Integrand size = 21, antiderivative size = 119 \[ \int (a+a \sec (c+d x))^2 \tan ^4(c+d x) \, dx=a^2 x+\frac {3 a^2 \text {arctanh}(\sin (c+d x))}{4 d}-\frac {a^2 \tan (c+d x)}{d}-\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{4 d}+\frac {a^2 \tan ^3(c+d x)}{3 d}+\frac {a^2 \sec (c+d x) \tan ^3(c+d x)}{2 d}+\frac {a^2 \tan ^5(c+d x)}{5 d} \] Output:
a^2*x+3/4*a^2*arctanh(sin(d*x+c))/d-a^2*tan(d*x+c)/d-3/4*a^2*sec(d*x+c)*ta n(d*x+c)/d+1/3*a^2*tan(d*x+c)^3/d+1/2*a^2*sec(d*x+c)*tan(d*x+c)^3/d+1/5*a^ 2*tan(d*x+c)^5/d
Time = 0.30 (sec) , antiderivative size = 150, normalized size of antiderivative = 1.26 \[ \int (a+a \sec (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {a^2 \arctan (\tan (c+d x))}{d}+\frac {3 a^2 \text {arctanh}(\sin (c+d x))}{4 d}-\frac {a^2 \tan (c+d x)}{d}+\frac {3 a^2 \sec (c+d x) \tan (c+d x)}{4 d}-\frac {3 a^2 \sec ^3(c+d x) \tan (c+d x)}{2 d}+\frac {a^2 \tan ^3(c+d x)}{3 d}+\frac {2 a^2 \sec (c+d x) \tan ^3(c+d x)}{d}+\frac {a^2 \tan ^5(c+d x)}{5 d} \] Input:
Integrate[(a + a*Sec[c + d*x])^2*Tan[c + d*x]^4,x]
Output:
(a^2*ArcTan[Tan[c + d*x]])/d + (3*a^2*ArcTanh[Sin[c + d*x]])/(4*d) - (a^2* Tan[c + d*x])/d + (3*a^2*Sec[c + d*x]*Tan[c + d*x])/(4*d) - (3*a^2*Sec[c + d*x]^3*Tan[c + d*x])/(2*d) + (a^2*Tan[c + d*x]^3)/(3*d) + (2*a^2*Sec[c + d*x]*Tan[c + d*x]^3)/d + (a^2*Tan[c + d*x]^5)/(5*d)
Time = 0.37 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {3042, 4374, 2009}
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.
| \(\displaystyle \int \tan ^4(c+d x) (a \sec (c+d x)+a)^2 \, dx\) |
| \(\Big \downarrow \) 3042 |
| \(\displaystyle \int \cot \left (c+d x+\frac {\pi }{2}\right )^4 \left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^2dx\) |
| \(\Big \downarrow \) 4374 |
| \(\displaystyle \int \left (a^2 \tan ^4(c+d x)+a^2 \tan ^4(c+d x) \sec ^2(c+d x)+2 a^2 \tan ^4(c+d x) \sec (c+d x)\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {3 a^2 \text {arctanh}(\sin (c+d x))}{4 d}+\frac {a^2 \tan ^5(c+d x)}{5 d}+\frac {a^2 \tan ^3(c+d x)}{3 d}-\frac {a^2 \tan (c+d x)}{d}+\frac {a^2 \tan ^3(c+d x) \sec (c+d x)}{2 d}-\frac {3 a^2 \tan (c+d x) \sec (c+d x)}{4 d}+a^2 x\) |
Input:
Int[(a + a*Sec[c + d*x])^2*Tan[c + d*x]^4,x]
Output:
a^2*x + (3*a^2*ArcTanh[Sin[c + d*x]])/(4*d) - (a^2*Tan[c + d*x])/d - (3*a^ 2*Sec[c + d*x]*Tan[c + d*x])/(4*d) + (a^2*Tan[c + d*x]^3)/(3*d) + (a^2*Sec [c + d*x]*Tan[c + d*x]^3)/(2*d) + (a^2*Tan[c + d*x]^5)/(5*d)
Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + ( a_))^(n_), x_Symbol] :> Int[ExpandIntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[ c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]
Time = 0.31 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.09
| method | result | size |
| derivativedivides | \(\frac {\frac {a^{2} \sin \left (d x +c \right )^{5}}{5 \cos \left (d x +c \right )^{5}}+2 a^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )}{d}\) | \(130\) |
| default | \(\frac {\frac {a^{2} \sin \left (d x +c \right )^{5}}{5 \cos \left (d x +c \right )^{5}}+2 a^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )+a^{2} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+d x +c \right )}{d}\) | \(130\) |
| parts | \(\frac {a^{2} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {a^{2} \tan \left (d x +c \right )^{5}}{5 d}+\frac {2 a^{2} \left (\frac {\sin \left (d x +c \right )^{5}}{4 \cos \left (d x +c \right )^{4}}-\frac {\sin \left (d x +c \right )^{5}}{8 \cos \left (d x +c \right )^{2}}-\frac {\sin \left (d x +c \right )^{3}}{8}-\frac {3 \sin \left (d x +c \right )}{8}+\frac {3 \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8}\right )}{d}\) | \(130\) |
| risch | \(a^{2} x +\frac {i a^{2} \left (75 \,{\mathrm e}^{9 i \left (d x +c \right )}-60 \,{\mathrm e}^{8 i \left (d x +c \right )}+30 \,{\mathrm e}^{7 i \left (d x +c \right )}-360 \,{\mathrm e}^{6 i \left (d x +c \right )}-320 \,{\mathrm e}^{4 i \left (d x +c \right )}-30 \,{\mathrm e}^{3 i \left (d x +c \right )}-280 \,{\mathrm e}^{2 i \left (d x +c \right )}-75 \,{\mathrm e}^{i \left (d x +c \right )}-68\right )}{30 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{4 d}+\frac {3 a^{2} \ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{4 d}\) | \(161\) |
Input:
int((a+a*sec(d*x+c))^2*tan(d*x+c)^4,x,method=_RETURNVERBOSE)
Output:
1/d*(1/5*a^2*sin(d*x+c)^5/cos(d*x+c)^5+2*a^2*(1/4*sin(d*x+c)^5/cos(d*x+c)^ 4-1/8*sin(d*x+c)^5/cos(d*x+c)^2-1/8*sin(d*x+c)^3-3/8*sin(d*x+c)+3/8*ln(sec (d*x+c)+tan(d*x+c)))+a^2*(1/3*tan(d*x+c)^3-tan(d*x+c)+d*x+c))
Time = 0.13 (sec) , antiderivative size = 139, normalized size of antiderivative = 1.17 \[ \int (a+a \sec (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {120 \, a^{2} d x \cos \left (d x + c\right )^{5} + 45 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (\sin \left (d x + c\right ) + 1\right ) - 45 \, a^{2} \cos \left (d x + c\right )^{5} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (68 \, a^{2} \cos \left (d x + c\right )^{4} + 75 \, a^{2} \cos \left (d x + c\right )^{3} + 4 \, a^{2} \cos \left (d x + c\right )^{2} - 30 \, a^{2} \cos \left (d x + c\right ) - 12 \, a^{2}\right )} \sin \left (d x + c\right )}{120 \, d \cos \left (d x + c\right )^{5}} \] Input:
integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^4,x, algorithm="fricas")
Output:
1/120*(120*a^2*d*x*cos(d*x + c)^5 + 45*a^2*cos(d*x + c)^5*log(sin(d*x + c) + 1) - 45*a^2*cos(d*x + c)^5*log(-sin(d*x + c) + 1) - 2*(68*a^2*cos(d*x + c)^4 + 75*a^2*cos(d*x + c)^3 + 4*a^2*cos(d*x + c)^2 - 30*a^2*cos(d*x + c) - 12*a^2)*sin(d*x + c))/(d*cos(d*x + c)^5)
\[ \int (a+a \sec (c+d x))^2 \tan ^4(c+d x) \, dx=a^{2} \left (\int 2 \tan ^{4}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int \tan ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int \tan ^{4}{\left (c + d x \right )}\, dx\right ) \] Input:
integrate((a+a*sec(d*x+c))**2*tan(d*x+c)**4,x)
Output:
a**2*(Integral(2*tan(c + d*x)**4*sec(c + d*x), x) + Integral(tan(c + d*x)* *4*sec(c + d*x)**2, x) + Integral(tan(c + d*x)**4, x))
Time = 0.12 (sec) , antiderivative size = 119, normalized size of antiderivative = 1.00 \[ \int (a+a \sec (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {24 \, a^{2} \tan \left (d x + c\right )^{5} + 40 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} + 15 \, a^{2} {\left (\frac {2 \, {\left (5 \, \sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1} + 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, \log \left (\sin \left (d x + c\right ) - 1\right )\right )}}{120 \, d} \] Input:
integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^4,x, algorithm="maxima")
Output:
1/120*(24*a^2*tan(d*x + c)^5 + 40*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d* x + c))*a^2 + 15*a^2*(2*(5*sin(d*x + c)^3 - 3*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
Time = 0.44 (sec) , antiderivative size = 148, normalized size of antiderivative = 1.24 \[ \int (a+a \sec (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {60 \, {\left (d x + c\right )} a^{2} + 45 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) - 45 \, a^{2} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) + \frac {2 \, {\left (15 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{9} - 110 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{7} + 328 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 530 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 105 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{5}}}{60 \, d} \] Input:
integrate((a+a*sec(d*x+c))^2*tan(d*x+c)^4,x, algorithm="giac")
Output:
1/60*(60*(d*x + c)*a^2 + 45*a^2*log(abs(tan(1/2*d*x + 1/2*c) + 1)) - 45*a^ 2*log(abs(tan(1/2*d*x + 1/2*c) - 1)) + 2*(15*a^2*tan(1/2*d*x + 1/2*c)^9 - 110*a^2*tan(1/2*d*x + 1/2*c)^7 + 328*a^2*tan(1/2*d*x + 1/2*c)^5 - 530*a^2* tan(1/2*d*x + 1/2*c)^3 + 105*a^2*tan(1/2*d*x + 1/2*c))/(tan(1/2*d*x + 1/2* c)^2 - 1)^5)/d
Time = 14.13 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.46 \[ \int (a+a \sec (c+d x))^2 \tan ^4(c+d x) \, dx=a^2\,x+\frac {3\,a^2\,\mathrm {atanh}\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\right )}{2\,d}+\frac {\frac {a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{2}-\frac {11\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{3}+\frac {164\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{15}-\frac {53\,a^2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{3}+\frac {7\,a^2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{2}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8+10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )} \] Input:
int(tan(c + d*x)^4*(a + a/cos(c + d*x))^2,x)
Output:
a^2*x + (3*a^2*atanh(tan(c/2 + (d*x)/2)))/(2*d) + ((164*a^2*tan(c/2 + (d*x )/2)^5)/15 - (53*a^2*tan(c/2 + (d*x)/2)^3)/3 - (11*a^2*tan(c/2 + (d*x)/2)^ 7)/3 + (a^2*tan(c/2 + (d*x)/2)^9)/2 + (7*a^2*tan(c/2 + (d*x)/2))/2)/(d*(5* tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^4 + 10*tan(c/2 + (d*x)/2)^6 - 5*tan(c/2 + (d*x)/2)^8 + tan(c/2 + (d*x)/2)^10 - 1))
Time = 0.17 (sec) , antiderivative size = 399, normalized size of antiderivative = 3.35 \[ \int (a+a \sec (c+d x))^2 \tan ^4(c+d x) \, dx=\frac {a^{2} \left (-45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{4}+90 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right ) \sin \left (d x +c \right )^{2}-45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )+45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{4}-90 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right ) \sin \left (d x +c \right )^{2}+45 \cos \left (d x +c \right ) \mathrm {log}\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )+20 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} \tan \left (d x +c \right )^{3}-60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} \tan \left (d x +c \right )+60 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{4} d x +75 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{3}-40 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )^{3}+120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} \tan \left (d x +c \right )-120 \cos \left (d x +c \right ) \sin \left (d x +c \right )^{2} d x -45 \cos \left (d x +c \right ) \sin \left (d x +c \right )+20 \cos \left (d x +c \right ) \tan \left (d x +c \right )^{3}-60 \cos \left (d x +c \right ) \tan \left (d x +c \right )+60 \cos \left (d x +c \right ) d x +12 \sin \left (d x +c \right )^{5}\right )}{60 \cos \left (d x +c \right ) d \left (\sin \left (d x +c \right )^{4}-2 \sin \left (d x +c \right )^{2}+1\right )} \] Input:
int((a+a*sec(d*x+c))^2*tan(d*x+c)^4,x)
Output:
(a**2*( - 45*cos(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**4 + 90*c os(c + d*x)*log(tan((c + d*x)/2) - 1)*sin(c + d*x)**2 - 45*cos(c + d*x)*lo g(tan((c + d*x)/2) - 1) + 45*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**4 - 90*cos(c + d*x)*log(tan((c + d*x)/2) + 1)*sin(c + d*x)**2 + 45 *cos(c + d*x)*log(tan((c + d*x)/2) + 1) + 20*cos(c + d*x)*sin(c + d*x)**4* tan(c + d*x)**3 - 60*cos(c + d*x)*sin(c + d*x)**4*tan(c + d*x) + 60*cos(c + d*x)*sin(c + d*x)**4*d*x + 75*cos(c + d*x)*sin(c + d*x)**3 - 40*cos(c + d*x)*sin(c + d*x)**2*tan(c + d*x)**3 + 120*cos(c + d*x)*sin(c + d*x)**2*ta n(c + d*x) - 120*cos(c + d*x)*sin(c + d*x)**2*d*x - 45*cos(c + d*x)*sin(c + d*x) + 20*cos(c + d*x)*tan(c + d*x)**3 - 60*cos(c + d*x)*tan(c + d*x) + 60*cos(c + d*x)*d*x + 12*sin(c + d*x)**5))/(60*cos(c + d*x)*d*(sin(c + d*x )**4 - 2*sin(c + d*x)**2 + 1))