Integrand size = 21, antiderivative size = 105 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\text {arctanh}(\sin (c+d x))}{a^3 d}-\frac {\sec ^2(c+d x) \tan (c+d x)}{5 d (a+a \sec (c+d x))^3}+\frac {7 \tan (c+d x)}{15 a d (a+a \sec (c+d x))^2}-\frac {29 \tan (c+d x)}{15 d \left (a^3+a^3 \sec (c+d x)\right )} \]
arctanh(sin(d*x+c))/a^3/d-1/5*sec(d*x+c)^2*tan(d*x+c)/d/(a+a*sec(d*x+c))^3 +7/15*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^2-29/15*tan(d*x+c)/d/(a^3+a^3*sec(d* x+c))
Time = 0.29 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.91 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x) \left (120 \text {arctanh}(\sin (c+d x)) \cos ^5\left (\frac {1}{2} (c+d x)\right )-35 \sin \left (\frac {1}{2} (c+d x)\right )-40 \sin \left (\frac {3}{2} (c+d x)\right )-11 \sin \left (\frac {5}{2} (c+d x)\right )\right )}{15 a^3 d (1+\sec (c+d x))^3} \]
(Cos[(c + d*x)/2]*Sec[c + d*x]^3*(120*ArcTanh[Sin[c + d*x]]*Cos[(c + d*x)/ 2]^5 - 35*Sin[(c + d*x)/2] - 40*Sin[(3*(c + d*x))/2] - 11*Sin[(5*(c + d*x) )/2]))/(15*a^3*d*(1 + Sec[c + d*x])^3)
Time = 0.75 (sec) , antiderivative size = 115, normalized size of antiderivative = 1.10, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.476, Rules used = {3042, 4303, 3042, 4496, 25, 3042, 4486, 3042, 4257, 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.
| \(\displaystyle \int \frac {\sec ^4(c+d x)}{(a \sec (c+d x)+a)^3} \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^3}dx\) |
| \(\Big \downarrow \) 4303 |
| \(\displaystyle -\frac {\int \frac {\sec ^2(c+d x) (2 a-5 a \sec (c+d x))}{(\sec (c+d x) a+a)^2}dx}{5 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (2 a-5 a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^2}dx}{5 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4496 |
| \(\displaystyle -\frac {-\frac {\int -\frac {\sec (c+d x) \left (14 a^2-15 a^2 \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {7 a \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 25 |
| \(\displaystyle -\frac {\frac {\int \frac {\sec (c+d x) \left (14 a^2-15 a^2 \sec (c+d x)\right )}{\sec (c+d x) a+a}dx}{3 a^2}-\frac {7 a \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (14 a^2-15 a^2 \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx}{3 a^2}-\frac {7 a \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4486 |
| \(\displaystyle -\frac {\frac {29 a^2 \int \frac {\sec (c+d x)}{\sec (c+d x) a+a}dx-15 a \int \sec (c+d x)dx}{3 a^2}-\frac {7 a \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle -\frac {\frac {29 a^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-15 a \int \csc \left (c+d x+\frac {\pi }{2}\right )dx}{3 a^2}-\frac {7 a \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle -\frac {\frac {29 a^2 \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-\frac {15 a \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {7 a \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
| \(\Big \downarrow \) 4281 |
| \(\displaystyle -\frac {\frac {\frac {29 a^2 \tan (c+d x)}{d (a \sec (c+d x)+a)}-\frac {15 a \text {arctanh}(\sin (c+d x))}{d}}{3 a^2}-\frac {7 a \tan (c+d x)}{3 d (a \sec (c+d x)+a)^2}}{5 a^2}-\frac {\tan (c+d x) \sec ^2(c+d x)}{5 d (a \sec (c+d x)+a)^3}\) |
-1/5*(Sec[c + d*x]^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^3) - ((-7*a*Tan [c + d*x])/(3*d*(a + a*Sec[c + d*x])^2) + ((-15*a*ArcTanh[Sin[c + d*x]])/d + (29*a^2*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])))/(3*a^2))/(5*a^2)
3.1.63.3.1 Defintions of rubi rules used
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
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_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_), x_Symbol] :> Simp[(-d^2)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d *Csc[e + f*x])^(n - 2)/(f*(2*m + 1))), x] + Simp[d^2/(a*b*(2*m + 1)) Int[ (a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(b*(n - 2) + a*(m - n + 2)*Csc[e + f*x]), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[n, 2] && (IntegersQ[2*m, 2*n] || IntegerQ[m])
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/(csc[( e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[B/b Int[Csc[e + f*x], x], x] + Simp[(A*b - a*B)/b Int[Csc[e + f*x]/(a + b*Csc[e + f*x]), x], x ] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[A*b - a*B, 0]
Int[csc[(e_.) + (f_.)*(x_)]^2*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*( csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-(A*b - a*B))*Cot [e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Simp[1/(b^2*(2*m + 1)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*m - a*B*m + b *B*(2*m + 1)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, e, f, A, B}, x] && Ne Q[A*b - a*B, 0] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Time = 0.27 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.71
| method | result | size |
| derivativedivides | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(75\) |
| default | \(\frac {-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{5}-\frac {4 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{3}-7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )-4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+4 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{4 d \,a^{3}}\) | \(75\) |
| parallelrisch | \(\frac {-3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}-20 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}-60 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+60 \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )-105 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{60 a^{3} d}\) | \(75\) |
| risch | \(-\frac {2 i \left (15 \,{\mathrm e}^{4 i \left (d x +c \right )}+75 \,{\mathrm e}^{3 i \left (d x +c \right )}+145 \,{\mathrm e}^{2 i \left (d x +c \right )}+95 \,{\mathrm e}^{i \left (d x +c \right )}+22\right )}{15 d \,a^{3} \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{5}}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{a^{3} d}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{a^{3} d}\) | \(111\) |
| norman | \(\frac {\frac {7 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 a d}-\frac {59 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 a d}+\frac {43 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{10 a d}-\frac {9 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{10 a d}-\frac {11 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{9}}{60 a d}-\frac {\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{11}}{20 a d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{3} a^{2}}+\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{a^{3} d}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{a^{3} d}\) | \(173\) |
1/4/d/a^3*(-1/5*tan(1/2*d*x+1/2*c)^5-4/3*tan(1/2*d*x+1/2*c)^3-7*tan(1/2*d* x+1/2*c)-4*ln(tan(1/2*d*x+1/2*c)-1)+4*ln(tan(1/2*d*x+1/2*c)+1))
Time = 0.29 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.50 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {15 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 15 \, {\left (\cos \left (d x + c\right )^{3} + 3 \, \cos \left (d x + c\right )^{2} + 3 \, \cos \left (d x + c\right ) + 1\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (22 \, \cos \left (d x + c\right )^{2} + 51 \, \cos \left (d x + c\right ) + 32\right )} \sin \left (d x + c\right )}{30 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \]
1/30*(15*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1)*log(sin( d*x + c) + 1) - 15*(cos(d*x + c)^3 + 3*cos(d*x + c)^2 + 3*cos(d*x + c) + 1 )*log(-sin(d*x + c) + 1) - 2*(22*cos(d*x + c)^2 + 51*cos(d*x + c) + 32)*si n(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)
\[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\int \frac {\sec ^{4}{\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \]
Time = 0.34 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.13 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {\frac {\frac {105 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {20 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {3 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}}}{a^{3}} - \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}{a^{3}} + \frac {60 \, \log \left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}{a^{3}}}{60 \, d} \]
-1/60*((105*sin(d*x + c)/(cos(d*x + c) + 1) + 20*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 - 60*log(sin(d*x + c)/(cos(d*x + c) + 1) + 1)/a^3 + 60*log(sin(d*x + c)/(cos(d*x + c) + 1) - 1)/a^3)/d
Time = 0.33 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.90 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=\frac {\frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right )}{a^{3}} - \frac {60 \, \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right )}{a^{3}} - \frac {3 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 20 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, a^{12} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{a^{15}}}{60 \, d} \]
1/60*(60*log(abs(tan(1/2*d*x + 1/2*c) + 1))/a^3 - 60*log(abs(tan(1/2*d*x + 1/2*c) - 1))/a^3 - (3*a^12*tan(1/2*d*x + 1/2*c)^5 + 20*a^12*tan(1/2*d*x + 1/2*c)^3 + 105*a^12*tan(1/2*d*x + 1/2*c))/a^15)/d
Time = 12.39 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.55 \[ \int \frac {\sec ^4(c+d x)}{(a+a \sec (c+d x))^3} \, dx=-\frac {105\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-120\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )+20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+3\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{60\,a^3\,d} \]