3.52 \(\int x^3 (a+b \tanh ^{-1}(c x^2)) \, dx\)

Optimal. Leaf size=43 \[ \frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac{b \tanh ^{-1}\left (c x^2\right )}{4 c^2}+\frac{b x^2}{4 c} \]

[Out]

(b*x^2)/(4*c) - (b*ArcTanh[c*x^2])/(4*c^2) + (x^4*(a + b*ArcTanh[c*x^2]))/4

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Rubi [A]  time = 0.0297085, antiderivative size = 43, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {6097, 275, 321, 206} \[ \frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac{b \tanh ^{-1}\left (c x^2\right )}{4 c^2}+\frac{b x^2}{4 c} \]

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c*x^2]),x]

[Out]

(b*x^2)/(4*c) - (b*ArcTanh[c*x^2])/(4*c^2) + (x^4*(a + b*ArcTanh[c*x^2]))/4

Rule 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 321

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 206

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

Rubi steps

\begin{align*} \int x^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \, dx &=\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac{1}{2} (b c) \int \frac{x^5}{1-c^2 x^4} \, dx\\ &=\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac{1}{4} (b c) \operatorname{Subst}\left (\int \frac{x^2}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\frac{b x^2}{4 c}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac{b \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,x^2\right )}{4 c}\\ &=\frac{b x^2}{4 c}-\frac{b \tanh ^{-1}\left (c x^2\right )}{4 c^2}+\frac{1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0129561, size = 67, normalized size = 1.56 \[ \frac{a x^4}{4}+\frac{b \log \left (1-c x^2\right )}{8 c^2}-\frac{b \log \left (c x^2+1\right )}{8 c^2}+\frac{b x^2}{4 c}+\frac{1}{4} b x^4 \tanh ^{-1}\left (c x^2\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcTanh[c*x^2]),x]

[Out]

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

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Maple [A]  time = 0.012, size = 57, normalized size = 1.3 \begin{align*}{\frac{{x}^{4}a}{4}}+{\frac{b{x}^{4}{\it Artanh} \left ( c{x}^{2} \right ) }{4}}+{\frac{b{x}^{2}}{4\,c}}+{\frac{b\ln \left ( c{x}^{2}-1 \right ) }{8\,{c}^{2}}}-{\frac{b\ln \left ( c{x}^{2}+1 \right ) }{8\,{c}^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctanh(c*x^2)),x)

[Out]

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

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Maxima [A]  time = 0.955376, size = 78, normalized size = 1.81 \begin{align*} \frac{1}{4} \, a x^{4} + \frac{1}{8} \,{\left (2 \, x^{4} \operatorname{artanh}\left (c x^{2}\right ) + c{\left (\frac{2 \, x^{2}}{c^{2}} - \frac{\log \left (c x^{2} + 1\right )}{c^{3}} + \frac{\log \left (c x^{2} - 1\right )}{c^{3}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x^2)),x, algorithm="maxima")

[Out]

1/4*a*x^4 + 1/8*(2*x^4*arctanh(c*x^2) + c*(2*x^2/c^2 - log(c*x^2 + 1)/c^3 + log(c*x^2 - 1)/c^3))*b

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Fricas [A]  time = 2.02008, size = 112, normalized size = 2.6 \begin{align*} \frac{2 \, a c^{2} x^{4} + 2 \, b c x^{2} +{\left (b c^{2} x^{4} - b\right )} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right )}{8 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x^2)),x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 14.0799, size = 48, normalized size = 1.12 \begin{align*} \begin{cases} \frac{a x^{4}}{4} + \frac{b x^{4} \operatorname{atanh}{\left (c x^{2} \right )}}{4} + \frac{b x^{2}}{4 c} - \frac{b \operatorname{atanh}{\left (c x^{2} \right )}}{4 c^{2}} & \text{for}\: c \neq 0 \\\frac{a x^{4}}{4} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atanh(c*x**2)),x)

[Out]

Piecewise((a*x**4/4 + b*x**4*atanh(c*x**2)/4 + b*x**2/(4*c) - b*atanh(c*x**2)/(4*c**2), Ne(c, 0)), (a*x**4/4,
True))

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Giac [A]  time = 1.18379, size = 93, normalized size = 2.16 \begin{align*} \frac{1}{8} \, b x^{4} \log \left (-\frac{c x^{2} + 1}{c x^{2} - 1}\right ) + \frac{1}{4} \, a x^{4} + \frac{b x^{2}}{4 \, c} - \frac{b \log \left (c x^{2} + 1\right )}{8 \, c^{2}} + \frac{b \log \left (c x^{2} - 1\right )}{8 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x^2)),x, algorithm="giac")

[Out]

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