Integrand size = 19, antiderivative size = 85 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {3 b \text {arctanh}(\sin (c+d x))}{8 d}+\frac {a \tan (c+d x)}{d}+\frac {3 b \sec (c+d x) \tan (c+d x)}{8 d}+\frac {b \sec ^3(c+d x) \tan (c+d x)}{4 d}+\frac {a \tan ^3(c+d x)}{3 d} \] Output:
3/8*b*arctanh(sin(d*x+c))/d+a*tan(d*x+c)/d+3/8*b*sec(d*x+c)*tan(d*x+c)/d+1 /4*b*sec(d*x+c)^3*tan(d*x+c)/d+1/3*a*tan(d*x+c)^3/d
Time = 0.18 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.69 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {9 b \text {arctanh}(\sin (c+d x))+\tan (c+d x) \left (9 b \sec (c+d x)+6 b \sec ^3(c+d x)+8 a \left (3+\tan ^2(c+d x)\right )\right )}{24 d} \] Input:
Integrate[Sec[c + d*x]^4*(a + b*Sec[c + d*x]),x]
Output:
(9*b*ArcTanh[Sin[c + d*x]] + Tan[c + d*x]*(9*b*Sec[c + d*x] + 6*b*Sec[c + d*x]^3 + 8*a*(3 + Tan[c + d*x]^2)))/(24*d)
Time = 0.51 (sec) , antiderivative size = 90, normalized size of antiderivative = 1.06, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {3042, 4274, 3042, 4254, 2009, 4255, 3042, 4255, 3042, 4257}
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.
| \(\displaystyle \int \sec ^4(c+d x) (a+b \sec (c+d x)) \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )dx\) |
| \(\Big \downarrow \) 4274 |
| \(\displaystyle a \int \sec ^4(c+d x)dx+b \int \sec ^5(c+d x)dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle a \int \csc \left (c+d x+\frac {\pi }{2}\right )^4dx+b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx\) |
| \(\Big \downarrow \) 4254 |
| \(\displaystyle b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {a \int \left (\tan ^2(c+d x)+1\right )d(-\tan (c+d x))}{d}\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle b \int \csc \left (c+d x+\frac {\pi }{2}\right )^5dx-\frac {a \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle b \left (\frac {3}{4} \int \sec ^3(c+d x)dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {3}{4} \int \csc \left (c+d x+\frac {\pi }{2}\right )^3dx+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4255 |
| \(\displaystyle b \left (\frac {3}{4} \left (\frac {1}{2} \int \sec (c+d x)dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle b \left (\frac {3}{4} \left (\frac {1}{2} \int \csc \left (c+d x+\frac {\pi }{2}\right )dx+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
| \(\Big \downarrow \) 4257 |
| \(\displaystyle b \left (\frac {3}{4} \left (\frac {\text {arctanh}(\sin (c+d x))}{2 d}+\frac {\tan (c+d x) \sec (c+d x)}{2 d}\right )+\frac {\tan (c+d x) \sec ^3(c+d x)}{4 d}\right )-\frac {a \left (-\frac {1}{3} \tan ^3(c+d x)-\tan (c+d x)\right )}{d}\) |
Input:
Int[Sec[c + d*x]^4*(a + b*Sec[c + d*x]),x]
Output:
-((a*(-Tan[c + d*x] - Tan[c + d*x]^3/3))/d) + b*((Sec[c + d*x]^3*Tan[c + d *x])/(4*d) + (3*(ArcTanh[Sin[c + d*x]]/(2*d) + (Sec[c + d*x]*Tan[c + d*x]) /(2*d)))/4)
Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Simp[-d^(-1) Subst[Int[Exp andIntegrand[(1 + x^2)^(n/2 - 1), x], x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]
Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(-b)*Cos[c + d* x]*((b*Csc[c + d*x])^(n - 1)/(d*(n - 1))), x] + Simp[b^2*((n - 2)/(n - 1)) Int[(b*Csc[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && IntegerQ[2*n]
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[a Int[(d*Csc[e + f*x])^n, x], x] + Simp[b/d In t[(d*Csc[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, n}, x]
Time = 1.51 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.86
| method | result | size |
| derivativedivides | \(\frac {-a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(73\) |
| default | \(\frac {-a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )+b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(73\) |
| parts | \(-\frac {a \left (-\frac {2}{3}-\frac {\sec \left (d x +c \right )^{2}}{3}\right ) \tan \left (d x +c \right )}{d}+\frac {b \left (-\left (-\frac {\sec \left (d x +c \right )^{3}}{4}-\frac {3 \sec \left (d x +c \right )}{8}\right ) \tan \left (d x +c \right )+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(75\) |
| risch | \(-\frac {i \left (9 b \,{\mathrm e}^{7 i \left (d x +c \right )}+33 b \,{\mathrm e}^{5 i \left (d x +c \right )}-48 \,{\mathrm e}^{4 i \left (d x +c \right )} a -33 b \,{\mathrm e}^{3 i \left (d x +c \right )}-64 \,{\mathrm e}^{2 i \left (d x +c \right )} a -9 b \,{\mathrm e}^{i \left (d x +c \right )}-16 a \right )}{12 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{8 d}+\frac {3 b \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{8 d}\) | \(135\) |
| norman | \(\frac {-\frac {\left (8 a -5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{7}}{4 d}+\frac {\left (8 a +5 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{4 d}-\frac {\left (40 a -9 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{3}}{12 d}+\frac {\left (40 a +9 b \right ) \tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{5}}{12 d}}{\left (-1+\tan \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )^{4}}-\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{8 d}+\frac {3 b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{8 d}\) | \(145\) |
| parallelrisch | \(\frac {-18 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+18 \left (\frac {3}{4}+\frac {\cos \left (4 d x +4 c \right )}{4}+\cos \left (2 d x +2 c \right )\right ) b \ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+8 a \sin \left (4 d x +4 c \right )+32 a \sin \left (2 d x +2 c \right )+9 b \sin \left (3 d x +3 c \right )+33 b \sin \left (d x +c \right )}{12 d \left (3+\cos \left (4 d x +4 c \right )+4 \cos \left (2 d x +2 c \right )\right )}\) | \(150\) |
Input:
int(sec(d*x+c)^4*(a+b*sec(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(-a*(-2/3-1/3*sec(d*x+c)^2)*tan(d*x+c)+b*(-(-1/4*sec(d*x+c)^3-3/8*sec( d*x+c))*tan(d*x+c)+3/8*ln(sec(d*x+c)+tan(d*x+c))))
Time = 0.10 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.16 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {9 \, b \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) - 9 \, b \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (16 \, a \cos \left (d x + c\right )^{3} + 9 \, b \cos \left (d x + c\right )^{2} + 8 \, a \cos \left (d x + c\right ) + 6 \, b\right )} \sin \left (d x + c\right )}{48 \, d \cos \left (d x + c\right )^{4}} \] Input:
integrate(sec(d*x+c)^4*(a+b*sec(d*x+c)),x, algorithm="fricas")
Output:
1/48*(9*b*cos(d*x + c)^4*log(sin(d*x + c) + 1) - 9*b*cos(d*x + c)^4*log(-s in(d*x + c) + 1) + 2*(16*a*cos(d*x + c)^3 + 9*b*cos(d*x + c)^2 + 8*a*cos(d *x + c) + 6*b)*sin(d*x + c))/(d*cos(d*x + c)^4)
\[ \int \sec ^4(c+d x) (a+b \sec (c+d x)) \, dx=\int \left (a + b \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}\, dx \] Input:
integrate(sec(d*x+c)**4*(a+b*sec(d*x+c)),x)
Output:
Integral((a + b*sec(c + d*x))*sec(c + d*x)**4, x)
Time = 0.04 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.12 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {16 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a - 3 \, b {\left (\frac {2 \, {\left (3 \, \sin \left (d x + c\right )^{3} - 5 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{48 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+b*sec(d*x+c)),x, algorithm="maxima")
Output:
1/48*(16*(tan(d*x + c)^3 + 3*tan(d*x + c))*a - 3*b*(2*(3*sin(d*x + c)^3 - 5*sin(d*x + c))/(sin(d*x + c)^4 - 2*sin(d*x + c)^2 + 1) - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1)))/d
Leaf count of result is larger than twice the leaf count of optimal. 164 vs. \(2 (77) = 154\).
Time = 0.13 (sec) , antiderivative size = 164, normalized size of antiderivative = 1.93 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {9 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 9 \, b \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (24 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 15 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} - 40 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 9 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 40 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 9 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 24 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 15 \, b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{4}}}{24 \, d} \] Input:
integrate(sec(d*x+c)^4*(a+b*sec(d*x+c)),x, algorithm="giac")
Output:
1/24*(9*b*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 9*b*log(abs(tan(1/2*d*x + 1 /2*c) - 1)) - 2*(24*a*tan(1/2*d*x + 1/2*c)^7 - 15*b*tan(1/2*d*x + 1/2*c)^7 - 40*a*tan(1/2*d*x + 1/2*c)^5 - 9*b*tan(1/2*d*x + 1/2*c)^5 + 40*a*tan(1/2 *d*x + 1/2*c)^3 - 9*b*tan(1/2*d*x + 1/2*c)^3 - 24*a*tan(1/2*d*x + 1/2*c) - 15*b*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2*c)^2 - 1)^4)/d
Time = 12.55 (sec) , antiderivative size = 152, normalized size of antiderivative = 1.79 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {\left (\frac {5\,b}{4}-2\,a\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+\left (\frac {10\,a}{3}+\frac {3\,b}{4}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {3\,b}{4}-\frac {10\,a}{3}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (2\,a+\frac {5\,b}{4}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+6\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}+\frac {3\,b\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{4\,d} \] Input:
int((a + b/cos(c + d*x))/cos(c + d*x)^4,x)
Output:
(tan(c/2 + (d*x)/2)*(2*a + (5*b)/4) - tan(c/2 + (d*x)/2)^7*(2*a - (5*b)/4) - tan(c/2 + (d*x)/2)^3*((10*a)/3 - (3*b)/4) + tan(c/2 + (d*x)/2)^5*((10*a )/3 + (3*b)/4))/(d*(6*tan(c/2 + (d*x)/2)^4 - 4*tan(c/2 + (d*x)/2)^2 - 4*ta n(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 + 1)) + (3*b*atanh(tan(c/2 + (d* x)/2)))/(4*d)
Time = 0.16 (sec) , antiderivative size = 202, normalized size of antiderivative = 2.38 \[ \int \sec ^4(c+d x) (a+b \sec (c+d x)) \, dx=\frac {-16 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3} a +24 \cos \left (d x +c \right ) \sin \left (d x +c \right ) a -9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4} b +18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2} b -9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) b +9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4} b -18 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2} b +9 \,\mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) b -9 \sin \left (d x +c \right )^{3} b +15 \sin \left (d x +c \right ) b}{24 d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int(sec(d*x+c)^4*(a+b*sec(d*x+c)),x)
Output:
( - 16*cos(c + d*x)*sin(c + d*x)**3*a + 24*cos(c + d*x)*sin(c + d*x)*a - 9 *log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4*b + 18*log(tan((c + d*x)/2) - 1 )*sin(c + d*x)**2*b - 9*log(tan((c + d*x)/2) - 1)*b + 9*log(tan((c + d*x)/ 2) + 1)*sin(c + d*x)**4*b - 18*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2*b + 9*log(tan((c + d*x)/2) + 1)*b - 9*sin(c + d*x)**3*b + 15*sin(c + d*x)*b )/(24*d*(sin(c + d*x)**4 - 2*sin(c + d*x)**2 + 1))