Integrand size = 29, antiderivative size = 80 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {(A b-a B) x}{a^2+b^2}-\frac {B \log (\cos (c+d x))}{b d}-\frac {a (A b-a B) \log (a \cos (c+d x)+b \sin (c+d x))}{b \left (a^2+b^2\right ) d} \] Output:
(A*b-B*a)*x/(a^2+b^2)-B*ln(cos(d*x+c))/b/d-a*(A*b-B*a)*ln(a*cos(d*x+c)+b*s in(d*x+c))/b/(a^2+b^2)/d
Result contains complex when optimal does not.
Time = 0.13 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.22 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {(a-i b) b (A+i B) \log (i-\tan (c+d x))+(a+i b) b (A-i B) \log (i+\tan (c+d x))+2 a (-A b+a B) \log (a+b \tan (c+d x))}{2 b \left (a^2+b^2\right ) d} \] Input:
Integrate[(Tan[c + d*x]*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x]
Output:
((a - I*b)*b*(A + I*B)*Log[I - Tan[c + d*x]] + (a + I*b)*b*(A - I*B)*Log[I + Tan[c + d*x]] + 2*a*(-(A*b) + a*B)*Log[a + b*Tan[c + d*x]])/(2*b*(a^2 + b^2)*d)
Time = 0.51 (sec) , antiderivative size = 75, normalized size of antiderivative = 0.94, number of steps used = 8, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.276, Rules used = {3042, 4072, 27, 3042, 3956, 4014, 3042, 4013}
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.
\(\displaystyle \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4072 |
\(\displaystyle \frac {\int \frac {(A b-a B) \tan (c+d x)}{a+b \tan (c+d x)}dx}{b}+\frac {B \int \tan (c+d x)dx}{b}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {(A b-a B) \int \frac {\tan (c+d x)}{a+b \tan (c+d x)}dx}{b}+\frac {B \int \tan (c+d x)dx}{b}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \int \frac {\tan (c+d x)}{a+b \tan (c+d x)}dx}{b}+\frac {B \int \tan (c+d x)dx}{b}\) |
\(\Big \downarrow \) 3956 |
\(\displaystyle \frac {(A b-a B) \int \frac {\tan (c+d x)}{a+b \tan (c+d x)}dx}{b}-\frac {B \log (\cos (c+d x))}{b d}\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle \frac {(A b-a B) \left (\frac {b x}{a^2+b^2}-\frac {a \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}\right )}{b}-\frac {B \log (\cos (c+d x))}{b d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {(A b-a B) \left (\frac {b x}{a^2+b^2}-\frac {a \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}\right )}{b}-\frac {B \log (\cos (c+d x))}{b d}\) |
\(\Big \downarrow \) 4013 |
\(\displaystyle \frac {(A b-a B) \left (\frac {b x}{a^2+b^2}-\frac {a \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}\right )}{b}-\frac {B \log (\cos (c+d x))}{b d}\) |
Input:
Int[(Tan[c + d*x]*(A + B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x]
Output:
-((B*Log[Cos[c + d*x]])/(b*d)) + ((A*b - a*B)*((b*x)/(a^2 + b^2) - (a*Log[ a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)*d)))/b
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d *x], x]]/d, x] /; FreeQ[{c, d}, x]
Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)* (x_)]), x_Symbol] :> Simp[(c/(b*f))*Log[RemoveContent[a*Cos[e + f*x] + b*Si n[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] && EqQ[a*c + b*d, 0]
Int[((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_.) + (b_.)*tan[(e_.) + (f_. )*(x_)]), x_Symbol] :> Simp[(a*c + b*d)*(x/(a^2 + b^2)), x] + Simp[(b*c - a *d)/(a^2 + b^2) Int[(b - a*Tan[e + f*x])/(a + b*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] && N eQ[a*c + b*d, 0]
Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_ .)*(x_)]))/((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[B*(d/ b) Int[Tan[e + f*x], x], x] + Simp[1/b Int[Simp[A*b*c + (A*b*d + B*(b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d , e, f, A, B}, x] && NeQ[b*c - a*d, 0]
Time = 0.10 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.09
method | result | size |
derivativedivides | \(\frac {\frac {\frac {\left (a A +B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (A b -B a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {a \left (A b -B a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) b}}{d}\) | \(87\) |
default | \(\frac {\frac {\frac {\left (a A +B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+\left (A b -B a \right ) \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {a \left (A b -B a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) b}}{d}\) | \(87\) |
norman | \(\frac {\left (A b -B a \right ) x}{a^{2}+b^{2}}+\frac {\left (a A +B b \right ) \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \left (a^{2}+b^{2}\right )}-\frac {a \left (A b -B a \right ) \ln \left (a +b \tan \left (d x +c \right )\right )}{\left (a^{2}+b^{2}\right ) d b}\) | \(90\) |
parallelrisch | \(\frac {2 A \,b^{2} d x -2 B a b d x +A \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a b -2 A \ln \left (a +b \tan \left (d x +c \right )\right ) a b +B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) b^{2}+2 B \ln \left (a +b \tan \left (d x +c \right )\right ) a^{2}}{2 \left (a^{2}+b^{2}\right ) d b}\) | \(98\) |
risch | \(\frac {x B}{i b -a}+\frac {i x A}{i b -a}+\frac {2 i a A x}{a^{2}+b^{2}}+\frac {2 i a A c}{d \left (a^{2}+b^{2}\right )}-\frac {2 i a^{2} B x}{b \left (a^{2}+b^{2}\right )}-\frac {2 i a^{2} B c}{b d \left (a^{2}+b^{2}\right )}+\frac {2 i B x}{b}+\frac {2 i B c}{b d}-\frac {a \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) A}{d \left (a^{2}+b^{2}\right )}+\frac {a^{2} \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B}{b d \left (a^{2}+b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right ) B}{b d}\) | \(240\) |
Input:
int(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)
Output:
1/d*(1/(a^2+b^2)*(1/2*(A*a+B*b)*ln(1+tan(d*x+c)^2)+(A*b-B*a)*arctan(tan(d* x+c)))-a*(A*b-B*a)/(a^2+b^2)/b*ln(a+b*tan(d*x+c)))
Time = 0.13 (sec) , antiderivative size = 110, normalized size of antiderivative = 1.38 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {2 \, {\left (B a b - A b^{2}\right )} d x - {\left (B a^{2} - A a b\right )} \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + {\left (B a^{2} + B b^{2}\right )} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, {\left (a^{2} b + b^{3}\right )} d} \] 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*(2*(B*a*b - A*b^2)*d*x - (B*a^2 - A*a*b)*log((b^2*tan(d*x + c)^2 + 2* a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) + (B*a^2 + B*b^2)*log(1/(tan (d*x + c)^2 + 1)))/((a^2*b + b^3)*d)
Result contains complex when optimal does not.
Time = 0.55 (sec) , antiderivative size = 700, normalized size of antiderivative = 8.75 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx =\text {Too large to display} \] Input:
integrate(tan(d*x+c)*(A+B*tan(d*x+c))/(a+b*tan(d*x+c)),x)
Output:
Piecewise((zoo*x*(A + B*tan(c)), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((A*log( tan(c + d*x)**2 + 1)/(2*d) - B*x + B*tan(c + d*x)/d)/a, Eq(b, 0)), (A*d*x* tan(c + d*x)/(2*b*d*tan(c + d*x) - 2*I*b*d) - I*A*d*x/(2*b*d*tan(c + d*x) - 2*I*b*d) - A/(2*b*d*tan(c + d*x) - 2*I*b*d) + I*B*d*x*tan(c + d*x)/(2*b* d*tan(c + d*x) - 2*I*b*d) + B*d*x/(2*b*d*tan(c + d*x) - 2*I*b*d) + B*log(t an(c + d*x)**2 + 1)*tan(c + d*x)/(2*b*d*tan(c + d*x) - 2*I*b*d) - I*B*log( tan(c + d*x)**2 + 1)/(2*b*d*tan(c + d*x) - 2*I*b*d) - I*B/(2*b*d*tan(c + d *x) - 2*I*b*d), Eq(a, -I*b)), (A*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2* I*b*d) + I*A*d*x/(2*b*d*tan(c + d*x) + 2*I*b*d) - A/(2*b*d*tan(c + d*x) + 2*I*b*d) - I*B*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + B*d*x/(2* b*d*tan(c + d*x) + 2*I*b*d) + B*log(tan(c + d*x)**2 + 1)*tan(c + d*x)/(2*b *d*tan(c + d*x) + 2*I*b*d) + I*B*log(tan(c + d*x)**2 + 1)/(2*b*d*tan(c + d *x) + 2*I*b*d) + I*B/(2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, I*b)), (x*(A + B*tan(c))*tan(c)/(a + b*tan(c)), Eq(d, 0)), (-2*A*a*b*log(a/b + tan(c + d* x))/(2*a**2*b*d + 2*b**3*d) + A*a*b*log(tan(c + d*x)**2 + 1)/(2*a**2*b*d + 2*b**3*d) + 2*A*b**2*d*x/(2*a**2*b*d + 2*b**3*d) + 2*B*a**2*log(a/b + tan (c + d*x))/(2*a**2*b*d + 2*b**3*d) - 2*B*a*b*d*x/(2*a**2*b*d + 2*b**3*d) + B*b**2*log(tan(c + d*x)**2 + 1)/(2*a**2*b*d + 2*b**3*d), True))
Time = 0.11 (sec) , antiderivative size = 94, normalized size of antiderivative = 1.18 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {\frac {2 \, {\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2} + b^{2}} - \frac {2 \, {\left (B a^{2} - A a b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} b + b^{3}} - \frac {{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \] 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*(2*(B*a - A*b)*(d*x + c)/(a^2 + b^2) - 2*(B*a^2 - A*a*b)*log(b*tan(d* x + c) + a)/(a^2*b + b^3) - (A*a + B*b)*log(tan(d*x + c)^2 + 1)/(a^2 + b^2 ))/d
Time = 0.23 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {{\left (B a - A b\right )} {\left (d x + c\right )}}{a^{2} d + b^{2} d} + \frac {{\left (A a + B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, {\left (a^{2} d + b^{2} d\right )}} + \frac {{\left (B a^{2} - A a b\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b d + b^{3} d} \] 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)/(a^2*d + b^2*d) + 1/2*(A*a + B*b)*log(tan(d*x + c)^ 2 + 1)/(a^2*d + b^2*d) + (B*a^2 - A*a*b)*log(abs(b*tan(d*x + c) + a))/(a^2 *b*d + b^3*d)
Time = 3.81 (sec) , antiderivative size = 100, normalized size of antiderivative = 1.25 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-B+A\,1{}\mathrm {i}\right )}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (A-B\,1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}-\frac {a\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )\,\left (A\,b-B\,a\right )}{b\,d\,\left (a^2+b^2\right )} \] Input:
int((tan(c + d*x)*(A + B*tan(c + d*x)))/(a + b*tan(c + d*x)),x)
Output:
(log(tan(c + d*x) - 1i)*(A*1i - B))/(2*d*(a*1i - b)) + (log(tan(c + d*x) + 1i)*(A - B*1i))/(2*d*(a - b*1i)) - (a*log(a + b*tan(c + d*x))*(A*b - B*a) )/(b*d*(a^2 + b^2))
Time = 0.16 (sec) , antiderivative size = 16, normalized size of antiderivative = 0.20 \[ \int \frac {\tan (c+d x) (A+B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right )}{2 d} \] 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)/(2*d)