3.39 \(\int (c+d x) (a+b \cot (e+f x)) \, dx\)

Optimal. Leaf size=83 \[ \frac {a (c+d x)^2}{2 d}+\frac {b (c+d x) \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac {i b (c+d x)^2}{2 d}-\frac {i b d \text {Li}_2\left (e^{2 i (e+f x)}\right )}{2 f^2} \]

[Out]

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

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Rubi [A]  time = 0.12, antiderivative size = 83, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {3722, 3717, 2190, 2279, 2391} \[ -\frac {i b d \text {PolyLog}\left (2,e^{2 i (e+f x)}\right )}{2 f^2}+\frac {a (c+d x)^2}{2 d}+\frac {b (c+d x) \log \left (1-e^{2 i (e+f x)}\right )}{f}-\frac {i b (c+d x)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)*(a + b*Cot[e + f*x]),x]

[Out]

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

Rule 3722

Int[((c_.) + (d_.)*(x_))^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Int[ExpandIntegrand[
(c + d*x)^m, (a + b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[m, 0] && IGtQ[n, 0]

Rubi steps

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

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Mathematica [B]  time = 5.04, size = 204, normalized size = 2.46 \[ a c x+\frac {1}{2} a d x^2+\frac {b c (\log (\tan (e+f x))+\log (\cos (e+f x)))}{f}-\frac {b d \csc (e) \sec (e) \left (f^2 x^2 e^{i \tan ^{-1}(\tan (e))}+\frac {\tan (e) \left (i \text {Li}_2\left (e^{2 i \left (f x+\tan ^{-1}(\tan (e))\right )}\right )+i f x \left (2 \tan ^{-1}(\tan (e))-\pi \right )-2 \left (\tan ^{-1}(\tan (e))+f x\right ) \log \left (1-e^{2 i \left (\tan ^{-1}(\tan (e))+f x\right )}\right )+2 \tan ^{-1}(\tan (e)) \log \left (\sin \left (\tan ^{-1}(\tan (e))+f x\right )\right )-\pi \log \left (1+e^{-2 i f x}\right )+\pi \log (\cos (f x))\right )}{\sqrt {\tan ^2(e)+1}}\right )}{2 f^2 \sqrt {\sec ^2(e) \left (\sin ^2(e)+\cos ^2(e)\right )}}+\frac {1}{2} b d x^2 \cot (e) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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fricas [B]  time = 0.52, size = 224, normalized size = 2.70 \[ \frac {2 \, a d f^{2} x^{2} + 4 \, a c f^{2} x - i \, b d {\rm Li}_2\left (\cos \left (2 \, f x + 2 \, e\right ) + i \, \sin \left (2 \, f x + 2 \, e\right )\right ) + i \, b d {\rm Li}_2\left (\cos \left (2 \, f x + 2 \, e\right ) - i \, \sin \left (2 \, f x + 2 \, e\right )\right ) - 2 \, {\left (b d e - b c f\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, f x + 2 \, e\right ) + \frac {1}{2} i \, \sin \left (2 \, f x + 2 \, e\right ) + \frac {1}{2}\right ) - 2 \, {\left (b d e - b c f\right )} \log \left (-\frac {1}{2} \, \cos \left (2 \, f x + 2 \, e\right ) - \frac {1}{2} i \, \sin \left (2 \, f x + 2 \, e\right ) + \frac {1}{2}\right ) + 2 \, {\left (b d f x + b d e\right )} \log \left (-\cos \left (2 \, f x + 2 \, e\right ) + i \, \sin \left (2 \, f x + 2 \, e\right ) + 1\right ) + 2 \, {\left (b d f x + b d e\right )} \log \left (-\cos \left (2 \, f x + 2 \, e\right ) - i \, \sin \left (2 \, f x + 2 \, e\right ) + 1\right )}{4 \, f^{2}} \]

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

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

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)

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maple [B]  time = 1.11, size = 240, normalized size = 2.89 \[ i b c x -\frac {i b d \polylog \left (2, {\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {a d \,x^{2}}{2}+c a x -\frac {2 b c \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f}+\frac {b c \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{f}+\frac {b c \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right )}{f}-\frac {i b d \,x^{2}}{2}-\frac {i b d \,e^{2}}{f^{2}}-\frac {i b d \polylog \left (2, -{\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}+\frac {b d \ln \left (1-{\mathrm e}^{i \left (f x +e \right )}\right ) x}{f}+\frac {b d \ln \left (1-{\mathrm e}^{i \left (f x +e \right )}\right ) e}{f^{2}}-\frac {2 i b d e x}{f}+\frac {b d \ln \left ({\mathrm e}^{i \left (f x +e \right )}+1\right ) x}{f}+\frac {2 b d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}\right )}{f^{2}}-\frac {b d e \ln \left ({\mathrm e}^{i \left (f x +e \right )}-1\right )}{f^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [B]  time = 0.93, size = 208, normalized size = 2.51 \[ \frac {{\left (a - i \, b\right )} d f^{2} x^{2} + 2 \, {\left (a - i \, b\right )} c f^{2} x - 2 i \, b d f x \arctan \left (\sin \left (f x + e\right ), -\cos \left (f x + e\right ) + 1\right ) + 2 i \, b c f \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) - 1\right ) - 2 i \, b d {\rm Li}_2\left (-e^{\left (i \, f x + i \, e\right )}\right ) - 2 i \, b d {\rm Li}_2\left (e^{\left (i \, f x + i \, e\right )}\right ) + {\left (2 i \, b d f x + 2 i \, b c f\right )} \arctan \left (\sin \left (f x + e\right ), \cos \left (f x + e\right ) + 1\right ) + {\left (b d f x + b c f\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} + 2 \, \cos \left (f x + e\right ) + 1\right ) + {\left (b d f x + b c f\right )} \log \left (\cos \left (f x + e\right )^{2} + \sin \left (f x + e\right )^{2} - 2 \, \cos \left (f x + e\right ) + 1\right )}{2 \, f^{2}} \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*cot(e + f*x))*(c + d*x),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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