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

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

[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 = 0.03, 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 = {6356, 327, 223, 212} \begin {gather*} -\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}} \end {gather*}

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 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 223

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 327

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 6356

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 ) \text {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 = 0.04, size = 88, normalized size = 0.87 \begin {gather*} \frac {\sqrt {e} x \left (3 d-2 e x^2\right ) \sqrt {d+e x^2}+8 e^2 x^4 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-3 d^2 \log \left (\sqrt {e} x+\sqrt {d+e x^2}\right )}{32 e^2} \end {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. \(204\) vs. \(2(77)=154\).
time = 0.01, size = 205, normalized size = 2.03

method	result	size
default	\(\frac {x^{4} \arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{4}+\frac {e^{\frac {3}{2}} \left (\frac {x^{5} \sqrt {e \,x^{2}+d}}{6 e}-\frac {5 d \left (\frac {x^{3} \sqrt {e \,x^{2}+d}}{4 e}-\frac {3 d \left (\frac {x \sqrt {e \,x^{2}+d}}{2 e}-\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 e^{\frac {3}{2}}}\right )}{4 e}\right )}{6 e}\right )}{4 d}-\frac {\sqrt {e}\, \left (\frac {x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{6 e}-\frac {d \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4 e}-\frac {d \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{4 e}\right )}{2 e}\right )}{4 d}\)	\(205\)

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,method=_RETURNVERBOSE)

[Out]

1/4*x^4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/4*e^(3/2)/d*(1/6*x^5/e*(e*x^2+d)^(1/2)-5/6*d/e*(1/4*x^3/e*(e*x^2+
d)^(1/2)-3/4*d/e*(1/2*x/e*(e*x^2+d)^(1/2)-1/2*d/e^(3/2)*ln(x*e^(1/2)+(e*x^2+d)^(1/2)))))-1/4*e^(1/2)/d*(1/6*x^
3*(e*x^2+d)^(3/2)/e-1/2*d/e*(1/4*x*(e*x^2+d)^(3/2)/e-1/4*d/e*(1/2*x*(e*x^2+d)^(1/2)+1/2/e^(1/2)*d*ln(x*e^(1/2)
+(e*x^2+d)^(1/2)))))

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 263 vs. \(2 (76) = 152\).
time = 0.34, size = 263, normalized size = 2.60 \begin {gather*} \frac {{\left (8 \, x^{4} \cosh \left (\frac {1}{2}\right )^{4} + 32 \, x^{4} \cosh \left (\frac {1}{2}\right )^{3} \sinh \left (\frac {1}{2}\right ) + 48 \, x^{4} \cosh \left (\frac {1}{2}\right )^{2} \sinh \left (\frac {1}{2}\right )^{2} + 32 \, x^{4} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{3} + 8 \, x^{4} \sinh \left (\frac {1}{2}\right )^{4} - 3 \, d^{2}\right )} \log \left (\frac {2 \, x^{2} \cosh \left (\frac {1}{2}\right )^{2} + 4 \, x^{2} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right ) + 2 \, x^{2} \sinh \left (\frac {1}{2}\right )^{2} + 2 \, {\left (x \cosh \left (\frac {1}{2}\right ) + x \sinh \left (\frac {1}{2}\right )\right )} \sqrt {\frac {{\left (x^{2} + d\right )} \cosh \left (\frac {1}{2}\right ) + {\left (x^{2} - d\right )} \sinh \left (\frac {1}{2}\right )}{\cosh \left (\frac {1}{2}\right ) - \sinh \left (\frac {1}{2}\right )}} + d}{d}\right ) - 2 \, {\left (2 \, x^{3} \cosh \left (\frac {1}{2}\right )^{3} + 6 \, x^{3} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{2} + 2 \, x^{3} \sinh \left (\frac {1}{2}\right )^{3} - 3 \, d x \cosh \left (\frac {1}{2}\right ) + 3 \, {\left (2 \, x^{3} \cosh \left (\frac {1}{2}\right )^{2} - d x\right )} \sinh \left (\frac {1}{2}\right )\right )} \sqrt {\frac {{\left (x^{2} + d\right )} \cosh \left (\frac {1}{2}\right ) + {\left (x^{2} - d\right )} \sinh \left (\frac {1}{2}\right )}{\cosh \left (\frac {1}{2}\right ) - \sinh \left (\frac {1}{2}\right )}}}{64 \, {\left (\cosh \left (\frac {1}{2}\right )^{4} + 4 \, \cosh \left (\frac {1}{2}\right )^{3} \sinh \left (\frac {1}{2}\right ) + 6 \, \cosh \left (\frac {1}{2}\right )^{2} \sinh \left (\frac {1}{2}\right )^{2} + 4 \, \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{3} + \sinh \left (\frac {1}{2}\right )^{4}\right )}} \end {gather*}

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*((8*x^4*cosh(1/2)^4 + 32*x^4*cosh(1/2)^3*sinh(1/2) + 48*x^4*cosh(1/2)^2*sinh(1/2)^2 + 32*x^4*cosh(1/2)*si
nh(1/2)^3 + 8*x^4*sinh(1/2)^4 - 3*d^2)*log((2*x^2*cosh(1/2)^2 + 4*x^2*cosh(1/2)*sinh(1/2) + 2*x^2*sinh(1/2)^2
+ 2*(x*cosh(1/2) + x*sinh(1/2))*sqrt(((x^2 + d)*cosh(1/2) + (x^2 - d)*sinh(1/2))/(cosh(1/2) - sinh(1/2))) + d)
/d) - 2*(2*x^3*cosh(1/2)^3 + 6*x^3*cosh(1/2)*sinh(1/2)^2 + 2*x^3*sinh(1/2)^3 - 3*d*x*cosh(1/2) + 3*(2*x^3*cosh
(1/2)^2 - d*x)*sinh(1/2))*sqrt(((x^2 + d)*cosh(1/2) + (x^2 - d)*sinh(1/2))/(cosh(1/2) - sinh(1/2))))/(cosh(1/2
)^4 + 4*cosh(1/2)^3*sinh(1/2) + 6*cosh(1/2)^2*sinh(1/2)^2 + 4*cosh(1/2)*sinh(1/2)^3 + sinh(1/2)^4)

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Sympy [A]
time = 0.61, size = 95, normalized size = 0.94 \begin {gather*} \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 {gather*}

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(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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]

Timed out

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

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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