Integrand size = 27, antiderivative size = 65 \[ \int \tan (c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-((A b+a B) x)-\frac {(a A-b B) \log (\cos (c+d x))}{d}+\frac {(A b+a B) \tan (c+d x)}{d}+\frac {b B \tan ^2(c+d x)}{2 d} \] Output:
-(A*b+B*a)*x-(A*a-B*b)*ln(cos(d*x+c))/d+(A*b+B*a)*tan(d*x+c)/d+1/2*b*B*tan (d*x+c)^2/d
Time = 0.27 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.03 \[ \int \tan (c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {-2 (A b+a B) \arctan (\tan (c+d x))+2 (-a A+b B) \log (\cos (c+d x))+b B \sec ^2(c+d x)+2 (A b+a B) \tan (c+d x)}{2 d} \] Input:
Integrate[Tan[c + d*x]*(a + b*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]
Output:
(-2*(A*b + a*B)*ArcTan[Tan[c + d*x]] + 2*(-(a*A) + b*B)*Log[Cos[c + d*x]] + b*B*Sec[c + d*x]^2 + 2*(A*b + a*B)*Tan[c + d*x])/(2*d)
Time = 0.37 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3042, 4075, 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.
\(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x))dx\) |
\(\Big \downarrow \) 4075 |
\(\displaystyle \int \tan (c+d x) (a A-b B+(A b+a B) \tan (c+d x))dx+\frac {b B \tan ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \tan (c+d x) (a A-b B+(A b+a B) \tan (c+d x))dx+\frac {b B \tan ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 4008 |
\(\displaystyle (a A-b B) \int \tan (c+d x)dx+\frac {(a B+A b) \tan (c+d x)}{d}-x (a B+A b)+\frac {b B \tan ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (a A-b B) \int \tan (c+d x)dx+\frac {(a B+A b) \tan (c+d x)}{d}-x (a B+A b)+\frac {b B \tan ^2(c+d x)}{2 d}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {(a B+A b) \tan (c+d x)}{d}-\frac {(a A-b B) \log (\cos (c+d x))}{d}-x (a B+A b)+\frac {b B \tan ^2(c+d x)}{2 d}\) |
Input:
Int[Tan[c + d*x]*(a + b*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]
Output:
-((A*b + a*B)*x) - ((a*A - b*B)*Log[Cos[c + d*x]])/d + ((A*b + a*B)*Tan[c + d*x])/d + (b*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_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B *d*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1))), x] + Int[(a + b*Tan[e + f* x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && !LeQ[m, -1]
Time = 0.06 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06
method | result | size |
norman | \(\left (-A b -B a \right ) x +\frac {\left (A b +B a \right ) \tan \left (d x +c \right )}{d}+\frac {b B \tan \left (d x +c \right )^{2}}{2 d}+\frac {\left (a A -B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}\) | \(69\) |
derivativedivides | \(\frac {\frac {B \tan \left (d x +c \right )^{2} b}{2}+A \tan \left (d x +c \right ) b +B \tan \left (d x +c \right ) a +\frac {\left (a A -B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-A b -B a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(74\) |
default | \(\frac {\frac {B \tan \left (d x +c \right )^{2} b}{2}+A \tan \left (d x +c \right ) b +B \tan \left (d x +c \right ) a +\frac {\left (a A -B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (-A b -B a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{d}\) | \(74\) |
parts | \(\frac {\left (A b +B a \right ) \left (\tan \left (d x +c \right )-\arctan \left (\tan \left (d x +c \right )\right )\right )}{d}+\frac {a A \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d}+\frac {B b \left (\frac {\tan \left (d x +c \right )^{2}}{2}-\frac {\ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}\right )}{d}\) | \(77\) |
parallelrisch | \(\frac {-2 A b d x -2 B x a d +B \tan \left (d x +c \right )^{2} b +A \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a +2 A \tan \left (d x +c \right ) b -B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) b +2 B \tan \left (d x +c \right ) a}{2 d}\) | \(79\) |
risch | \(-A b x -B a x +i A a x -i B b x +\frac {2 i a A c}{d}-\frac {2 i B b c}{d}+\frac {2 i \left (-i B b \,{\mathrm e}^{2 i \left (d x +c \right )}+A b \,{\mathrm e}^{2 i \left (d x +c \right )}+B a \,{\mathrm e}^{2 i \left (d x +c \right )}+A b +B a \right )}{d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{2}}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) A}{d}+\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B b}{d}\) | \(143\) |
Input:
int(tan(d*x+c)*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
(-A*b-B*a)*x+(A*b+B*a)*tan(d*x+c)/d+1/2*b*B*tan(d*x+c)^2/d+1/2*(A*a-B*b)/d *ln(1+tan(d*x+c)^2)
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int \tan (c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {B b \tan \left (d x + c\right )^{2} - 2 \, {\left (B a + A b\right )} d x - {\left (A a - B b\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, {\left (B a + A b\right )} \tan \left (d x + c\right )}{2 \, d} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="frica s")
Output:
1/2*(B*b*tan(d*x + c)^2 - 2*(B*a + A*b)*d*x - (A*a - B*b)*log(1/(tan(d*x + c)^2 + 1)) + 2*(B*a + A*b)*tan(d*x + c))/d
Time = 0.11 (sec) , antiderivative size = 104, normalized size of antiderivative = 1.60 \[ \int \tan (c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\begin {cases} \frac {A a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} - A b x + \frac {A b \tan {\left (c + d x \right )}}{d} - B a x + \frac {B a \tan {\left (c + d x \right )}}{d} - \frac {B b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B b \tan ^{2}{\left (c + d x \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (A + B \tan {\left (c \right )}\right ) \left (a + b \tan {\left (c \right )}\right ) \tan {\left (c \right )} & \text {otherwise} \end {cases} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x)
Output:
Piecewise((A*a*log(tan(c + d*x)**2 + 1)/(2*d) - A*b*x + A*b*tan(c + d*x)/d - B*a*x + B*a*tan(c + d*x)/d - B*b*log(tan(c + d*x)**2 + 1)/(2*d) + B*b*t an(c + d*x)**2/(2*d), Ne(d, 0)), (x*(A + B*tan(c))*(a + b*tan(c))*tan(c), True))
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.02 \[ \int \tan (c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {B b \tan \left (d x + c\right )^{2} - 2 \, {\left (B a + A b\right )} {\left (d x + c\right )} + {\left (A a - B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) + 2 \, {\left (B a + A b\right )} \tan \left (d x + c\right )}{2 \, d} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="maxim a")
Output:
1/2*(B*b*tan(d*x + c)^2 - 2*(B*a + A*b)*(d*x + c) + (A*a - B*b)*log(tan(d* x + c)^2 + 1) + 2*(B*a + A*b)*tan(d*x + c))/d
Time = 0.25 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.26 \[ \int \tan (c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=-\frac {{\left (B a + A b\right )} {\left (d x + c\right )}}{d} + \frac {{\left (A a - B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} + \frac {B b d \tan \left (d x + c\right )^{2} + 2 \, B a d \tan \left (d x + c\right ) + 2 \, A b d \tan \left (d x + c\right )}{2 \, d^{2}} \] Input:
integrate(tan(d*x+c)*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="giac" )
Output:
-(B*a + A*b)*(d*x + c)/d + 1/2*(A*a - B*b)*log(tan(d*x + c)^2 + 1)/d + 1/2 *(B*b*d*tan(d*x + c)^2 + 2*B*a*d*tan(d*x + c) + 2*A*b*d*tan(d*x + c))/d^2
Time = 3.56 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.97 \[ \int \tan (c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {\mathrm {tan}\left (c+d\,x\right )\,\left (A\,b+B\,a\right )+\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )\,\left (\frac {A\,a}{2}-\frac {B\,b}{2}\right )-d\,x\,\left (A\,b+B\,a\right )+\frac {B\,b\,{\mathrm {tan}\left (c+d\,x\right )}^2}{2}}{d} \] Input:
int(tan(c + d*x)*(A + B*tan(c + d*x))*(a + b*tan(c + d*x)),x)
Output:
(tan(c + d*x)*(A*b + B*a) + log(tan(c + d*x)^2 + 1)*((A*a)/2 - (B*b)/2) - d*x*(A*b + B*a) + (B*b*tan(c + d*x)^2)/2)/d
Time = 0.20 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00 \[ \int \tan (c+d x) (a+b \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a^{2}-\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) b^{2}+\tan \left (d x +c \right )^{2} b^{2}+4 \tan \left (d x +c \right ) a b -4 a b d x}{2 d} \] Input:
int(tan(d*x+c)*(a+b*tan(d*x+c))*(A+B*tan(d*x+c)),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)