Integrand size = 29, antiderivative size = 102 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {(c-d) \tan (e+f x)}{5 f (a+a \sec (e+f x))^3}+\frac {(2 c+3 d) \tan (e+f x)}{15 a f (a+a \sec (e+f x))^2}+\frac {(2 c+3 d) \tan (e+f x)}{15 f \left (a^3+a^3 \sec (e+f x)\right )} \] Output:
1/5*(c-d)*tan(f*x+e)/f/(a+a*sec(f*x+e))^3+1/15*(2*c+3*d)*tan(f*x+e)/a/f/(a +a*sec(f*x+e))^2+1/15*(2*c+3*d)*tan(f*x+e)/f/(a^3+a^3*sec(f*x+e))
Time = 0.85 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.32 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (e+f x)\right ) \sec \left (\frac {e}{2}\right ) \left (5 (8 c+3 d) \sin \left (\frac {f x}{2}\right )-15 (2 c+d) \sin \left (e+\frac {f x}{2}\right )+20 c \sin \left (e+\frac {3 f x}{2}\right )+15 d \sin \left (e+\frac {3 f x}{2}\right )-15 c \sin \left (2 e+\frac {3 f x}{2}\right )+7 c \sin \left (2 e+\frac {5 f x}{2}\right )+3 d \sin \left (2 e+\frac {5 f x}{2}\right )\right )}{30 a^3 f (1+\cos (e+f x))^3} \] Input:
Integrate[(Sec[e + f*x]*(c + d*Sec[e + f*x]))/(a + a*Sec[e + f*x])^3,x]
Output:
(Cos[(e + f*x)/2]*Sec[e/2]*(5*(8*c + 3*d)*Sin[(f*x)/2] - 15*(2*c + d)*Sin[ e + (f*x)/2] + 20*c*Sin[e + (3*f*x)/2] + 15*d*Sin[e + (3*f*x)/2] - 15*c*Si n[2*e + (3*f*x)/2] + 7*c*Sin[2*e + (5*f*x)/2] + 3*d*Sin[2*e + (5*f*x)/2])) /(30*a^3*f*(1 + Cos[e + f*x])^3)
Time = 0.48 (sec) , antiderivative size = 99, normalized size of antiderivative = 0.97, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.207, Rules used = {3042, 4488, 3042, 4283, 3042, 4281}
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.
| \(\displaystyle \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a \sec (e+f x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right ) \left (c+d \csc \left (e+f x+\frac {\pi }{2}\right )\right )}{\left (a \csc \left (e+f x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4488 |
| \(\displaystyle \frac {(2 c+3 d) \int \frac {\sec (e+f x)}{(\sec (e+f x) a+a)^2}dx}{5 a}+\frac {(c-d) \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(2 c+3 d) \int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\left (\csc \left (e+f x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}+\frac {(c-d) \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 4283 |
| \(\displaystyle \frac {(2 c+3 d) \left (\frac {\int \frac {\sec (e+f x)}{\sec (e+f x) a+a}dx}{3 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2}\right )}{5 a}+\frac {(c-d) \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \frac {(2 c+3 d) \left (\frac {\int \frac {\csc \left (e+f x+\frac {\pi }{2}\right )}{\csc \left (e+f x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2}\right )}{5 a}+\frac {(c-d) \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle \frac {(c-d) \tan (e+f x)}{5 f (a \sec (e+f x)+a)^3}+\frac {(2 c+3 d) \left (\frac {\tan (e+f x)}{3 a f (a \sec (e+f x)+a)}+\frac {\tan (e+f x)}{3 f (a \sec (e+f x)+a)^2}\right )}{5 a}\) |
Input:
Int[(Sec[e + f*x]*(c + d*Sec[e + f*x]))/(a + a*Sec[e + f*x])^3,x]
Output:
((c - d)*Tan[e + f*x])/(5*f*(a + a*Sec[e + f*x])^3) + ((2*c + 3*d)*(Tan[e + f*x]/(3*f*(a + a*Sec[e + f*x])^2) + Tan[e + f*x]/(3*a*f*(a + a*Sec[e + f *x]))))/(5*a)
Int[csc[(e_.) + (f_.)*(x_)]/(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbo l] :> Simp[-Cot[e + f*x]/(f*(b + a*Csc[e + f*x])), x] /; FreeQ[{a, b, e, f} , x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_ Symbol] :> Simp[b*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(m + 1)/(a*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1) ] && IntegerQ[2*m]
Int[csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(cs c[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(A*b - a*B)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[(a*B*m + A*b*(m + 1))/(a*b*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && NeQ[a*B*m + A*b*(m + 1), 0] && LtQ[m, -2^(-1)]
Time = 0.20 (sec) , antiderivative size = 56, normalized size of antiderivative = 0.55
| method | result | size |
| parallelrisch | \(\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4}-\frac {10 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}}{3}+5 c +5 d \right )}{20 a^{3} f}\) | \(56\) |
| derivativedivides | \(\frac {\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f \,a^{3}}\) | \(64\) |
| default | \(\frac {\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{5}-\frac {2 c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{3}+c \tan \left (\frac {f x}{2}+\frac {e}{2}\right )+d \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f \,a^{3}}\) | \(64\) |
| risch | \(\frac {2 i \left (15 c \,{\mathrm e}^{4 i \left (f x +e \right )}+30 c \,{\mathrm e}^{3 i \left (f x +e \right )}+15 d \,{\mathrm e}^{3 i \left (f x +e \right )}+40 c \,{\mathrm e}^{2 i \left (f x +e \right )}+15 d \,{\mathrm e}^{2 i \left (f x +e \right )}+20 \,{\mathrm e}^{i \left (f x +e \right )} c +15 d \,{\mathrm e}^{i \left (f x +e \right )}+7 c +3 d \right )}{15 f \,a^{3} \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )^{5}}\) | \(114\) |
| norman | \(\frac {\frac {\left (c -d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{7}}{20 f a}-\frac {\left (c +d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )}{4 f a}+\frac {\left (5 c +3 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{3}}{12 f a}-\frac {\left (13 c -3 d \right ) \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{5}}{60 f a}}{\left (\tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2}-1\right ) a^{2}}\) | \(117\) |
Input:
int(sec(f*x+e)*(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^3,x,method=_RETURNVERBOSE )
Output:
1/20*tan(1/2*f*x+1/2*e)*((c-d)*tan(1/2*f*x+1/2*e)^4-10/3*c*tan(1/2*f*x+1/2 *e)^2+5*c+5*d)/a^3/f
Time = 0.10 (sec) , antiderivative size = 93, normalized size of antiderivative = 0.91 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {{\left ({\left (7 \, c + 3 \, d\right )} \cos \left (f x + e\right )^{2} + 3 \, {\left (2 \, c + 3 \, d\right )} \cos \left (f x + e\right ) + 2 \, c + 3 \, d\right )} \sin \left (f x + e\right )}{15 \, {\left (a^{3} f \cos \left (f x + e\right )^{3} + 3 \, a^{3} f \cos \left (f x + e\right )^{2} + 3 \, a^{3} f \cos \left (f x + e\right ) + a^{3} f\right )}} \] Input:
integrate(sec(f*x+e)*(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^3,x, algorithm="fri cas")
Output:
1/15*((7*c + 3*d)*cos(f*x + e)^2 + 3*(2*c + 3*d)*cos(f*x + e) + 2*c + 3*d) *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)
\[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {\int \frac {c \sec {\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx + \int \frac {d \sec ^{2}{\left (e + f x \right )}}{\sec ^{3}{\left (e + f x \right )} + 3 \sec ^{2}{\left (e + f x \right )} + 3 \sec {\left (e + f x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sec(f*x+e)*(c+d*sec(f*x+e))/(a+a*sec(f*x+e))**3,x)
Output:
(Integral(c*sec(e + f*x)/(sec(e + f*x)**3 + 3*sec(e + f*x)**2 + 3*sec(e + f*x) + 1), x) + Integral(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
Time = 0.03 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.13 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {\frac {c {\left (\frac {15 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {10 \, \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}} + \frac {3 \, d {\left (\frac {5 \, \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} - \frac {\sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )}}{a^{3}}}{60 \, f} \] Input:
integrate(sec(f*x+e)*(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^3,x, algorithm="max ima")
Output:
1/60*(c*(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 + 3*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 = 0.16 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.74 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {3 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 3 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{5} - 10 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )^{3} + 15 \, c \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right ) + 15 \, d \tan \left (\frac {1}{2} \, f x + \frac {1}{2} \, e\right )}{60 \, a^{3} f} \] Input:
integrate(sec(f*x+e)*(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^3,x, algorithm="gia c")
Output:
1/60*(3*c*tan(1/2*f*x + 1/2*e)^5 - 3*d*tan(1/2*f*x + 1/2*e)^5 - 10*c*tan(1 /2*f*x + 1/2*e)^3 + 15*c*tan(1/2*f*x + 1/2*e) + 15*d*tan(1/2*f*x + 1/2*e)) /(a^3*f)
Time = 10.30 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )\,\left (15\,c+15\,d-10\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^2+3\,c\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4-3\,d\,{\mathrm {tan}\left (\frac {e}{2}+\frac {f\,x}{2}\right )}^4\right )}{60\,a^3\,f} \] Input:
int((c + d/cos(e + f*x))/(cos(e + f*x)*(a + a/cos(e + f*x))^3),x)
Output:
(tan(e/2 + (f*x)/2)*(15*c + 15*d - 10*c*tan(e/2 + (f*x)/2)^2 + 3*c*tan(e/2 + (f*x)/2)^4 - 3*d*tan(e/2 + (f*x)/2)^4))/(60*a^3*f)
Time = 0.21 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.65 \[ \int \frac {\sec (e+f x) (c+d \sec (e+f x))}{(a+a \sec (e+f x))^3} \, dx=\frac {\tan \left (\frac {f x}{2}+\frac {e}{2}\right ) \left (3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} c -3 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{4} d -10 \tan \left (\frac {f x}{2}+\frac {e}{2}\right )^{2} c +15 c +15 d \right )}{60 a^{3} f} \] Input:
int(sec(f*x+e)*(c+d*sec(f*x+e))/(a+a*sec(f*x+e))^3,x)
Output:
(tan((e + f*x)/2)*(3*tan((e + f*x)/2)**4*c - 3*tan((e + f*x)/2)**4*d - 10* tan((e + f*x)/2)**2*c + 15*c + 15*d))/(60*a**3*f)