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

Optimal. Leaf size=58 \[ \frac{2 b B x}{a^2+b^2}-\frac{B \left (a-\frac{b^2}{a}\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )} \]

[Out]

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

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Rubi [A]  time = 0.0776258, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 28, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.071, Rules used = {3531, 3530} \[ \frac{2 b B x}{a^2+b^2}-\frac{B \left (a-\frac{b^2}{a}\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 0.0969963, size = 65, normalized size = 1.12 \[ \frac{B \left (\left (a^2-b^2\right ) \left (\log \left (\sec ^2(c+d x)\right )-2 \log (a+b \tan (c+d x))\right )+4 a b \tan ^{-1}(\tan (c+d x))\right )}{2 a d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(B*(4*a*b*ArcTan[Tan[c + d*x]] + (a^2 - b^2)*(Log[Sec[c + d*x]^2] - 2*Log[a + b*Tan[c + d*x]])))/(2*a*(a^2 + b
^2)*d)

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Maple [B]  time = 0.033, size = 142, normalized size = 2.5 \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\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){b}^{2}}{2\,ad \left ({a}^{2}+{b}^{2} \right ) }}+2\,{\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 ) }}+{\frac{{b}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{ad \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.71733, size = 128, normalized size = 2.21 \begin{align*} \frac{\frac{4 \,{\left (d x + c\right )} B b}{a^{2} + b^{2}} - \frac{2 \,{\left (B a^{2} - B b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{3} + a b^{2}} + \frac{{\left (B a^{2} - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{3} + a b^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*(d*x + c)*B*b/(a^2 + b^2) - 2*(B*a^2 - B*b^2)*log(b*tan(d*x + c) + a)/(a^3 + a*b^2) + (B*a^2 - B*b^2)*l
og(tan(d*x + c)^2 + 1)/(a^3 + a*b^2))/d

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Fricas [A]  time = 1.73846, size = 174, normalized size = 3. \begin{align*} \frac{4 \, B a b d x -{\left (B a^{2} - B b^{2}\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 )}{2 \,{\left (a^{3} + a b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*B*a*b*d*x - (B*a^2 - B*b^2)*log((b^2*tan(d*x + c)^2 + 2*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)))/
((a^3 + a*b^2)*d)

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Sympy [A]  time = 2.40913, size = 233, normalized size = 4.02 \begin{align*} \begin{cases} \text{NaN} & \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}{- b d \tan{\left (c + d x \right )} + i b d} & \text{for}\: a = - i b \\- \frac{B}{b d \tan{\left (c + d x \right )} + i b d} & \text{for}\: a = i b \\\frac{x \left (B \tan{\left (c \right )} + \frac{B b}{a}\right )}{a + b \tan{\left (c \right )}} & \text{for}\: d = 0 \\- \frac{2 B a^{2} \log{\left (\frac{a}{b} + \tan{\left (c + d x \right )} \right )}}{2 a^{3} d + 2 a b^{2} d} + \frac{B a^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{3} d + 2 a b^{2} d} + \frac{4 B a b d x}{2 a^{3} d + 2 a b^{2} d} + \frac{2 B b^{2} \log{\left (\frac{a}{b} + \tan{\left (c + d x \right )} \right )}}{2 a^{3} d + 2 a b^{2} d} - \frac{B b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{3} d + 2 a b^{2} d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((nan, Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), (B*log(tan(c + d*x)**2 + 1)/(2*a*d), Eq(b, 0)), (B/(-b*d*tan(
c + d*x) + I*b*d), Eq(a, -I*b)), (-B/(b*d*tan(c + d*x) + I*b*d), Eq(a, I*b)), (x*(B*tan(c) + B*b/a)/(a + b*tan
(c)), Eq(d, 0)), (-2*B*a**2*log(a/b + tan(c + d*x))/(2*a**3*d + 2*a*b**2*d) + B*a**2*log(tan(c + d*x)**2 + 1)/
(2*a**3*d + 2*a*b**2*d) + 4*B*a*b*d*x/(2*a**3*d + 2*a*b**2*d) + 2*B*b**2*log(a/b + tan(c + d*x))/(2*a**3*d + 2
*a*b**2*d) - B*b**2*log(tan(c + d*x)**2 + 1)/(2*a**3*d + 2*a*b**2*d), True))

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Giac [A]  time = 1.16534, size = 134, normalized size = 2.31 \begin{align*} \frac{\frac{4 \,{\left (d x + c\right )} B b}{a^{2} + b^{2}} + \frac{{\left (B a^{2} - B b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{3} + a b^{2}} - \frac{2 \,{\left (B a^{2} b - B b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{3} b + a b^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(4*(d*x + c)*B*b/(a^2 + b^2) + (B*a^2 - B*b^2)*log(tan(d*x + c)^2 + 1)/(a^3 + a*b^2) - 2*(B*a^2*b - B*b^3)
*log(abs(b*tan(d*x + c) + a))/(a^3*b + a*b^3))/d