3.1.45 \(\int \frac {x (a+b \tanh ^{-1}(c x))}{d+c d x} \, dx\) [45]

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

[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 = 0.08, 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 = {6077, 6021, 266, 6055, 2449, 2352} \begin {gather*} \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} \end {gather*}

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 266

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 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 6021

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

Rule 6055

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 6077

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 \text {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 = 0.10, size = 75, normalized size = 0.80 \begin {gather*} \frac {2 a c x+2 b \tanh ^{-1}(c x) \left (c x+\log \left (1+e^{-2 \tanh ^{-1}(c x)}\right )\right )-2 a \log (1+c x)+b \log \left (1-c^2 x^2\right )-b \text {PolyLog}\left (2,-e^{-2 \tanh ^{-1}(c x)}\right )}{2 c^2 d} \end {gather*}

Antiderivative was successfully verified.

[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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Maple [A]
time = 0.28, size = 136, normalized size = 1.45

method	result	size
derivativedivides	\(\frac {\frac {a c x}{d}-\frac {a \ln \left (c x +1\right )}{d}-\frac {b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{d}+\frac {b \arctanh \left (c x \right ) c x}{d}+\frac {b \ln \left (c x +1\right )^{2}}{4 d}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 d}+\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 d}+\frac {b \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 d}+\frac {b \ln \left (\left (c x +1\right ) \left (c x -1\right )\right )}{2 d}}{c^{2}}\)	\(136\)
default	\(\frac {\frac {a c x}{d}-\frac {a \ln \left (c x +1\right )}{d}-\frac {b \arctanh \left (c x \right ) \ln \left (c x +1\right )}{d}+\frac {b \arctanh \left (c x \right ) c x}{d}+\frac {b \ln \left (c x +1\right )^{2}}{4 d}-\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{2 d}+\frac {b \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 d}+\frac {b \dilog \left (\frac {c x}{2}+\frac {1}{2}\right )}{2 d}+\frac {b \ln \left (\left (c x +1\right ) \left (c x -1\right )\right )}{2 d}}{c^{2}}\)	\(136\)
risch	\(-\frac {b \ln \left (c x +1\right )^{2}}{4 d \,c^{2}}+\frac {b x \ln \left (c x +1\right )}{2 d c}+\frac {b \ln \left (c x +1\right )}{2 d \,c^{2}}-\frac {b \ln \left (-c x +1\right ) x}{2 d c}-\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d \,c^{2}}+\frac {b \ln \left (\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (-c x +1\right )}{2 d \,c^{2}}+\frac {a x}{d c}-\frac {a \ln \left (-c x -1\right )}{d \,c^{2}}-\frac {b \dilog \left (-\frac {c x}{2}+\frac {1}{2}\right )}{2 d \,c^{2}}+\frac {b \ln \left (-c x +1\right )}{2 d \,c^{2}}-\frac {a}{d \,c^{2}}-\frac {b}{2 d \,c^{2}}\)	\(188\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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

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

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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