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

Optimal. Leaf size=37 \[ \frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {b \tanh ^{-1}(c x)}{2 c^2}+\frac {b x}{2 c} \]

[Out]

1/2*b*x/c-1/2*b*arctanh(c*x)/c^2+1/2*x^2*(a+b*arctanh(c*x))

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Rubi [A]  time = 0.02, antiderivative size = 37, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {5916, 321, 206} \[ \frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {b \tanh ^{-1}(c x)}{2 c^2}+\frac {b x}{2 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTanh[c*x]),x]

[Out]

(b*x)/(2*c) - (b*ArcTanh[c*x])/(2*c^2) + (x^2*(a + b*ArcTanh[c*x]))/2

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 321

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

Rule 5916

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

Rubi steps

\begin {align*} \int x \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{2} (b c) \int \frac {x^2}{1-c^2 x^2} \, dx\\ &=\frac {b x}{2 c}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )-\frac {b \int \frac {1}{1-c^2 x^2} \, dx}{2 c}\\ &=\frac {b x}{2 c}-\frac {b \tanh ^{-1}(c x)}{2 c^2}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 59, normalized size = 1.59 \[ \frac {a x^2}{2}+\frac {b \log (1-c x)}{4 c^2}-\frac {b \log (c x+1)}{4 c^2}+\frac {1}{2} b x^2 \tanh ^{-1}(c x)+\frac {b x}{2 c} \]

Antiderivative was successfully verified.

[In]

Integrate[x*(a + b*ArcTanh[c*x]),x]

[Out]

(b*x)/(2*c) + (a*x^2)/2 + (b*x^2*ArcTanh[c*x])/2 + (b*Log[1 - c*x])/(4*c^2) - (b*Log[1 + c*x])/(4*c^2)

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fricas [A]  time = 0.58, size = 48, normalized size = 1.30 \[ \frac {2 \, a c^{2} x^{2} + 2 \, b c x + {\left (b c^{2} x^{2} - b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{4 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/4*(2*a*c^2*x^2 + 2*b*c*x + (b*c^2*x^2 - b)*log(-(c*x + 1)/(c*x - 1)))/c^2

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giac [B]  time = 0.29, size = 148, normalized size = 4.00 \[ c {\left (\frac {{\left (c x + 1\right )} b \log \left (-\frac {c x + 1}{c x - 1}\right )}{{\left (\frac {{\left (c x + 1\right )}^{2} c^{3}}{{\left (c x - 1\right )}^{2}} - \frac {2 \, {\left (c x + 1\right )} c^{3}}{c x - 1} + c^{3}\right )} {\left (c x - 1\right )}} + \frac {\frac {2 \, {\left (c x + 1\right )} a}{c x - 1} + \frac {{\left (c x + 1\right )} b}{c x - 1} - b}{\frac {{\left (c x + 1\right )}^{2} c^{3}}{{\left (c x - 1\right )}^{2}} - \frac {2 \, {\left (c x + 1\right )} c^{3}}{c x - 1} + c^{3}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c*((c*x + 1)*b*log(-(c*x + 1)/(c*x - 1))/(((c*x + 1)^2*c^3/(c*x - 1)^2 - 2*(c*x + 1)*c^3/(c*x - 1) + c^3)*(c*x
- 1)) + (2*(c*x + 1)*a/(c*x - 1) + (c*x + 1)*b/(c*x - 1) - b)/((c*x + 1)^2*c^3/(c*x - 1)^2 - 2*(c*x + 1)*c^3/
(c*x - 1) + c^3))

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maple [A]  time = 0.01, size = 49, normalized size = 1.32 \[ \frac {a \,x^{2}}{2}+\frac {b \,x^{2} \arctanh \left (c x \right )}{2}+\frac {b x}{2 c}+\frac {b \ln \left (c x -1\right )}{4 c^{2}}-\frac {b \ln \left (c x +1\right )}{4 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.30, size = 50, normalized size = 1.35 \[ \frac {1}{2} \, a x^{2} + \frac {1}{4} \, {\left (2 \, x^{2} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, x}{c^{2}} - \frac {\log \left (c x + 1\right )}{c^{3}} + \frac {\log \left (c x - 1\right )}{c^{3}}\right )}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.73, size = 35, normalized size = 0.95 \[ \frac {a\,x^2}{2}-\frac {\frac {b\,\mathrm {atanh}\left (c\,x\right )}{2}-\frac {b\,c\,x}{2}}{c^2}+\frac {b\,x^2\,\mathrm {atanh}\left (c\,x\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atanh(c*x)),x)

[Out]

(a*x^2)/2 - ((b*atanh(c*x))/2 - (b*c*x)/2)/c^2 + (b*x^2*atanh(c*x))/2

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sympy [A]  time = 0.49, size = 42, normalized size = 1.14 \[ \begin {cases} \frac {a x^{2}}{2} + \frac {b x^{2} \operatorname {atanh}{\left (c x \right )}}{2} + \frac {b x}{2 c} - \frac {b \operatorname {atanh}{\left (c x \right )}}{2 c^{2}} & \text {for}\: c \neq 0 \\\frac {a x^{2}}{2} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**2/2 + b*x**2*atanh(c*x)/2 + b*x/(2*c) - b*atanh(c*x)/(2*c**2), Ne(c, 0)), (a*x**2/2, True))

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