Integrand size = 19, antiderivative size = 83 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {2 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}+\frac {2 \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \] Output:
1/5*tan(d*x+c)/d/(a+a*sec(d*x+c))^3+2/15*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2 +2/15*tan(d*x+c)/d/(a^3+a^3*sec(d*x+c))
Time = 0.38 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.04 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\sec \left (\frac {c}{2}\right ) \sec ^5\left (\frac {1}{2} (c+d x)\right ) \left (40 \sin \left (\frac {d x}{2}\right )-30 \sin \left (c+\frac {d x}{2}\right )+20 \sin \left (c+\frac {3 d x}{2}\right )-15 \sin \left (2 c+\frac {3 d x}{2}\right )+7 \sin \left (2 c+\frac {5 d x}{2}\right )\right )}{240 a^3 d} \] Input:
Integrate[Sec[c + d*x]/(a + a*Sec[c + d*x])^3,x]
Output:
(Sec[c/2]*Sec[(c + d*x)/2]^5*(40*Sin[(d*x)/2] - 30*Sin[c + (d*x)/2] + 20*S in[c + (3*d*x)/2] - 15*Sin[2*c + (3*d*x)/2] + 7*Sin[2*c + (5*d*x)/2]))/(24 0*a^3*d)
Time = 0.43 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {3042, 4283, 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.
\(\displaystyle \int \frac {\sec (c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
\(\Big \downarrow \) 4283 |
\(\displaystyle \frac {2 \int \frac {\sec (c+d x)}{(\sec (c+d x) a+a)^2}dx}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4283 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {2 \left (\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}+\frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
\(\Big \downarrow \) 4281 |
\(\displaystyle \frac {\tan (c+d x)}{5 d (a \sec (c+d x)+a)^3}+\frac {2 \left (\frac {\tan (c+d x)}{3 a d (a \sec (c+d x)+a)}+\frac {\tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}\right )}{5 a}\) |
Input:
Int[Sec[c + d*x]/(a + a*Sec[c + d*x])^3,x]
Output:
Tan[c + d*x]/(5*d*(a + a*Sec[c + d*x])^3) + (2*(Tan[c + d*x]/(3*d*(a + a*S ec[c + d*x])^2) + Tan[c + d*x]/(3*a*d*(a + a*Sec[c + d*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]
Time = 0.35 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.54
method | result | size |
derivativedivides | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(45\) |
default | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {2 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d \,a^{3}}\) | \(45\) |
parallelrisch | \(\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}+15 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{60 d \,a^{3}}\) | \(47\) |
norman | \(\frac {\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{6 a d}+\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{20 a d}}{a^{2}}\) | \(61\) |
risch | \(\frac {2 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+30 \,{\mathrm e}^{3 i \left (d x +c \right )}+40 \,{\mathrm e}^{2 i \left (d x +c \right )}+20 \,{\mathrm e}^{i \left (d x +c \right )}+7\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}\) | \(69\) |
Input:
int(sec(d*x+c)/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
1/4/d/a^3*(1/5*tan(1/2*d*x+1/2*c)^5-2/3*tan(1/2*d*x+1/2*c)^3+tan(1/2*d*x+1 /2*c))
Time = 0.07 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.90 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {{\left (7 \, \cos \left (d x + c\right )^{2} + 6 \, \cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right )}{15 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \] Input:
integrate(sec(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")
Output:
1/15*(7*cos(d*x + c)^2 + 6*cos(d*x + c) + 2)*sin(d*x + c)/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)
\[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\sec {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \] Input:
integrate(sec(d*x+c)/(a+a*sec(d*x+c))**3,x)
Output:
Integral(sec(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x ) + 1), x)/a**3
Time = 0.04 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.81 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {15 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{60 \, a^{3} d} \] Input:
integrate(sec(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")
Output:
1/60*(15*sin(d*x + c)/(cos(d*x + c) + 1) - 10*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 3*sin(d*x + c)^5/(cos(d*x + c) + 1)^5)/(a^3*d)
Time = 0.19 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.55 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {3 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 10 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 15 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{60 \, a^{3} d} \] Input:
integrate(sec(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="giac")
Output:
1/60*(3*tan(1/2*d*x + 1/2*c)^5 - 10*tan(1/2*d*x + 1/2*c)^3 + 15*tan(1/2*d* x + 1/2*c))/(a^3*d)
Time = 9.61 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.54 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+15\right )}{60\,a^3\,d} \] Input:
int(1/(cos(c + d*x)*(a + a/cos(c + d*x))^3),x)
Output:
(tan(c/2 + (d*x)/2)*(3*tan(c/2 + (d*x)/2)^4 - 10*tan(c/2 + (d*x)/2)^2 + 15 ))/(60*a^3*d)
Time = 0.16 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.54 \[ \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-10 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+15\right )}{60 a^{3} d} \] Input:
int(sec(d*x+c)/(a+a*sec(d*x+c))^3,x)
Output:
(tan((c + d*x)/2)*(3*tan((c + d*x)/2)**4 - 10*tan((c + d*x)/2)**2 + 15))/( 60*a**3*d)