Integrand size = 19, antiderivative size = 77 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x)) \, dx=a x-\frac {b \log (\cos (c+d x))}{d}-\frac {a \tan (c+d x)}{d}-\frac {b \tan ^2(c+d x)}{2 d}+\frac {a \tan ^3(c+d x)}{3 d}+\frac {b \tan ^4(c+d x)}{4 d} \] Output:
a*x-b*ln(cos(d*x+c))/d-a*tan(d*x+c)/d-1/2*b*tan(d*x+c)^2/d+1/3*a*tan(d*x+c )^3/d+1/4*b*tan(d*x+c)^4/d
Time = 0.03 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.09 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {a \arctan (\tan (c+d x))}{d}-\frac {b \log (\cos (c+d x))}{d}-\frac {b \sec ^2(c+d x)}{d}+\frac {b \sec ^4(c+d x)}{4 d}-\frac {a \tan (c+d x)}{d}+\frac {a \tan ^3(c+d x)}{3 d} \] Input:
Integrate[Tan[c + d*x]^4*(a + b*Tan[c + d*x]),x]
Output:
(a*ArcTan[Tan[c + d*x]])/d - (b*Log[Cos[c + d*x]])/d - (b*Sec[c + d*x]^2)/ d + (b*Sec[c + d*x]^4)/(4*d) - (a*Tan[c + d*x])/d + (a*Tan[c + d*x]^3)/(3* d)
Time = 0.53 (sec) , antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.526, Rules used = {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.
\(\displaystyle \int \tan ^4(c+d x) (a+b \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^4 (a+b \tan (c+d x))dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \tan ^3(c+d x) (a \tan (c+d x)-b)dx+\frac {b \tan ^4(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^3 (a \tan (c+d x)-b)dx+\frac {b \tan ^4(c+d x)}{4 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \tan ^2(c+d x) (-a-b \tan (c+d x))dx+\frac {a \tan ^3(c+d x)}{3 d}+\frac {b \tan ^4(c+d x)}{4 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x)^2 (-a-b \tan (c+d x))dx+\frac {a \tan ^3(c+d x)}{3 d}+\frac {b \tan ^4(c+d x)}{4 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int \tan (c+d x) (b-a \tan (c+d x))dx+\frac {a \tan ^3(c+d x)}{3 d}+\frac {b \tan ^4(c+d x)}{4 d}-\frac {b \tan ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x) (b-a \tan (c+d x))dx+\frac {a \tan ^3(c+d x)}{3 d}+\frac {b \tan ^4(c+d x)}{4 d}-\frac {b \tan ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle b \int \tan (c+d x)dx+\frac {a \tan ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+a x+\frac {b \tan ^4(c+d x)}{4 d}-\frac {b \tan ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle b \int \tan (c+d x)dx+\frac {a \tan ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+a x+\frac {b \tan ^4(c+d x)}{4 d}-\frac {b \tan ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {a \tan ^3(c+d x)}{3 d}-\frac {a \tan (c+d x)}{d}+a x+\frac {b \tan ^4(c+d x)}{4 d}-\frac {b \tan ^2(c+d x)}{2 d}-\frac {b \log (\cos (c+d x))}{d}\) |
Input:
Int[Tan[c + d*x]^4*(a + b*Tan[c + d*x]),x]
Output:
a*x - (b*Log[Cos[c + d*x]])/d - (a*Tan[c + d*x])/d - (b*Tan[c + d*x]^2)/(2 *d) + (a*Tan[c + d*x]^3)/(3*d) + (b*Tan[c + d*x]^4)/(4*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 = 0.44 (sec) , antiderivative size = 68, normalized size of antiderivative = 0.88
method | result | size |
parallelrisch | \(\frac {3 b \tan \left (d x +c \right )^{4}+4 a \tan \left (d x +c \right )^{3}+12 a x d -6 b \tan \left (d x +c \right )^{2}+6 b \ln \left (1+\tan \left (d x +c \right )^{2}\right )-12 a \tan \left (d x +c \right )}{12 d}\) | \(68\) |
derivativedivides | \(\frac {\frac {b \tan \left (d x +c \right )^{4}}{4}+\frac {a \tan \left (d x +c \right )^{3}}{3}-\frac {b \tan \left (d x +c \right )^{2}}{2}-a \tan \left (d x +c \right )+\frac {b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(71\) |
default | \(\frac {\frac {b \tan \left (d x +c \right )^{4}}{4}+\frac {a \tan \left (d x +c \right )^{3}}{3}-\frac {b \tan \left (d x +c \right )^{2}}{2}-a \tan \left (d x +c \right )+\frac {b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+a \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(71\) |
parts | \(\frac {a \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {b \left (\frac {\tan \left (d x +c \right )^{4}}{4}-\frac {\tan \left (d x +c \right )^{2}}{2}+\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}\) | \(72\) |
norman | \(a x -\frac {a \tan \left (d x +c \right )}{d}+\frac {a \tan \left (d x +c \right )^{3}}{3 d}-\frac {b \tan \left (d x +c \right )^{2}}{2 d}+\frac {b \tan \left (d x +c \right )^{4}}{4 d}+\frac {b \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(76\) |
risch | \(i b x +a x +\frac {2 i b c}{d}-\frac {4 \left (3 i a \,{\mathrm e}^{6 i \left (d x +c \right )}+3 b \,{\mathrm e}^{6 i \left (d x +c \right )}+6 i a \,{\mathrm e}^{4 i \left (d x +c \right )}+3 b \,{\mathrm e}^{4 i \left (d x +c \right )}+5 i a \,{\mathrm e}^{2 i \left (d x +c \right )}+3 b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 i a \right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{4}}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\) | \(134\) |
Input:
int(tan(d*x+c)^4*(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/12*(3*b*tan(d*x+c)^4+4*a*tan(d*x+c)^3+12*a*x*d-6*b*tan(d*x+c)^2+6*b*ln(1 +tan(d*x+c)^2)-12*a*tan(d*x+c))/d
Time = 0.10 (sec) , antiderivative size = 69, normalized size of antiderivative = 0.90 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 \, b \tan \left (d x + c\right )^{4} + 4 \, a \tan \left (d x + c\right )^{3} + 12 \, a d x - 6 \, b \tan \left (d x + c\right )^{2} - 6 \, b \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) - 12 \, a \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate(tan(d*x+c)^4*(a+b*tan(d*x+c)),x, algorithm="fricas")
Output:
1/12*(3*b*tan(d*x + c)^4 + 4*a*tan(d*x + c)^3 + 12*a*d*x - 6*b*tan(d*x + c )^2 - 6*b*log(1/(tan(d*x + c)^2 + 1)) - 12*a*tan(d*x + c))/d
Time = 0.11 (sec) , antiderivative size = 83, normalized size of antiderivative = 1.08 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x)) \, dx=\begin {cases} a x + \frac {a \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {a \tan {\left (c + d x \right )}}{d} + \frac {b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b \tan ^{4}{\left (c + d x \right )}}{4 d} - \frac {b \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right ) \tan ^{4}{\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(tan(d*x+c)**4*(a+b*tan(d*x+c)),x)
Output:
Piecewise((a*x + a*tan(c + d*x)**3/(3*d) - a*tan(c + d*x)/d + b*log(tan(c + d*x)**2 + 1)/(2*d) + b*tan(c + d*x)**4/(4*d) - b*tan(c + d*x)**2/(2*d), Ne(d, 0)), (x*(a + b*tan(c))*tan(c)**4, True))
Time = 0.11 (sec) , antiderivative size = 70, normalized size of antiderivative = 0.91 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {3 \, b \tan \left (d x + c\right )^{4} + 4 \, a \tan \left (d x + c\right )^{3} - 6 \, b \tan \left (d x + c\right )^{2} + 12 \, {\left (d x + c\right )} a + 6 \, b \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 12 \, a \tan \left (d x + c\right )}{12 \, d} \] Input:
integrate(tan(d*x+c)^4*(a+b*tan(d*x+c)),x, algorithm="maxima")
Output:
1/12*(3*b*tan(d*x + c)^4 + 4*a*tan(d*x + c)^3 - 6*b*tan(d*x + c)^2 + 12*(d *x + c)*a + 6*b*log(tan(d*x + c)^2 + 1) - 12*a*tan(d*x + c))/d
Time = 0.19 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.14 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {{\left (d x + c\right )} a}{d} + \frac {b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} + \frac {3 \, b d^{3} \tan \left (d x + c\right )^{4} + 4 \, a d^{3} \tan \left (d x + c\right )^{3} - 6 \, b d^{3} \tan \left (d x + c\right )^{2} - 12 \, a d^{3} \tan \left (d x + c\right )}{12 \, d^{4}} \] Input:
integrate(tan(d*x+c)^4*(a+b*tan(d*x+c)),x, algorithm="giac")
Output:
(d*x + c)*a/d + 1/2*b*log(tan(d*x + c)^2 + 1)/d + 1/12*(3*b*d^3*tan(d*x + c)^4 + 4*a*d^3*tan(d*x + c)^3 - 6*b*d^3*tan(d*x + c)^2 - 12*a*d^3*tan(d*x + c))/d^4
Time = 1.00 (sec) , antiderivative size = 65, normalized size of antiderivative = 0.84 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {\frac {b\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2}-a\,\mathrm {tan}\left (c+d\,x\right )+\frac {a\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3}-\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+\frac {b\,{\mathrm {tan}\left (c+d\,x\right )}^4}{4}+a\,d\,x}{d} \] Input:
int(tan(c + d*x)^4*(a + b*tan(c + d*x)),x)
Output:
((b*log(tan(c + d*x)^2 + 1))/2 - a*tan(c + d*x) + (a*tan(c + d*x)^3)/3 - ( b*tan(c + d*x)^2)/2 + (b*tan(c + d*x)^4)/4 + a*d*x)/d
Time = 0.20 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.87 \[ \int \tan ^4(c+d x) (a+b \tan (c+d x)) \, dx=\frac {6 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) b +3 \tan \left (d x +c \right )^{4} b +4 \tan \left (d x +c \right )^{3} a -6 \tan \left (d x +c \right )^{2} b -12 \tan \left (d x +c \right ) a +12 a d x}{12 d} \] Input:
int(tan(d*x+c)^4*(a+b*tan(d*x+c)),x)
Output:
(6*log(tan(c + d*x)**2 + 1)*b + 3*tan(c + d*x)**4*b + 4*tan(c + d*x)**3*a - 6*tan(c + d*x)**2*b - 12*tan(c + d*x)*a + 12*a*d*x)/(12*d)