3.1.28 \(\int \frac {a+b \text {ArcTan}(c+d x)}{e+f x} \, dx\) [28]

Optimal. Leaf size=162 \[ -\frac {(a+b \text {ArcTan}(c+d x)) \log \left (\frac {2}{1-i (c+d x)}\right )}{f}+\frac {(a+b \text {ArcTan}(c+d x)) \log \left (\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{f}+\frac {i b \text {PolyLog}\left (2,1-\frac {2}{1-i (c+d x)}\right )}{2 f}-\frac {i b \text {PolyLog}\left (2,1-\frac {2 d (e+f x)}{(d e+i f-c f) (1-i (c+d x))}\right )}{2 f} \]

[Out]

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

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Rubi [A]
time = 0.12, antiderivative size = 162, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.278, Rules used = {5155, 4966, 2449, 2352, 2497} \begin {gather*} \frac {(a+b \text {ArcTan}(c+d x)) \log \left (\frac {2 d (e+f x)}{(1-i (c+d x)) (-c f+d e+i f)}\right )}{f}-\frac {\log \left (\frac {2}{1-i (c+d x)}\right ) (a+b \text {ArcTan}(c+d x))}{f}-\frac {i b \text {Li}_2\left (1-\frac {2 d (e+f x)}{(d e-c f+i f) (1-i (c+d x))}\right )}{2 f}+\frac {i b \text {Li}_2\left (1-\frac {2}{1-i (c+d x)}\right )}{2 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTan[c + d*x])/(e + f*x),x]

[Out]

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

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2497

Int[Log[u_]*(Pq_)^(m_.), x_Symbol] :> With[{C = FullSimplify[Pq^m*((1 - u)/D[u, x])]}, Simp[C*PolyLog[2, 1 - u
], x] /; FreeQ[C, x]] /; IntegerQ[m] && PolyQ[Pq, x] && RationalFunctionQ[u, x] && LeQ[RationalFunctionExponen
ts[u, x][[2]], Expon[Pq, x]]

Rule 4966

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x]))*(Log[2/(1
- I*c*x)]/e), x] + (Dist[b*(c/e), Int[Log[2/(1 - I*c*x)]/(1 + c^2*x^2), x], x] - Dist[b*(c/e), Int[Log[2*c*((
d + e*x)/((c*d + I*e)*(1 - I*c*x)))]/(1 + c^2*x^2), x], x] + Simp[(a + b*ArcTan[c*x])*(Log[2*c*((d + e*x)/((c*
d + I*e)*(1 - I*c*x)))]/e), x]) /; FreeQ[{a, b, c, d, e}, x] && NeQ[c^2*d^2 + e^2, 0]

Rule 5155

Int[((a_.) + ArcTan[(c_) + (d_.)*(x_)]*(b_.))^(p_.)*((e_.) + (f_.)*(x_))^(m_.), x_Symbol] :> Dist[1/d, Subst[I
nt[((d*e - c*f)/d + f*(x/d))^m*(a + b*ArcTan[x])^p, x], x, c + d*x], x] /; FreeQ[{a, b, c, d, e, f, m, p}, x]
&& IGtQ[p, 0]

Rubi steps

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

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Mathematica [A]
time = 0.18, size = 315, normalized size = 1.94 \begin {gather*} \frac {a \log (e+f x)+b \text {ArcTan}(c+d x) \left (-\log \left (\frac {1}{\sqrt {1+(c+d x)^2}}\right )+\log \left (\sin \left (\text {ArcTan}\left (\frac {d e-c f}{f}\right )+\text {ArcTan}(c+d x)\right )\right )\right )+\frac {1}{2} b \left (-\frac {1}{4} i (\pi -2 \text {ArcTan}(c+d x))^2-i \left (\text {ArcTan}\left (\frac {d e-c f}{f}\right )+\text {ArcTan}(c+d x)\right )^2+(\pi -2 \text {ArcTan}(c+d x)) \log \left (1+e^{-2 i \text {ArcTan}(c+d x)}\right )+2 \left (\text {ArcTan}\left (\frac {d e-c f}{f}\right )+\text {ArcTan}(c+d x)\right ) \log \left (1-e^{2 i \left (\text {ArcTan}\left (\frac {d e-c f}{f}\right )+\text {ArcTan}(c+d x)\right )}\right )-(\pi -2 \text {ArcTan}(c+d x)) \log \left (\frac {2}{\sqrt {1+(c+d x)^2}}\right )-2 \left (\text {ArcTan}\left (\frac {d e-c f}{f}\right )+\text {ArcTan}(c+d x)\right ) \log \left (2 \sin \left (\text {ArcTan}\left (\frac {d e-c f}{f}\right )+\text {ArcTan}(c+d x)\right )\right )-i \text {PolyLog}\left (2,-e^{-2 i \text {ArcTan}(c+d x)}\right )-i \text {PolyLog}\left (2,e^{2 i \left (\text {ArcTan}\left (\frac {d e-c f}{f}\right )+\text {ArcTan}(c+d x)\right )}\right )\right )}{f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Log[e + f*x] + b*ArcTan[c + d*x]*(-Log[1/Sqrt[1 + (c + d*x)^2]] + Log[Sin[ArcTan[(d*e - c*f)/f] + ArcTan[c
+ d*x]]]) + (b*((-1/4*I)*(Pi - 2*ArcTan[c + d*x])^2 - I*(ArcTan[(d*e - c*f)/f] + ArcTan[c + d*x])^2 + (Pi - 2*
ArcTan[c + d*x])*Log[1 + E^((-2*I)*ArcTan[c + d*x])] + 2*(ArcTan[(d*e - c*f)/f] + ArcTan[c + d*x])*Log[1 - E^(
(2*I)*(ArcTan[(d*e - c*f)/f] + ArcTan[c + d*x]))] - (Pi - 2*ArcTan[c + d*x])*Log[2/Sqrt[1 + (c + d*x)^2]] - 2*
(ArcTan[(d*e - c*f)/f] + ArcTan[c + d*x])*Log[2*Sin[ArcTan[(d*e - c*f)/f] + ArcTan[c + d*x]]] - I*PolyLog[2, -
E^((-2*I)*ArcTan[c + d*x])] - I*PolyLog[2, E^((2*I)*(ArcTan[(d*e - c*f)/f] + ArcTan[c + d*x]))]))/2)/f

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Maple [A]
time = 0.10, size = 238, normalized size = 1.47

method	result	size
derivativedivides	\(\frac {\frac {a d \ln \left (c f -d e -f \left (d x +c \right )\right )}{f}+\frac {b d \ln \left (c f -d e -f \left (d x +c \right )\right ) \arctan \left (d x +c \right )}{f}-\frac {i b d \ln \left (c f -d e -f \left (d x +c \right )\right ) \ln \left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )}{2 f}+\frac {i b d \ln \left (c f -d e -f \left (d x +c \right )\right ) \ln \left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )}{2 f}-\frac {i b d \dilog \left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )}{2 f}+\frac {i b d \dilog \left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )}{2 f}}{d}\)	\(238\)
default	\(\frac {\frac {a d \ln \left (c f -d e -f \left (d x +c \right )\right )}{f}+\frac {b d \ln \left (c f -d e -f \left (d x +c \right )\right ) \arctan \left (d x +c \right )}{f}-\frac {i b d \ln \left (c f -d e -f \left (d x +c \right )\right ) \ln \left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )}{2 f}+\frac {i b d \ln \left (c f -d e -f \left (d x +c \right )\right ) \ln \left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )}{2 f}-\frac {i b d \dilog \left (\frac {i f +f \left (d x +c \right )}{c f -d e +i f}\right )}{2 f}+\frac {i b d \dilog \left (\frac {i f -f \left (d x +c \right )}{-c f +d e +i f}\right )}{2 f}}{d}\)	\(238\)
risch	\(\frac {a \ln \left (i c f -i d e +\left (-i d x -i c +1\right ) f -f \right )}{f}+\frac {i b \dilog \left (\frac {i c f -i d e +\left (-i d x -i c +1\right ) f -f}{i c f -i d e -f}\right )}{2 f}+\frac {i b \ln \left (-i d x -i c +1\right ) \ln \left (\frac {i c f -i d e +\left (-i d x -i c +1\right ) f -f}{i c f -i d e -f}\right )}{2 f}-\frac {i b \dilog \left (\frac {-i c f +i d e +\left (i d x +i c +1\right ) f -f}{-i c f +i d e -f}\right )}{2 f}-\frac {i b \ln \left (i d x +i c +1\right ) \ln \left (\frac {-i c f +i d e +\left (i d x +i c +1\right ) f -f}{-i c f +i d e -f}\right )}{2 f}\)	\(267\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctan(d*x+c))/(f*x+e),x,method=_RETURNVERBOSE)

[Out]

1/d*(a*d*ln(c*f-d*e-f*(d*x+c))/f+b*d*ln(c*f-d*e-f*(d*x+c))/f*arctan(d*x+c)-1/2*I*b*d*ln(c*f-d*e-f*(d*x+c))/f*l
n((I*f+f*(d*x+c))/(c*f-d*e+I*f))+1/2*I*b*d*ln(c*f-d*e-f*(d*x+c))/f*ln((I*f-f*(d*x+c))/(d*e+I*f-c*f))-1/2*I*b*d
/f*dilog((I*f+f*(d*x+c))/(c*f-d*e+I*f))+1/2*I*b*d/f*dilog((I*f-f*(d*x+c))/(d*e+I*f-c*f)))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2*b*integrate(1/2*arctan(d*x + c)/(f*x + e), x) + a*log(f*x + e)/f

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*arctan(d*x + c) + a)/(f*x + e), x)

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {a+b\,\mathrm {atan}\left (c+d\,x\right )}{e+f\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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