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

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

[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 = 0.01, 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 = {6037, 327, 212} \begin {gather*} \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} \end {gather*}

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 212

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

Rule 327

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 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
]

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 \begin {gather*} \frac {b x}{2 c}+\frac {a x^2}{2}+\frac {1}{2} b x^2 \tanh ^{-1}(c x)+\frac {b \log (1-c x)}{4 c^2}-\frac {b \log (1+c x)}{4 c^2} \end {gather*}

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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Maple [A]
time = 0.01, size = 51, normalized size = 1.38

method	result	size
derivativedivides	\(\frac {\frac {a \,c^{2} x^{2}}{2}+\frac {\arctanh \left (c x \right ) b \,c^{2} x^{2}}{2}+\frac {b c x}{2}+\frac {b \ln \left (c x -1\right )}{4}-\frac {b \ln \left (c x +1\right )}{4}}{c^{2}}\)	\(51\)
default	\(\frac {\frac {a \,c^{2} x^{2}}{2}+\frac {\arctanh \left (c x \right ) b \,c^{2} x^{2}}{2}+\frac {b c x}{2}+\frac {b \ln \left (c x -1\right )}{4}-\frac {b \ln \left (c x +1\right )}{4}}{c^{2}}\)	\(51\)
risch	\(\frac {b \,x^{2} \ln \left (c x +1\right )}{4}-\frac {b \,x^{2} \ln \left (-c x +1\right )}{4}+\frac {a \,x^{2}}{2}+\frac {b x}{2 c}+\frac {b \ln \left (-c x +1\right )}{4 c^{2}}-\frac {b \ln \left (c x +1\right )}{4 c^{2}}\)	\(65\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 50, normalized size = 1.35 \begin {gather*} \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 \end {gather*}

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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Fricas [A]
time = 0.34, size = 48, normalized size = 1.30 \begin {gather*} \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}} \end {gather*}

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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Sympy [A]
time = 0.18, size = 42, normalized size = 1.14 \begin {gather*} \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} \end {gather*}

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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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 148 vs. \(2 (31) = 62\).
time = 0.41, size = 148, normalized size = 4.00 \begin {gather*} 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 )} \end {gather*}

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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Mupad [B]
time = 0.73, size = 35, normalized size = 0.95 \begin {gather*} \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} \end {gather*}

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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