3.1.96 \(\int (a+b x) \coth ^{-1}(a+b x) \, dx\) [96]

Optimal. Leaf size=39 \[ \frac {x}{2}+\frac {(a+b x)^2 \coth ^{-1}(a+b x)}{2 b}-\frac {\tanh ^{-1}(a+b x)}{2 b} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {6243, 6038, 327, 212} \begin {gather*} -\frac {\tanh ^{-1}(a+b x)}{2 b}+\frac {(a+b x)^2 \coth ^{-1}(a+b x)}{2 b}+\frac {x}{2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)*ArcCoth[a + b*x],x]

[Out]

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

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 6038

Int[((a_.) + ArcCoth[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcCoth[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcCoth[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 6243

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

Rubi steps

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

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Mathematica [A]
time = 0.03, size = 66, normalized size = 1.69 \begin {gather*} \frac {2 b x+2 b x (2 a+b x) \coth ^{-1}(a+b x)-\left (-1+a^2\right ) \log (1-a-b x)-\log (1+a+b x)+a^2 \log (1+a+b x)}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)*ArcCoth[a + b*x],x]

[Out]

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

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Maple [A]
time = 0.05, size = 46, normalized size = 1.18

method	result	size
derivativedivides	\(\frac {\frac {\left (b x +a \right )^{2} \mathrm {arccoth}\left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}+\frac {\ln \left (b x +a -1\right )}{4}-\frac {\ln \left (b x +a +1\right )}{4}}{b}\)	\(46\)
default	\(\frac {\frac {\left (b x +a \right )^{2} \mathrm {arccoth}\left (b x +a \right )}{2}+\frac {b x}{2}+\frac {a}{2}+\frac {\ln \left (b x +a -1\right )}{4}-\frac {\ln \left (b x +a +1\right )}{4}}{b}\)	\(46\)
risch	\(\left (\frac {1}{4} b \,x^{2}+\frac {1}{2} a x \right ) \ln \left (b x +a +1\right )-\frac {b \,x^{2} \ln \left (b x +a -1\right )}{4}-\frac {a x \ln \left (b x +a -1\right )}{2}-\frac {\ln \left (b x +a -1\right ) a^{2}}{4 b}+\frac {\ln \left (-b x -a -1\right ) a^{2}}{4 b}+\frac {x}{2}+\frac {\ln \left (b x +a -1\right )}{4 b}-\frac {\ln \left (-b x -a -1\right )}{4 b}\)	\(108\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.26, size = 62, normalized size = 1.59 \begin {gather*} \frac {1}{4} \, b {\left (\frac {2 \, x}{b} + \frac {{\left (a^{2} - 1\right )} \log \left (b x + a + 1\right )}{b^{2}} - \frac {{\left (a^{2} - 1\right )} \log \left (b x + a - 1\right )}{b^{2}}\right )} + \frac {1}{2} \, {\left (b x^{2} + 2 \, a x\right )} \operatorname {arcoth}\left (b x + a\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.36, size = 44, normalized size = 1.13 \begin {gather*} \frac {2 \, b x + {\left (b^{2} x^{2} + 2 \, a b x + a^{2} - 1\right )} \log \left (\frac {b x + a + 1}{b x + a - 1}\right )}{4 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.25, size = 56, normalized size = 1.44 \begin {gather*} \begin {cases} \frac {a^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 b} + a x \operatorname {acoth}{\left (a + b x \right )} + \frac {b x^{2} \operatorname {acoth}{\left (a + b x \right )}}{2} + \frac {x}{2} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{2 b} & \text {for}\: b \neq 0 \\a x \operatorname {acoth}{\left (a \right )} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)*acoth(b*x+a),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 188 vs. \(2 (33) = 66\).
time = 0.40, size = 188, normalized size = 4.82 \begin {gather*} \frac {1}{2} \, {\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {1}{b^{2} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}} + \frac {{\left (b x + a + 1\right )} \log \left (-\frac {\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} + 1}{\frac {1}{a - \frac {{\left (\frac {{\left (b x + a + 1\right )} {\left (a - 1\right )}}{b x + a - 1} - a - 1\right )} b}{\frac {{\left (b x + a + 1\right )} b}{b x + a - 1} - b}} - 1}\right )}{{\left (b x + a - 1\right )} b^{2} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}^{2}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.02, size = 50, normalized size = 1.28 \begin {gather*} \frac {x}{2}-\frac {\frac {\mathrm {acoth}\left (a+b\,x\right )}{2}-\frac {a^2\,\mathrm {acoth}\left (a+b\,x\right )}{2}}{b}+a\,x\,\mathrm {acoth}\left (a+b\,x\right )+\frac {b\,x^2\,\mathrm {acoth}\left (a+b\,x\right )}{2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(acoth(a + b*x)*(a + b*x),x)

[Out]

x/2 - (acoth(a + b*x)/2 - (a^2*acoth(a + b*x))/2)/b + a*x*acoth(a + b*x) + (b*x^2*acoth(a + b*x))/2

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