3.2 \(\int x^3 \tanh ^{-1}(\frac{\sqrt{e} x}{\sqrt{d+e x^2}}) \, dx\)

Optimal. Leaf size=101 \[ -\frac{3 d^2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{32 e^2}+\frac{3 d x \sqrt{d+e x^2}}{32 e^{3/2}}-\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{e}}+\frac{1}{4} x^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \]

[Out]

(3*d*x*Sqrt[d + e*x^2])/(32*e^(3/2)) - (x^3*Sqrt[d + e*x^2])/(16*Sqrt[e]) - (3*d^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d
+ e*x^2]])/(32*e^2) + (x^4*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/4

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Rubi [A]  time = 0.0388202, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {6221, 321, 217, 206} \[ -\frac{3 d^2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{32 e^2}+\frac{3 d x \sqrt{d+e x^2}}{32 e^{3/2}}-\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{e}}+\frac{1}{4} x^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^3*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(3*d*x*Sqrt[d + e*x^2])/(32*e^(3/2)) - (x^3*Sqrt[d + e*x^2])/(16*Sqrt[e]) - (3*d^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d
+ e*x^2]])/(32*e^2) + (x^4*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/4

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]

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 217

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

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 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right ) \, dx &=\frac{1}{4} x^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{1}{4} \sqrt{e} \int \frac{x^4}{\sqrt{d+e x^2}} \, dx\\ &=-\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{e}}+\frac{1}{4} x^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )+\frac{(3 d) \int \frac{x^2}{\sqrt{d+e x^2}} \, dx}{16 \sqrt{e}}\\ &=\frac{3 d x \sqrt{d+e x^2}}{32 e^{3/2}}-\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{e}}+\frac{1}{4} x^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{\left (3 d^2\right ) \int \frac{1}{\sqrt{d+e x^2}} \, dx}{32 e^{3/2}}\\ &=\frac{3 d x \sqrt{d+e x^2}}{32 e^{3/2}}-\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{e}}+\frac{1}{4} x^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )-\frac{\left (3 d^2\right ) \operatorname{Subst}\left (\int \frac{1}{1-e x^2} \, dx,x,\frac{x}{\sqrt{d+e x^2}}\right )}{32 e^{3/2}}\\ &=\frac{3 d x \sqrt{d+e x^2}}{32 e^{3/2}}-\frac{x^3 \sqrt{d+e x^2}}{16 \sqrt{e}}-\frac{3 d^2 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )}{32 e^2}+\frac{1}{4} x^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )\\ \end{align*}

Mathematica [A]  time = 0.0481845, size = 88, normalized size = 0.87 \[ \frac{-3 d^2 \log \left (\sqrt{d+e x^2}+\sqrt{e} x\right )+8 e^2 x^4 \tanh ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d+e x^2}}\right )+\sqrt{e} x \left (3 d-2 e x^2\right ) \sqrt{d+e x^2}}{32 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[x^3*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(Sqrt[e]*x*(3*d - 2*e*x^2)*Sqrt[d + e*x^2] + 8*e^2*x^4*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]] - 3*d^2*Log[Sqrt[e
]*x + Sqrt[d + e*x^2]])/(32*e^2)

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Maple [A]  time = 0.031, size = 134, normalized size = 1.3 \begin{align*}{\frac{{x}^{4}}{4}{\it Artanh} \left ({x\sqrt{e}{\frac{1}{\sqrt{e{x}^{2}+d}}}} \right ) }+{\frac{{x}^{5}}{24\,d}\sqrt{e}\sqrt{e{x}^{2}+d}}-{\frac{5\,{x}^{3}}{96}\sqrt{e{x}^{2}+d}{\frac{1}{\sqrt{e}}}}+{\frac{dx}{16}\sqrt{e{x}^{2}+d}{e}^{-{\frac{3}{2}}}}-{\frac{3\,{d}^{2}}{32\,{e}^{2}}\ln \left ( x\sqrt{e}+\sqrt{e{x}^{2}+d} \right ) }-{\frac{{x}^{3}}{24\,d} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}{\frac{1}{\sqrt{e}}}}+{\frac{x}{32} \left ( e{x}^{2}+d \right ) ^{{\frac{3}{2}}}{e}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x)

[Out]

1/4*x^4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/24*e^(1/2)/d*x^5*(e*x^2+d)^(1/2)-5/96*x^3*(e*x^2+d)^(1/2)/e^(1/2)
+1/16*d*x*(e*x^2+d)^(1/2)/e^(3/2)-3/32/e^2*d^2*ln(x*e^(1/2)+(e*x^2+d)^(1/2))-1/24/e^(1/2)/d*x^3*(e*x^2+d)^(3/2
)+1/32/e^(3/2)*x*(e*x^2+d)^(3/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{1}{8} \, x^{4} \log \left (\sqrt{e} x + \sqrt{e x^{2} + d}\right ) - \frac{1}{8} \, x^{4} \log \left (-\sqrt{e} x + \sqrt{e x^{2} + d}\right ) - \frac{1}{2} \, d \sqrt{e} \int -\frac{\sqrt{e x^{2} + d} x^{4}}{2 \,{\left (e^{2} x^{4} + d e x^{2} -{\left (e x^{2} + d\right )}^{2}\right )}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="maxima")

[Out]

1/8*x^4*log(sqrt(e)*x + sqrt(e*x^2 + d)) - 1/8*x^4*log(-sqrt(e)*x + sqrt(e*x^2 + d)) - 1/2*d*sqrt(e)*integrate
(-1/2*sqrt(e*x^2 + d)*x^4/(e^2*x^4 + d*e*x^2 - (e*x^2 + d)^2), x)

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Fricas [A]  time = 2.10243, size = 176, normalized size = 1.74 \begin{align*} -\frac{2 \,{\left (2 \, e x^{3} - 3 \, d x\right )} \sqrt{e x^{2} + d} \sqrt{e} -{\left (8 \, e^{2} x^{4} - 3 \, d^{2}\right )} \log \left (\frac{2 \, e x^{2} + 2 \, \sqrt{e x^{2} + d} \sqrt{e} x + d}{d}\right )}{64 \, e^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

-1/64*(2*(2*e*x^3 - 3*d*x)*sqrt(e*x^2 + d)*sqrt(e) - (8*e^2*x^4 - 3*d^2)*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt
(e)*x + d)/d))/e^2

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Sympy [A]  time = 4.84909, size = 95, normalized size = 0.94 \begin{align*} \begin{cases} - \frac{3 d^{2} \operatorname{atanh}{\left (\frac{\sqrt{e} x}{\sqrt{d + e x^{2}}} \right )}}{32 e^{2}} + \frac{3 d x \sqrt{d + e x^{2}}}{32 e^{\frac{3}{2}}} + \frac{x^{4} \operatorname{atanh}{\left (\frac{\sqrt{e} x}{\sqrt{d + e x^{2}}} \right )}}{4} - \frac{x^{3} \sqrt{d + e x^{2}}}{16 \sqrt{e}} & \text{for}\: e \neq 0 \\0 & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*atanh(x*e**(1/2)/(e*x**2+d)**(1/2)),x)

[Out]

Piecewise((-3*d**2*atanh(sqrt(e)*x/sqrt(d + e*x**2))/(32*e**2) + 3*d*x*sqrt(d + e*x**2)/(32*e**(3/2)) + x**4*a
tanh(sqrt(e)*x/sqrt(d + e*x**2))/4 - x**3*sqrt(d + e*x**2)/(16*sqrt(e)), Ne(e, 0)), (0, True))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \left [\mathit{undef}, \mathit{undef}, \mathit{undef}, -\frac{1}{2} \, d e^{\frac{1}{2}}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="giac")

[Out]

[undef, undef, undef, -1/2*d*e^(1/2)]