3.1.6 \(\int \frac {(d+c d x) (a+b \tanh ^{-1}(c x))}{x^2} \, dx\) [6]

Optimal. Leaf size=70 \[ -\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)+b c d \log (x)-\frac {1}{2} b c d \log \left (1-c^2 x^2\right )-\frac {1}{2} b c d \text {PolyLog}(2,-c x)+\frac {1}{2} b c d \text {PolyLog}(2,c x) \]

[Out]

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

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Rubi [A]
time = 0.07, antiderivative size = 70, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.389, Rules used = {6087, 6037, 272, 36, 29, 31, 6031} \begin {gather*} -\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)-\frac {1}{2} b c d \log \left (1-c^2 x^2\right )-\frac {1}{2} b c d \text {Li}_2(-c x)+\frac {1}{2} b c d \text {Li}_2(c x)+b c d \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((d + c*d*x)*(a + b*ArcTanh[c*x]))/x^2,x]

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

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

Rule 36

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

Rule 272

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 6031

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

Rule 6037

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

Rule 6087

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Int[E
xpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x)^q, x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && IGtQ[
p, 0] && IntegerQ[q] && (GtQ[q, 0] || NeQ[a, 0] || IntegerQ[m])

Rubi steps

\begin {align*} \int \frac {(d+c d x) \left (a+b \tanh ^{-1}(c x)\right )}{x^2} \, dx &=\int \left (\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x^2}+\frac {c d \left (a+b \tanh ^{-1}(c x)\right )}{x}\right ) \, dx\\ &=d \int \frac {a+b \tanh ^{-1}(c x)}{x^2} \, dx+(c d) \int \frac {a+b \tanh ^{-1}(c x)}{x} \, dx\\ &=-\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)-\frac {1}{2} b c d \text {Li}_2(-c x)+\frac {1}{2} b c d \text {Li}_2(c x)+(b c d) \int \frac {1}{x \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)-\frac {1}{2} b c d \text {Li}_2(-c x)+\frac {1}{2} b c d \text {Li}_2(c x)+\frac {1}{2} (b c d) \text {Subst}\left (\int \frac {1}{x \left (1-c^2 x\right )} \, dx,x,x^2\right )\\ &=-\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)-\frac {1}{2} b c d \text {Li}_2(-c x)+\frac {1}{2} b c d \text {Li}_2(c x)+\frac {1}{2} (b c d) \text {Subst}\left (\int \frac {1}{x} \, dx,x,x^2\right )+\frac {1}{2} \left (b c^3 d\right ) \text {Subst}\left (\int \frac {1}{1-c^2 x} \, dx,x,x^2\right )\\ &=-\frac {d \left (a+b \tanh ^{-1}(c x)\right )}{x}+a c d \log (x)+b c d \log (x)-\frac {1}{2} b c d \log \left (1-c^2 x^2\right )-\frac {1}{2} b c d \text {Li}_2(-c x)+\frac {1}{2} b c d \text {Li}_2(c x)\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 71, normalized size = 1.01 \begin {gather*} -\frac {a d}{x}+a c d \log (x)+b c d \left (-\frac {\tanh ^{-1}(c x)}{c x}+\log (c x)-\frac {1}{2} \log \left (1-c^2 x^2\right )\right )+\frac {1}{2} b c d (-\text {PolyLog}(2,-c x)+\text {PolyLog}(2,c x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((d + c*d*x)*(a + b*ArcTanh[c*x]))/x^2,x]

[Out]

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

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Maple [A]
time = 0.16, size = 105, normalized size = 1.50

method	result	size
derivativedivides	\(c \left (-\frac {d a}{c x}+d a \ln \left (c x \right )-\frac {d b \arctanh \left (c x \right )}{c x}+d b \arctanh \left (c x \right ) \ln \left (c x \right )+d b \ln \left (c x \right )-\frac {d b \ln \left (c x +1\right )}{2}-\frac {d b \ln \left (c x -1\right )}{2}-\frac {d b \dilog \left (c x \right )}{2}-\frac {d b \dilog \left (c x +1\right )}{2}-\frac {d b \ln \left (c x \right ) \ln \left (c x +1\right )}{2}\right )\)	\(105\)
default	\(c \left (-\frac {d a}{c x}+d a \ln \left (c x \right )-\frac {d b \arctanh \left (c x \right )}{c x}+d b \arctanh \left (c x \right ) \ln \left (c x \right )+d b \ln \left (c x \right )-\frac {d b \ln \left (c x +1\right )}{2}-\frac {d b \ln \left (c x -1\right )}{2}-\frac {d b \dilog \left (c x \right )}{2}-\frac {d b \dilog \left (c x +1\right )}{2}-\frac {d b \ln \left (c x \right ) \ln \left (c x +1\right )}{2}\right )\)	\(105\)
risch	\(\frac {d c b \ln \left (-c x \right )}{2}-\frac {\ln \left (-c x +1\right ) b c d}{2}+\frac {d b \ln \left (-c x +1\right )}{2 x}+\frac {d c \dilog \left (-c x +1\right ) b}{2}-\frac {d a}{x}+d c a \ln \left (-c x \right )+\frac {b c d \ln \left (c x \right )}{2}-\frac {\ln \left (c x +1\right ) b c d}{2}-\frac {b d \ln \left (c x +1\right )}{2 x}-\frac {b c d \dilog \left (c x +1\right )}{2}\)	\(110\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*d*x+d)*(a+b*arctanh(c*x))/x^2,x,method=_RETURNVERBOSE)

[Out]

c*(-d*a/c/x+d*a*ln(c*x)-d*b*arctanh(c*x)/c/x+d*b*arctanh(c*x)*ln(c*x)+d*b*ln(c*x)-1/2*d*b*ln(c*x+1)-1/2*d*b*ln
(c*x-1)-1/2*d*b*dilog(c*x)-1/2*d*b*dilog(c*x+1)-1/2*d*b*ln(c*x)*ln(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((c*d*x+d)*(a+b*arctanh(c*x))/x^2,x, algorithm="maxima")

[Out]

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

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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((c*d*x+d)*(a+b*arctanh(c*x))/x^2,x, algorithm="fricas")

[Out]

integral((a*c*d*x + a*d + (b*c*d*x + b*d)*arctanh(c*x))/x^2, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} d \left (\int \frac {a}{x^{2}}\, dx + \int \frac {a c}{x}\, dx + \int \frac {b \operatorname {atanh}{\left (c x \right )}}{x^{2}}\, dx + \int \frac {b c \operatorname {atanh}{\left (c x \right )}}{x}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*d*x+d)*(a+b*atanh(c*x))/x**2,x)

[Out]

d*(Integral(a/x**2, x) + Integral(a*c/x, x) + Integral(b*atanh(c*x)/x**2, x) + Integral(b*c*atanh(c*x)/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((c*d*x+d)*(a+b*arctanh(c*x))/x^2,x, algorithm="giac")

[Out]

integrate((c*d*x + d)*(b*arctanh(c*x) + a)/x^2, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,\left (d+c\,d\,x\right )}{x^2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atanh(c*x))*(d + c*d*x))/x^2,x)

[Out]

int(((a + b*atanh(c*x))*(d + c*d*x))/x^2, x)

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