3.1.5 \(\int \frac {a+b \tanh ^{-1}(c x)}{d+e x} \, dx\) [5]

Optimal. Leaf size=114 \[ -\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{1+c x}\right )}{e}+\frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{e}+\frac {b \text {PolyLog}\left (2,1-\frac {2}{1+c x}\right )}{2 e}-\frac {b \text {PolyLog}\left (2,1-\frac {2 c (d+e x)}{(c d+e) (1+c x)}\right )}{2 e} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.06, antiderivative size = 114, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6057, 2449, 2352, 2497} \begin {gather*} \frac {\left (a+b \tanh ^{-1}(c x)\right ) \log \left (\frac {2 c (d+e x)}{(c x+1) (c d+e)}\right )}{e}-\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{e}-\frac {b \text {Li}_2\left (1-\frac {2 c (d+e x)}{(c d+e) (c x+1)}\right )}{2 e}+\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 e} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/(d + e*x),x]

[Out]

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

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 6057

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

Rubi steps

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

________________________________________________________________________________________

Mathematica [C] Result contains complex when optimal does not.
time = 0.24, size = 257, normalized size = 2.25 \begin {gather*} \frac {a \log (d+e x)+b \tanh ^{-1}(c x) \left (\frac {1}{2} \log \left (1-c^2 x^2\right )+\log \left (i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )\right )-\frac {1}{2} i b \left (-\frac {1}{4} i \left (\pi -2 i \tanh ^{-1}(c x)\right )^2+i \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )^2+\left (\pi -2 i \tanh ^{-1}(c x)\right ) \log \left (1+e^{2 \tanh ^{-1}(c x)}\right )+2 i \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right ) \log \left (1-e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )-\left (\pi -2 i \tanh ^{-1}(c x)\right ) \log \left (\frac {2}{\sqrt {1-c^2 x^2}}\right )-2 i \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right ) \log \left (2 i \sinh \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )\right )-i \text {PolyLog}\left (2,-e^{2 \tanh ^{-1}(c x)}\right )-i \text {PolyLog}\left (2,e^{-2 \left (\tanh ^{-1}\left (\frac {c d}{e}\right )+\tanh ^{-1}(c x)\right )}\right )\right )}{e} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]
time = 0.75, size = 158, normalized size = 1.39

method	result	size
derivativedivides	\(\frac {\frac {a c \ln \left (c e x +d c \right )}{e}+\frac {b c \ln \left (c e x +d c \right ) \arctanh \left (c x \right )}{e}+\frac {b c \ln \left (c e x +d c \right ) \ln \left (\frac {c e x -e}{-d c -e}\right )}{2 e}+\frac {b c \dilog \left (\frac {c e x -e}{-d c -e}\right )}{2 e}-\frac {b c \ln \left (c e x +d c \right ) \ln \left (\frac {c e x +e}{-d c +e}\right )}{2 e}-\frac {b c \dilog \left (\frac {c e x +e}{-d c +e}\right )}{2 e}}{c}\)	\(158\)
default	\(\frac {\frac {a c \ln \left (c e x +d c \right )}{e}+\frac {b c \ln \left (c e x +d c \right ) \arctanh \left (c x \right )}{e}+\frac {b c \ln \left (c e x +d c \right ) \ln \left (\frac {c e x -e}{-d c -e}\right )}{2 e}+\frac {b c \dilog \left (\frac {c e x -e}{-d c -e}\right )}{2 e}-\frac {b c \ln \left (c e x +d c \right ) \ln \left (\frac {c e x +e}{-d c +e}\right )}{2 e}-\frac {b c \dilog \left (\frac {c e x +e}{-d c +e}\right )}{2 e}}{c}\)	\(158\)
risch	\(-\frac {b \dilog \left (\frac {\left (-c x +1\right ) e -d c -e}{-d c -e}\right )}{2 e}-\frac {b \ln \left (-c x +1\right ) \ln \left (\frac {\left (-c x +1\right ) e -d c -e}{-d c -e}\right )}{2 e}+\frac {a \ln \left (\left (-c x +1\right ) e -d c -e \right )}{e}+\frac {b \dilog \left (\frac {\left (c x +1\right ) e +d c -e}{d c -e}\right )}{2 e}+\frac {b \ln \left (c x +1\right ) \ln \left (\frac {\left (c x +1\right ) e +d c -e}{d c -e}\right )}{2 e}\)	\(167\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*arctanh(c*x))/(e*x+d),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

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*arctanh(c*x))/(e*x+d),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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*arctanh(c*x))/(e*x+d),x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/(d + e*x),x)

[Out]

int((a + b*atanh(c*x))/(d + e*x), x)

________________________________________________________________________________________