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

Optimal. Leaf size=81 \[ \frac{b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 d \left (a^2+b^2\right )}-\frac{a x}{a^2+b^2}-\frac{b \log (\sin (c+d x))}{a^2 d}-\frac{\cot (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.17318, antiderivative size = 81, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {3569, 3651, 3530, 3475} \[ \frac{b^3 \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 d \left (a^2+b^2\right )}-\frac{a x}{a^2+b^2}-\frac{b \log (\sin (c+d x))}{a^2 d}-\frac{\cot (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.430862, size = 96, normalized size = 1.19 \[ -\frac{-\frac{b^3 \log (a \cot (c+d x)+b)}{a^2 \left (a^2+b^2\right )}-\frac{\log (-\cot (c+d x)+i)}{2 (b+i a)}+\frac{\log (\cot (c+d x)+i)}{2 (-b+i a)}+\frac{\cot (c+d x)}{a}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.063, size = 112, normalized size = 1.4 \begin{align*}{\frac{b\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) }{2\,d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{a\arctan \left ( \tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{1}{ad\tan \left ( dx+c \right ) }}-{\frac{b\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{2}d}}+{\frac{{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) }{d \left ({a}^{2}+{b}^{2} \right ){a}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.62395, size = 135, normalized size = 1.67 \begin{align*} \frac{\frac{2 \, b^{3} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{4} + a^{2} b^{2}} - \frac{2 \,{\left (d x + c\right )} a}{a^{2} + b^{2}} + \frac{b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, b \log \left (\tan \left (d x + c\right )\right )}{a^{2}} - \frac{2}{a \tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.18599, size = 331, normalized size = 4.09 \begin{align*} -\frac{2 \, a^{3} d x \tan \left (d x + c\right ) - b^{3} \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 ) \tan \left (d x + c\right ) + 2 \, a^{3} + 2 \, a b^{2} +{\left (a^{2} b + b^{3}\right )} \log \left (\frac{\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )}{2 \,{\left (a^{4} + a^{2} b^{2}\right )} d \tan \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]  time = 28.4359, size = 1080, normalized size = 13.33 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28445, size = 157, normalized size = 1.94 \begin{align*} \frac{\frac{2 \, b^{4} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{4} b + a^{2} b^{3}} - \frac{2 \,{\left (d x + c\right )} a}{a^{2} + b^{2}} + \frac{b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac{2 \,{\left (b \tan \left (d x + c\right ) - a\right )}}{a^{2} \tan \left (d x + c\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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