3.1.54 \(\int \frac {\text {ArcTan}(a+b x)}{c+d x} \, dx\) [54]

Optimal. Leaf size=152 \[ -\frac {\text {ArcTan}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}+\frac {\text {ArcTan}(a+b x) \log \left (\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{d}+\frac {i \text {PolyLog}\left (2,1-\frac {2}{1-i (a+b x)}\right )}{2 d}-\frac {i \text {PolyLog}\left (2,1-\frac {2 b (c+d x)}{(b c+i d-a d) (1-i (a+b x))}\right )}{2 d} \]

[Out]

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

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Rubi [A]
time = 0.11, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {5155, 4966, 2449, 2352, 2497} \begin {gather*} \frac {\text {ArcTan}(a+b x) \log \left (\frac {2 b (c+d x)}{(1-i (a+b x)) (-a d+b c+i d)}\right )}{d}-\frac {\text {ArcTan}(a+b x) \log \left (\frac {2}{1-i (a+b x)}\right )}{d}-\frac {i \text {Li}_2\left (1-\frac {2 b (c+d x)}{(b c-a d+i d) (1-i (a+b x))}\right )}{2 d}+\frac {i \text {Li}_2\left (1-\frac {2}{1-i (a+b x)}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTan[a + b*x]/(c + d*x),x]

[Out]

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

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

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Mathematica [A]
time = 0.02, size = 231, normalized size = 1.52 \begin {gather*} \frac {i \log (1-i (a+b x)) \log \left (-\frac {i \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{-\frac {d}{b}-\frac {i (b c-a d)}{b}}\right )}{2 d}-\frac {i \log (1+i (a+b x)) \log \left (\frac {i \left (\frac {b c-a d}{b}+\frac {d (a+b x)}{b}\right )}{-\frac {d}{b}+\frac {i (b c-a d)}{b}}\right )}{2 d}+\frac {i \text {PolyLog}\left (2,-\frac {i d (1-i (a+b x))}{b c-i d-a d}\right )}{2 d}-\frac {i \text {PolyLog}\left (2,\frac {i d (1+i (a+b x))}{b c+i d-a d}\right )}{2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.08, size = 186, normalized size = 1.22

method	result	size
derivativedivides	\(\frac {\frac {b \ln \left (a d -b c -d \left (b x +a \right )\right ) \arctan \left (b x +a \right )}{d}+b \left (-\frac {i \ln \left (a d -b c -d \left (b x +a \right )\right ) \left (\ln \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\ln \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}-\frac {i \left (\dilog \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\dilog \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}\right )}{b}\)	\(186\)
default	\(\frac {\frac {b \ln \left (a d -b c -d \left (b x +a \right )\right ) \arctan \left (b x +a \right )}{d}+b \left (-\frac {i \ln \left (a d -b c -d \left (b x +a \right )\right ) \left (\ln \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\ln \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}-\frac {i \left (\dilog \left (\frac {i d +d \left (b x +a \right )}{a d -b c +i d}\right )-\dilog \left (\frac {i d -d \left (b x +a \right )}{-a d +b c +i d}\right )\right )}{2 d}\right )}{b}\)	\(186\)
risch	\(\frac {i \dilog \left (\frac {i a d -i c b +\left (-i b x -i a +1\right ) d -d}{i a d -i c b -d}\right )}{2 d}+\frac {i \ln \left (-i b x -i a +1\right ) \ln \left (\frac {i a d -i c b +\left (-i b x -i a +1\right ) d -d}{i a d -i c b -d}\right )}{2 d}-\frac {i \dilog \left (\frac {-i a d +i c b +\left (i b x +i a +1\right ) d -d}{-i a d +i c b -d}\right )}{2 d}-\frac {i \ln \left (i b x +i a +1\right ) \ln \left (\frac {-i a d +i c b +\left (i b x +i a +1\right ) d -d}{-i a d +i c b -d}\right )}{2 d}\)	\(230\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 284 vs. \(2 (130) = 260\).
time = 0.56, size = 284, normalized size = 1.87 \begin {gather*} \frac {\arctan \left (b x + a\right ) \log \left (d x + c\right )}{d} - \frac {\arctan \left (\frac {b^{2} x + a b}{b}\right ) \log \left (d x + c\right )}{d} - \frac {\arctan \left (\frac {b d^{2} x + b c d}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}, \frac {b^{2} c^{2} - a b c d + {\left (b^{2} c d - a b d^{2}\right )} x}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}\right ) \log \left (b^{2} x^{2} + 2 \, a b x + a^{2} + 1\right ) - \arctan \left (b x + a\right ) \log \left (\frac {b^{2} d^{2} x^{2} + 2 \, b^{2} c d x + b^{2} c^{2}}{b^{2} c^{2} - 2 \, a b c d + {\left (a^{2} + 1\right )} d^{2}}\right ) + i \, {\rm Li}_2\left (\frac {i \, b d x + {\left (i \, a + 1\right )} d}{-i \, b c + {\left (i \, a + 1\right )} d}\right ) - i \, {\rm Li}_2\left (\frac {i \, b d x + {\left (i \, a - 1\right )} d}{-i \, b c + {\left (i \, a - 1\right )} d}\right )}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\operatorname {atan}{\left (a + b x \right )}}{c + d x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(atan(a + b*x)/(c + d*x), x)

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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(arctan(b*x+a)/(d*x+c),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 {\mathrm {atan}\left (a+b\,x\right )}{c+d\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atan(a + b*x)/(c + d*x),x)

[Out]

int(atan(a + b*x)/(c + d*x), x)

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