3.4 \(\int \cot ^2(c+d x) (a+b \tan (c+d x)) (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=37 \[ x (a C+b B)+\frac{a B \log (\sin (c+d x))}{d}-\frac{b C \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.109605, antiderivative size = 37, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.105, Rules used = {3632, 3589, 3475, 3531} \[ x (a C+b B)+\frac{a B \log (\sin (c+d x))}{d}-\frac{b C \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*B + a*C)*x - (b*C*Log[Cos[c + d*x]])/d + (a*B*Log[Sin[c + d*x]])/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 3589

Int[(((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]))/((a_.) + (b_.)*tan[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> Dist[(B*d)/b, Int[Tan[e + f*x], x], x] + Dist[1/b, Int[Simp[A*b*c + (A*b*d + B*(
b*c - a*d))*Tan[e + f*x], x]/(a + b*Tan[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a
*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]

Rule 3531

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

Rubi steps

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

Mathematica [A]  time = 0.0689106, size = 44, normalized size = 1.19 \[ \frac{a B (\log (\tan (c+d x))+\log (\cos (c+d x)))}{d}+a C x+b B x-\frac{b C \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.068, size = 51, normalized size = 1.4 \begin{align*} Bxb+Cxa+{\frac{aB\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}+{\frac{Bbc}{d}}-{\frac{Cb\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{Cac}{d}} \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))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x)

[Out]

B*x*b+C*x*a+1/d*a*B*ln(sin(d*x+c))+1/d*B*b*c-b*C*ln(cos(d*x+c))/d+1/d*C*a*c

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Maxima [A]  time = 1.66812, size = 70, normalized size = 1.89 \begin{align*} \frac{2 \, B a \log \left (\tan \left (d x + c\right )\right ) + 2 \,{\left (C a + B b\right )}{\left (d x + c\right )} -{\left (B a - C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\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))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.41548, size = 146, normalized size = 3.95 \begin{align*} \frac{2 \,{\left (C a + B b\right )} d x + B a \log \left (\frac{\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - C b \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\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))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 4.86821, size = 85, normalized size = 2.3 \begin{align*} \begin{cases} - \frac{B a \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{B a \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + B b x + C a x + \frac{C b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text{for}\: d \neq 0 \\x \left (a + b \tan{\left (c \right )}\right ) \left (B \tan{\left (c \right )} + C \tan ^{2}{\left (c \right )}\right ) \cot ^{2}{\left (c \right )} & \text{otherwise} \end{cases} \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))*(B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

Piecewise((-B*a*log(tan(c + d*x)**2 + 1)/(2*d) + B*a*log(tan(c + d*x))/d + B*b*x + C*a*x + C*b*log(tan(c + d*x
)**2 + 1)/(2*d), Ne(d, 0)), (x*(a + b*tan(c))*(B*tan(c) + C*tan(c)**2)*cot(c)**2, True))

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Giac [A]  time = 1.49365, size = 72, normalized size = 1.95 \begin{align*} \frac{2 \, B a \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + 2 \,{\left (C a + B b\right )}{\left (d x + c\right )} -{\left (B a - C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\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))*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

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