3.1.28 \(\int x^3 \tanh ^{-1}(a+b x^4) \, dx\) [28]

Optimal. Leaf size=44 \[ \frac {\left (a+b x^4\right ) \tanh ^{-1}\left (a+b x^4\right )}{4 b}+\frac {\log \left (1-\left (a+b x^4\right )^2\right )}{8 b} \]

[Out]

1/4*(b*x^4+a)*arctanh(b*x^4+a)/b+1/8*ln(1-(b*x^4+a)^2)/b

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Rubi [A]
time = 0.03, antiderivative size = 44, 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 = {6847, 6238, 6021, 266} \begin {gather*} \frac {\log \left (1-\left (a+b x^4\right )^2\right )}{8 b}+\frac {\left (a+b x^4\right ) \tanh ^{-1}\left (a+b x^4\right )}{4 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTanh[a + b*x^4],x]

[Out]

((a + b*x^4)*ArcTanh[a + b*x^4])/(4*b) + Log[1 - (a + b*x^4)^2]/(8*b)

Rule 266

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

Rule 6021

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

Rule 6238

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

Rule 6847

Int[(u_)*(x_)^(m_.), x_Symbol] :> Dist[1/(m + 1), Subst[Int[SubstFor[x^(m + 1), u, x], x], x, x^(m + 1)], x] /
; FreeQ[m, x] && NeQ[m, -1] && FunctionOfQ[x^(m + 1), u, x]

Rubi steps

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

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Mathematica [A]
time = 0.01, size = 39, normalized size = 0.89 \begin {gather*} \frac {2 \left (a+b x^4\right ) \tanh ^{-1}\left (a+b x^4\right )+\log \left (1-\left (a+b x^4\right )^2\right )}{8 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcTanh[a + b*x^4],x]

[Out]

(2*(a + b*x^4)*ArcTanh[a + b*x^4] + Log[1 - (a + b*x^4)^2])/(8*b)

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Maple [A]
time = 0.04, size = 39, normalized size = 0.89

method	result	size
derivativedivides	\(\frac {\left (b \,x^{4}+a \right ) \arctanh \left (b \,x^{4}+a \right )+\frac {\ln \left (1-\left (b \,x^{4}+a \right )^{2}\right )}{2}}{4 b}\)	\(39\)
default	\(\frac {\left (b \,x^{4}+a \right ) \arctanh \left (b \,x^{4}+a \right )+\frac {\ln \left (1-\left (b \,x^{4}+a \right )^{2}\right )}{2}}{4 b}\)	\(39\)
risch	\(\frac {x^{4} \ln \left (b \,x^{4}+a +1\right )}{8}-\frac {x^{4} \ln \left (-b \,x^{4}-a +1\right )}{8}+\frac {\ln \left (b \,x^{4}+a +1\right ) a}{8 b}-\frac {\ln \left (-b \,x^{4}-a +1\right ) a}{8 b}+\frac {\ln \left (b \,x^{4}+a +1\right )}{8 b}+\frac {\ln \left (-b \,x^{4}-a +1\right )}{8 b}\)	\(97\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctanh(b*x^4+a),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.25, size = 37, normalized size = 0.84 \begin {gather*} \frac {2 \, {\left (b x^{4} + a\right )} \operatorname {artanh}\left (b x^{4} + a\right ) + \log \left (-{\left (b x^{4} + a\right )}^{2} + 1\right )}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(b*x^4+a),x, algorithm="maxima")

[Out]

1/8*(2*(b*x^4 + a)*arctanh(b*x^4 + a) + log(-(b*x^4 + a)^2 + 1))/b

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Fricas [A]
time = 0.37, size = 59, normalized size = 1.34 \begin {gather*} \frac {b x^{4} \log \left (-\frac {b x^{4} + a + 1}{b x^{4} + a - 1}\right ) + {\left (a + 1\right )} \log \left (b x^{4} + a + 1\right ) - {\left (a - 1\right )} \log \left (b x^{4} + a - 1\right )}{8 \, b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(b*x^4+a),x, algorithm="fricas")

[Out]

1/8*(b*x^4*log(-(b*x^4 + a + 1)/(b*x^4 + a - 1)) + (a + 1)*log(b*x^4 + a + 1) - (a - 1)*log(b*x^4 + a - 1))/b

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Sympy [A]
time = 0.98, size = 60, normalized size = 1.36 \begin {gather*} \begin {cases} \frac {a \operatorname {atanh}{\left (a + b x^{4} \right )}}{4 b} + \frac {x^{4} \operatorname {atanh}{\left (a + b x^{4} \right )}}{4} + \frac {\log {\left (a + b x^{4} + 1 \right )}}{4 b} - \frac {\operatorname {atanh}{\left (a + b x^{4} \right )}}{4 b} & \text {for}\: b \neq 0 \\\frac {x^{4} \operatorname {atanh}{\left (a \right )}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atanh(b*x**4+a),x)

[Out]

Piecewise((a*atanh(a + b*x**4)/(4*b) + x**4*atanh(a + b*x**4)/4 + log(a + b*x**4 + 1)/(4*b) - atanh(a + b*x**4
)/(4*b), Ne(b, 0)), (x**4*atanh(a)/4, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 223 vs. \(2 (40) = 80\).
time = 0.41, size = 223, normalized size = 5.07 \begin {gather*} \frac {1}{8} \, {\left ({\left (a + 1\right )} b - {\left (a - 1\right )} b\right )} {\left (\frac {\log \left (\frac {{\left | -b x^{4} - a - 1 \right |}}{{\left | b x^{4} + a - 1 \right |}}\right )}{b^{2}} - \frac {\log \left ({\left | -\frac {b x^{4} + a + 1}{b x^{4} + a - 1} + 1 \right |}\right )}{b^{2}} + \frac {\log \left (-\frac {a - \frac {{\left (\frac {{\left (b x^{4} + a + 1\right )} {\left (a - 1\right )}}{b x^{4} + a - 1} - a - 1\right )} b}{\frac {{\left (b x^{4} + a + 1\right )} b}{b x^{4} + a - 1} - b} + 1}{a - \frac {{\left (\frac {{\left (b x^{4} + a + 1\right )} {\left (a - 1\right )}}{b x^{4} + a - 1} - a - 1\right )} b}{\frac {{\left (b x^{4} + a + 1\right )} b}{b x^{4} + a - 1} - b} - 1}\right )}{b^{2} {\left (\frac {b x^{4} + a + 1}{b x^{4} + a - 1} - 1\right )}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(b*x^4+a),x, algorithm="giac")

[Out]

1/8*((a + 1)*b - (a - 1)*b)*(log(abs(-b*x^4 - a - 1)/abs(b*x^4 + a - 1))/b^2 - log(abs(-(b*x^4 + a + 1)/(b*x^4
+ a - 1) + 1))/b^2 + log(-(a - ((b*x^4 + a + 1)*(a - 1)/(b*x^4 + a - 1) - a - 1)*b/((b*x^4 + a + 1)*b/(b*x^4
+ a - 1) - b) + 1)/(a - ((b*x^4 + a + 1)*(a - 1)/(b*x^4 + a - 1) - a - 1)*b/((b*x^4 + a + 1)*b/(b*x^4 + a - 1)
- b) - 1))/(b^2*((b*x^4 + a + 1)/(b*x^4 + a - 1) - 1)))

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Mupad [B]
time = 0.32, size = 90, normalized size = 2.05 \begin {gather*} \frac {\ln \left (b\,x^4+a-1\right )}{8\,b}-\frac {x^4\,\ln \left (-b\,x^4-a+1\right )}{8}+\frac {\ln \left (b\,x^4+a+1\right )}{8\,b}+\frac {x^4\,\ln \left (b\,x^4+a+1\right )}{8}-\frac {a\,\ln \left (b\,x^4+a-1\right )}{8\,b}+\frac {a\,\ln \left (b\,x^4+a+1\right )}{8\,b} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*atanh(a + b*x^4),x)

[Out]

log(a + b*x^4 - 1)/(8*b) - (x^4*log(1 - b*x^4 - a))/8 + log(a + b*x^4 + 1)/(8*b) + (x^4*log(a + b*x^4 + 1))/8
- (a*log(a + b*x^4 - 1))/(8*b) + (a*log(a + b*x^4 + 1))/(8*b)

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