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]

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

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Rubi [A]  time = 0.21, antiderivative size = 72, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 40, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, 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 3475

Int[tan[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Log[RemoveContent[Cos[c + d*x], x]]/d, x] /; FreeQ[{c, d}, x]

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 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]

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*}

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Mathematica [C]  time = 0.25, 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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fricas [A]  time = 0.77, size = 112, normalized size = 1.56 \[ -\frac {C b^{2} \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right ) + 2 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} d x \tan \left (d x + c\right ) + 2 \, B a^{2} - {\left (C a^{2} + 2 \, B a b\right )} \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) \tan \left (d x + c\right )}{2 \, d \tan \left (d x + c\right )} \]

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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giac [A]  time = 6.39, size = 118, normalized size = 1.64 \[ -\frac {2 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} {\left (d x + c\right )} + {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (C a^{2} + 2 \, B a b\right )} \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + \frac {2 \, {\left (C a^{2} \tan \left (d x + c\right ) + 2 \, B a b \tan \left (d x + c\right ) + B a^{2}\right )}}{\tan \left (d x + c\right )}}{2 \, d} \]

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

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maple [A]  time = 0.52, size = 110, normalized size = 1.53 \[ -a^{2} B x +B x \,b^{2}+2 a b C x -\frac {a^{2} B \cot \left (d x +c \right )}{d}+\frac {2 B a b \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {B \,a^{2} c}{d}+\frac {B \,b^{2} c}{d}+\frac {a^{2} C \ln \left (\sin \left (d x +c \right )\right )}{d}-\frac {b^{2} C \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {2 C a b c}{d} \]

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*x*b^2+2*a*b*C*x-a^2*B*cot(d*x+c)/d+2/d*B*a*b*ln(sin(d*x+c))-1/d*B*a^2*c+1/d*B*b^2*c+1/d*a^2*C*ln(si
n(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 = 0.74, size = 93, normalized size = 1.29 \[ -\frac {2 \, {\left (B a^{2} - 2 \, C a b - B b^{2}\right )} {\left (d x + c\right )} + {\left (C a^{2} + 2 \, B a b - C b^{2}\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right ) - 2 \, {\left (C a^{2} + 2 \, B a b\right )} \log \left (\tan \left (d x + c\right )\right ) + \frac {2 \, B a^{2}}{\tan \left (d x + c\right )}}{2 \, d} \]

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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mupad [B]  time = 9.00, size = 100, normalized size = 1.39 \[ \frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )\,\left (C\,a^2+2\,B\,b\,a\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (-C+B\,1{}\mathrm {i}\right )\,{\left (-b+a\,1{}\mathrm {i}\right )}^2}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (C+B\,1{}\mathrm {i}\right )\,{\left (b+a\,1{}\mathrm {i}\right )}^2}{2\,d}-\frac {B\,a^2\,\mathrm {cot}\left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 2.30, size = 158, normalized size = 2.19 \[ \begin {cases} \text {NaN} & \text {for}\: c = 0 \wedge d = 0 \\x \left (a + b \tan {\relax (c )}\right )^{2} \left (B \tan {\relax (c )} + C \tan ^{2}{\relax (c )}\right ) \cot ^{3}{\relax (c )} & \text {for}\: d = 0 \\\text {NaN} & \text {for}\: c = - d x \\- B a^{2} x - \frac {B a^{2}}{d \tan {\left (c + d x \right )}} - \frac {B a b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{d} + \frac {2 B a b \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + B b^{2} x - \frac {C a^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {C a^{2} \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + 2 C a b x + \frac {C b^{2} \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text {otherwise} \end {cases} \]

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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