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

Optimal. Leaf size=137 \[ \frac{b (A b-a B)}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{b \left (3 a^2 A b-2 a^3 B+A b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 d \left (a^2+b^2\right )^2}-\frac{x \left (a^2 (-B)+2 a A b+b^2 B\right )}{\left (a^2+b^2\right )^2}+\frac{A \log (\sin (c+d x))}{a^2 d} \]

[Out]

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

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Rubi [A]  time = 0.31891, antiderivative size = 137, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {3609, 3651, 3530, 3475} \[ \frac{b (A b-a B)}{a d \left (a^2+b^2\right ) (a+b \tan (c+d x))}-\frac{b \left (3 a^2 A b-2 a^3 B+A b^3\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 d \left (a^2+b^2\right )^2}-\frac{x \left (a^2 (-B)+2 a A b+b^2 B\right )}{\left (a^2+b^2\right )^2}+\frac{A \log (\sin (c+d x))}{a^2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3609

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(A*b - a*B)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n
+ 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e +
f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[b*B*(b*c*(m + 1) + a*d*(n + 1)) + A*(a*(b*c - a*d)*(m + 1) - b^2*d*(
m + n + 2)) - (A*b - a*B)*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b*d*(A*b - a*B)*(m + n + 2)*Tan[e + f*x]^2, x], x
], x] /; FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0]
&& LtQ[m, -1] && (IntegerQ[m] || IntegersQ[2*m, 2*n]) &&  !(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))}{(a+b \tan (c+d x))^2} \, dx &=\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{\int \frac{\cot (c+d x) \left (A \left (a^2+b^2\right )-a (A b-a B) \tan (c+d x)+b (A b-a B) \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx}{a \left (a^2+b^2\right )}\\ &=-\frac{\left (2 a A b-a^2 B+b^2 B\right ) x}{\left (a^2+b^2\right )^2}+\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}+\frac{A \int \cot (c+d x) \, dx}{a^2}-\frac{\left (b \left (3 a^2 A b+A b^3-2 a^3 B\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{\left (2 a A b-a^2 B+b^2 B\right ) x}{\left (a^2+b^2\right )^2}+\frac{A \log (\sin (c+d x))}{a^2 d}-\frac{b \left (3 a^2 A b+A b^3-2 a^3 B\right ) \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 \left (a^2+b^2\right )^2 d}+\frac{b (A b-a B)}{a \left (a^2+b^2\right ) d (a+b \tan (c+d x))}\\ \end{align*}

Mathematica [C]  time = 0.788527, size = 183, normalized size = 1.34 \[ \frac{\frac{b \left (-3 a^2 A b+2 a^3 B-A b^3\right ) \log (a+b \tan (c+d x))}{a \left (a^2+b^2\right )}+\frac{A \left (a^2+b^2\right ) \log (\tan (c+d x))}{a}+\frac{b (A b-a B)}{a+b \tan (c+d x)}-\frac{a (a-i b) (A+i B) \log (-\tan (c+d x)+i)}{2 (a+i b)}-\frac{a (a+i b) (A-i B) \log (\tan (c+d x)+i)}{2 (a-i b)}}{a d \left (a^2+b^2\right )} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.126, size = 325, normalized size = 2.4 \begin{align*} -{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ){a}^{2}A}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) A{b}^{2}}{2\,d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Bab}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-2\,{\frac{A\arctan \left ( \tan \left ( dx+c \right ) \right ) ab}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ){a}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ){b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{A\ln \left ( \tan \left ( dx+c \right ) \right ) }{{a}^{2}d}}-3\,{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A{b}^{2}}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}-{\frac{{b}^{4}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) A}{{a}^{2}d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+2\,{\frac{\ln \left ( a+b\tan \left ( dx+c \right ) \right ) Bab}{d \left ({a}^{2}+{b}^{2} \right ) ^{2}}}+{\frac{A{b}^{2}}{ad \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) }}-{\frac{Bb}{d \left ({a}^{2}+{b}^{2} \right ) \left ( a+b\tan \left ( dx+c \right ) \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)*(A+B*tan(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*A+1/2/d/(a^2+b^2)^2*ln(1+tan(d*x+c)^2)*A*b^2-1/d/(a^2+b^2)^2*ln(1+ta
n(d*x+c)^2)*B*a*b-2/d/(a^2+b^2)^2*A*arctan(tan(d*x+c))*a*b+1/d/(a^2+b^2)^2*B*arctan(tan(d*x+c))*a^2-1/d/(a^2+b
^2)^2*B*arctan(tan(d*x+c))*b^2+1/d/a^2*A*ln(tan(d*x+c))-3/d/(a^2+b^2)^2*ln(a+b*tan(d*x+c))*A*b^2-1/d*b^4/a^2/(
a^2+b^2)^2*ln(a+b*tan(d*x+c))*A+2/d/(a^2+b^2)^2*ln(a+b*tan(d*x+c))*B*a*b+1/d*b^2/a/(a^2+b^2)/(a+b*tan(d*x+c))*
A-1/d*b/(a^2+b^2)/(a+b*tan(d*x+c))*B

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Maxima [A]  time = 1.49221, size = 281, normalized size = 2.05 \begin{align*} \frac{\frac{2 \,{\left (B a^{2} - 2 \, A a b - B b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} + \frac{2 \,{\left (2 \, B a^{3} b - 3 \, A a^{2} b^{2} - A 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 a^{2} + 2 \, B a b - A 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 (B a b - A b^{2}\right )}}{a^{4} + a^{2} b^{2} +{\left (a^{3} b + a b^{3}\right )} \tan \left (d x + c\right )} + \frac{2 \, A \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))/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 2.07281, size = 701, normalized size = 5.12 \begin{align*} -\frac{2 \, B a^{2} b^{3} - 2 \, A a b^{4} - 2 \,{\left (B a^{5} - 2 \, A a^{4} b - B a^{3} b^{2}\right )} d x -{\left (A a^{5} + 2 \, A a^{3} b^{2} + A a b^{4} +{\left (A a^{4} b + 2 \, A a^{2} b^{3} + A 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 (2 \, B a^{4} b - 3 \, A a^{3} b^{2} - A a b^{4} +{\left (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3} - A 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 (B a^{3} b^{2} - A a^{2} b^{3} +{\left (B a^{4} b - 2 \, A a^{3} b^{2} - B a^{2} b^{3}\right )} d x\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))/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/2*(2*B*a^2*b^3 - 2*A*a*b^4 - 2*(B*a^5 - 2*A*a^4*b - B*a^3*b^2)*d*x - (A*a^5 + 2*A*a^3*b^2 + A*a*b^4 + (A*a^
4*b + 2*A*a^2*b^3 + A*b^5)*tan(d*x + c))*log(tan(d*x + c)^2/(tan(d*x + c)^2 + 1)) - (2*B*a^4*b - 3*A*a^3*b^2 -
A*a*b^4 + (2*B*a^3*b^2 - 3*A*a^2*b^3 - A*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*(B*a^3*b^2 - A*a^2*b^3 + (B*a^4*b - 2*A*a^3*b^2 - B*a^2*b^3)*d*x)*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(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.26438, size = 377, normalized size = 2.75 \begin{align*} \frac{\frac{2 \,{\left (B a^{2} - 2 \, A a b - B b^{2}\right )}{\left (d x + c\right )}}{a^{4} + 2 \, a^{2} b^{2} + b^{4}} - \frac{{\left (A a^{2} + 2 \, B a b - A 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 (2 \, B a^{3} b^{2} - 3 \, A a^{2} b^{3} - A 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 \, A \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2}} - \frac{2 \,{\left (2 \, B a^{3} b^{2} \tan \left (d x + c\right ) - 3 \, A a^{2} b^{3} \tan \left (d x + c\right ) - A b^{5} \tan \left (d x + c\right ) + 3 \, B a^{4} b - 4 \, A a^{3} b^{2} + B a^{2} b^{3} - 2 \, A 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 )}}}{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))/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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