3.13 \(\int \cot ^3(c+d x) (a+b \tan (c+d x))^2 (B \tan (c+d x)+C \tan ^2(c+d x)) \, dx\)

Optimal. Leaf size=72 \[ -x \left (a^2 B-2 a b C-b^2 B\right )-\frac{a^2 B \cot (c+d x)}{d}+\frac{a (a C+2 b B) \log (\sin (c+d x))}{d}-\frac{b^2 C \log (\cos (c+d x))}{d} \]

[Out]

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

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Rubi [A]  time = 0.206626, antiderivative size = 72, 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, 3604, 3624, 3475} \[ -x \left (a^2 B-2 a b C-b^2 B\right )-\frac{a^2 B \cot (c+d x)}{d}+\frac{a (a C+2 b B) \log (\sin (c+d x))}{d}-\frac{b^2 C \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^2*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

-((a^2*B - b^2*B - 2*a*b*C)*x) - (a^2*B*Cot[c + d*x])/d - (b^2*C*Log[Cos[c + d*x]])/d + (a*(2*b*B + a*C)*Log[S
in[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 3604

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

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 ^3(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 ^2(c+d x) (a+b \tan (c+d x))^2 (B+C \tan (c+d x)) \, dx\\ &=-\frac{a^2 B \cot (c+d x)}{d}+\int \cot (c+d x) \left (a (2 b B+a C)-\left (a^2 B-b^2 B-2 a b C\right ) \tan (c+d x)+b^2 C \tan ^2(c+d x)\right ) \, dx\\ &=-\left (a^2 B-b^2 B-2 a b C\right ) x-\frac{a^2 B \cot (c+d x)}{d}+\left (b^2 C\right ) \int \tan (c+d x) \, dx+(a (2 b B+a C)) \int \cot (c+d x) \, dx\\ &=-\left (a^2 B-b^2 B-2 a b C\right ) x-\frac{a^2 B \cot (c+d x)}{d}-\frac{b^2 C \log (\cos (c+d x))}{d}+\frac{a (2 b B+a C) \log (\sin (c+d x))}{d}\\ \end{align*}

Mathematica [C]  time = 0.247485, size = 100, normalized size = 1.39 \[ \frac{-2 a^2 B \cot (c+d x)+2 a (a C+2 b B) \log (\tan (c+d x))+i (a+i b)^2 (B+i C) \log (-\tan (c+d x)+i)-(a-i b)^2 (C+i B) \log (\tan (c+d x)+i)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cot[c + d*x]^3*(a + b*Tan[c + d*x])^2*(B*Tan[c + d*x] + C*Tan[c + d*x]^2),x]

[Out]

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

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Maple [A]  time = 0.076, size = 110, normalized size = 1.5 \begin{align*} -{a}^{2}Bx+{b}^{2}Bx+2\,Cabx-{\frac{B\cot \left ( dx+c \right ){a}^{2}}{d}}+2\,{\frac{Bab\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{B{a}^{2}c}{d}}+{\frac{B{b}^{2}c}{d}}+{\frac{C{a}^{2}\ln \left ( \sin \left ( dx+c \right ) \right ) }{d}}-{\frac{{b}^{2}C\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+2\,{\frac{Cabc}{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^3*(a+b*tan(d*x+c))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x)

[Out]

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

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Maxima [A]  time = 1.75614, size = 126, normalized size = 1.75 \begin{align*} -\frac{2 \,{\left (B a^{2} - 2 \, C a b - B b^{2}\right )}{\left (d x + c\right )} +{\left (C a^{2} + 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \,{\left (C a^{2} + 2 \, B a b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac{2 \, B 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)^3*(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 - 2*C*a*b - B*b^2)*(d*x + c) + (C*a^2 + 2*B*a*b - C*b^2)*log(tan(d*x + c)^2 + 1) - 2*(C*a^2 + 2
*B*a*b)*log(tan(d*x + c)) + 2*B*a^2/tan(d*x + c))/d

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Fricas [A]  time = 1.4045, size = 274, normalized size = 3.81 \begin{align*} -\frac{C b^{2} \log \left (\frac{1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) + 2 \,{\left (B a^{2} - 2 \, C a b - B b^{2}\right )} d x \tan \left (d x + c\right ) + 2 \, B a^{2} -{\left (C a^{2} + 2 \, B a b\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 \, 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)^3*(a+b*tan(d*x+c))^2*(B*tan(d*x+c)+C*tan(d*x+c)^2),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 18.9267, size = 158, normalized size = 2.19 \begin{align*} \begin{cases} \text{NaN} & \text{for}\: c = 0 \wedge d = 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 ^{3}{\left (c \right )} & \text{for}\: d = 0 \\\text{NaN} & \text{for}\: c = - d x \\- B a^{2} x - \frac{B a^{2}}{d \tan{\left (c + d x \right )}} - \frac{B a b \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac{2 B a b \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + B b^{2} x - \frac{C a^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac{C a^{2} \log{\left (\tan{\left (c + d x \right )} \right )}}{d} + 2 C a b x + \frac{C b^{2} \log{\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**3*(a+b*tan(d*x+c))**2*(B*tan(d*x+c)+C*tan(d*x+c)**2),x)

[Out]

Piecewise((nan, Eq(c, 0) & Eq(d, 0)), (x*(a + b*tan(c))**2*(B*tan(c) + C*tan(c)**2)*cot(c)**3, Eq(d, 0)), (nan
, Eq(c, -d*x)), (-B*a**2*x - B*a**2/(d*tan(c + d*x)) - B*a*b*log(tan(c + d*x)**2 + 1)/d + 2*B*a*b*log(tan(c +
d*x))/d + B*b**2*x - C*a**2*log(tan(c + d*x)**2 + 1)/(2*d) + C*a**2*log(tan(c + d*x))/d + 2*C*a*b*x + C*b**2*l
og(tan(c + d*x)**2 + 1)/(2*d), True))

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Giac [A]  time = 1.83617, size = 159, normalized size = 2.21 \begin{align*} -\frac{2 \,{\left (B a^{2} - 2 \, C a b - B b^{2}\right )}{\left (d x + c\right )} +{\left (C a^{2} + 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \,{\left (C a^{2} + 2 \, B a b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + \frac{2 \,{\left (C a^{2} \tan \left (d x + c\right ) + 2 \, B a b \tan \left (d x + c\right ) + B a^{2}\right )}}{\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)^3*(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 - 2*C*a*b - B*b^2)*(d*x + c) + (C*a^2 + 2*B*a*b - C*b^2)*log(tan(d*x + c)^2 + 1) - 2*(C*a^2 + 2
*B*a*b)*log(abs(tan(d*x + c))) + 2*(C*a^2*tan(d*x + c) + 2*B*a*b*tan(d*x + c) + B*a^2)/tan(d*x + c))/d