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

Optimal. Leaf size=65 \[ -\frac {i d \text {Li}_2\left (e^{2 i (a+b x)}\right )}{2 b^2}+\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {i (c+d x)^2}{2 d} \]

[Out]

-1/2*I*(d*x+c)^2/d+(d*x+c)*ln(1-exp(2*I*(b*x+a)))/b-1/2*I*d*polylog(2,exp(2*I*(b*x+a)))/b^2

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Rubi [A]  time = 0.10, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3717, 2190, 2279, 2391} \[ -\frac {i d \text {PolyLog}\left (2,e^{2 i (a+b x)}\right )}{2 b^2}+\frac {(c+d x) \log \left (1-e^{2 i (a+b x)}\right )}{b}-\frac {i (c+d x)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*Cot[a + b*x],x]

[Out]

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

Rule 2190

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m*Log[1 + (b*(F^(g*(e + f*x)))^n)/a])/(b*f*g*n*Log[F]), x]
- Dist[(d*m)/(b*f*g*n*Log[F]), Int[(c + d*x)^(m - 1)*Log[1 + (b*(F^(g*(e + f*x)))^n)/a], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

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 3717

Int[((c_.) + (d_.)*(x_))^(m_.)*tan[(e_.) + Pi*(k_.) + (f_.)*(x_)], x_Symbol] :> Simp[(I*(c + d*x)^(m + 1))/(d*
(m + 1)), x] - Dist[2*I, Int[((c + d*x)^m*E^(2*I*k*Pi)*E^(2*I*(e + f*x)))/(1 + E^(2*I*k*Pi)*E^(2*I*(e + f*x)))
, x], x] /; FreeQ[{c, d, e, f}, x] && IntegerQ[4*k] && IGtQ[m, 0]

Rubi steps

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

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Mathematica [B]  time = 5.11, size = 188, normalized size = 2.89 \[ -\frac {d \csc (a) \sec (a) \left (b^2 x^2 e^{i \tan ^{-1}(\tan (a))}+\frac {\tan (a) \left (i \text {Li}_2\left (e^{2 i \left (b x+\tan ^{-1}(\tan (a))\right )}\right )+i b x \left (2 \tan ^{-1}(\tan (a))-\pi \right )-2 \left (\tan ^{-1}(\tan (a))+b x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (a))+b x\right )}\right )+2 \tan ^{-1}(\tan (a)) \log \left (\sin \left (\tan ^{-1}(\tan (a))+b x\right )\right )-\pi \log \left (1+e^{-2 i b x}\right )+\pi \log (\cos (b x))\right )}{\sqrt {\tan ^2(a)+1}}\right )}{2 b^2 \sqrt {\sec ^2(a) \left (\sin ^2(a)+\cos ^2(a)\right )}}+\frac {c (\log (\tan (a+b x))+\log (\cos (a+b x)))}{b}+\frac {1}{2} d x^2 \cot (a) \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(c + d*x)*Cot[a + b*x],x]

[Out]

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

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fricas [B]  time = 0.50, size = 250, normalized size = 3.85 \[ \frac {-i \, d {\rm Li}_2\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) + i \, d {\rm Li}_2\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + i \, d {\rm Li}_2\left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right )\right ) - i \, d {\rm Li}_2\left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right )\right ) + {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right ) + {\left (b c - a d\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) + \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) + {\left (b c - a d\right )} \log \left (-\frac {1}{2} \, \cos \left (b x + a\right ) - \frac {1}{2} i \, \sin \left (b x + a\right ) + \frac {1}{2}\right ) + {\left (b d x + a d\right )} \log \left (-\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ) + 1\right ) + {\left (b d x + a d\right )} \log \left (-\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ) + 1\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-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))/b^2

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 0.07, size = 215, normalized size = 3.31 \[ -\frac {i d \,x^{2}}{2}+i c x +\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b}+\frac {c \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right )}{b}-\frac {2 c \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b}-\frac {2 i d a x}{b}-\frac {i d \,a^{2}}{b^{2}}+\frac {d \ln \left ({\mathrm e}^{i \left (b x +a \right )}+1\right ) x}{b}-\frac {i d \polylog \left (2, -{\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}+\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) x}{b}+\frac {d \ln \left (1-{\mathrm e}^{i \left (b x +a \right )}\right ) a}{b^{2}}-\frac {i d \polylog \left (2, {\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}}-\frac {d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}-1\right )}{b^{2}}+\frac {2 d a \ln \left ({\mathrm e}^{i \left (b x +a \right )}\right )}{b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*I*d*x^2+I*c*x+1/b*c*ln(exp(I*(b*x+a))-1)+1/b*c*ln(exp(I*(b*x+a))+1)-2/b*c*ln(exp(I*(b*x+a)))-2*I/b*d*a*x-
I/b^2*d*a^2+1/b*d*ln(exp(I*(b*x+a))+1)*x-I*d*polylog(2,-exp(I*(b*x+a)))/b^2+1/b*d*ln(1-exp(I*(b*x+a)))*x+1/b^2
*d*ln(1-exp(I*(b*x+a)))*a-I*d*polylog(2,exp(I*(b*x+a)))/b^2-1/b^2*d*a*ln(exp(I*(b*x+a))-1)+2/b^2*d*a*ln(exp(I*
(b*x+a)))

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maxima [B]  time = 0.46, size = 189, normalized size = 2.91 \[ \frac {-i \, b^{2} d x^{2} - 2 i \, b^{2} c x - 2 i \, b d x \arctan \left (\sin \left (b x + a\right ), -\cos \left (b x + a\right ) + 1\right ) + 2 i \, b c \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) - 1\right ) + {\left (2 i \, b d x + 2 i \, b c\right )} \arctan \left (\sin \left (b x + a\right ), \cos \left (b x + a\right ) + 1\right ) - 2 i \, d {\rm Li}_2\left (-e^{\left (i \, b x + i \, a\right )}\right ) - 2 i \, d {\rm Li}_2\left (e^{\left (i \, b x + i \, a\right )}\right ) + {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} + 2 \, \cos \left (b x + a\right ) + 1\right ) + {\left (b d x + b c\right )} \log \left (\cos \left (b x + a\right )^{2} + \sin \left (b x + a\right )^{2} - 2 \, \cos \left (b x + a\right ) + 1\right )}{2 \, b^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(-I*b^2*d*x^2 - 2*I*b^2*c*x - 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*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))/b^2

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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {\cos \left (a+b\,x\right )\,\left (c+d\,x\right )}{\sin \left (a+b\,x\right )} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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