Integrand size = 19, antiderivative size = 58 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=-2 a b x-\frac {\left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {a b \tan (c+d x)}{d}+\frac {(a+b \tan (c+d x))^2}{2 d} \] Output:
-2*a*b*x-(a^2-b^2)*ln(cos(d*x+c))/d+a*b*tan(d*x+c)/d+1/2*(a+b*tan(d*x+c))^ 2/d
Result contains complex when optimal does not.
Time = 0.07 (sec) , antiderivative size = 162, normalized size of antiderivative = 2.79 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {a^2 \log (i-\tan (c+d x))}{2 d}+\frac {i a b \log (i-\tan (c+d x))}{d}-\frac {b^2 \log (i-\tan (c+d x))}{2 d}+\frac {a^2 \log (i+\tan (c+d x))}{2 d}-\frac {i a b \log (i+\tan (c+d x))}{d}-\frac {b^2 \log (i+\tan (c+d x))}{2 d}+\frac {2 a b \tan (c+d x)}{d}+\frac {b^2 \tan ^2(c+d x)}{2 d} \] Input:
Integrate[Tan[c + d*x]*(a + b*Tan[c + d*x])^2,x]
Output:
(a^2*Log[I - Tan[c + d*x]])/(2*d) + (I*a*b*Log[I - Tan[c + d*x]])/d - (b^2 *Log[I - Tan[c + d*x]])/(2*d) + (a^2*Log[I + Tan[c + d*x]])/(2*d) - (I*a*b *Log[I + Tan[c + d*x]])/d - (b^2*Log[I + Tan[c + d*x]])/(2*d) + (2*a*b*Tan [c + d*x])/d + (b^2*Tan[c + d*x]^2)/(2*d)
Time = 0.37 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {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.
\(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x))^2dx\) |
\(\Big \downarrow \) 4011 |
\(\displaystyle \int (a \tan (c+d x)-b) (a+b \tan (c+d x))dx+\frac {(a+b \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a \tan (c+d x)-b) (a+b \tan (c+d x))dx+\frac {(a+b \tan (c+d x))^2}{2 d}\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle \left (a^2-b^2\right ) \int \tan (c+d x)dx+\frac {(a+b \tan (c+d x))^2}{2 d}+\frac {a b \tan (c+d x)}{d}-2 a b x\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \left (a^2-b^2\right ) \int \tan (c+d x)dx+\frac {(a+b \tan (c+d x))^2}{2 d}+\frac {a b \tan (c+d x)}{d}-2 a b x\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle -\frac {\left (a^2-b^2\right ) \log (\cos (c+d x))}{d}+\frac {(a+b \tan (c+d x))^2}{2 d}+\frac {a b \tan (c+d x)}{d}-2 a b x\) |
Input:
Int[Tan[c + d*x]*(a + b*Tan[c + d*x])^2,x]
Output:
-2*a*b*x - ((a^2 - b^2)*Log[Cos[c + d*x]])/d + (a*b*Tan[c + d*x])/d + (a + b*Tan[c + d*x])^2/(2*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 = 0.40 (sec) , antiderivative size = 61, normalized size of antiderivative = 1.05
method | result | size |
norman | \(-2 a b x +\frac {b^{2} \tan \left (d x +c \right )^{2}}{2 d}+\frac {2 a b \tan \left (d x +c \right )}{d}+\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(61\) |
derivativedivides | \(\frac {\frac {\tan \left (d x +c \right )^{2} b^{2}}{2}+2 \tan \left (d x +c \right ) a b +\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}-2 a b \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(62\) |
default | \(\frac {\frac {\tan \left (d x +c \right )^{2} b^{2}}{2}+2 \tan \left (d x +c \right ) a b +\frac {\left (a^{2}-b^{2}\right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}-2 a b \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(62\) |
parallelrisch | \(\frac {-4 a b d x +\tan \left (d x +c \right )^{2} b^{2}+\ln \left (1+\tan \left (d x +c \right )^{2}\right ) a^{2}-\ln \left (1+\tan \left (d x +c \right )^{2}\right ) b^{2}+4 \tan \left (d x +c \right ) a b}{2 d}\) | \(66\) |
parts | \(\frac {a^{2} \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}+\frac {b^{2} \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}+\frac {2 a b \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}\) | \(75\) |
risch | \(-2 a b x +i a^{2} x -i b^{2} x +\frac {2 i a^{2} c}{d}-\frac {2 i b^{2} c}{d}+\frac {2 i b \left (-i b \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) a^{2}}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) b^{2}}{d}\) | \(129\) |
Input:
int(tan(d*x+c)*(a+b*tan(d*x+c))^2,x,method=_RETURNVERBOSE)
Output:
-2*a*b*x+1/2*b^2/d*tan(d*x+c)^2+2*a*b*tan(d*x+c)/d+1/2*(a^2-b^2)/d*ln(1+ta n(d*x+c)^2)
Time = 0.08 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {4 \, a b d x - b^{2} \tan \left (d x + c\right )^{2} - 4 \, a b \tan \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, d} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="fricas")
Output:
-1/2*(4*a*b*d*x - b^2*tan(d*x + c)^2 - 4*a*b*tan(d*x + c) + (a^2 - b^2)*lo g(1/(tan(d*x + c)^2 + 1)))/d
Time = 0.10 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.47 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=\begin {cases} \frac {a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - 2 a b x + \frac {2 a b \tan {\left (c + d x \right )}}{d} - \frac {b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {b^{2} \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\left (c \right )}\right )^{2} \tan {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))**2,x)
Output:
Piecewise((a**2*log(tan(c + d*x)**2 + 1)/(2*d) - 2*a*b*x + 2*a*b*tan(c + d *x)/d - b**2*log(tan(c + d*x)**2 + 1)/(2*d) + b**2*tan(c + d*x)**2/(2*d), Ne(d, 0)), (x*(a + b*tan(c))**2*tan(c), True))
Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.00 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {b^{2} \tan \left (d x + c\right )^{2} - 4 \, {\left (d x + c\right )} a b + 4 \, a b \tan \left (d x + c\right ) + {\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="maxima")
Output:
1/2*(b^2*tan(d*x + c)^2 - 4*(d*x + c)*a*b + 4*a*b*tan(d*x + c) + (a^2 - b^ 2)*log(tan(d*x + c)^2 + 1))/d
Time = 0.20 (sec) , antiderivative size = 68, normalized size of antiderivative = 1.17 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=-\frac {2 \, {\left (d x + c\right )} a b}{d} + \frac {{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} + \frac {b^{2} d \tan \left (d x + c\right )^{2} + 4 \, a b d \tan \left (d x + c\right )}{2 \, d^{2}} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))^2,x, algorithm="giac")
Output:
-2*(d*x + c)*a*b/d + 1/2*(a^2 - b^2)*log(tan(d*x + c)^2 + 1)/d + 1/2*(b^2* d*tan(d*x + c)^2 + 4*a*b*d*tan(d*x + c))/d^2
Time = 1.05 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.98 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {a^2}{2}-\frac {b^2}{2}\right )+\frac {b^2\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}+2\,a\,b\,\mathrm {tan}\left (c+d\,x\right )-2\,a\,b\,d\,x}{d} \] Input:
int(tan(c + d*x)*(a + b*tan(c + d*x))^2,x)
Output:
(log(tan(c + d*x)^2 + 1)*(a^2/2 - b^2/2) + (b^2*tan(c + d*x)^2)/2 + 2*a*b* tan(c + d*x) - 2*a*b*d*x)/d
Time = 0.22 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.12 \[ \int \tan (c+d x) (a+b \tan (c+d x))^2 \, dx=\frac {\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a^{2}-\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) b^{2}+\tan \left (d x +c \right )^{2} b^{2}+4 \tan \left (d x +c \right ) a b -4 a b d x}{2 d} \] Input:
int(tan(d*x+c)*(a+b*tan(d*x+c))^2,x)
Output:
(log(tan(c + d*x)**2 + 1)*a**2 - log(tan(c + d*x)**2 + 1)*b**2 + tan(c + d *x)**2*b**2 + 4*tan(c + d*x)*a*b - 4*a*b*d*x)/(2*d)