3.45 \(\int \frac {x (a+b \tanh ^{-1}(c x))}{d+c d x} \, dx\)

Optimal. Leaf size=94 \[ \frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2 d}+\frac {a x}{c d}-\frac {b \text {Li}_2\left (1-\frac {2}{c x+1}\right )}{2 c^2 d}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^2 d}+\frac {b x \tanh ^{-1}(c x)}{c d} \]

[Out]

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

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Rubi [A]  time = 0.10, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {5930, 5910, 260, 5918, 2402, 2315} \[ -\frac {b \text {PolyLog}\left (2,1-\frac {2}{c x+1}\right )}{2 c^2 d}+\frac {\log \left (\frac {2}{c x+1}\right ) \left (a+b \tanh ^{-1}(c x)\right )}{c^2 d}+\frac {a x}{c d}+\frac {b \log \left (1-c^2 x^2\right )}{2 c^2 d}+\frac {b x \tanh ^{-1}(c x)}{c d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 2315

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

Rule 2402

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 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 5918

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

Rule 5930

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)), x_Symbol] :> Dist[f/e,
Int[(f*x)^(m - 1)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(d*f)/e, Int[((f*x)^(m - 1)*(a + b*ArcTanh[c*x])^p)/(
d + e*x), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 - e^2, 0] && GtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.16, size = 75, normalized size = 0.80 \[ \frac {2 a c x-2 a \log (c x+1)+b \log \left (1-c^2 x^2\right )-b \text {Li}_2\left (-e^{-2 \tanh ^{-1}(c x)}\right )+2 b \tanh ^{-1}(c x) \left (c x+\log \left (e^{-2 \tanh ^{-1}(c x)}+1\right )\right )}{2 c^2 d} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(2*a*c*x + 2*b*ArcTanh[c*x]*(c*x + Log[1 + E^(-2*ArcTanh[c*x])]) - 2*a*Log[1 + c*x] + b*Log[1 - c^2*x^2] - b*P
olyLog[2, -E^(-2*ArcTanh[c*x])])/(2*c^2*d)

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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x \operatorname {artanh}\left (c x\right ) + a x}{c d x + d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.05, size = 157, normalized size = 1.67 \[ \frac {a x}{c d}-\frac {a \ln \left (c x +1\right )}{c^{2} d}-\frac {b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{c^{2} d}+\frac {b x \arctanh \left (c x \right )}{c d}+\frac {b \ln \left (c x +1\right )^{2}}{4 c^{2} d}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 c^{2} d}+\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c^{2} d}+\frac {b \dilog \left (\frac {1}{2}+\frac {c x}{2}\right )}{2 c^{2} d}+\frac {b \ln \left (\left (c x -1\right ) \left (c x +1\right )\right )}{2 c^{2} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a*x/c/d-1/c^2*a/d*ln(c*x+1)-1/c^2*b/d*arctanh(c*x)*ln(c*x+1)+b*x*arctanh(c*x)/c/d+1/4/c^2*b/d*ln(c*x+1)^2-1/2/
c^2*b/d*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/2/c^2*b/d*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/2/c^2*b/d*dilog(1/2+1/2*c*x)
+1/2/c^2*b/d*ln((c*x-1)*(c*x+1))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, {\left (c^{2} {\left (\frac {2 \, x}{c^{3} d} - \frac {\log \left (c x + 1\right )}{c^{4} d} + \frac {\log \left (c x - 1\right )}{c^{4} d}\right )} + 2 \, c^{2} \int \frac {x^{2} \log \left (c x + 1\right )}{c^{3} d x^{2} - c d}\,{d x} - 4 \, c \int \frac {x \log \left (c x + 1\right )}{c^{3} d x^{2} - c d}\,{d x} - \frac {2 \, {\left (c x - \log \left (c x + 1\right )\right )} \log \left (-c x + 1\right )}{c^{2} d} + \frac {\log \left (c^{3} d x^{2} - c d\right )}{c^{2} d} - 2 \, \int \frac {\log \left (c x + 1\right )}{c^{3} d x^{2} - c d}\,{d x}\right )} b + a {\left (\frac {x}{c d} - \frac {\log \left (c x + 1\right )}{c^{2} d}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(c^2*(2*x/(c^3*d) - log(c*x + 1)/(c^4*d) + log(c*x - 1)/(c^4*d)) + 2*c^2*integrate(x^2*log(c*x + 1)/(c^3*d
*x^2 - c*d), x) - 4*c*integrate(x*log(c*x + 1)/(c^3*d*x^2 - c*d), x) - 2*(c*x - log(c*x + 1))*log(-c*x + 1)/(c
^2*d) + log(c^3*d*x^2 - c*d)/(c^2*d) - 2*integrate(log(c*x + 1)/(c^3*d*x^2 - c*d), x))*b + a*(x/(c*d) - log(c*
x + 1)/(c^2*d))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {\int \frac {a x}{c x + 1}\, dx + \int \frac {b x \operatorname {atanh}{\left (c x \right )}}{c x + 1}\, dx}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(Integral(a*x/(c*x + 1), x) + Integral(b*x*atanh(c*x)/(c*x + 1), x))/d

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