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

Optimal. Leaf size=80 \[ -\frac{b (b B-a C) \log (a \cos (c+d x)+b \sin (c+d x))}{a d \left (a^2+b^2\right )}-\frac{x (b B-a C)}{a^2+b^2}+\frac{B \log (\sin (c+d x))}{a d} \]

[Out]

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

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Rubi [A]  time = 0.200835, antiderivative size = 80, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3632, 3611, 3530, 3475} \[ -\frac{b (b B-a C) \log (a \cos (c+d x)+b \sin (c+d x))}{a d \left (a^2+b^2\right )}-\frac{x (b B-a C)}{a^2+b^2}+\frac{B \log (\sin (c+d x))}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((b*B - a*C)*x)/(a^2 + b^2)) + (B*Log[Sin[c + d*x]])/(a*d) - (b*(b*B - a*C)*Log[a*Cos[c + d*x] + b*Sin[c + d
*x]])/(a*(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 3611

Int[((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])/(((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) +
(f_.)*(x_)])), x_Symbol] :> Simp[((B*(b*c + a*d) + A*(a*c - b*d))*x)/((a^2 + b^2)*(c^2 + d^2)), x] + (Dist[(b
*(A*b - a*B))/((b*c - a*d)*(a^2 + b^2)), Int[(b - a*Tan[e + f*x])/(a + b*Tan[e + f*x]), x], x] + Dist[(d*(B*c
- A*d))/((b*c - a*d)*(c^2 + d^2)), Int[(d - c*Tan[e + f*x])/(c + d*Tan[e + f*x]), x], x]) /; FreeQ[{a, b, c, d
, e, f, A, B}, 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) \left (B \tan (c+d x)+C \tan ^2(c+d x)\right )}{a+b \tan (c+d x)} \, dx &=\int \frac{\cot (c+d x) (B+C \tan (c+d x))}{a+b \tan (c+d x)} \, dx\\ &=-\frac{(b B-a C) x}{a^2+b^2}+\frac{B \int \cot (c+d x) \, dx}{a}-\frac{(b (b B-a C)) \int \frac{b-a \tan (c+d x)}{a+b \tan (c+d x)} \, dx}{a \left (a^2+b^2\right )}\\ &=-\frac{(b B-a C) x}{a^2+b^2}+\frac{B \log (\sin (c+d x))}{a d}-\frac{b (b B-a C) \log (a \cos (c+d x)+b \sin (c+d x))}{a \left (a^2+b^2\right ) d}\\ \end{align*}

Mathematica [C]  time = 0.312373, size = 113, normalized size = 1.41 \[ -\frac{\frac{2 b (b B-a C) \log (a+b \tan (c+d x))}{a \left (a^2+b^2\right )}+\frac{(B+i C) \log (-\tan (c+d x)+i)}{a+i b}+\frac{(B-i C) \log (\tan (c+d x)+i)}{a-i b}-\frac{2 B \log (\tan (c+d x))}{a}}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.13, size = 174, normalized size = 2.2 \begin{align*} -{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) aB}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{\ln \left ( 1+ \left ( \tan \left ( dx+c \right ) \right ) ^{2} \right ) Cb}{2\,d \left ({a}^{2}+{b}^{2} \right ) }}-{\frac{B\arctan \left ( \tan \left ( dx+c \right ) \right ) b}{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{C\arctan \left ( \tan \left ( dx+c \right ) \right ) a}{d \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{B\ln \left ( \tan \left ( dx+c \right ) \right ) }{ad}}-{\frac{{b}^{2}\ln \left ( a+b\tan \left ( dx+c \right ) \right ) B}{ad \left ({a}^{2}+{b}^{2} \right ) }}+{\frac{b\ln \left ( a+b\tan \left ( dx+c \right ) \right ) C}{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*(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)*a*B-1/2/d/(a^2+b^2)*ln(1+tan(d*x+c)^2)*C*b-1/d/(a^2+b^2)*B*arctan(tan(d*x+
c))*b+1/d/(a^2+b^2)*C*arctan(tan(d*x+c))*a+1/d/a*B*ln(tan(d*x+c))-1/d*b^2/a/(a^2+b^2)*ln(a+b*tan(d*x+c))*B+1/d
*b/(a^2+b^2)*ln(a+b*tan(d*x+c))*C

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Maxima [A]  time = 1.75587, size = 144, normalized size = 1.8 \begin{align*} \frac{\frac{2 \,{\left (C a - B b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} + \frac{2 \,{\left (C a b - B b^{2}\right )} \log \left (b \tan \left (d x + c\right ) + a\right )}{a^{3} + a b^{2}} - \frac{{\left (B a + C b\right )} \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 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.26378, size = 267, normalized size = 3.34 \begin{align*} \frac{2 \,{\left (C a^{2} - B a b\right )} d x +{\left (B a^{2} + B b^{2}\right )} \log \left (\frac{\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) +{\left (C a b - B b^{2}\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^{3} + a b^{2}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.66261, size = 153, normalized size = 1.91 \begin{align*} \frac{\frac{2 \,{\left (C a - B b\right )}{\left (d x + c\right )}}{a^{2} + b^{2}} - \frac{{\left (B a + C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{a^{2} + b^{2}} + \frac{2 \,{\left (C a b^{2} - B b^{3}\right )} \log \left ({\left | b \tan \left (d x + c\right ) + a \right |}\right )}{a^{3} b + a b^{3}} + \frac{2 \, B \log \left ({\left | \tan \left (d x + c\right ) \right |}\right )}{a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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