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

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Rubi [A]  time = 0.11, antiderivative size = 37, normalized size of antiderivative = 1.00, 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 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]

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 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 ^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*}

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Mathematica [A]  time = 0.07, 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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fricas [A]  time = 0.46, size = 59, normalized size = 1.59 \[ \frac {2 \, {\left (C a + B b\right )} d x + B a \log \left (\frac {\tan \left (d x + c\right )^{2}}{\tan \left (d x + c\right )^{2} + 1}\right ) - C b \log \left (\frac {1}{\tan \left (d x + c\right )^{2} + 1}\right )}{2 \, d} \]

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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giac [A]  time = 3.67, size = 53, normalized size = 1.43 \[ \frac {2 \, B a \log \left ({\left | \tan \left (d x + c\right ) \right |}\right ) + 2 \, {\left (C a + B b\right )} {\left (d x + c\right )} - {\left (B a - C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

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

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maple [A]  time = 0.57, size = 51, normalized size = 1.38 \[ B x b +a C x +\frac {a B \ln \left (\sin \left (d x +c \right )\right )}{d}+\frac {B b c}{d}-\frac {b C \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {C a c}{d} \]

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+a*C*x+a*B*ln(sin(d*x+c))/d+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 = 0.59, size = 52, normalized size = 1.41 \[ \frac {2 \, B a \log \left (\tan \left (d x + c\right )\right ) + 2 \, {\left (C a + B b\right )} {\left (d x + c\right )} - {\left (B a - C b\right )} \log \left (\tan \left (d x + c\right )^{2} + 1\right )}{2 \, d} \]

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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mupad [B]  time = 8.96, size = 69, normalized size = 1.86 \[ \frac {B\,a\,\ln \left (\mathrm {tan}\left (c+d\,x\right )\right )}{d}-\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )-\mathrm {i}\right )\,\left (B+C\,1{}\mathrm {i}\right )\,\left (a+b\,1{}\mathrm {i}\right )}{2\,d}+\frac {\ln \left (\mathrm {tan}\left (c+d\,x\right )+1{}\mathrm {i}\right )\,\left (B-C\,1{}\mathrm {i}\right )\,\left (b+a\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{2\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[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]

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

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sympy [A]  time = 0.98, size = 85, normalized size = 2.30 \[ \begin {cases} - \frac {B a \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} + \frac {B a \log {\left (\tan {\left (c + d x \right )} \right )}}{d} + B b x + C a x + \frac {C b \log {\left (\tan ^{2}{\left (c + d x \right )} + 1 \right )}}{2 d} & \text {for}\: d \neq 0 \\x \left (a + b \tan {\relax (c )}\right ) \left (B \tan {\relax (c )} + C \tan ^{2}{\relax (c )}\right ) \cot ^{2}{\relax (c )} & \text {otherwise} \end {cases} \]

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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