Integrand size = 19, antiderivative size = 93 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=-b x-\frac {a \log (\cos (c+d x))}{d}+\frac {b \tan (c+d x)}{d}-\frac {a \tan ^2(c+d x)}{2 d}-\frac {b \tan ^3(c+d x)}{3 d}+\frac {a \tan ^4(c+d x)}{4 d}+\frac {b \tan ^5(c+d x)}{5 d} \] Output:
-b*x-a*ln(cos(d*x+c))/d+b*tan(d*x+c)/d-1/2*a*tan(d*x+c)^2/d-1/3*b*tan(d*x+ c)^3/d+1/4*a*tan(d*x+c)^4/d+1/5*b*tan(d*x+c)^5/d
Time = 0.05 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.08 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {b \arctan (\tan (c+d x))}{d}-\frac {a \log (\cos (c+d x))}{d}-\frac {a \sec ^2(c+d x)}{d}+\frac {a \sec ^4(c+d x)}{4 d}+\frac {b \tan (c+d x)}{d}-\frac {b \tan ^3(c+d x)}{3 d}+\frac {b \tan ^5(c+d x)}{5 d} \] Input:
Integrate[Tan[c + d*x]^5*(a + b*Tan[c + d*x]),x]
Output:
-((b*ArcTan[Tan[c + d*x]])/d) - (a*Log[Cos[c + d*x]])/d - (a*Sec[c + d*x]^ 2)/d + (a*Sec[c + d*x]^4)/(4*d) + (b*Tan[c + d*x])/d - (b*Tan[c + d*x]^3)/ (3*d) + (b*Tan[c + d*x]^5)/(5*d)
Time = 0.63 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.632, Rules used = {3042, 4011, 3042, 4011, 3042, 4011, 3042, 4011, 3042, 4008, 3042, 3956}
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.
\(\displaystyle \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^5 (a+b \tan (c+d x))dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \tan ^4(c+d x) (a \tan (c+d x)-b)dx+\frac {b \tan ^5(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^4 (a \tan (c+d x)-b)dx+\frac {b \tan ^5(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \tan ^3(c+d x) (-a-b \tan (c+d x))dx+\frac {a \tan ^4(c+d x)}{4 d}+\frac {b \tan ^5(c+d x)}{5 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^3 (-a-b \tan (c+d x))dx+\frac {a \tan ^4(c+d x)}{4 d}+\frac {b \tan ^5(c+d x)}{5 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \tan ^2(c+d x) (b-a \tan (c+d x))dx+\frac {a \tan ^4(c+d x)}{4 d}+\frac {b \tan ^5(c+d x)}{5 d}-\frac {b \tan ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^2 (b-a \tan (c+d x))dx+\frac {a \tan ^4(c+d x)}{4 d}+\frac {b \tan ^5(c+d x)}{5 d}-\frac {b \tan ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x))dx+\frac {a \tan ^4(c+d x)}{4 d}-\frac {a \tan ^2(c+d x)}{2 d}+\frac {b \tan ^5(c+d x)}{5 d}-\frac {b \tan ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x))dx+\frac {a \tan ^4(c+d x)}{4 d}-\frac {a \tan ^2(c+d x)}{2 d}+\frac {b \tan ^5(c+d x)}{5 d}-\frac {b \tan ^3(c+d x)}{3 d}\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle a \int \tan (c+d x)dx+\frac {a \tan ^4(c+d x)}{4 d}-\frac {a \tan ^2(c+d x)}{2 d}+\frac {b \tan ^5(c+d x)}{5 d}-\frac {b \tan ^3(c+d x)}{3 d}+\frac {b \tan (c+d x)}{d}-b x\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \int \tan (c+d x)dx+\frac {a \tan ^4(c+d x)}{4 d}-\frac {a \tan ^2(c+d x)}{2 d}+\frac {b \tan ^5(c+d x)}{5 d}-\frac {b \tan ^3(c+d x)}{3 d}+\frac {b \tan (c+d x)}{d}-b x\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {a \tan ^4(c+d x)}{4 d}-\frac {a \tan ^2(c+d x)}{2 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {b \tan ^5(c+d x)}{5 d}-\frac {b \tan ^3(c+d x)}{3 d}+\frac {b \tan (c+d x)}{d}-b x\) |
Input:
Int[Tan[c + d*x]^5*(a + b*Tan[c + d*x]),x]
Output:
-(b*x) - (a*Log[Cos[c + d*x]])/d + (b*Tan[c + d*x])/d - (a*Tan[c + d*x]^2) /(2*d) - (b*Tan[c + d*x]^3)/(3*d) + (a*Tan[c + d*x]^4)/(4*d) + (b*Tan[c + d*x]^5)/(5*d)
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.) *(x_)]), x_Symbol] :> Simp[(a*c - b*d)*x, x] + (Simp[b*d*(Tan[e + f*x]/f), x] + Simp[(b*c + a*d) Int[Tan[e + f*x], x], x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[b*c + a*d, 0]
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[d*((a + b*Tan[e + f*x])^m/(f*m)), x] + Int [(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
Time = 0.48 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.85
method | result | size |
parallelrisch | \(\frac {12 b \tan \left (d x +c \right )^{5}+15 a \tan \left (d x +c \right )^{4}-20 b \tan \left (d x +c \right )^{3}-60 b d x -30 a \tan \left (d x +c \right )^{2}+30 a \ln \left (1+\tan \left (d x +c \right )^{2}\right )+60 b \tan \left (d x +c \right )}{60 d}\) | \(79\) |
derivativedivides | \(\frac {\frac {b \tan \left (d x +c \right )^{5}}{5}+\frac {a \tan \left (d x +c \right )^{4}}{4}-\frac {b \tan \left (d x +c \right )^{3}}{3}-\frac {a \tan \left (d x +c \right )^{2}}{2}+b \tan \left (d x +c \right )+\frac {a \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}-b \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(82\) |
default | \(\frac {\frac {b \tan \left (d x +c \right )^{5}}{5}+\frac {a \tan \left (d x +c \right )^{4}}{4}-\frac {b \tan \left (d x +c \right )^{3}}{3}-\frac {a \tan \left (d x +c \right )^{2}}{2}+b \tan \left (d x +c \right )+\frac {a \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}-b \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(82\) |
parts | \(\frac {a \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {b \left (\frac {\tan \left (d x +c \right )^{5}}{5}-\frac {\tan \left (d x +c \right )^{3}}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(82\) |
norman | \(\frac {b \tan \left (d x +c \right )}{d}-b x -\frac {a \tan \left (d x +c \right )^{2}}{2 d}+\frac {a \tan \left (d x +c \right )^{4}}{4 d}-\frac {b \tan \left (d x +c \right )^{3}}{3 d}+\frac {b \tan \left (d x +c \right )^{5}}{5 d}+\frac {a \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(90\) |
risch | \(-b x +i a x +\frac {2 i a c}{d}+\frac {2 i \left (30 i a \,{\mathrm e}^{8 i \left (d x +c \right )}+45 b \,{\mathrm e}^{8 i \left (d x +c \right )}+60 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+90 b \,{\mathrm e}^{6 i \left (d x +c \right )}+60 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+140 b \,{\mathrm e}^{4 i \left (d x +c \right )}+30 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+70 b \,{\mathrm e}^{2 i \left (d x +c \right )}+23 b \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(160\) |
Input:
int(tan(d*x+c)^5*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/60*(12*b*tan(d*x+c)^5+15*a*tan(d*x+c)^4-20*b*tan(d*x+c)^3-60*b*d*x-30*a* tan(d*x+c)^2+30*a*ln(1+tan(d*x+c)^2)+60*b*tan(d*x+c))/d
Time = 0.09 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {12 \, b \tan \left (d x + c\right )^{5} + 15 \, a \tan \left (d x + c\right )^{4} - 20 \, b \tan \left (d x + c\right )^{3} - 60 \, b d x - 30 \, a \tan \left (d x + c\right )^{2} - 30 \, a \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 60 \, b \tan \left (d x + c\right )}{60 \, d} \] Input:
integrate(tan(d*x+c)^5*(a+b*tan(d*x+c)),x, algorithm="fricas")
Output:
1/60*(12*b*tan(d*x + c)^5 + 15*a*tan(d*x + c)^4 - 20*b*tan(d*x + c)^3 - 60 *b*d*x - 30*a*tan(d*x + c)^2 - 30*a*log(1/(tan(d*x + c)^2 + 1)) + 60*b*tan (d*x + c))/d
Time = 0.14 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.04 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=\begin {cases} \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} - b x + \frac {b \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac {b \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac {b \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right ) \tan ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(tan(d*x+c)**5*(a+b*tan(d*x+c)),x)
Output:
Piecewise((a*log(tan(c + d*x)**2 + 1)/(2*d) + a*tan(c + d*x)**4/(4*d) - a* tan(c + d*x)**2/(2*d) - b*x + b*tan(c + d*x)**5/(5*d) - b*tan(c + d*x)**3/ (3*d) + b*tan(c + d*x)/d, Ne(d, 0)), (x*(a + b*tan(c))*tan(c)**5, True))
Time = 0.11 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.87 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {12 \, b \tan \left (d x + c\right )^{5} + 15 \, a \tan \left (d x + c\right )^{4} - 20 \, b \tan \left (d x + c\right )^{3} - 30 \, a \tan \left (d x + c\right )^{2} - 60 \, {\left (d x + c\right )} b + 30 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 60 \, b \tan \left (d x + c\right )}{60 \, d} \] Input:
integrate(tan(d*x+c)^5*(a+b*tan(d*x+c)),x, algorithm="maxima")
Output:
1/60*(12*b*tan(d*x + c)^5 + 15*a*tan(d*x + c)^4 - 20*b*tan(d*x + c)^3 - 30 *a*tan(d*x + c)^2 - 60*(d*x + c)*b + 30*a*log(tan(d*x + c)^2 + 1) + 60*b*t an(d*x + c))/d
Time = 0.20 (sec) , antiderivative size = 103, normalized size of antiderivative = 1.11 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {{\left (d x + c\right )} b}{d} + \frac {a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} + \frac {12 \, b d^{4} \tan \left (d x + c\right )^{5} + 15 \, a d^{4} \tan \left (d x + c\right )^{4} - 20 \, b d^{4} \tan \left (d x + c\right )^{3} - 30 \, a d^{4} \tan \left (d x + c\right )^{2} + 60 \, b d^{4} \tan \left (d x + c\right )}{60 \, d^{5}} \] Input:
integrate(tan(d*x+c)^5*(a+b*tan(d*x+c)),x, algorithm="giac")
Output:
-(d*x + c)*b/d + 1/2*a*log(tan(d*x + c)^2 + 1)/d + 1/60*(12*b*d^4*tan(d*x + c)^5 + 15*a*d^4*tan(d*x + c)^4 - 20*b*d^4*tan(d*x + c)^3 - 30*a*d^4*tan( d*x + c)^2 + 60*b*d^4*tan(d*x + c))/d^5
Time = 1.05 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {b\,\mathrm {tan}\left (c+d\,x\right )+\frac {a\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}-\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}-b\,d\,x}{d} \] Input:
int(tan(c + d*x)^5*(a + b*tan(c + d*x)),x)
Output:
(b*tan(c + d*x) + (a*log(tan(c + d*x)^2 + 1))/2 - (a*tan(c + d*x)^2)/2 + ( a*tan(c + d*x)^4)/4 - (b*tan(c + d*x)^3)/3 + (b*tan(c + d*x)^5)/5 - b*d*x) /d
Time = 0.19 (sec) , antiderivative size = 78, normalized size of antiderivative = 0.84 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {30 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a +12 \tan \left (d x +c \right )^{5} b +15 \tan \left (d x +c \right )^{4} a -20 \tan \left (d x +c \right )^{3} b -30 \tan \left (d x +c \right )^{2} a +60 \tan \left (d x +c \right ) b -60 b d x}{60 d} \] Input:
int(tan(d*x+c)^5*(a+b*tan(d*x+c)),x)
Output:
(30*log(tan(c + d*x)**2 + 1)*a + 12*tan(c + d*x)**5*b + 15*tan(c + d*x)**4 *a - 20*tan(c + d*x)**3*b - 30*tan(c + d*x)**2*a + 60*tan(c + d*x)*b - 60* b*d*x)/(60*d)