3.1.10 \(\int x^4 \tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}}) \, dx\) [10]

Optimal. Leaf size=91 \[ -\frac {d^2 \sqrt {d+e x^2}}{5 e^{5/2}}+\frac {2 d \left (d+e x^2\right )^{3/2}}{15 e^{5/2}}-\frac {\left (d+e x^2\right )^{5/2}}{25 e^{5/2}}+\frac {1}{5} x^5 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]

[Out]

2/15*d*(e*x^2+d)^(3/2)/e^(5/2)-1/25*(e*x^2+d)^(5/2)/e^(5/2)+1/5*x^5*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))-1/5*d^2
*(e*x^2+d)^(1/2)/e^(5/2)

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Rubi [A]
time = 0.03, antiderivative size = 91, 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 = {6356, 272, 45} \begin {gather*} -\frac {d^2 \sqrt {d+e x^2}}{5 e^{5/2}}-\frac {\left (d+e x^2\right )^{5/2}}{25 e^{5/2}}+\frac {2 d \left (d+e x^2\right )^{3/2}}{15 e^{5/2}}+\frac {1}{5} x^5 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^4*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

-1/5*(d^2*Sqrt[d + e*x^2])/e^(5/2) + (2*d*(d + e*x^2)^(3/2))/(15*e^(5/2)) - (d + e*x^2)^(5/2)/(25*e^(5/2)) + (
x^5*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/5

Rule 45

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 272

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 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^4 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx &=\frac {1}{5} x^5 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{5} \sqrt {e} \int \frac {x^5}{\sqrt {d+e x^2}} \, dx\\ &=\frac {1}{5} x^5 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{10} \sqrt {e} \text {Subst}\left (\int \frac {x^2}{\sqrt {d+e x}} \, dx,x,x^2\right )\\ &=\frac {1}{5} x^5 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{10} \sqrt {e} \text {Subst}\left (\int \left (\frac {d^2}{e^2 \sqrt {d+e x}}-\frac {2 d \sqrt {d+e x}}{e^2}+\frac {(d+e x)^{3/2}}{e^2}\right ) \, dx,x,x^2\right )\\ &=-\frac {d^2 \sqrt {d+e x^2}}{5 e^{5/2}}+\frac {2 d \left (d+e x^2\right )^{3/2}}{15 e^{5/2}}-\frac {\left (d+e x^2\right )^{5/2}}{25 e^{5/2}}+\frac {1}{5} x^5 \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 = 68, normalized size = 0.75 \begin {gather*} -\frac {\sqrt {d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{75 e^{5/2}}+\frac {1}{5} x^5 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^4*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

-1/75*(Sqrt[d + e*x^2]*(8*d^2 - 4*d*e*x^2 + 3*e^2*x^4))/e^(5/2) + (x^5*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/5

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(175\) vs. \(2(67)=134\).
time = 0.01, size = 176, normalized size = 1.93

method	result	size
default	\(\frac {x^{5} \arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{5}+\frac {e^{\frac {3}{2}} \left (\frac {x^{6} \sqrt {e \,x^{2}+d}}{7 e}-\frac {6 d \left (\frac {x^{4} \sqrt {e \,x^{2}+d}}{5 e}-\frac {4 d \left (\frac {x^{2} \sqrt {e \,x^{2}+d}}{3 e}-\frac {2 d \sqrt {e \,x^{2}+d}}{3 e^{2}}\right )}{5 e}\right )}{7 e}\right )}{5 d}-\frac {\sqrt {e}\, \left (\frac {x^{4} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 e}-\frac {4 d \left (\frac {x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 e}-\frac {2 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 e^{2}}\right )}{7 e}\right )}{5 d}\)	\(176\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

1/5*x^5*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/5*e^(3/2)/d*(1/7*x^6/e*(e*x^2+d)^(1/2)-6/7*d/e*(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))))-1/5*e^(1/2)/d*(1/7*x^4*(e*x^2+d)^(3/2
)/e-4/7*d/e*(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 = 0.28, size = 132, normalized size = 1.45 \begin {gather*} \frac {1}{5} \, x^{5} \operatorname {artanh}\left (\frac {x e^{\frac {1}{2}}}{\sqrt {x^{2} e + d}}\right ) - \frac {{\left (15 \, {\left (x^{2} e + d\right )}^{\frac {7}{2}} - 42 \, {\left (x^{2} e + d\right )}^{\frac {5}{2}} d + 35 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} d^{2}\right )} e^{\left (-\frac {5}{2}\right )}}{525 \, d} + \frac {{\left (5 \, {\left (x^{2} e + d\right )}^{\frac {7}{2}} - 21 \, {\left (x^{2} e + d\right )}^{\frac {5}{2}} d + 35 \, {\left (x^{2} e + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {x^{2} e + d} d^{3}\right )} e^{\left (-\frac {5}{2}\right )}}{175 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="maxima")

[Out]

1/5*x^5*arctanh(x*e^(1/2)/sqrt(x^2*e + d)) - 1/525*(15*(x^2*e + d)^(7/2) - 42*(x^2*e + d)^(5/2)*d + 35*(x^2*e
+ d)^(3/2)*d^2)*e^(-5/2)/d + 1/175*(5*(x^2*e + d)^(7/2) - 21*(x^2*e + d)^(5/2)*d + 35*(x^2*e + d)^(3/2)*d^2 -
35*sqrt(x^2*e + d)*d^3)*e^(-5/2)/d

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 315 vs. \(2 (67) = 134\).
time = 0.36, size = 315, normalized size = 3.46 \begin {gather*} \frac {15 \, {\left (x^{5} \cosh \left (\frac {1}{2}\right )^{5} + 5 \, x^{5} \cosh \left (\frac {1}{2}\right )^{4} \sinh \left (\frac {1}{2}\right ) + 10 \, x^{5} \cosh \left (\frac {1}{2}\right )^{3} \sinh \left (\frac {1}{2}\right )^{2} + 10 \, x^{5} \cosh \left (\frac {1}{2}\right )^{2} \sinh \left (\frac {1}{2}\right )^{3} + 5 \, x^{5} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{4} + x^{5} \sinh \left (\frac {1}{2}\right )^{5}\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 (3 \, x^{4} \cosh \left (\frac {1}{2}\right )^{4} + 12 \, x^{4} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{3} + 3 \, x^{4} \sinh \left (\frac {1}{2}\right )^{4} - 4 \, d x^{2} \cosh \left (\frac {1}{2}\right )^{2} + 2 \, {\left (9 \, x^{4} \cosh \left (\frac {1}{2}\right )^{2} - 2 \, d x^{2}\right )} \sinh \left (\frac {1}{2}\right )^{2} + 8 \, d^{2} + 4 \, {\left (3 \, x^{4} \cosh \left (\frac {1}{2}\right )^{3} - 2 \, d x^{2} \cosh \left (\frac {1}{2}\right )\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 )}}}{150 \, {\left (\cosh \left (\frac {1}{2}\right )^{5} + 5 \, \cosh \left (\frac {1}{2}\right )^{4} \sinh \left (\frac {1}{2}\right ) + 10 \, \cosh \left (\frac {1}{2}\right )^{3} \sinh \left (\frac {1}{2}\right )^{2} + 10 \, \cosh \left (\frac {1}{2}\right )^{2} \sinh \left (\frac {1}{2}\right )^{3} + 5 \, \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{4} + \sinh \left (\frac {1}{2}\right )^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^4*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

1/150*(15*(x^5*cosh(1/2)^5 + 5*x^5*cosh(1/2)^4*sinh(1/2) + 10*x^5*cosh(1/2)^3*sinh(1/2)^2 + 10*x^5*cosh(1/2)^2
*sinh(1/2)^3 + 5*x^5*cosh(1/2)*sinh(1/2)^4 + x^5*sinh(1/2)^5)*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*(3*x^4*cosh(1/2)^4 + 12*x^4*cosh(1/2)*sinh(1/2)^3 + 3*x^4*sinh(1/2)^4 - 4*d*x^2
*cosh(1/2)^2 + 2*(9*x^4*cosh(1/2)^2 - 2*d*x^2)*sinh(1/2)^2 + 8*d^2 + 4*(3*x^4*cosh(1/2)^3 - 2*d*x^2*cosh(1/2))
*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)^5 + 5*cosh(1
/2)^4*sinh(1/2) + 10*cosh(1/2)^3*sinh(1/2)^2 + 10*cosh(1/2)^2*sinh(1/2)^3 + 5*cosh(1/2)*sinh(1/2)^4 + sinh(1/2
)^5)

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Sympy [A]
time = 0.81, size = 90, normalized size = 0.99 \begin {gather*} \begin {cases} - \frac {8 d^{2} \sqrt {d + e x^{2}}}{75 e^{\frac {5}{2}}} + \frac {4 d x^{2} \sqrt {d + e x^{2}}}{75 e^{\frac {3}{2}}} + \frac {x^{5} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{5} - \frac {x^{4} \sqrt {d + e x^{2}}}{25 \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**4*atanh(x*e**(1/2)/(e*x**2+d)**(1/2)),x)

[Out]

Piecewise((-8*d**2*sqrt(d + e*x**2)/(75*e**(5/2)) + 4*d*x**2*sqrt(d + e*x**2)/(75*e**(3/2)) + x**5*atanh(sqrt(
e)*x/sqrt(d + e*x**2))/5 - x**4*sqrt(d + e*x**2)/(25*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^4*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^4\,\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^4*atanh((e^(1/2)*x)/(d + e*x^2)^(1/2)),x)

[Out]

int(x^4*atanh((e^(1/2)*x)/(d + e*x^2)^(1/2)), x)

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