3.11 \(\int x^2 \tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}}) \, dx\)

Optimal. Leaf size=68 \[ -\frac {\left (d+e x^2\right )^{3/2}}{9 e^{3/2}}+\frac {d \sqrt {d+e x^2}}{3 e^{3/2}}+\frac {1}{3} x^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]

[Out]

-1/9*(e*x^2+d)^(3/2)/e^(3/2)+1/3*x^3*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/3*d*(e*x^2+d)^(1/2)/e^(3/2)

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Rubi [A]  time = 0.03, antiderivative size = 68, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {6221, 266, 43} \[ -\frac {\left (d+e x^2\right )^{3/2}}{9 e^{3/2}}+\frac {d \sqrt {d+e x^2}}{3 e^{3/2}}+\frac {1}{3} x^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(d*Sqrt[d + e*x^2])/(3*e^(3/2)) - (d + e*x^2)^(3/2)/(9*e^(3/2)) + (x^3*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/3

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 266

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

Rule 6221

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

Rubi steps

\begin {align*} \int x^2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx &=\frac {1}{3} x^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{3} \sqrt {e} \int \frac {x^3}{\sqrt {d+e x^2}} \, dx\\ &=\frac {1}{3} x^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{6} \sqrt {e} \operatorname {Subst}\left (\int \frac {x}{\sqrt {d+e x}} \, dx,x,x^2\right )\\ &=\frac {1}{3} x^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{6} \sqrt {e} \operatorname {Subst}\left (\int \left (-\frac {d}{e \sqrt {d+e x}}+\frac {\sqrt {d+e x}}{e}\right ) \, dx,x,x^2\right )\\ &=\frac {d \sqrt {d+e x^2}}{3 e^{3/2}}-\frac {\left (d+e x^2\right )^{3/2}}{9 e^{3/2}}+\frac {1}{3} x^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 56, normalized size = 0.82 \[ \frac {1}{9} \left (\frac {\left (2 d-e x^2\right ) \sqrt {d+e x^2}}{e^{3/2}}+3 x^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[x^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(((2*d - e*x^2)*Sqrt[d + e*x^2])/e^(3/2) + 3*x^3*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/9

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fricas [A]  time = 0.49, size = 65, normalized size = 0.96 \[ \frac {3 \, e^{2} x^{3} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) - 2 \, \sqrt {e x^{2} + d} {\left (e x^{2} - 2 \, d\right )} \sqrt {e}}{18 \, e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

1/18*(3*e^2*x^3*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - 2*sqrt(e*x^2 + d)*(e*x^2 - 2*d)*sqrt(e))/
e^2

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: 1/2*(x^3/3*ln((1+(sqrt(exp(1)*x^2+d))^-1
*exp(1/2)*x)/(1-(sqrt(exp(1)*x^2+d))^-1*exp(1/2)*x))-2*d*exp(1/2)/exp(1)*(sqrt(d+x^2*exp(1))/(3*exp(1)-3*exp(1
/2)^2)-2*d*exp(1)/2/(3*exp(1)-3*exp(1/2)^2)/sqrt(-d*exp(1/2)^2+d*exp(1))/exp(1/2)*atan((sqrt(d+x^2*exp(1))*exp
(1)-sqrt(d+x^2*exp(1))*exp(1/2)^2)/sqrt(-d*exp(1/2)^2+d*exp(1))/exp(1/2))))

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maple [B]  time = 0.03, size = 128, normalized size = 1.88 \[ \frac {x^{3} \arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{3}+\frac {e^{\frac {3}{2}} \left (\frac {x^{4} \sqrt {e \,x^{2}+d}}{5 e}-\frac {4 d \left (\frac {x^{2} \sqrt {e \,x^{2}+d}}{3 e}-\frac {2 d \sqrt {e \,x^{2}+d}}{3 e^{2}}\right )}{5 e}\right )}{3 d}-\frac {\sqrt {e}\, \left (\frac {x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 e}-\frac {2 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 e^{2}}\right )}{3 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x)

[Out]

1/3*x^3*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/3*e^(3/2)/d*(1/5*x^4/e*(e*x^2+d)^(1/2)-4/5*d/e*(1/3*x^2/e*(e*x^2+
d)^(1/2)-2/3*d/e^2*(e*x^2+d)^(1/2)))-1/3*e^(1/2)/d*(1/5*x^2*(e*x^2+d)^(3/2)/e-2/15*d/e^2*(e*x^2+d)^(3/2))

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maxima [A]  time = 0.33, size = 99, normalized size = 1.46 \[ \frac {1}{3} \, x^{3} \operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right ) - \frac {3 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} - 5 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d}{45 \, d e^{\frac {3}{2}}} + \frac {3 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} - 10 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d + 15 \, \sqrt {e x^{2} + d} d^{2}}{45 \, d e^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="maxima")

[Out]

1/3*x^3*arctanh(sqrt(e)*x/sqrt(e*x^2 + d)) - 1/45*(3*(e*x^2 + d)^(5/2) - 5*(e*x^2 + d)^(3/2)*d)/(d*e^(3/2)) +
1/45*(3*(e*x^2 + d)^(5/2) - 10*(e*x^2 + d)^(3/2)*d + 15*sqrt(e*x^2 + d)*d^2)/(d*e^(3/2))

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^2\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*atanh((e^(1/2)*x)/(d + e*x^2)^(1/2)),x)

[Out]

int(x^2*atanh((e^(1/2)*x)/(d + e*x^2)^(1/2)), x)

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sympy [A]  time = 1.03, size = 65, normalized size = 0.96 \[ \begin {cases} \frac {2 d \sqrt {d + e x^{2}}}{9 e^{\frac {3}{2}}} + \frac {x^{3} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{3} - \frac {x^{2} \sqrt {d + e x^{2}}}{9 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*atanh(x*e**(1/2)/(e*x**2+d)**(1/2)),x)

[Out]

Piecewise((2*d*sqrt(d + e*x**2)/(9*e**(3/2)) + x**3*atanh(sqrt(e)*x/sqrt(d + e*x**2))/3 - x**2*sqrt(d + e*x**2
)/(9*sqrt(e)), Ne(e, 0)), (0, True))

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