3.1.4 \(\int \frac {a+b \text {ArcTan}(c+d x)}{c e+d e x} \, dx\) [4]

Optimal. Leaf size=63 \[ \frac {a \log (c+d x)}{d e}+\frac {i b \text {PolyLog}(2,-i (c+d x))}{2 d e}-\frac {i b \text {PolyLog}(2,i (c+d x))}{2 d e} \]

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {5151, 12, 4940, 2438} \begin {gather*} \frac {a \log (c+d x)}{d e}+\frac {i b \text {Li}_2(-i (c+d x))}{2 d e}-\frac {i b \text {Li}_2(i (c+d x))}{2 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2438

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 4940

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

Rule 5151

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

Rubi steps

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

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(189\) vs. \(2(63)=126\).
time = 0.07, size = 189, normalized size = 3.00 \begin {gather*} -\frac {i b \pi ^2-4 i b \pi \text {ArcTan}(c+d x)+8 i b \text {ArcTan}(c+d x)^2+b \pi \log (16)-4 b \pi \log \left (1+e^{-2 i \text {ArcTan}(c+d x)}\right )+8 b \text {ArcTan}(c+d x) \log \left (1+e^{-2 i \text {ArcTan}(c+d x)}\right )-8 b \text {ArcTan}(c+d x) \log \left (1-e^{2 i \text {ArcTan}(c+d x)}\right )-8 a \log (c+d x)-2 b \pi \log \left (1+c^2+2 c d x+d^2 x^2\right )+4 i b \text {PolyLog}\left (2,-e^{-2 i \text {ArcTan}(c+d x)}\right )+4 i b \text {PolyLog}\left (2,e^{2 i \text {ArcTan}(c+d x)}\right )}{8 d e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/8*(I*b*Pi^2 - (4*I)*b*Pi*ArcTan[c + d*x] + (8*I)*b*ArcTan[c + d*x]^2 + b*Pi*Log[16] - 4*b*Pi*Log[1 + E^((-2
*I)*ArcTan[c + d*x])] + 8*b*ArcTan[c + d*x]*Log[1 + E^((-2*I)*ArcTan[c + d*x])] - 8*b*ArcTan[c + d*x]*Log[1 -
E^((2*I)*ArcTan[c + d*x])] - 8*a*Log[c + d*x] - 2*b*Pi*Log[1 + c^2 + 2*c*d*x + d^2*x^2] + (4*I)*b*PolyLog[2, -
E^((-2*I)*ArcTan[c + d*x])] + (4*I)*b*PolyLog[2, E^((2*I)*ArcTan[c + d*x])])/(d*e)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 117 vs. \(2 (55 ) = 110\).
time = 0.08, size = 118, normalized size = 1.87

method	result	size
risch	\(-\frac {i b \dilog \left (-i d x -i c +1\right )}{2 e d}+\frac {a \ln \left (-i d x -i c \right )}{e d}+\frac {i b \dilog \left (i d x +i c +1\right )}{2 e d}\)	\(65\)
derivativedivides	\(\frac {\frac {a \ln \left (d x +c \right )}{e}+\frac {b \ln \left (d x +c \right ) \arctan \left (d x +c \right )}{e}+\frac {i b \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )}{2 e}-\frac {i b \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )}{2 e}+\frac {i b \dilog \left (1+i \left (d x +c \right )\right )}{2 e}-\frac {i b \dilog \left (1-i \left (d x +c \right )\right )}{2 e}}{d}\)	\(118\)
default	\(\frac {\frac {a \ln \left (d x +c \right )}{e}+\frac {b \ln \left (d x +c \right ) \arctan \left (d x +c \right )}{e}+\frac {i b \ln \left (d x +c \right ) \ln \left (1+i \left (d x +c \right )\right )}{2 e}-\frac {i b \ln \left (d x +c \right ) \ln \left (1-i \left (d x +c \right )\right )}{2 e}+\frac {i b \dilog \left (1+i \left (d x +c \right )\right )}{2 e}-\frac {i b \dilog \left (1-i \left (d x +c \right )\right )}{2 e}}{d}\)	\(118\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))/(d*e*x+c*e),x, algorithm="maxima")

[Out]

2*b*integrate(1/2*arctan(d*x + c)/(d*x*e + c*e), x) + a*e^(-1)*log(d*x*e + c*e)/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((a+b*arctan(d*x+c))/(d*e*x+c*e),x, algorithm="fricas")

[Out]

integral((b*arctan(d*x + c) + a)*e^(-1)/(d*x + c), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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))/(d*e*x+c*e),x, algorithm="giac")

[Out]

sage0*x

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {a+b\,\mathrm {atan}\left (c+d\,x\right )}{c\,e+d\,e\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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