3.1.3 \(\int x^3 (a+b \tanh ^{-1}(c x)) \, dx\) [3]

Optimal. Leaf size=48 \[ \frac {b x}{4 c^3}+\frac {b x^3}{12 c}-\frac {b \tanh ^{-1}(c x)}{4 c^4}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right ) \]

[Out]

1/4*b*x/c^3+1/12*b*x^3/c-1/4*b*arctanh(c*x)/c^4+1/4*x^4*(a+b*arctanh(c*x))

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Rubi [A]
time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6037, 308, 212} \begin {gather*} \frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {b \tanh ^{-1}(c x)}{4 c^4}+\frac {b x}{4 c^3}+\frac {b x^3}{12 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c*x]),x]

[Out]

(b*x)/(4*c^3) + (b*x^3)/(12*c) - (b*ArcTanh[c*x])/(4*c^4) + (x^4*(a + b*ArcTanh[c*x]))/4

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 308

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

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^3 \left (a+b \tanh ^{-1}(c x)\right ) \, dx &=\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{4} (b c) \int \frac {x^4}{1-c^2 x^2} \, dx\\ &=\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {1}{4} (b c) \int \left (-\frac {1}{c^4}-\frac {x^2}{c^2}+\frac {1}{c^4 \left (1-c^2 x^2\right )}\right ) \, dx\\ &=\frac {b x}{4 c^3}+\frac {b x^3}{12 c}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )-\frac {b \int \frac {1}{1-c^2 x^2} \, dx}{4 c^3}\\ &=\frac {b x}{4 c^3}+\frac {b x^3}{12 c}-\frac {b \tanh ^{-1}(c x)}{4 c^4}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}(c x)\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 70, normalized size = 1.46 \begin {gather*} \frac {b x}{4 c^3}+\frac {b x^3}{12 c}+\frac {a x^4}{4}+\frac {1}{4} b x^4 \tanh ^{-1}(c x)+\frac {b \log (1-c x)}{8 c^4}-\frac {b \log (1+c x)}{8 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcTanh[c*x]),x]

[Out]

(b*x)/(4*c^3) + (b*x^3)/(12*c) + (a*x^4)/4 + (b*x^4*ArcTanh[c*x])/4 + (b*Log[1 - c*x])/(8*c^4) - (b*Log[1 + c*
x])/(8*c^4)

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Maple [A]
time = 0.01, size = 60, normalized size = 1.25

method	result	size
derivativedivides	\(\frac {\frac {c^{4} x^{4} a}{4}+\frac {c^{4} x^{4} b \arctanh \left (c x \right )}{4}+\frac {b \,c^{3} x^{3}}{12}+\frac {b c x}{4}+\frac {b \ln \left (c x -1\right )}{8}-\frac {b \ln \left (c x +1\right )}{8}}{c^{4}}\)	\(60\)
default	\(\frac {\frac {c^{4} x^{4} a}{4}+\frac {c^{4} x^{4} b \arctanh \left (c x \right )}{4}+\frac {b \,c^{3} x^{3}}{12}+\frac {b c x}{4}+\frac {b \ln \left (c x -1\right )}{8}-\frac {b \ln \left (c x +1\right )}{8}}{c^{4}}\)	\(60\)
risch	\(\frac {x^{4} b \ln \left (c x +1\right )}{8}-\frac {x^{4} b \ln \left (-c x +1\right )}{8}+\frac {x^{4} a}{4}+\frac {b \,x^{3}}{12 c}+\frac {b x}{4 c^{3}}+\frac {b \ln \left (-c x +1\right )}{8 c^{4}}-\frac {b \ln \left (c x +1\right )}{8 c^{4}}\)	\(74\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/c^4*(1/4*c^4*x^4*a+1/4*c^4*x^4*b*arctanh(c*x)+1/12*b*c^3*x^3+1/4*b*c*x+1/8*b*ln(c*x-1)-1/8*b*ln(c*x+1))

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Maxima [A]
time = 0.27, size = 61, normalized size = 1.27 \begin {gather*} \frac {1}{4} \, a x^{4} + \frac {1}{24} \, {\left (6 \, x^{4} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{3} + 3 \, x\right )}}{c^{4}} - \frac {3 \, \log \left (c x + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x - 1\right )}{c^{5}}\right )}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]
time = 0.34, size = 58, normalized size = 1.21 \begin {gather*} \frac {6 \, a c^{4} x^{4} + 2 \, b c^{3} x^{3} + 6 \, b c x + 3 \, {\left (b c^{4} x^{4} - b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{24 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [A]
time = 0.24, size = 53, normalized size = 1.10 \begin {gather*} \begin {cases} \frac {a x^{4}}{4} + \frac {b x^{4} \operatorname {atanh}{\left (c x \right )}}{4} + \frac {b x^{3}}{12 c} + \frac {b x}{4 c^{3}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{4 c^{4}} & \text {for}\: c \neq 0 \\\frac {a x^{4}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Piecewise((a*x**4/4 + b*x**4*atanh(c*x)/4 + b*x**3/(12*c) + b*x/(4*c**3) - b*atanh(c*x)/(4*c**4), Ne(c, 0)), (
a*x**4/4, True))

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 296 vs. \(2 (40) = 80\).
time = 0.42, size = 296, normalized size = 6.17 \begin {gather*} \frac {1}{3} \, c {\left (\frac {3 \, {\left (\frac {{\left (c x + 1\right )}^{3} b}{{\left (c x - 1\right )}^{3}} + \frac {{\left (c x + 1\right )} b}{c x - 1}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{5}}{c x - 1} + c^{5}} + \frac {\frac {6 \, {\left (c x + 1\right )}^{3} a}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )} a}{c x - 1} + \frac {3 \, {\left (c x + 1\right )}^{3} b}{{\left (c x - 1\right )}^{3}} - \frac {6 \, {\left (c x + 1\right )}^{2} b}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} b}{c x - 1} - 2 \, b}{\frac {{\left (c x + 1\right )}^{4} c^{5}}{{\left (c x - 1\right )}^{4}} - \frac {4 \, {\left (c x + 1\right )}^{3} c^{5}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} c^{5}}{{\left (c x - 1\right )}^{2}} - \frac {4 \, {\left (c x + 1\right )} c^{5}}{c x - 1} + c^{5}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 0.77, size = 43, normalized size = 0.90 \begin {gather*} \frac {a\,x^4}{4}+\frac {\frac {b\,c^3\,x^3}{12}-\frac {b\,\mathrm {atanh}\left (c\,x\right )}{4}+\frac {b\,c\,x}{4}}{c^4}+\frac {b\,x^4\,\mathrm {atanh}\left (c\,x\right )}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atanh(c*x)),x)

[Out]

(a*x^4)/4 + ((b*c^3*x^3)/12 - (b*atanh(c*x))/4 + (b*c*x)/4)/c^4 + (b*x^4*atanh(c*x))/4

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