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

Optimal. Leaf size=85 \[ \frac{b^3 B \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 d \left (a^2+b^2\right )}-\frac{a B x}{a^2+b^2}-\frac{b B \log (\sin (c+d x))}{a^2 d}-\frac{B \cot (c+d x)}{a d} \]

[Out]

-((a*B*x)/(a^2 + b^2)) - (B*Cot[c + d*x])/(a*d) - (b*B*Log[Sin[c + d*x]])/(a^2*d) + (b^3*B*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 = 0.182393, antiderivative size = 85, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.147, Rules used = {21, 3569, 3651, 3530, 3475} \[ \frac{b^3 B \log (a \cos (c+d x)+b \sin (c+d x))}{a^2 d \left (a^2+b^2\right )}-\frac{a B x}{a^2+b^2}-\frac{b B \log (\sin (c+d x))}{a^2 d}-\frac{B \cot (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*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 ^2(c+d x) (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^2} \, dx &=B \int \frac{\cot ^2(c+d x)}{a+b \tan (c+d x)} \, dx\\ &=-\frac{B \cot (c+d x)}{a d}-\frac{B \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 B x}{a^2+b^2}-\frac{B \cot (c+d x)}{a d}-\frac{(b B) \int \cot (c+d x) \, dx}{a^2}+\frac{\left (b^3 B\right ) \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 B x}{a^2+b^2}-\frac{B \cot (c+d x)}{a d}-\frac{b B \log (\sin (c+d x))}{a^2 d}+\frac{b^3 B \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 = 0.387123, size = 97, normalized size = 1.14 \[ -\frac{B \left (-\frac{b^3 \log (a \cot (c+d x)+b)}{a^2 \left (a^2+b^2\right )}-\frac{\log (-\cot (c+d x)+i)}{2 (b+i a)}+\frac{\log (\cot (c+d x)+i)}{2 (-b+i a)}+\frac{\cot (c+d x)}{a}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((B*(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 = 0.081, size = 117, normalized size = 1.4 \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{B\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{B}{ad\tan \left ( dx+c \right ) }}-{\frac{\ln \left ( \tan \left ( dx+c \right ) \right ) Bb}{{a}^{2}d}}+{\frac{{b}^{3}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{{a}^{2}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)^2*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.75692, size = 142, normalized size = 1.67 \begin{align*} \frac{\frac{2 \, B 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 )} B a}{a^{2} + b^{2}} + \frac{B b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, B b \log \left (\tan \left (d x + c\right )\right )}{a^{2}} - \frac{2 \, B}{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+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.84326, size = 347, normalized size = 4.08 \begin{align*} -\frac{2 \, B a^{3} d x \tan \left (d x + c\right ) - B 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 \, B a^{3} + 2 \, B a b^{2} +{\left (B a^{2} b + 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+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.28116, size = 165, normalized size = 1.94 \begin{align*} \frac{\frac{2 \, B 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 )} B a}{a^{2} + b^{2}} + \frac{B b \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} - \frac{2 \, B b \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a^{2}} + \frac{2 \,{\left (B b \tan \left (d x + c\right ) - B 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+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^2,x, algorithm="giac")

[Out]

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