\(\int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx\) [300]

Optimal result

Integrand size = 32, antiderivative size = 13 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \log (\cos (c+d x))}{d} \]

[Out]

-B*ln(cos(d*x+c))/d

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.062, Rules used = {21, 3556} \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \log (\cos (c+d x))}{d} \]

[In]

Int[(Tan[c + d*x]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x]

[Out]

-((B*Log[Cos[c + d*x]])/d)

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(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])

Rule 3556

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> Simp[-Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps \begin{align*} \text {integral}& = B \int \tan (c+d x) \, dx \\ & = -\frac {B \log (\cos (c+d x))}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.01 (sec) , antiderivative size = 13, normalized size of antiderivative = 1.00 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \log (\cos (c+d x))}{d} \]

[In]

Integrate[(Tan[c + d*x]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x]),x]

[Out]

-((B*Log[Cos[c + d*x]])/d)

Maple [A] (verified)

Time = 0.02 (sec) , antiderivative size = 18, normalized size of antiderivative = 1.38

method	result	size
derivativedivides	\(\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(18\)
default	\(\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(18\)
norman	\(\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(18\)
parallelrisch	\(\frac {B \ln \left (1+\tan ^{2}\left (d x +c \right )\right )}{2 d}\)	\(18\)
risch	\(i B x +\frac {2 i B c}{d}-\frac {B \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{d}\)	\(33\)

[In]

int(tan(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/2*B/d*ln(1+tan(d*x+c)^2)

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 19, normalized size of antiderivative = 1.46 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=-\frac {B \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, d} \]

[In]

integrate(tan(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*B*log(1/(tan(d*x + c)^2 + 1))/d

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 37 vs. \(2 (12) = 24\).

Time = 0.33 (sec) , antiderivative size = 37, normalized size of antiderivative = 2.85 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\begin {cases} \frac {B \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text {for}\: d \neq 0 \\\frac {x \left (B a + B b \tan {\left (c \right )}\right ) \tan {\left (c \right )}}{a + b \tan {\left (c \right )}} & \text {otherwise} \end {cases} \]

[In]

integrate(tan(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x)

[Out]

Piecewise((B*log(tan(c + d*x)**2 + 1)/(2*d), Ne(d, 0)), (x*(B*a + B*b*tan(c))*tan(c)/(a + b*tan(c)), True))

Maxima [A] (verification not implemented)

none

Time = 0.34 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {B \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

[In]

integrate(tan(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/2*B*log(tan(d*x + c)^2 + 1)/d

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 99 vs. \(2 (13) = 26\).

Time = 0.33 (sec) , antiderivative size = 99, normalized size of antiderivative = 7.62 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {B \log \left ({\left | -\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} - \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 2 \right |}\right ) - B \log \left ({\left | -\frac {\cos \left (d x + c\right ) + 1}{\cos \left (d x + c\right ) - 1} - \frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 2 \right |}\right )}{2 \, d} \]

[In]

integrate(tan(d*x+c)*(B*a+b*B*tan(d*x+c))/(a+b*tan(d*x+c)),x, algorithm="giac")

[Out]

1/2*(B*log(abs(-(cos(d*x + c) + 1)/(cos(d*x + c) - 1) - (cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 2)) - B*log(ab
s(-(cos(d*x + c) + 1)/(cos(d*x + c) - 1) - (cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 2)))/d

Mupad [B] (verification not implemented)

Time = 7.23 (sec) , antiderivative size = 17, normalized size of antiderivative = 1.31 \[ \int \frac {\tan (c+d x) (a B+b B \tan (c+d x))}{a+b \tan (c+d x)} \, dx=\frac {B\,\ln \left ({\mathrm {tan}\left (c+d\,x\right )}^2+1\right )}{2\,d} \]

[In]

int((tan(c + d*x)*(B*a + B*b*tan(c + d*x)))/(a + b*tan(c + d*x)),x)

[Out]

(B*log(tan(c + d*x)^2 + 1))/(2*d)