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

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

[Out]

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

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Rubi [A]  time = 0.143978, antiderivative size = 58, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.079, Rules used = {3632, 3531, 3530} \[ \frac{(b B-a C) \log (a \cos (c+d x)+b \sin (c+d x))}{d \left (a^2+b^2\right )}+\frac{x (a B+b C)}{a^2+b^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3632

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_.)*((A_.) + (B_.)
*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Dist[1/b^2, Int[(a + b*Tan[e + f*x])
^(m + 1)*(c + d*Tan[e + f*x])^n*(b*B - a*C + b*C*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C, m,
n}, x] && NeQ[b*c - a*d, 0] && EqQ[A*b^2 - a*b*B + a^2*C, 0]

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

Mathematica [A]  time = 0.115209, size = 67, normalized size = 1.16 \[ \frac{(b B-a C) \left (2 \log (a \cot (c+d x)+b)-\log \left (\csc ^2(c+d x)\right )\right )-2 (a B+b C) \tan ^{-1}(\cot (c+d x))}{2 d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*(a*B + b*C)*ArcTan[Cot[c + d*x]] + (b*B - a*C)*(2*Log[b + a*Cot[c + d*x]] - Log[Csc[c + d*x]^2]))/(2*(a^2
+ b^2)*d)

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Maple [B]  time = 0.109, size = 153, normalized size = 2.6 \begin{align*} -{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Bb}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Ca}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{C\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 ) Bb}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) Ca}{d \left ({a}^{2}+{b}^{2} \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.7382, size = 119, normalized size = 2.05 \begin{align*} \frac{\frac{2 \,{\left (B a + C b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} - \frac{2 \,{\left (C a - B b\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{2} + b^{2}} + \frac{{\left (C a - B b\right )} \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(cot(d*x+c)*(B*tan(d*x+c)+C*tan(d*x+c)^2)/(a+b*tan(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.08596, size = 174, normalized size = 3. \begin{align*} \frac{2 \,{\left (B a + C b\right )} d x -{\left (C a - B 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 )}{2 \,{\left (a^{2} + b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*(B*a + C*b)*d*x - (C*a - B*b)*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)

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Sympy [A]  time = 89.4981, size = 541, normalized size = 9.33 \begin{align*} \begin{cases} \frac{\tilde{\infty } x \left (B \tan{\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot{\left (c \right )}}{\tan{\left (c \right )}} & \text{for}\: a = 0 \wedge b = 0 \wedge d = 0 \\\frac{B x + \frac{C \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d}}{a} & \text{for}\: b = 0 \\- \frac{i B d x \tan{\left (c + d x \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{B d x}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{i B}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{C d x \tan{\left (c + d x \right )}}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{i C d x}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{C}{- 2 b d \tan{\left (c + d x \right )} + 2 i b d} & \text{for}\: a = - i b \\- \frac{i B d x \tan{\left (c + d x \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{B d x}{2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{i B}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{C d x \tan{\left (c + d x \right )}}{2 b d \tan{\left (c + d x \right )} + 2 i b d} + \frac{i C d x}{2 b d \tan{\left (c + d x \right )} + 2 i b d} - \frac{C}{2 b d \tan{\left (c + d x \right )} + 2 i b d} & \text{for}\: a = i b \\\frac{x \left (B \tan{\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot{\left (c \right )}}{a + b \tan{\left (c \right )}} & \text{for}\: d = 0 \\\frac{2 B a d x}{2 a^{2} d + 2 b^{2} d} + \frac{2 B b \log{\left (\frac{a}{b} + \tan{\left (c + d x \right )} \right )}}{2 a^{2} d + 2 b^{2} d} - \frac{B b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d + 2 b^{2} d} - \frac{2 C a \log{\left (\frac{a}{b} + \tan{\left (c + d x \right )} \right )}}{2 a^{2} d + 2 b^{2} d} + \frac{C a \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 a^{2} d + 2 b^{2} d} + \frac{2 C b d x}{2 a^{2} d + 2 b^{2} d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((zoo*x*(B*tan(c) + C*tan(c)**2)*cot(c)/tan(c), Eq(a, 0) & Eq(b, 0) & Eq(d, 0)), ((B*x + C*log(tan(c
+ d*x)**2 + 1)/(2*d))/a, Eq(b, 0)), (-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) - I*B/(-2*b*d*tan(c + d*x) + 2*I*b*d) - C*d*x*tan(c + d*x)/(-2*b*d*tan(c + d*x) + 2*I*b*d
) + I*C*d*x/(-2*b*d*tan(c + d*x) + 2*I*b*d) + C/(-2*b*d*tan(c + d*x) + 2*I*b*d), Eq(a, -I*b)), (-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) - I*B/(2*b*d*tan(c + d*x) + 2*I*
b*d) + C*d*x*tan(c + d*x)/(2*b*d*tan(c + d*x) + 2*I*b*d) + I*C*d*x/(2*b*d*tan(c + d*x) + 2*I*b*d) - C/(2*b*d*t
an(c + d*x) + 2*I*b*d), Eq(a, I*b)), (x*(B*tan(c) + C*tan(c)**2)*cot(c)/(a + b*tan(c)), Eq(d, 0)), (2*B*a*d*x/
(2*a**2*d + 2*b**2*d) + 2*B*b*log(a/b + tan(c + d*x))/(2*a**2*d + 2*b**2*d) - B*b*log(tan(c + d*x)**2 + 1)/(2*
a**2*d + 2*b**2*d) - 2*C*a*log(a/b + tan(c + d*x))/(2*a**2*d + 2*b**2*d) + C*a*log(tan(c + d*x)**2 + 1)/(2*a**
2*d + 2*b**2*d) + 2*C*b*d*x/(2*a**2*d + 2*b**2*d), True))

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Giac [A]  time = 1.55263, size = 127, normalized size = 2.19 \begin{align*} \frac{\frac{2 \,{\left (B a + C b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} + \frac{{\left (C a - B b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \,{\left (C a b - B b^{2}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{2} b + b^{3}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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