3.101 \(\int (c+d x) \cos (a+b x) \cot (a+b x) \, dx\)

Optimal. Leaf size=94 \[ \frac {i d \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {i d \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {d \sin (a+b x)}{b^2}+\frac {(c+d x) \cos (a+b x)}{b}-\frac {2 (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

[Out]

-2*(d*x+c)*arctanh(exp(I*(b*x+a)))/b+(d*x+c)*cos(b*x+a)/b+I*d*polylog(2,-exp(I*(b*x+a)))/b^2-I*d*polylog(2,exp
(I*(b*x+a)))/b^2-d*sin(b*x+a)/b^2

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Rubi [A]  time = 0.06, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {4408, 3296, 2637, 4183, 2279, 2391} \[ \frac {i d \text {PolyLog}\left (2,-e^{i (a+b x)}\right )}{b^2}-\frac {i d \text {PolyLog}\left (2,e^{i (a+b x)}\right )}{b^2}-\frac {d \sin (a+b x)}{b^2}+\frac {(c+d x) \cos (a+b x)}{b}-\frac {2 (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Cos[a + b*x]*Cot[a + b*x],x]

[Out]

(-2*(c + d*x)*ArcTanh[E^(I*(a + b*x))])/b + ((c + d*x)*Cos[a + b*x])/b + (I*d*PolyLog[2, -E^(I*(a + b*x))])/b^
2 - (I*d*PolyLog[2, E^(I*(a + b*x))])/b^2 - (d*Sin[a + b*x])/b^2

Rule 2279

Int[Log[(a_) + (b_.)*((F_)^((e_.)*((c_.) + (d_.)*(x_))))^(n_.)], x_Symbol] :> Dist[1/(d*e*n*Log[F]), Subst[Int
[Log[a + b*x]/x, x], x, (F^(e*(c + d*x)))^n], x] /; FreeQ[{F, a, b, c, d, e, n}, x] && GtQ[a, 0]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2637

Int[sin[Pi/2 + (c_.) + (d_.)*(x_)], x_Symbol] :> Simp[Sin[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 3296

Int[((c_.) + (d_.)*(x_))^(m_.)*sin[(e_.) + (f_.)*(x_)], x_Symbol] :> -Simp[((c + d*x)^m*Cos[e + f*x])/f, x] +
Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Cos[e + f*x], x], x] /; FreeQ[{c, d, e, f}, x] && GtQ[m, 0]

Rule 4183

Int[csc[(e_.) + (f_.)*(x_)]*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> Simp[(-2*(c + d*x)^m*ArcTanh[E^(I*(e + f*
x))])/f, x] + (-Dist[(d*m)/f, Int[(c + d*x)^(m - 1)*Log[1 - E^(I*(e + f*x))], x], x] + Dist[(d*m)/f, Int[(c +
d*x)^(m - 1)*Log[1 + E^(I*(e + f*x))], x], x]) /; FreeQ[{c, d, e, f}, x] && IGtQ[m, 0]

Rule 4408

Int[Cos[(a_.) + (b_.)*(x_)]^(n_.)*Cot[(a_.) + (b_.)*(x_)]^(p_.)*((c_.) + (d_.)*(x_))^(m_.), x_Symbol] :> -Int[
(c + d*x)^m*Cos[a + b*x]^n*Cot[a + b*x]^(p - 2), x] + Int[(c + d*x)^m*Cos[a + b*x]^(n - 2)*Cot[a + b*x]^p, x]
/; FreeQ[{a, b, c, d, m}, x] && IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

\begin {align*} \int (c+d x) \cos (a+b x) \cot (a+b x) \, dx &=\int (c+d x) \csc (a+b x) \, dx-\int (c+d x) \sin (a+b x) \, dx\\ &=-\frac {2 (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x) \cos (a+b x)}{b}-\frac {d \int \cos (a+b x) \, dx}{b}-\frac {d \int \log \left (1-e^{i (a+b x)}\right ) \, dx}{b}+\frac {d \int \log \left (1+e^{i (a+b x)}\right ) \, dx}{b}\\ &=-\frac {2 (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x) \cos (a+b x)}{b}-\frac {d \sin (a+b x)}{b^2}+\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1-x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^2}-\frac {(i d) \operatorname {Subst}\left (\int \frac {\log (1+x)}{x} \, dx,x,e^{i (a+b x)}\right )}{b^2}\\ &=-\frac {2 (c+d x) \tanh ^{-1}\left (e^{i (a+b x)}\right )}{b}+\frac {(c+d x) \cos (a+b x)}{b}+\frac {i d \text {Li}_2\left (-e^{i (a+b x)}\right )}{b^2}-\frac {i d \text {Li}_2\left (e^{i (a+b x)}\right )}{b^2}-\frac {d \sin (a+b x)}{b^2}\\ \end {align*}

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Mathematica [A]  time = 0.17, size = 176, normalized size = 1.87 \[ \frac {d \left (i \left (\text {Li}_2\left (-e^{i (a+b x)}\right )-\text {Li}_2\left (e^{i (a+b x)}\right )\right )+(a+b x) \left (\log \left (1-e^{i (a+b x)}\right )-\log \left (1+e^{i (a+b x)}\right )\right )-a \log \left (\tan \left (\frac {1}{2} (a+b x)\right )\right )\right )}{b^2}+\frac {d \cos (b x) (b x \cos (a)-\sin (a))}{b^2}-\frac {d \sin (b x) (b x \sin (a)+\cos (a))}{b^2}+\frac {c \cos (a+b x)}{b}+\frac {c \log \left (\sin \left (\frac {1}{2} (a+b x)\right )\right )}{b}-\frac {c \log \left (\cos \left (\frac {1}{2} (a+b x)\right )\right )}{b} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)*Cos[a + b*x]*Cot[a + b*x],x]

[Out]

(c*Cos[a + b*x])/b - (c*Log[Cos[(a + b*x)/2]])/b + (c*Log[Sin[(a + b*x)/2]])/b + (d*((a + b*x)*(Log[1 - E^(I*(
a + b*x))] - Log[1 + E^(I*(a + b*x))]) - a*Log[Tan[(a + b*x)/2]] + I*(PolyLog[2, -E^(I*(a + b*x))] - PolyLog[2
, E^(I*(a + b*x))])))/b^2 + (d*Cos[b*x]*(b*x*Cos[a] - Sin[a]))/b^2 - (d*(Cos[a] + b*x*Sin[a])*Sin[b*x])/b^2

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fricas [B]  time = 0.65, size = 277, normalized size = 2.95 \[ \frac {2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right ) - i \, d {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, d {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - i \, d {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, d {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) - {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) - {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) + {\left (b c - a d\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) + {\left (b c - a d\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) + {\left (b d x + a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) - 2 \, d \sin \left (b x + a\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*cos(b*x+a)*cot(b*x+a),x, algorithm="fricas")

[Out]

1/2*(2*(b*d*x + b*c)*cos(b*x + a) - I*d*dilog(cos(b*x + a) + I*sin(b*x + a)) + I*d*dilog(cos(b*x + a) - I*sin(
b*x + a)) - I*d*dilog(-cos(b*x + a) + I*sin(b*x + a)) + I*d*dilog(-cos(b*x + a) - I*sin(b*x + a)) - (b*d*x + b
*c)*log(cos(b*x + a) + I*sin(b*x + a) + 1) - (b*d*x + b*c)*log(cos(b*x + a) - I*sin(b*x + a) + 1) + (b*c - a*d
)*log(-1/2*cos(b*x + a) + 1/2*I*sin(b*x + a) + 1/2) + (b*c - a*d)*log(-1/2*cos(b*x + a) - 1/2*I*sin(b*x + a) +
1/2) + (b*d*x + a*d)*log(-cos(b*x + a) + I*sin(b*x + a) + 1) + (b*d*x + a*d)*log(-cos(b*x + a) - I*sin(b*x +
a) + 1) - 2*d*sin(b*x + a))/b^2

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )} \cos \left (b x + a\right ) \cot \left (b x + a\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*cos(b*x+a)*cot(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)*cos(b*x + a)*cot(b*x + a), x)

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maple [B]  time = 0.08, size = 199, normalized size = 2.12 \[ \frac {d \cos \left (b x +a \right ) x}{b}-\frac {d \sin \left (b x +a \right )}{b^{2}}+\frac {c \cos \left (b x +a \right )}{b}+\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}-\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}+\frac {i d \dilog \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b^{2}}-\frac {i d \dilog \left (1-{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) a}{b^{2}}-\frac {d a \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{b^{2}}+\frac {c \ln \left (\csc \left (b x +a \right )-\cot \left (b x +a \right )\right )}{b} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*cos(b*x+a)*cot(b*x+a),x)

[Out]

1/b*d*cos(b*x+a)*x-d*sin(b*x+a)/b^2+1/b*c*cos(b*x+a)+1/b*d*ln(1-exp(I*(b*x+a)))*x-1/b*d*ln(exp(I*(b*x+a))+1)*x
+I/b^2*d*dilog(exp(I*(b*x+a))+1)-I/b^2*d*dilog(1-exp(I*(b*x+a)))+1/b^2*d*ln(1-exp(I*(b*x+a)))*a-1/b^2*d*ln(exp
(I*(b*x+a))+1)*a-1/b^2*d*a*ln(csc(b*x+a)-cot(b*x+a))+1/b*c*ln(csc(b*x+a)-cot(b*x+a))

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maxima [B]  time = 0.45, size = 199, normalized size = 2.12 \[ -\frac {2 i \, b d x \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) - 2 i \, b c \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) + {\left (2 i \, b d x + 2 i \, b c\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - 2 \, {\left (b d x + b c\right )} \cos \left (b x + a\right ) - 2 i \, d {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) + 2 i \, d {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) - {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right ) + 2 \, d \sin \left (b x + a\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*cos(b*x+a)*cot(b*x+a),x, algorithm="maxima")

[Out]

-1/2*(2*I*b*d*x*arctan2(sin(b*x + a), -cos(b*x + a) + 1) - 2*I*b*c*arctan2(sin(b*x + a), cos(b*x + a) - 1) + (
2*I*b*d*x + 2*I*b*c)*arctan2(sin(b*x + a), cos(b*x + a) + 1) - 2*(b*d*x + b*c)*cos(b*x + a) - 2*I*d*dilog(-e^(
I*b*x + I*a)) + 2*I*d*dilog(e^(I*b*x + I*a)) + (b*d*x + b*c)*log(cos(b*x + a)^2 + sin(b*x + a)^2 + 2*cos(b*x +
a) + 1) - (b*d*x + b*c)*log(cos(b*x + a)^2 + sin(b*x + a)^2 - 2*cos(b*x + a) + 1) + 2*d*sin(b*x + a))/b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \cos \left (a+b\,x\right )\,\mathrm {cot}\left (a+b\,x\right )\,\left (c+d\,x\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(a + b*x)*cot(a + b*x)*(c + d*x),x)

[Out]

int(cos(a + b*x)*cot(a + b*x)*(c + d*x), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right ) \cos {\left (a + b x \right )} \cot {\left (a + b x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*cos(b*x+a)*cot(b*x+a),x)

[Out]

Integral((c + d*x)*cos(a + b*x)*cot(a + b*x), x)

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