3.309 \(\int \frac{\tan (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx\)

Optimal. Leaf size=48 \[ \frac{b B x}{a^2+b^2}-\frac{a B \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )} \]

[Out]

(b*B*x)/(a^2 + b^2) - (a*B*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)*d)

________________________________________________________________________________________

Rubi [A]  time = 0.0666127, antiderivative size = 48, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 32, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.094, Rules used = {21, 3531, 3530} \[ \frac{b B x}{a^2+b^2}-\frac{a B \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Int[(Tan[c + d*x]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

(b*B*x)/(a^2 + b^2) - (a*B*Log[a*Cos[c + d*x] + b*Sin[c + d*x]])/((a^2 + b^2)*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 3531

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] + Dist[(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] && NeQ[a*c + b*d, 0]

Rule 3530

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(c*Log[Re
moveContent[a*Cos[e + f*x] + b*Sin[e + f*x], x]])/(b*f), 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]

Rubi steps

\begin{align*} \int \frac{\tan (c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx &=B \int \frac{\tan (c+d x)}{a+b \tan (c+d x)} \, dx\\ &=\frac{b B x}{a^2+b^2}-\frac{(a B) \int \frac{b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^2+b^2}\\ &=\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}\\ \end{align*}

Mathematica [C]  time = 0.107548, size = 67, normalized size = 1.4 \[ \frac{B \left (2 (b-i a) (c+d x)-a \log \left ((a \cos (c+d x)+b \sin (c+d x))^2\right )+2 i a \tan ^{-1}(\tan (c+d x))\right )}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

Integrate[(Tan[c + d*x]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^2,x]

[Out]

(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)

________________________________________________________________________________________

Maple [A]  time = 0.029, size = 78, normalized size = 1.6 \begin{align*}{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) aB}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) aB}{d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x)

[Out]

1/2/d/(a^2+b^2)*ln(1+tan(d*x+c)^2)*a*B+1/d/(a^2+b^2)*B*arctan(tan(d*x+c))*b-1/d/(a^2+b^2)*ln(a+b*tan(d*x+c))*a
*B

________________________________________________________________________________________

Maxima [A]  time = 1.72114, size = 96, normalized size = 2. \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 1.63656, size = 153, normalized size = 3.19 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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
)

________________________________________________________________________________________

Sympy [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: AttributeError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))**2,x)

[Out]

Exception raised: AttributeError

________________________________________________________________________________________

Giac [A]  time = 1.23563, size = 103, normalized size = 2.15 \begin{align*} -\frac{\frac{2 \, B a b \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}} - \frac{2 \,{\left (d x + c\right )} B b}{a^{2} + b^{2}} - \frac{B a \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

-1/2*(2*B*a*b*log(abs(b*tan(d*x + c) + a))/(a^2*b + b^3) - 2*(d*x + c)*B*b/(a^2 + b^2) - B*a*log(tan(d*x + c)^
2 + 1)/(a^2 + b^2))/d