3.54 \(\int \frac{c+d x}{a+b \cot (e+f x)} \, dx\)

Optimal. Leaf size=126 \[ \frac{i b d \text{PolyLog}\left (2,\frac{(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f^2 \left (a^2+b^2\right )}-\frac{b (c+d x) \log \left (1-\frac{(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{f \left (a^2+b^2\right )}+\frac{(c+d x)^2}{2 d (a-i b)} \]

[Out]

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

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Rubi [A]  time = 0.162368, antiderivative size = 126, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.222, Rules used = {3731, 2190, 2279, 2391} \[ \frac{i b d \text{PolyLog}\left (2,\frac{(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{2 f^2 \left (a^2+b^2\right )}-\frac{b (c+d x) \log \left (1-\frac{(a+i b) e^{2 i (e+f x)}}{a-i b}\right )}{f \left (a^2+b^2\right )}+\frac{(c+d x)^2}{2 d (a-i b)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)/(a + b*Cot[e + f*x]),x]

[Out]

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

Rule 3731

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

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]

Rubi steps

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

Mathematica [A]  time = 1.51226, size = 182, normalized size = 1.44 \[ \frac{x \sin (e) (2 c+d x)}{2 (a \sin (e)+b \cos (e))}+\frac{1}{2} b \left (-\frac{i d \text{PolyLog}\left (2,\frac{(a-i b) e^{-2 i (e+f x)}}{a+i b}\right )}{f^2 \left (a^2+b^2\right )}-\frac{2 (c+d x) \log \left (1+\frac{(-a+i b) e^{-2 i (e+f x)}}{a+i b}\right )}{f \left (a^2+b^2\right )}+\frac{2 i (c+d x)^2}{d (a+i b) \left (a \left (-1+e^{2 i e}\right )+i b \left (1+e^{2 i e}\right )\right )}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)/(a + b*Cot[e + f*x]),x]

[Out]

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

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Maple [B]  time = 0.404, size = 445, normalized size = 3.5 \begin{align*}{\frac{d{x}^{2}}{2\,a+2\,ib}}+{\frac{cx}{a+ib}}-2\,{\frac{bc\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{ \left ( a+ib \right ) f \left ( ib-a \right ) }}+{\frac{bc\ln \left ( a{{\rm e}^{2\,i \left ( fx+e \right ) }}+i{{\rm e}^{2\,i \left ( fx+e \right ) }}b-a+ib \right ) }{ \left ( a+ib \right ) f \left ( ib-a \right ) }}-{\frac{bdx}{ \left ( a+ib \right ) f \left ( a-ib \right ) }\ln \left ( 1-{\frac{ \left ( a+ib \right ){{\rm e}^{2\,i \left ( fx+e \right ) }}}{a-ib}} \right ) }-{\frac{bde}{ \left ( a+ib \right ){f}^{2} \left ( a-ib \right ) }\ln \left ( 1-{\frac{ \left ( a+ib \right ){{\rm e}^{2\,i \left ( fx+e \right ) }}}{a-ib}} \right ) }+{\frac{ibd{x}^{2}}{ \left ( a+ib \right ) \left ( a-ib \right ) }}+{\frac{2\,ibdex}{ \left ( a+ib \right ) f \left ( a-ib \right ) }}+{\frac{ibd{e}^{2}}{ \left ( a+ib \right ){f}^{2} \left ( a-ib \right ) }}+{\frac{{\frac{i}{2}}bd}{ \left ( a+ib \right ){f}^{2} \left ( a-ib \right ) }{\it polylog} \left ( 2,{\frac{ \left ( a+ib \right ){{\rm e}^{2\,i \left ( fx+e \right ) }}}{a-ib}} \right ) }+2\,{\frac{bde\ln \left ({{\rm e}^{i \left ( fx+e \right ) }} \right ) }{ \left ( a+ib \right ){f}^{2} \left ( ib-a \right ) }}-{\frac{bde\ln \left ( a{{\rm e}^{2\,i \left ( fx+e \right ) }}+i{{\rm e}^{2\,i \left ( fx+e \right ) }}b-a+ib \right ) }{ \left ( a+ib \right ){f}^{2} \left ( ib-a \right ) }} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/(a+b*cot(f*x+e)),x)

[Out]

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

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Maxima [B]  time = 2.3692, size = 548, normalized size = 4.35 \begin{align*} \frac{{\left (a + i \, b\right )} d f^{2} x^{2} + 2 \,{\left (a + i \, b\right )} c f^{2} x - 2 i \, b d f x \arctan \left (-\frac{2 \, a b \cos \left (2 \, f x + 2 \, e\right ) +{\left (a^{2} - b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}, \frac{2 \, a b \sin \left (2 \, f x + 2 \, e\right ) + a^{2} + b^{2} -{\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - b d f x \log \left (\frac{{\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} - 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - 2 i \, b c f \arctan \left (b \cos \left (2 \, f x + 2 \, e\right ) + a \sin \left (2 \, f x + 2 \, e\right ) + b, a \cos \left (2 \, f x + 2 \, e\right ) - b \sin \left (2 \, f x + 2 \, e\right ) - a\right ) - b c f \log \left ({\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )^{2} + 4 \, a b \sin \left (2 \, f x + 2 \, e\right ) +{\left (a^{2} + b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )^{2} + a^{2} + b^{2} - 2 \,{\left (a^{2} - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right )\right ) + i \, b d{\rm Li}_2\left (\frac{{\left (i \, a - b\right )} e^{\left (2 i \, f x + 2 i \, e\right )}}{i \, a + b}\right )}{2 \,{\left (a^{2} + b^{2}\right )} f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*((a + I*b)*d*f^2*x^2 + 2*(a + I*b)*c*f^2*x - 2*I*b*d*f*x*arctan2(-(2*a*b*cos(2*f*x + 2*e) + (a^2 - b^2)*si
n(2*f*x + 2*e))/(a^2 + b^2), (2*a*b*sin(2*f*x + 2*e) + a^2 + b^2 - (a^2 - b^2)*cos(2*f*x + 2*e))/(a^2 + b^2))
- b*d*f*x*log(((a^2 + b^2)*cos(2*f*x + 2*e)^2 + 4*a*b*sin(2*f*x + 2*e) + (a^2 + b^2)*sin(2*f*x + 2*e)^2 + a^2
+ b^2 - 2*(a^2 - b^2)*cos(2*f*x + 2*e))/(a^2 + b^2)) - 2*I*b*c*f*arctan2(b*cos(2*f*x + 2*e) + a*sin(2*f*x + 2*
e) + b, a*cos(2*f*x + 2*e) - b*sin(2*f*x + 2*e) - a) - b*c*f*log((a^2 + b^2)*cos(2*f*x + 2*e)^2 + 4*a*b*sin(2*
f*x + 2*e) + (a^2 + b^2)*sin(2*f*x + 2*e)^2 + a^2 + b^2 - 2*(a^2 - b^2)*cos(2*f*x + 2*e)) + I*b*d*dilog((I*a -
b)*e^(2*I*f*x + 2*I*e)/(I*a + b)))/((a^2 + b^2)*f^2)

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Fricas [B]  time = 2.16641, size = 1123, normalized size = 8.91 \begin{align*} \frac{2 \, a d f^{2} x^{2} + 4 \, a c f^{2} x + i \, b d{\rm Li}_2\left (-\frac{a^{2} + b^{2} -{\left (a^{2} + 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) +{\left (-i \, a^{2} + 2 \, a b + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}} + 1\right ) - i \, b d{\rm Li}_2\left (-\frac{a^{2} + b^{2} -{\left (a^{2} - 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) +{\left (i \, a^{2} + 2 \, a b - i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}} + 1\right ) + 2 \,{\left (b d e - b c f\right )} \log \left (\frac{1}{2} \, a^{2} + i \, a b - \frac{1}{2} \, b^{2} - \frac{1}{2} \,{\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + \frac{1}{2} \,{\left (i \, a^{2} + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )\right ) + 2 \,{\left (b d e - b c f\right )} \log \left (-\frac{1}{2} \, a^{2} + i \, a b + \frac{1}{2} \, b^{2} + \frac{1}{2} \,{\left (a^{2} + b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) + \frac{1}{2} \,{\left (i \, a^{2} + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )\right ) - 2 \,{\left (b d f x + b d e\right )} \log \left (\frac{a^{2} + b^{2} -{\left (a^{2} + 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) +{\left (-i \, a^{2} + 2 \, a b + i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right ) - 2 \,{\left (b d f x + b d e\right )} \log \left (\frac{a^{2} + b^{2} -{\left (a^{2} - 2 i \, a b - b^{2}\right )} \cos \left (2 \, f x + 2 \, e\right ) +{\left (i \, a^{2} + 2 \, a b - i \, b^{2}\right )} \sin \left (2 \, f x + 2 \, e\right )}{a^{2} + b^{2}}\right )}{4 \,{\left (a^{2} + b^{2}\right )} f^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*d*f^2*x^2 + 4*a*c*f^2*x + I*b*d*dilog(-(a^2 + b^2 - (a^2 + 2*I*a*b - b^2)*cos(2*f*x + 2*e) + (-I*a^2
+ 2*a*b + I*b^2)*sin(2*f*x + 2*e))/(a^2 + b^2) + 1) - I*b*d*dilog(-(a^2 + b^2 - (a^2 - 2*I*a*b - b^2)*cos(2*f*
x + 2*e) + (I*a^2 + 2*a*b - I*b^2)*sin(2*f*x + 2*e))/(a^2 + b^2) + 1) + 2*(b*d*e - b*c*f)*log(1/2*a^2 + I*a*b
- 1/2*b^2 - 1/2*(a^2 + b^2)*cos(2*f*x + 2*e) + 1/2*(I*a^2 + I*b^2)*sin(2*f*x + 2*e)) + 2*(b*d*e - b*c*f)*log(-
1/2*a^2 + I*a*b + 1/2*b^2 + 1/2*(a^2 + b^2)*cos(2*f*x + 2*e) + 1/2*(I*a^2 + I*b^2)*sin(2*f*x + 2*e)) - 2*(b*d*
f*x + b*d*e)*log((a^2 + b^2 - (a^2 + 2*I*a*b - b^2)*cos(2*f*x + 2*e) + (-I*a^2 + 2*a*b + I*b^2)*sin(2*f*x + 2*
e))/(a^2 + b^2)) - 2*(b*d*f*x + b*d*e)*log((a^2 + b^2 - (a^2 - 2*I*a*b - b^2)*cos(2*f*x + 2*e) + (I*a^2 + 2*a*
b - I*b^2)*sin(2*f*x + 2*e))/(a^2 + b^2)))/((a^2 + b^2)*f^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{c + d x}{a + b \cot{\left (e + f x \right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(a+b*cot(f*x+e)),x)

[Out]

Integral((c + d*x)/(a + b*cot(e + f*x)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{b \cot \left (f x + e\right ) + a}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)/(b*cot(f*x + e) + a), x)