Integrand size = 32, antiderivative size = 48 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {b B x}{a^2+b^2}-\frac {a B \log (a \cos (c+d x)+b \sin (c+d x))}{\left (a^2+b^2\right ) d} \] Output:
b*B*x/(a^2+b^2)-a*B*ln(a*cos(d*x+c)+b*sin(d*x+c))/(a^2+b^2)/d
Result contains complex when optimal does not.
Time = 0.08 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.40 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {B \left (2 (-i a+b) (c+d x)+2 i a \arctan (\tan (c+d x))-a \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )\right )}{2 \left (a^2+b^2\right ) d} \] Input:
Integrate[(Tan[c + d*x]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x ]
Output:
(B*(2*((-I)*a + b)*(c + d*x) + (2*I)*a*ArcTan[Tan[c + d*x]] - a*Log[(a*Cos [c + d*x] + b*Sin[c + d*x])^2]))/(2*(a^2 + b^2)*d)
Time = 0.33 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.156, Rules used = {2011, 3042, 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.
\(\displaystyle \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\) |
\(\Big \downarrow \) 2011 |
\(\displaystyle B \int \frac {\tan (c+d x)}{a+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \int \frac {\tan (c+d x)}{a+b \tan (c+d x)}dx\) |
\(\Big \downarrow \) 4014 |
\(\displaystyle B \left (\frac {b x}{a^2+b^2}-\frac {a \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle B \left (\frac {b x}{a^2+b^2}-\frac {a \int \frac {b-a \tan (c+d x)}{a+b \tan (c+d x)}dx}{a^2+b^2}\right )\) |
\(\Big \downarrow \) 4013 |
\(\displaystyle B \left (\frac {b x}{a^2+b^2}-\frac {a \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}\right )\) |
Input:
Int[(Tan[c + d*x]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]
Output:
B*((b*x)/(a^2 + b^2) - (a*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^ 2)*d))
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Simp[(b/d)^m Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x ] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x , a + b*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]
Time = 0.06 (sec) , antiderivative size = 51, normalized size of antiderivative = 1.06
method | result | size |
parallelrisch | \(\frac {2 B b d x +B \ln \left (1+\tan \left (d x +c \right )^{2}\right ) a -2 B \ln \left (a +b \tan \left (d x +c \right )\right ) a}{2 d \left (a^{2}+b^{2}\right )}\) | \(51\) |
derivativedivides | \(\frac {B \left (\frac {\frac {a \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+b \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {a \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2}+b^{2}}\right )}{d}\) | \(64\) |
default | \(\frac {B \left (\frac {\frac {a \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2}+b \arctan \left (\tan \left (d x +c \right )\right )}{a^{2}+b^{2}}-\frac {a \ln \left (a +b \tan \left (d x +c \right )\right )}{a^{2}+b^{2}}\right )}{d}\) | \(64\) |
risch | \(\frac {i x B}{i b -a}+\frac {2 i B x a}{a^{2}+b^{2}}+\frac {2 i B a c}{d \left (a^{2}+b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{2 i \left (d x +c \right )}-\frac {i b +a}{i b -a}\right ) B a}{d \left (a^{2}+b^{2}\right )}\) | \(95\) |
norman | \(\frac {\frac {b B a x}{a^{2}+b^{2}}+\frac {b^{2} B x \tan \left (d x +c \right )}{a^{2}+b^{2}}}{a +b \tan \left (d x +c \right )}+\frac {B a \ln \left (1+\tan \left (d x +c \right )^{2}\right )}{2 d \left (a^{2}+b^{2}\right )}-\frac {B a \ln \left (a +b \tan \left (d x +c \right )\right )}{d \left (a^{2}+b^{2}\right )}\) | \(105\) |
Input:
int(tan(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x,method=_RETURNVER BOSE)
Output:
1/2*(2*B*b*d*x+B*ln(1+tan(d*x+c)^2)*a-2*B*ln(a+b*tan(d*x+c))*a)/d/(a^2+b^2 )
Time = 0.11 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.35 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {2 \, B b d x - B a \log \left (\frac {b^{2} \tan \left (d x + c\right )^{2} + 2 \, a b \tan \left (d x + c\right ) + a^{2}}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, {\left (a^{2} + b^{2}\right )} d} \] Input:
integrate(tan(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm= "fricas")
Output:
1/2*(2*B*b*d*x - B*a*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/( tan(d*x + c)^2 + 1)))/((a^2 + b^2)*d)
Result contains complex when optimal does not.
Time = 0.60 (sec) , antiderivative size = 282, normalized size of antiderivative = 5.88 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\begin {cases} \tilde {\infty } B x & \text {for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac {B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a d} & \text {for}\: b = 0 \\\frac {B d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {i B d x}{2 b d \tan {\left (c + d x \right )} - 2 i b d} - \frac {B}{2 b d \tan {\left (c + d x \right )} - 2 i b d} & \text {for}\: a = - i b \\\frac {B d x \tan {\left (c + d x \right )}}{2 b d \tan {\left (c + d x \right )} + 2 i b d} + \frac {i B d x}{2 b d \tan {\left (c + d x \right )} + 2 i b d} - \frac {B}{2 b d \tan {\left (c + d x \right )} + 2 i b d} & \text {for}\: a = i b \\\frac {x \left (B a + B b \tan {\left (c \right )}\right ) \tan {\left (c \right )}}{\left (a + b \tan {\left (c \right )}\right )^{2}} & \text {for}\: d = 0 \\- \frac {2 B a \log {\left (\frac {a}{b} + \tan {\left (c + d x \right )} \right )}}{2 a^{2} d + 2 b^{2} d} + \frac {B a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d + 2 b^{2} d} + \frac {2 B b d x}{2 a^{2} d + 2 b^{2} d} & \text {otherwise} \end {cases} \] Input:
integrate(tan(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))**2,x)
Output:
Piecewise((zoo*B*x, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (B*log(tan(c + d*x)** 2 + 1)/(2*a*d), Eq(b, 0)), (B*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) - 2*I*b *d) - I*B*d*x/(2*b*d*tan(c + d*x) - 2*I*b*d) - B/(2*b*d*tan(c + d*x) - 2*I *b*d), Eq(a, -I*b)), (B*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + I*B*d*x/(2*b*d*tan(c + d*x) + 2*I*b*d) - B/(2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, I*b)), (x*(B*a + B*b*tan(c))*tan(c)/(a + b*tan(c))**2, Eq(d, 0)), ( -2*B*a*log(a/b + tan(c + d*x))/(2*a**2*d + 2*b**2*d) + B*a*log(tan(c + d*x )**2 + 1)/(2*a**2*d + 2*b**2*d) + 2*B*b*d*x/(2*a**2*d + 2*b**2*d), True))
Time = 0.11 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.48 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {\frac {2 \, {\left (d x + c\right )} B b}{a^{2} + b^{2}} - \frac {2 \, B a \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} + b^{2}} + \frac {B a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \] Input:
integrate(tan(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm= "maxima")
Output:
1/2*(2*(d*x + c)*B*b/(a^2 + b^2) - 2*B*a*log(b*tan(d*x + c) + a)/(a^2 + b^ 2) + B*a*log(tan(d*x + c)^2 + 1)/(a^2 + b^2))/d
Time = 0.22 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.71 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=-\frac {1}{2} \, {\left (\frac {2 \, a b \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b d + b^{3} d} - \frac {2 \, {\left (d x + c\right )} b}{a^{2} d + b^{2} d} - \frac {a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} d + b^{2} d}\right )} B \] Input:
integrate(tan(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm= "giac")
Output:
-1/2*(2*a*b*log(abs(b*tan(d*x + c) + a))/(a^2*b*d + b^3*d) - 2*(d*x + c)*b /(a^2*d + b^2*d) - a*log(tan(d*x + c)^2 + 1)/(a^2*d + b^2*d))*B
Time = 3.53 (sec) , antiderivative size = 79, normalized size of antiderivative = 1.65 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {B\,\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )}{2\,d\,\left (a-b\,1{}\mathrm {i}\right )}-\frac {B\,a\,\ln \left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}{d\,\left (a^2+b^2\right )}+\frac {B\,\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d\,\left (-b+a\,1{}\mathrm {i}\right )} \] Input:
int((tan(c + d*x)*(B*a + B*b*tan(c + d*x)))/(a + b*tan(c + d*x))^2,x)
Output:
(B*log(tan(c + d*x) + 1i))/(2*d*(a - b*1i)) + (B*log(tan(c + d*x) - 1i)*1i )/(2*d*(a*1i - b)) - (B*a*log(a + b*tan(c + d*x)))/(d*(a^2 + b^2))
Time = 0.16 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx=\frac {b \left (\mathrm {log}\left (\tan \left (d x +c \right )^{2}+1\right ) a -2 \,\mathrm {log}\left (a +\tan \left (d x +c \right ) b \right ) a +2 b d x \right )}{2 d \left (a^{2}+b^{2}\right )} \] Input:
int(tan(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x)
Output:
(b*(log(tan(c + d*x)**2 + 1)*a - 2*log(tan(c + d*x)*b + a)*a + 2*b*d*x))/( 2*d*(a**2 + b**2))