Integrand size = 19, antiderivative size = 93 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=-b x-\frac {a \log (\cos (c+d x))}{d}+\frac {b \tan (c+d x)}{d}-\frac {a \tan ^2(c+d x)}{2 d}-\frac {b \tan ^3(c+d x)}{3 d}+\frac {a \tan ^4(c+d x)}{4 d}+\frac {b \tan ^5(c+d x)}{5 d} \]
-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 = 0.34 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.02 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=-\frac {b \arctan (\tan (c+d x))}{d}+\frac {b \tan (c+d x)}{d}-\frac {b \tan ^3(c+d x)}{3 d}+\frac {b \tan ^5(c+d x)}{5 d}-\frac {a \left (4 \log (\cos (c+d x))+2 \tan ^2(c+d x)-\tan ^4(c+d x)\right )}{4 d} \]
-((b*ArcTan[Tan[c + d*x]])/d) + (b*Tan[c + d*x])/d - (b*Tan[c + d*x]^3)/(3 *d) + (b*Tan[c + d*x]^5)/(5*d) - (a*(4*Log[Cos[c + d*x]] + 2*Tan[c + d*x]^ 2 - Tan[c + d*x]^4))/(4*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 |
\(\displaystyle \frac {a \tan ^4(c+d x)}{4 d}-\frac {a \tan ^2(c+d x)}{2 d}-\frac {a \log (\cos (c+d x))}{d}+\frac {b \tan ^5(c+d x)}{5 d}-\frac {b \tan ^3(c+d x)}{3 d}+\frac {b \tan (c+d x)}{d}-b x\) |
-(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)
3.5.12.3.1 Defintions of rubi rules used
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 = 0.27 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.85
method | result | size |
parallelrisch | \(\frac {12 b \left (\tan ^{5}\left (d x +c \right )\right )+15 a \left (\tan ^{4}\left (d x +c \right )\right )-20 b \left (\tan ^{3}\left (d x +c \right )\right )-60 b d x -30 \left (\tan ^{2}\left (d x +c \right )\right ) a +30 a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )+60 b \tan \left (d x +c \right )}{60 d}\) | \(79\) |
derivativedivides | \(\frac {\frac {b \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {a \left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {b \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\tan ^{2}\left (d x +c \right )\right ) a}{2}+b \tan \left (d x +c \right )+\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}-b \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(82\) |
default | \(\frac {\frac {b \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {a \left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {b \left (\tan ^{3}\left (d x +c \right )\right )}{3}-\frac {\left (\tan ^{2}\left (d x +c \right )\right ) a}{2}+b \tan \left (d x +c \right )+\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}-b \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(82\) |
parts | \(\frac {a \left (\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{4}-\frac {\left (\tan ^{2}\left (d x +c \right )\right )}{2}+\frac {\ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d}+\frac {b \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(82\) |
norman | \(\frac {b \tan \left (d x +c \right )}{d}-b x -\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}+\frac {a \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {b \left (\tan ^{3}\left (d x +c \right )\right )}{3 d}+\frac {b \left (\tan ^{5}\left (d x +c \right )\right )}{5 d}+\frac {a \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\) | \(90\) |
risch | \(-b x +i a x +\frac {2 i a c}{d}+\frac {2 i \left (30 i a \,{\mathrm e}^{8 i \left (d x +c \right )}+45 b \,{\mathrm e}^{8 i \left (d x +c \right )}+60 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+90 b \,{\mathrm e}^{6 i \left (d x +c \right )}+60 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+140 b \,{\mathrm e}^{4 i \left (d x +c \right )}+30 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+70 b \,{\mathrm e}^{2 i \left (d x +c \right )}+23 b \right )}{15 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{5}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(160\) |
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*ta n(d*x+c)^2*a+30*a*ln(1+tan(d*x+c)^2)+60*b*tan(d*x+c))/d
Time = 0.25 (sec) , antiderivative size = 80, normalized size of antiderivative = 0.86 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {12 \, b \tan \left (d x + c\right )^{5} + 15 \, a \tan \left (d x + c\right )^{4} - 20 \, b \tan \left (d x + c\right )^{3} - 60 \, b d x - 30 \, a \tan \left (d x + c\right )^{2} - 30 \, a \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 60 \, b \tan \left (d x + c\right )}{60 \, d} \]
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 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=\begin {cases} \frac {a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {a \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {a \tan ^{2}{\left (c + d x \right )}}{2 d} - b x + \frac {b \tan ^{5}{\left (c + d x \right )}}{5 d} - \frac {b \tan ^{3}{\left (c + d x \right )}}{3 d} + \frac {b \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right ) \tan ^{5}{\left (c \right )} & \text {otherwise} \end {cases} \]
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 = 0.31 (sec) , antiderivative size = 81, normalized size of antiderivative = 0.87 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {12 \, b \tan \left (d x + c\right )^{5} + 15 \, a \tan \left (d x + c\right )^{4} - 20 \, b \tan \left (d x + c\right )^{3} - 30 \, a \tan \left (d x + c\right )^{2} - 60 \, {\left (d x + c\right )} b + 30 \, a \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 60 \, b \tan \left (d x + c\right )}{60 \, d} \]
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
Leaf count of result is larger than twice the leaf count of optimal. 887 vs. \(2 (85) = 170\).
Time = 2.50 (sec) , antiderivative size = 887, normalized size of antiderivative = 9.54 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=\text {Too large to display} \]
-1/60*(60*b*d*x*tan(d*x)^5*tan(c)^5 + 30*a*log(4*(tan(d*x)^2*tan(c)^2 - 2* tan(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*ta n(d*x)^5*tan(c)^5 - 300*b*d*x*tan(d*x)^4*tan(c)^4 + 45*a*tan(d*x)^5*tan(c) ^5 - 150*a*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/(tan(d*x)^2 *tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^4*tan(c)^4 + 60*b*tan(d*x )^5*tan(c)^4 + 60*b*tan(d*x)^4*tan(c)^5 + 600*b*d*x*tan(d*x)^3*tan(c)^3 + 30*a*tan(d*x)^5*tan(c)^3 - 165*a*tan(d*x)^4*tan(c)^4 + 30*a*tan(d*x)^3*tan (c)^5 - 20*b*tan(d*x)^5*tan(c)^2 + 300*a*log(4*(tan(d*x)^2*tan(c)^2 - 2*ta n(d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan( d*x)^3*tan(c)^3 - 300*b*tan(d*x)^4*tan(c)^3 - 300*b*tan(d*x)^3*tan(c)^4 - 20*b*tan(d*x)^2*tan(c)^5 - 15*a*tan(d*x)^5*tan(c) - 600*b*d*x*tan(d*x)^2*t an(c)^2 - 150*a*tan(d*x)^4*tan(c)^2 + 180*a*tan(d*x)^3*tan(c)^3 - 150*a*ta n(d*x)^2*tan(c)^4 - 15*a*tan(d*x)*tan(c)^5 + 12*b*tan(d*x)^5 + 100*b*tan(d *x)^4*tan(c) - 300*a*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan(d*x)*tan(c) + 1)/( tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*tan(d*x)^2*tan(c)^2 + 60 0*b*tan(d*x)^3*tan(c)^2 + 600*b*tan(d*x)^2*tan(c)^3 + 100*b*tan(d*x)*tan(c )^4 + 12*b*tan(c)^5 + 15*a*tan(d*x)^4 + 300*b*d*x*tan(d*x)*tan(c) + 150*a* tan(d*x)^3*tan(c) - 180*a*tan(d*x)^2*tan(c)^2 + 150*a*tan(d*x)*tan(c)^3 + 15*a*tan(c)^4 - 20*b*tan(d*x)^3 + 150*a*log(4*(tan(d*x)^2*tan(c)^2 - 2*tan (d*x)*tan(c) + 1)/(tan(d*x)^2*tan(c)^2 + tan(d*x)^2 + tan(c)^2 + 1))*ta...
Time = 4.54 (sec) , antiderivative size = 76, normalized size of antiderivative = 0.82 \[ \int \tan ^5(c+d x) (a+b \tan (c+d x)) \, dx=\frac {b\,\mathrm {tan}\left (c+d\,x\right )+\frac {a\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2}-\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}-\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^5}{5}-b\,d\,x}{d} \]