3.1.62 \(\int x^3 \coth ^{-1}(a+b x) \, dx\) [62]

Optimal. Leaf size=101 \[ \frac {\left (1+6 a^2\right ) x}{4 b^3}-\frac {a (a+b x)^2}{2 b^4}+\frac {(a+b x)^3}{12 b^4}+\frac {1}{4} x^4 \coth ^{-1}(a+b x)+\frac {(1-a)^4 \log (1-a-b x)}{8 b^4}-\frac {(1+a)^4 \log (1+a+b x)}{8 b^4} \]

[Out]

1/4*(6*a^2+1)*x/b^3-1/2*a*(b*x+a)^2/b^4+1/12*(b*x+a)^3/b^4+1/4*x^4*arccoth(b*x+a)+1/8*(1-a)^4*ln(-b*x-a+1)/b^4
-1/8*(1+a)^4*ln(b*x+a+1)/b^4

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Rubi [A]
time = 0.09, antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 5, integrand size = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6247, 6064, 716, 647, 31} \begin {gather*} \frac {\left (6 a^2+1\right ) x}{4 b^3}+\frac {(a+b x)^3}{12 b^4}-\frac {a (a+b x)^2}{2 b^4}+\frac {(1-a)^4 \log (-a-b x+1)}{8 b^4}-\frac {(a+1)^4 \log (a+b x+1)}{8 b^4}+\frac {1}{4} x^4 \coth ^{-1}(a+b x) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*ArcCoth[a + b*x],x]

[Out]

((1 + 6*a^2)*x)/(4*b^3) - (a*(a + b*x)^2)/(2*b^4) + (a + b*x)^3/(12*b^4) + (x^4*ArcCoth[a + b*x])/4 + ((1 - a)
^4*Log[1 - a - b*x])/(8*b^4) - ((1 + a)^4*Log[1 + a + b*x])/(8*b^4)

Rule 31

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

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 716

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[PolynomialDivide[(d + e*x)^m, a + c*x^2,
x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && IGtQ[m, 1] && (NeQ[d, 0] || GtQ[m, 2])

Rule 6064

Int[((a_.) + ArcCoth[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_))^(q_.), x_Symbol] :> Simp[(d + e*x)^(q + 1)*((a + b
*ArcCoth[c*x])/(e*(q + 1))), x] - Dist[b*(c/(e*(q + 1))), Int[(d + e*x)^(q + 1)/(1 - c^2*x^2), x], x] /; FreeQ
[{a, b, c, d, e, q}, x] && NeQ[q, -1]

Rule 6247

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

Rubi steps

\begin {align*} \int x^3 \coth ^{-1}(a+b x) \, dx &=\frac {\text {Subst}\left (\int \left (-\frac {a}{b}+\frac {x}{b}\right )^3 \coth ^{-1}(x) \, dx,x,a+b x\right )}{b}\\ &=\frac {1}{4} x^4 \coth ^{-1}(a+b x)-\frac {1}{4} \text {Subst}\left (\int \frac {\left (-\frac {a}{b}+\frac {x}{b}\right )^4}{1-x^2} \, dx,x,a+b x\right )\\ &=\frac {1}{4} x^4 \coth ^{-1}(a+b x)-\frac {1}{4} \text {Subst}\left (\int \left (-\frac {1+6 a^2}{b^4}+\frac {4 a x}{b^4}-\frac {x^2}{b^4}+\frac {1+6 a^2+a^4-4 a \left (1+a^2\right ) x}{b^4 \left (1-x^2\right )}\right ) \, dx,x,a+b x\right )\\ &=\frac {\left (1+6 a^2\right ) x}{4 b^3}-\frac {a (a+b x)^2}{2 b^4}+\frac {(a+b x)^3}{12 b^4}+\frac {1}{4} x^4 \coth ^{-1}(a+b x)-\frac {\text {Subst}\left (\int \frac {1+6 a^2+a^4-4 a \left (1+a^2\right ) x}{1-x^2} \, dx,x,a+b x\right )}{4 b^4}\\ &=\frac {\left (1+6 a^2\right ) x}{4 b^3}-\frac {a (a+b x)^2}{2 b^4}+\frac {(a+b x)^3}{12 b^4}+\frac {1}{4} x^4 \coth ^{-1}(a+b x)-\frac {(1-a)^4 \text {Subst}\left (\int \frac {1}{1-x} \, dx,x,a+b x\right )}{8 b^4}+\frac {(1+a)^4 \text {Subst}\left (\int \frac {1}{-1-x} \, dx,x,a+b x\right )}{8 b^4}\\ &=\frac {\left (1+6 a^2\right ) x}{4 b^3}-\frac {a (a+b x)^2}{2 b^4}+\frac {(a+b x)^3}{12 b^4}+\frac {1}{4} x^4 \coth ^{-1}(a+b x)+\frac {(1-a)^4 \log (1-a-b x)}{8 b^4}-\frac {(1+a)^4 \log (1+a+b x)}{8 b^4}\\ \end {align*}

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Mathematica [A]
time = 0.03, size = 81, normalized size = 0.80 \begin {gather*} \frac {6 \left (1+3 a^2\right ) b x-6 a b^2 x^2+2 b^3 x^3+6 b^4 x^4 \coth ^{-1}(a+b x)+3 (-1+a)^4 \log (1-a-b x)-3 (1+a)^4 \log (1+a+b x)}{24 b^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcCoth[a + b*x],x]

[Out]

(6*(1 + 3*a^2)*b*x - 6*a*b^2*x^2 + 2*b^3*x^3 + 6*b^4*x^4*ArcCoth[a + b*x] + 3*(-1 + a)^4*Log[1 - a - b*x] - 3*
(1 + a)^4*Log[1 + a + b*x])/(24*b^4)

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Maple [A]
time = 0.06, size = 172, normalized size = 1.70

method	result	size
derivativedivides	\(\frac {\frac {\mathrm {arccoth}\left (b x +a \right ) a^{4}}{4}-\mathrm {arccoth}\left (b x +a \right ) a^{3} \left (b x +a \right )+\frac {3 \,\mathrm {arccoth}\left (b x +a \right ) a^{2} \left (b x +a \right )^{2}}{2}-\mathrm {arccoth}\left (b x +a \right ) a \left (b x +a \right )^{3}+\frac {\mathrm {arccoth}\left (b x +a \right ) \left (b x +a \right )^{4}}{4}+\frac {3 a^{2} \left (b x +a \right )}{2}-\frac {\left (b x +a \right )^{2} a}{2}+\frac {\left (b x +a \right )^{3}}{12}+\frac {b x}{4}+\frac {a}{4}+\frac {\left (a^{4}-4 a^{3}+6 a^{2}-4 a +1\right ) \ln \left (b x +a -1\right )}{8}-\frac {\left (a^{4}+4 a^{3}+6 a^{2}+4 a +1\right ) \ln \left (b x +a +1\right )}{8}}{b^{4}}\)	\(172\)
default	\(\frac {\frac {\mathrm {arccoth}\left (b x +a \right ) a^{4}}{4}-\mathrm {arccoth}\left (b x +a \right ) a^{3} \left (b x +a \right )+\frac {3 \,\mathrm {arccoth}\left (b x +a \right ) a^{2} \left (b x +a \right )^{2}}{2}-\mathrm {arccoth}\left (b x +a \right ) a \left (b x +a \right )^{3}+\frac {\mathrm {arccoth}\left (b x +a \right ) \left (b x +a \right )^{4}}{4}+\frac {3 a^{2} \left (b x +a \right )}{2}-\frac {\left (b x +a \right )^{2} a}{2}+\frac {\left (b x +a \right )^{3}}{12}+\frac {b x}{4}+\frac {a}{4}+\frac {\left (a^{4}-4 a^{3}+6 a^{2}-4 a +1\right ) \ln \left (b x +a -1\right )}{8}-\frac {\left (a^{4}+4 a^{3}+6 a^{2}+4 a +1\right ) \ln \left (b x +a +1\right )}{8}}{b^{4}}\)	\(172\)
risch	\(\frac {x^{4} \ln \left (b x +a +1\right )}{8}-\frac {x^{4} \ln \left (b x +a -1\right )}{8}+\frac {x^{3}}{12 b}+\frac {\ln \left (-b x -a +1\right ) a^{4}}{8 b^{4}}-\frac {\ln \left (b x +a +1\right ) a^{4}}{8 b^{4}}-\frac {a \,x^{2}}{4 b^{2}}-\frac {\ln \left (-b x -a +1\right ) a^{3}}{2 b^{4}}-\frac {\ln \left (b x +a +1\right ) a^{3}}{2 b^{4}}+\frac {3 a^{2} x}{4 b^{3}}+\frac {3 \ln \left (-b x -a +1\right ) a^{2}}{4 b^{4}}-\frac {3 \ln \left (b x +a +1\right ) a^{2}}{4 b^{4}}-\frac {\ln \left (-b x -a +1\right ) a}{2 b^{4}}-\frac {\ln \left (b x +a +1\right ) a}{2 b^{4}}+\frac {x}{4 b^{3}}+\frac {\ln \left (-b x -a +1\right )}{8 b^{4}}-\frac {\ln \left (b x +a +1\right )}{8 b^{4}}\)	\(213\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arccoth(b*x+a),x,method=_RETURNVERBOSE)

[Out]

1/b^4*(1/4*arccoth(b*x+a)*a^4-arccoth(b*x+a)*a^3*(b*x+a)+3/2*arccoth(b*x+a)*a^2*(b*x+a)^2-arccoth(b*x+a)*a*(b*
x+a)^3+1/4*arccoth(b*x+a)*(b*x+a)^4+3/2*a^2*(b*x+a)-1/2*(b*x+a)^2*a+1/12*(b*x+a)^3+1/4*b*x+1/4*a+1/8*(a^4-4*a^
3+6*a^2-4*a+1)*ln(b*x+a-1)-1/8*(a^4+4*a^3+6*a^2+4*a+1)*ln(b*x+a+1))

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Maxima [A]
time = 0.25, size = 106, normalized size = 1.05 \begin {gather*} \frac {1}{4} \, x^{4} \operatorname {arcoth}\left (b x + a\right ) + \frac {1}{24} \, b {\left (\frac {2 \, {\left (b^{2} x^{3} - 3 \, a b x^{2} + 3 \, {\left (3 \, a^{2} + 1\right )} x\right )}}{b^{4}} - \frac {3 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right )}{b^{5}} + \frac {3 \, {\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )}{b^{5}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccoth(b*x+a),x, algorithm="maxima")

[Out]

1/4*x^4*arccoth(b*x + a) + 1/24*b*(2*(b^2*x^3 - 3*a*b*x^2 + 3*(3*a^2 + 1)*x)/b^4 - 3*(a^4 + 4*a^3 + 6*a^2 + 4*
a + 1)*log(b*x + a + 1)/b^5 + 3*(a^4 - 4*a^3 + 6*a^2 - 4*a + 1)*log(b*x + a - 1)/b^5)

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Fricas [A]
time = 0.36, size = 112, normalized size = 1.11 \begin {gather*} \frac {3 \, b^{4} x^{4} \log \left (\frac {b x + a + 1}{b x + a - 1}\right ) + 2 \, b^{3} x^{3} - 6 \, a b^{2} x^{2} + 6 \, {\left (3 \, a^{2} + 1\right )} b x - 3 \, {\left (a^{4} + 4 \, a^{3} + 6 \, a^{2} + 4 \, a + 1\right )} \log \left (b x + a + 1\right ) + 3 \, {\left (a^{4} - 4 \, a^{3} + 6 \, a^{2} - 4 \, a + 1\right )} \log \left (b x + a - 1\right )}{24 \, b^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccoth(b*x+a),x, algorithm="fricas")

[Out]

1/24*(3*b^4*x^4*log((b*x + a + 1)/(b*x + a - 1)) + 2*b^3*x^3 - 6*a*b^2*x^2 + 6*(3*a^2 + 1)*b*x - 3*(a^4 + 4*a^
3 + 6*a^2 + 4*a + 1)*log(b*x + a + 1) + 3*(a^4 - 4*a^3 + 6*a^2 - 4*a + 1)*log(b*x + a - 1))/b^4

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Sympy [A]
time = 0.53, size = 153, normalized size = 1.51 \begin {gather*} \begin {cases} - \frac {a^{4} \operatorname {acoth}{\left (a + b x \right )}}{4 b^{4}} - \frac {a^{3} \log {\left (a + b x + 1 \right )}}{b^{4}} + \frac {a^{3} \operatorname {acoth}{\left (a + b x \right )}}{b^{4}} + \frac {3 a^{2} x}{4 b^{3}} - \frac {3 a^{2} \operatorname {acoth}{\left (a + b x \right )}}{2 b^{4}} - \frac {a x^{2}}{4 b^{2}} - \frac {a \log {\left (a + b x + 1 \right )}}{b^{4}} + \frac {a \operatorname {acoth}{\left (a + b x \right )}}{b^{4}} + \frac {x^{4} \operatorname {acoth}{\left (a + b x \right )}}{4} + \frac {x^{3}}{12 b} + \frac {x}{4 b^{3}} - \frac {\operatorname {acoth}{\left (a + b x \right )}}{4 b^{4}} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {acoth}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*acoth(b*x+a),x)

[Out]

Piecewise((-a**4*acoth(a + b*x)/(4*b**4) - a**3*log(a + b*x + 1)/b**4 + a**3*acoth(a + b*x)/b**4 + 3*a**2*x/(4
*b**3) - 3*a**2*acoth(a + b*x)/(2*b**4) - a*x**2/(4*b**2) - a*log(a + b*x + 1)/b**4 + a*acoth(a + b*x)/b**4 +
x**4*acoth(a + b*x)/4 + x**3/(12*b) + x/(4*b**3) - acoth(a + b*x)/(4*b**4), Ne(b, 0)), (x**4*acoth(a)/4, True)
)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 512 vs. \(2 (87) = 174\).
time = 0.41, size = 512, normalized size = 5.07 \begin {gather*} -\frac {1}{6} \, {\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {3 \, {\left (a^{3} + a\right )} \log \left (\frac {{\left | b x + a + 1 \right |}}{{\left | b x + a - 1 \right |}}\right )}{b^{5}} - \frac {3 \, {\left (a^{3} + a\right )} \log \left ({\left | \frac {b x + a + 1}{b x + a - 1} - 1 \right |}\right )}{b^{5}} - \frac {9 \, a^{2} + \frac {3 \, {\left (3 \, a^{2} - 2 \, a + 1\right )} {\left (b x + a + 1\right )}^{2}}{{\left (b x + a - 1\right )}^{2}} - \frac {3 \, {\left (6 \, a^{2} - 2 \, a + 1\right )} {\left (b x + a + 1\right )}}{b x + a - 1} + 2}{b^{5} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}^{3}} + \frac {3 \, {\left (\frac {{\left (b x + a + 1\right )}^{3} a^{3}}{{\left (b x + a - 1\right )}^{3}} - \frac {3 \, {\left (b x + a + 1\right )}^{2} a^{3}}{{\left (b x + a - 1\right )}^{2}} + \frac {3 \, {\left (b x + a + 1\right )} a^{3}}{b x + a - 1} - a^{3} - \frac {3 \, {\left (b x + a + 1\right )}^{3} a^{2}}{{\left (b x + a - 1\right )}^{3}} + \frac {6 \, {\left (b x + a + 1\right )}^{2} a^{2}}{{\left (b x + a - 1\right )}^{2}} - \frac {3 \, {\left (b x + a + 1\right )} a^{2}}{b x + a - 1} + \frac {3 \, {\left (b x + a + 1\right )}^{3} a}{{\left (b x + a - 1\right )}^{3}} - \frac {3 \, {\left (b x + a + 1\right )}^{2} a}{{\left (b x + a - 1\right )}^{2}} + \frac {{\left (b x + a + 1\right )} a}{b x + a - 1} - a - \frac {{\left (b x + a + 1\right )}^{3}}{{\left (b x + a - 1\right )}^{3}} - \frac {b x + a + 1}{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 )}{b^{5} {\left (\frac {b x + a + 1}{b x + a - 1} - 1\right )}^{4}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arccoth(b*x+a),x, algorithm="giac")

[Out]

-1/6*((a + 1)*b - (a - 1)*b)*(3*(a^3 + a)*log(abs(b*x + a + 1)/abs(b*x + a - 1))/b^5 - 3*(a^3 + a)*log(abs((b*
x + a + 1)/(b*x + a - 1) - 1))/b^5 - (9*a^2 + 3*(3*a^2 - 2*a + 1)*(b*x + a + 1)^2/(b*x + a - 1)^2 - 3*(6*a^2 -
2*a + 1)*(b*x + a + 1)/(b*x + a - 1) + 2)/(b^5*((b*x + a + 1)/(b*x + a - 1) - 1)^3) + 3*((b*x + a + 1)^3*a^3/
(b*x + a - 1)^3 - 3*(b*x + a + 1)^2*a^3/(b*x + a - 1)^2 + 3*(b*x + a + 1)*a^3/(b*x + a - 1) - a^3 - 3*(b*x + a
+ 1)^3*a^2/(b*x + a - 1)^3 + 6*(b*x + a + 1)^2*a^2/(b*x + a - 1)^2 - 3*(b*x + a + 1)*a^2/(b*x + a - 1) + 3*(b
*x + a + 1)^3*a/(b*x + a - 1)^3 - 3*(b*x + a + 1)^2*a/(b*x + a - 1)^2 + (b*x + a + 1)*a/(b*x + a - 1) - a - (b
*x + a + 1)^3/(b*x + a - 1)^3 - (b*x + a + 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^5*((b*x + a + 1)/(b*x + a - 1) - 1)^4))

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Mupad [B]
time = 1.47, size = 134, normalized size = 1.33 \begin {gather*} \frac {x^4\,\ln \left (\frac {1}{a+b\,x}+1\right )}{8}-x\,\left (\frac {4\,a^2-4}{16\,b^3}-\frac {a^2}{b^3}\right )-\frac {x^4\,\ln \left (1-\frac {1}{a+b\,x}\right )}{8}+\frac {x^3}{12\,b}-\frac {a\,x^2}{4\,b^2}+\frac {\ln \left (a+b\,x-1\right )\,\left (a^4-4\,a^3+6\,a^2-4\,a+1\right )}{8\,b^4}-\frac {\ln \left (a+b\,x+1\right )\,\left (a^4+4\,a^3+6\,a^2+4\,a+1\right )}{8\,b^4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*acoth(a + b*x),x)

[Out]

(x^4*log(1/(a + b*x) + 1))/8 - x*((4*a^2 - 4)/(16*b^3) - a^2/b^3) - (x^4*log(1 - 1/(a + b*x)))/8 + x^3/(12*b)
- (a*x^2)/(4*b^2) + (log(a + b*x - 1)*(6*a^2 - 4*a - 4*a^3 + a^4 + 1))/(8*b^4) - (log(a + b*x + 1)*(4*a + 6*a^
2 + 4*a^3 + a^4 + 1))/(8*b^4)

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