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

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

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 101, normalized size of antiderivative = 1.00, 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 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])

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 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 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^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*}

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Mathematica [A]  time = 0.05, 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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fricas [A]  time = 0.69, size = 75, normalized size = 0.74 \[ -\frac {2 \, {\left (2 \, e x^{3} - 3 \, d x\right )} \sqrt {e x^{2} + d} \sqrt {e} - {\left (8 \, e^{2} x^{4} - 3 \, d^{2}\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{64 \, e^{2}} \]

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

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]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to transpose Error: Bad Argument Value

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maple [A]  time = 0.04, size = 134, normalized size = 1.33 \[ \frac {x^{4} \arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{4}+\frac {\sqrt {e}\, x^{5} \sqrt {e \,x^{2}+d}}{24 d}-\frac {5 x^{3} \sqrt {e \,x^{2}+d}}{96 \sqrt {e}}+\frac {d x \sqrt {e \,x^{2}+d}}{16 e^{\frac {3}{2}}}-\frac {3 d^{2} \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{32 e^{2}}-\frac {x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{24 \sqrt {e}\, d}+\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{32 e^{\frac {3}{2}}} \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{8} \, x^{4} \log \left (\sqrt {e} x + \sqrt {e x^{2} + d}\right ) - \frac {1}{8} \, x^{4} \log \left (-\sqrt {e} x + \sqrt {e x^{2} + d}\right ) - \frac {1}{2} \, d \sqrt {e} \int -\frac {\sqrt {e x^{2} + d} x^{4}}{2 \, {\left (e^{2} x^{4} + d e x^{2} - {\left (e x^{2} + d\right )}^{2}\right )}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^3\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [A]  time = 1.72, size = 95, normalized size = 0.94 \[ \begin {cases} - \frac {3 d^{2} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{32 e^{2}} + \frac {3 d x \sqrt {d + e x^{2}}}{32 e^{\frac {3}{2}}} + \frac {x^{4} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{4} - \frac {x^{3} \sqrt {d + e x^{2}}}{16 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]

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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