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

Optimal. Leaf size=70 \[ x \left (a^2 C+2 a b B-b^2 C\right )+\frac{a^2 B \log (\sin (c+d x))}{d}-\frac{b (2 a C+b B) \log (\cos (c+d x))}{d}+\frac{b^2 C \tan (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.184683, antiderivative size = 70, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3632, 3606, 3624, 3475} \[ x \left (a^2 C+2 a b B-b^2 C\right )+\frac{a^2 B \log (\sin (c+d x))}{d}-\frac{b (2 a C+b B) \log (\cos (c+d x))}{d}+\frac{b^2 C \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*b*B + a^2*C - b^2*C)*x - (b*(b*B + 2*a*C)*Log[Cos[c + d*x]])/d + (a^2*B*Log[Sin[c + d*x]])/d + (b^2*C*Tan
[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 3606

Int[(((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^2*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)]))/((c_.) + (d_.)*tan[(e_
.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b^2*B*Tan[e + f*x])/(d*f), x] + Dist[1/d, Int[(a^2*A*d - b^2*B*c + (2*a*
A*b + B*(a^2 - b^2))*d*Tan[e + f*x] + (A*b^2*d - b*B*(b*c - 2*a*d))*Tan[e + f*x]^2)/(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 3624

Int[((A_) + (B_.)*tan[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2)/tan[(e_.) + (f_.)*(x_)], x_Symbol
] :> Simp[B*x, x] + (Dist[A, Int[1/Tan[e + f*x], x], x] + Dist[C, Int[Tan[e + f*x], x], x]) /; FreeQ[{e, f, A,
B, C}, x] && NeQ[A, C]

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

Mathematica [C]  time = 0.26808, size = 91, normalized size = 1.3 \[ -\frac{-2 a^2 B \log (\tan (c+d x))+(a+i b)^2 (B+i C) \log (-\tan (c+d x)+i)+(a-i b)^2 (B-i C) \log (\tan (c+d x)+i)-2 b^2 C \tan (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.076, size = 109, normalized size = 1.6 \begin{align*} 2\,Babx+Cx{a}^{2}-{b}^{2}Cx+{\frac{{a}^{2}B\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{b}^{2}B\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{Babc}{d}}+{\frac{{b}^{2}C\tan \left ( dx+c \right ) }{d}}-2\,{\frac{Cab\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{C{a}^{2}c}{d}}-{\frac{C{b}^{2}c}{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))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 1.74779, size = 115, normalized size = 1.64 \begin{align*} \frac{2 \, B a^{2} \log \left (\tan \left (d x + c\right )\right ) + 2 \, C b^{2} \tan \left (d x + c\right ) + 2 \,{\left (C a^{2} + 2 \, B a b - C b^{2}\right )}{\left (d x + c\right )} -{\left (B a^{2} - 2 \, C a b - B b^{2}\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))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.42113, size = 217, normalized size = 3.1 \begin{align*} \frac{B a^{2} \log \left (\frac{\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) + 2 \, C b^{2} \tan \left (d x + c\right ) + 2 \,{\left (C a^{2} + 2 \, B a b - C b^{2}\right )} d x -{\left (2 \, C a b + B b^{2}\right )} \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))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 11.2536, size = 136, normalized size = 1.94 \begin{align*} \begin{cases} - \frac{B a^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{B a^{2} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + 2 B a b x + \frac{B b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + C a^{2} x + \frac{C a b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} - C b^{2} x + \frac{C b^{2} \tan{\left (c + d x \right )}}{d} & \text{for}\: d \neq 0 \\x \left (a + b \tan{\left (c \right )}\right )^{2} \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))**2*(B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

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

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Giac [A]  time = 1.87633, size = 116, normalized size = 1.66 \begin{align*} \frac{2 \, B a^{2} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + 2 \, C b^{2} \tan \left (d x + c\right ) + 2 \,{\left (C a^{2} + 2 \, B a b - C b^{2}\right )}{\left (d x + c\right )} -{\left (B a^{2} - 2 \, C a b - B b^{2}\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))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="giac")

[Out]

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