e \right )}+23\right )}{105 f \,c^{4} \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )^{7}}\) | \(92\) |
norman | \(\frac {\frac {a}{28 c f}-\frac {19 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{140 c f}+\frac {11 a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}}{60 c f}-\frac {a \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{6}}{12 c f}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) c^{3} \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}\) | \(101\) |
Time = 0.26 (sec) , antiderivative size = 104, normalized size of antiderivative = 0.90 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^4} \, dx=\frac {23 \, a \cos \left (f x + e\right )^{4} + 36 \, a \cos \left (f x + e\right )^{3} + 5 \, a \cos \left (f x + e\right )^{2} - 6 \, a \cos \left (f x + e\right ) + 2 \, a}{105 \, {\left (c^{4} f \cos \left (f x + e\right )^{3} - 3 \, c^{4} f \cos \left (f x + e\right )^{2} + 3 \, c^{4} f \cos \left (f x + e\right ) - c^{4} f\right )} \sin \left (f x + e\right )} \]
1/105*(23*a*cos(f*x + e)^4 + 36*a*cos(f*x + e)^3 + 5*a*cos(f*x + e)^2 - 6* a*cos(f*x + e) + 2*a)/((c^4*f*cos(f*x + e)^3 - 3*c^4*f*cos(f*x + e)^2 + 3* c^4*f*cos(f*x + e) - c^4*f)*sin(f*x + e))
\[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^4} \, dx=\frac {a \left (\int \frac {\sec {\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {\sec ^{2}{\left (e + f x \right )}}{\sec ^{4}{\left (e + f x \right )} - 4 \sec ^{3}{\left (e + f x \right )} + 6 \sec ^{2}{\left (e + f x \right )} - 4 \sec {\left (e + f x \right )} + 1}\, dx\right )}{c^{4}} \]
a*(Integral(sec(e + f*x)/(sec(e + f*x)**4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1), x) + Integral(sec(e + f*x)**2/(sec(e + f*x) **4 - 4*sec(e + f*x)**3 + 6*sec(e + f*x)**2 - 4*sec(e + f*x) + 1), x))/c** 4
Time = 0.23 (sec) , antiderivative size = 177, normalized size of antiderivative = 1.53 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^4} \, dx=\frac {\frac {a {\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + \frac {35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {105 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 15\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}} + \frac {3 \, a {\left (\frac {21 \, \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {35 \, \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} + \frac {35 \, \sin \left (f x + e\right )^{6}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{6}} - 5\right )} {\left (\cos \left (f x + e\right ) + 1\right )}^{7}}{c^{4} \sin \left (f x + e\right )^{7}}}{840 \, f} \]
1/840*(a*(21*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 35*sin(f*x + e)^4/(cos( f*x + e) + 1)^4 - 105*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 15)*(cos(f*x + e) + 1)^7/(c^4*sin(f*x + e)^7) + 3*a*(21*sin(f*x + e)^2/(cos(f*x + e) + 1 )^2 - 35*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 + 35*sin(f*x + e)^6/(cos(f*x + e) + 1)^6 - 5)*(cos(f*x + e) + 1)^7/(c^4*sin(f*x + e)^7))/f
Time = 0.32 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.44 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^4} \, dx=-\frac {35 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{4} - 42 \, a \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{2} + 15 \, a}{420 \, c^{4} f \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{7}} \]
-1/420*(35*a*tan(1/2*f*x + 1/2*e)^4 - 42*a*tan(1/2*f*x + 1/2*e)^2 + 15*a)/ (c^4*f*tan(1/2*f*x + 1/2*e)^7)
Time = 14.12 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.53 \[ \int \frac {\sec (e+f x) (a+a \sec (e+f x))}{(c-c \sec (e+f x))^4} \, dx=\frac {a\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^5}{10\,c^4\,f}-\frac {a\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^3}{12\,c^4\,f}-\frac {a\,{\mathrm {cot}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^7}{28\,c^4\,f} \]