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

Optimal. Leaf size=107 \[ \frac{b^2}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{b^2 \left (3 a^2+b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 d \left (a^2+b^2\right )^2}-\frac{2 a b x}{\left (a^2+b^2\right )^2}+\frac{\log (\sin (c+d x))}{a^2 d} \]

[Out]

(-2*a*b*x)/(a^2 + b^2)^2 + Log[Sin[c + d*x]]/(a^2*d) - (b^2*(3*a^2 + b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]
)/(a^2*(a^2 + b^2)^2*d) + b^2/(a*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))

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Rubi [A]  time = 0.215382, antiderivative size = 107, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.21, Rules used = {3569, 3651, 3530, 3475} \[ \frac{b^2}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{b^2 \left (3 a^2+b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 d \left (a^2+b^2\right )^2}-\frac{2 a b x}{\left (a^2+b^2\right )^2}+\frac{\log (\sin (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a*b*x)/(a^2 + b^2)^2 + Log[Sin[c + d*x]]/(a^2*d) - (b^2*(3*a^2 + b^2)*Log[a*Cos[c + d*x] + b*Sin[c + d*x]]
)/(a^2*(a^2 + b^2)^2*d) + b^2/(a*(a^2 + b^2)*d*(a + b*Tan[c + d*x]))

Rule 3569

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[(b^2*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d)), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) &&  !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))

Rule 3651

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

Rule 3475

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*} \int \frac{\cot (c+d x)}{(a+b \tan (c+d x))^2} \, dx &=\frac{b^2}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \frac{\cot (c+d x) \left (a^2+b^2-a b \tan (c+d x)+b^2 \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{a \left (a^2+b^2\right )}\\ &=-\frac{2 a b x}{\left (a^2+b^2\right )^2}+\frac{b^2}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \cot (c+d x) \, dx}{a^2}-\frac{\left (b^2 \left (3 a^2+b^2\right )\right ) \int \frac{b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a^2 \left (a^2+b^2\right )^2}\\ &=-\frac{2 a b x}{\left (a^2+b^2\right )^2}+\frac{\log (\sin (c+d x))}{a^2 d}-\frac{b^2 \left (3 a^2+b^2\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 \left (a^2+b^2\right )^2 d}+\frac{b^2}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.616245, size = 154, normalized size = 1.44 \[ \frac{-\frac{b^2 \left (3 a^2+b^2\right ) \log (a+b \tan (c+d x))}{a \left (a^2+b^2\right )}+\frac{\left (a^2+b^2\right ) \log (\tan (c+d x))}{a}+\frac{b^2}{a+b \tan (c+d x)}-\frac{a (a-i b) \log (-\tan (c+d x)+i)}{2 (a+i b)}-\frac{a (a+i b) \log (\tan (c+d x)+i)}{2 (a-i b)}}{a d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a*(a - I*b)*Log[I - Tan[c + d*x]])/(2*(a + I*b)) + ((a^2 + b^2)*Log[Tan[c + d*x]])/a - (a*(a + I*b)*Log[I +
Tan[c + d*x]])/(2*(a - I*b)) - (b^2*(3*a^2 + b^2)*Log[a + b*Tan[c + d*x]])/(a*(a^2 + b^2)) + b^2/(a + b*Tan[c
+ d*x]))/(a*(a^2 + b^2)*d)

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Maple [A]  time = 0.082, size = 185, normalized size = 1.7 \begin{align*} -{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){a}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-2\,{\frac{ab\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{2}d}}+{\frac{{b}^{2}}{ \left ({a}^{2}+{b}^{2} \right ) ad \left ( a+b\tan \left ( dx+c \right ) \right ) }}-3\,{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ){b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{{b}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) ^{2}{a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.5317, size = 221, normalized size = 2.07 \begin{align*} -\frac{\frac{4 \,{\left (d x + c\right )} a b}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \, b^{2}}{a^{4} + a^{2} b^{2} +{\left (a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )} + \frac{2 \,{\left (3 \, a^{2} b^{2} + b^{4}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \, \log \left (\tan \left (d x + c\right )\right )}{a^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*(d*x + c)*a*b/(a^4 + 2*a^2*b^2 + b^4) - 2*b^2/(a^4 + a^2*b^2 + (a^3*b + a*b^3)*tan(d*x + c)) + 2*(3*a^
2*b^2 + b^4)*log(b*tan(d*x + c) + a)/(a^6 + 2*a^4*b^2 + a^2*b^4) + (a^2 - b^2)*log(tan(d*x + c)^2 + 1)/(a^4 +
2*a^2*b^2 + b^4) - 2*log(tan(d*x + c))/a^2)/d

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Fricas [B]  time = 2.33887, size = 528, normalized size = 4.93 \begin{align*} -\frac{4 \, a^{4} b d x - 2 \, a b^{4} -{\left (a^{5} + 2 \, a^{3} b^{2} + a b^{4} +{\left (a^{4} b + 2 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )\right )} \log \left (\frac{\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) +{\left (3 \, a^{3} b^{2} + a b^{4} +{\left (3 \, a^{2} b^{3} + b^{5}\right )} \tan \left (d x + c\right )\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 (2 \, a^{3} b^{2} d x + a^{2} b^{3}\right )} \tan \left (d x + c\right )}{2 \,{\left ({\left (a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5}\right )} d \tan \left (d x + c\right ) +{\left (a^{7} + 2 \, a^{5} b^{2} + a^{3} b^{4}\right )} d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*a^4*b*d*x - 2*a*b^4 - (a^5 + 2*a^3*b^2 + a*b^4 + (a^4*b + 2*a^2*b^3 + b^5)*tan(d*x + c))*log(tan(d*x +
c)^2/(tan(d*x + c)^2 + 1)) + (3*a^3*b^2 + a*b^4 + (3*a^2*b^3 + b^5)*tan(d*x + c))*log((b^2*tan(d*x + c)^2 + 2
*a*b*tan(d*x + c) + a^2)/(tan(d*x + c)^2 + 1)) + 2*(2*a^3*b^2*d*x + a^2*b^3)*tan(d*x + c))/((a^6*b + 2*a^4*b^3
+ a^2*b^5)*d*tan(d*x + c) + (a^7 + 2*a^5*b^2 + a^3*b^4)*d)

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

[Out]

Exception raised: AttributeError

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Giac [A]  time = 1.19943, size = 278, normalized size = 2.6 \begin{align*} -\frac{\frac{4 \,{\left (d x + c\right )} a b}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (a^{2} - b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (3 \, a^{2} b^{3} + b^{5}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{6} b + 2 \, a^{4} b^{3} + a^{2} b^{5}} - \frac{2 \,{\left (3 \, a^{2} b^{3} \tan \left (d x + c\right ) + b^{5} \tan \left (d x + c\right ) + 4 \, a^{3} b^{2} + 2 \, a b^{4}\right )}}{{\left (a^{6} + 2 \, a^{4} b^{2} + a^{2} b^{4}\right )}{\left (b \tan \left (d x + c\right ) + a\right )}} - \frac{2 \, \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(4*(d*x + c)*a*b/(a^4 + 2*a^2*b^2 + b^4) + (a^2 - b^2)*log(tan(d*x + c)^2 + 1)/(a^4 + 2*a^2*b^2 + b^4) +
2*(3*a^2*b^3 + b^5)*log(abs(b*tan(d*x + c) + a))/(a^6*b + 2*a^4*b^3 + a^2*b^5) - 2*(3*a^2*b^3*tan(d*x + c) + b
^5*tan(d*x + c) + 4*a^3*b^2 + 2*a*b^4)/((a^6 + 2*a^4*b^2 + a^2*b^4)*(b*tan(d*x + c) + a)) - 2*log(abs(tan(d*x
+ c)))/a^2)/d