Integrand size = 19, antiderivative size = 97 \[ \int \tan (c+d x) (a+b \tan (c+d x))^3 \, dx=-b \left (3 a^2-b^2\right ) x-\frac {a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}+\frac {b \left (a^2-b^2\right ) \tan (c+d x)}{d}+\frac {a (a+b \tan (c+d x))^2}{2 d}+\frac {(a+b \tan (c+d x))^3}{3 d} \] Output:
-b*(3*a^2-b^2)*x-a*(a^2-3*b^2)*ln(cos(d*x+c))/d+b*(a^2-b^2)*tan(d*x+c)/d+1 /2*a*(a+b*tan(d*x+c))^2/d+1/3*(a+b*tan(d*x+c))^3/d
Result contains complex when optimal does not.
Time = 0.40 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.03 \[ \int \tan (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {3 \left ((a+i b)^3 \log (i-\tan (c+d x))+(a-i b)^3 \log (i+\tan (c+d x))\right )-6 b \left (-3 a^2+b^2\right ) \tan (c+d x)+9 a b^2 \tan ^2(c+d x)+2 b^3 \tan ^3(c+d x)}{6 d} \] Input:
Integrate[Tan[c + d*x]*(a + b*Tan[c + d*x])^3,x]
Output:
(3*((a + I*b)^3*Log[I - Tan[c + d*x]] + (a - I*b)^3*Log[I + Tan[c + d*x]]) - 6*b*(-3*a^2 + b^2)*Tan[c + d*x] + 9*a*b^2*Tan[c + d*x]^2 + 2*b^3*Tan[c + d*x]^3)/(6*d)
Time = 0.49 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.421, Rules used = {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.
\(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x))^3 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x))^3dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int (a \tan (c+d x)-b) (a+b \tan (c+d x))^2dx+\frac {(a+b \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \tan (c+d x)-b) (a+b \tan (c+d x))^2dx+\frac {(a+b \tan (c+d x))^3}{3 d}\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int (a+b \tan (c+d x)) \left (\left (a^2-b^2\right ) \tan (c+d x)-2 a b\right )dx+\frac {(a+b \tan (c+d x))^3}{3 d}+\frac {a (a+b \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+b \tan (c+d x)) \left (\left (a^2-b^2\right ) \tan (c+d x)-2 a b\right )dx+\frac {(a+b \tan (c+d x))^3}{3 d}+\frac {a (a+b \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle a \left (a^2-3 b^2\right ) \int \tan (c+d x)dx+\frac {b \left (a^2-b^2\right ) \tan (c+d x)}{d}-b x \left (3 a^2-b^2\right )+\frac {(a+b \tan (c+d x))^3}{3 d}+\frac {a (a+b \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle a \left (a^2-3 b^2\right ) \int \tan (c+d x)dx+\frac {b \left (a^2-b^2\right ) \tan (c+d x)}{d}-b x \left (3 a^2-b^2\right )+\frac {(a+b \tan (c+d x))^3}{3 d}+\frac {a (a+b \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {b \left (a^2-b^2\right ) \tan (c+d x)}{d}-\frac {a \left (a^2-3 b^2\right ) \log (\cos (c+d x))}{d}-b x \left (3 a^2-b^2\right )+\frac {(a+b \tan (c+d x))^3}{3 d}+\frac {a (a+b \tan (c+d x))^2}{2 d}\) |
Input:
Int[Tan[c + d*x]*(a + b*Tan[c + d*x])^3,x]
Output:
-(b*(3*a^2 - b^2)*x) - (a*(a^2 - 3*b^2)*Log[Cos[c + d*x]])/d + (b*(a^2 - b ^2)*Tan[c + d*x])/d + (a*(a + b*Tan[c + d*x])^2)/(2*d) + (a + b*Tan[c + d* x])^3/(3*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 = 0.43 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98
method | result | size |
norman | \(\left (-3 a^{2} b +b^{3}\right ) x +\frac {b \left (3 a^{2}-b^{2}\right ) \tan \left (d x +c \right )}{d}+\frac {b^{3} \tan \left (d x +c \right )^{3}}{3 d}+\frac {3 a \,b^{2} \tan \left (d x +c \right )^{2}}{2 d}+\frac {a \left (a^{2}-3 b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(95\) |
derivativedivides | \(\frac {\frac {\tan \left (d x +c \right )^{3} b^{3}}{3}+\frac {3 \tan \left (d x +c \right )^{2} a \,b^{2}}{2}+3 \tan \left (d x +c \right ) a^{2} b -b^{3} \tan \left (d x +c \right )+\frac {\left (a^{3}-3 a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-3 a^{2} b +b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(97\) |
default | \(\frac {\frac {\tan \left (d x +c \right )^{3} b^{3}}{3}+\frac {3 \tan \left (d x +c \right )^{2} a \,b^{2}}{2}+3 \tan \left (d x +c \right ) a^{2} b -b^{3} \tan \left (d x +c \right )+\frac {\left (a^{3}-3 a \,b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-3 a^{2} b +b^{3}\right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(97\) |
parallelrisch | \(\frac {2 \tan \left (d x +c \right )^{3} b^{3}-18 a^{2} b d x +6 b^{3} d x +9 \tan \left (d x +c \right )^{2} a \,b^{2}+3 a^{3} \ln \left (1+\tan \left (d x +c \right )^{2}\right )-9 \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a \,b^{2}+18 \tan \left (d x +c \right ) a^{2} b -6 b^{3} \tan \left (d x +c \right )}{6 d}\) | \(105\) |
parts | \(\frac {a^{3} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}+\frac {b^{3} \left (\frac {\tan \left (d x +c \right )^{3}}{3}-\tan \left (d x +c \right )+\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {3 a \,b^{2} \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {3 a^{2} b \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(112\) |
risch | \(-3 a^{2} b x +b^{3} x +i a^{3} x -3 i a \,b^{2} x +\frac {2 i a^{3} c}{d}-\frac {6 i a \,b^{2} c}{d}-\frac {2 i b \left (9 i a b \,{\mathrm e}^{4 i \left (d x +c \right )}-9 a^{2} {\mathrm e}^{4 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{4 i \left (d x +c \right )}+9 i a b \,{\mathrm e}^{2 i \left (d x +c \right )}-18 a^{2} {\mathrm e}^{2 i \left (d x +c \right )}+6 b^{2} {\mathrm e}^{2 i \left (d x +c \right )}-9 a^{2}+4 b^{2}\right )}{3 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{3}}-\frac {a^{3} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}+\frac {3 a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b^{2}}{d}\) | \(206\) |
Input:
int(tan(d*x+c)*(a+b*tan(d*x+c))^3,x,method=_RETURNVERBOSE)
Output:
(-3*a^2*b+b^3)*x+b*(3*a^2-b^2)/d*tan(d*x+c)+1/3*b^3/d*tan(d*x+c)^3+3/2*a*b ^2/d*tan(d*x+c)^2+1/2*a*(a^2-3*b^2)/d*ln(1+tan(d*x+c)^2)
Time = 0.09 (sec) , antiderivative size = 94, normalized size of antiderivative = 0.97 \[ \int \tan (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {2 \, b^{3} \tan \left (d x + c\right )^{3} + 9 \, a b^{2} \tan \left (d x + c\right )^{2} - 6 \, {\left (3 \, a^{2} b - b^{3}\right )} d x - 3 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 6 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )}{6 \, d} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))^3,x, algorithm="fricas")
Output:
1/6*(2*b^3*tan(d*x + c)^3 + 9*a*b^2*tan(d*x + c)^2 - 6*(3*a^2*b - b^3)*d*x - 3*(a^3 - 3*a*b^2)*log(1/(tan(d*x + c)^2 + 1)) + 6*(3*a^2*b - b^3)*tan(d *x + c))/d
Time = 0.13 (sec) , antiderivative size = 128, normalized size of antiderivative = 1.32 \[ \int \tan (c+d x) (a+b \tan (c+d x))^3 \, dx=\begin {cases} \frac {a^{3} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 3 a^{2} b x + \frac {3 a^{2} b \tan {\left (c + d x \right )}}{d} - \frac {3 a b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {3 a b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} + b^{3} x + \frac {b^{3} \tan ^{3}{\left (c + d x \right )}}{3 d} - \frac {b^{3} \tan {\left (c + d x \right )}}{d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right )^{3} \tan {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))**3,x)
Output:
Piecewise((a**3*log(tan(c + d*x)**2 + 1)/(2*d) - 3*a**2*b*x + 3*a**2*b*tan (c + d*x)/d - 3*a*b**2*log(tan(c + d*x)**2 + 1)/(2*d) + 3*a*b**2*tan(c + d *x)**2/(2*d) + b**3*x + b**3*tan(c + d*x)**3/(3*d) - b**3*tan(c + d*x)/d, Ne(d, 0)), (x*(a + b*tan(c))**3*tan(c), True))
Time = 0.12 (sec) , antiderivative size = 95, normalized size of antiderivative = 0.98 \[ \int \tan (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {2 \, b^{3} \tan \left (d x + c\right )^{3} + 9 \, a b^{2} \tan \left (d x + c\right )^{2} - 6 \, {\left (3 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )} + 3 \, {\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 6 \, {\left (3 \, a^{2} b - b^{3}\right )} \tan \left (d x + c\right )}{6 \, d} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))^3,x, algorithm="maxima")
Output:
1/6*(2*b^3*tan(d*x + c)^3 + 9*a*b^2*tan(d*x + c)^2 - 6*(3*a^2*b - b^3)*(d* x + c) + 3*(a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1) + 6*(3*a^2*b - b^3)*tan (d*x + c))/d
Time = 0.21 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.21 \[ \int \tan (c+d x) (a+b \tan (c+d x))^3 \, dx=-\frac {{\left (3 \, a^{2} b - b^{3}\right )} {\left (d x + c\right )}}{d} + \frac {{\left (a^{3} - 3 \, a b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} + \frac {2 \, b^{3} d^{2} \tan \left (d x + c\right )^{3} + 9 \, a b^{2} d^{2} \tan \left (d x + c\right )^{2} + 18 \, a^{2} b d^{2} \tan \left (d x + c\right ) - 6 \, b^{3} d^{2} \tan \left (d x + c\right )}{6 \, d^{3}} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))^3,x, algorithm="giac")
Output:
-(3*a^2*b - b^3)*(d*x + c)/d + 1/2*(a^3 - 3*a*b^2)*log(tan(d*x + c)^2 + 1) /d + 1/6*(2*b^3*d^2*tan(d*x + c)^3 + 9*a*b^2*d^2*tan(d*x + c)^2 + 18*a^2*b *d^2*tan(d*x + c) - 6*b^3*d^2*tan(d*x + c))/d^3
Time = 1.12 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.39 \[ \int \tan (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a^2\,b-b^3\right )}{d}+\frac {b^3\,{\mathrm {tan}\left (c+d\,x\right )}^3}{3\,d}-\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {3\,a\,b^2}{2}-\frac {a^3}{2}\right )}{d}+\frac {3\,a\,b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2\,d}-\frac {b\,\mathrm {atan}\left (\frac {b\,\mathrm {tan}\left (c+d\,x\right )\,\left (3\,a^2-b^2\right )}{3\,a^2\,b-b^3}\right )\,\left (3\,a^2-b^2\right )}{d} \] Input:
int(tan(c + d*x)*(a + b*tan(c + d*x))^3,x)
Output:
(tan(c + d*x)*(3*a^2*b - b^3))/d + (b^3*tan(c + d*x)^3)/(3*d) - (log(tan(c + d*x)^2 + 1)*((3*a*b^2)/2 - a^3/2))/d + (3*a*b^2*tan(c + d*x)^2)/(2*d) - (b*atan((b*tan(c + d*x)*(3*a^2 - b^2))/(3*a^2*b - b^3))*(3*a^2 - b^2))/d
Time = 0.20 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.07 \[ \int \tan (c+d x) (a+b \tan (c+d x))^3 \, dx=\frac {3 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a^{3}-9 \,\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a \,b^{2}+2 \tan \left (d x +c \right )^{3} b^{3}+9 \tan \left (d x +c \right )^{2} a \,b^{2}+18 \tan \left (d x +c \right ) a^{2} b -6 \tan \left (d x +c \right ) b^{3}-18 a^{2} b d x +6 b^{3} d x}{6 d} \] Input:
int(tan(d*x+c)*(a+b*tan(d*x+c))^3,x)
Output:
(3*log(tan(c + d*x)**2 + 1)*a**3 - 9*log(tan(c + d*x)**2 + 1)*a*b**2 + 2*t an(c + d*x)**3*b**3 + 9*tan(c + d*x)**2*a*b**2 + 18*tan(c + d*x)*a**2*b - 6*tan(c + d*x)*b**3 - 18*a**2*b*d*x + 6*b**3*d*x)/(6*d)