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

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

[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 = 0.05, 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 = {6221, 266, 43} \[ -\frac {d^2 \sqrt {d+e x^2}}{5 e^{5/2}}-\frac {\left (d+e x^2\right )^{5/2}}{25 e^{5/2}}+\frac {2 d \left (d+e x^2\right )^{3/2}}{15 e^{5/2}}+\frac {1}{5} x^5 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(d^2*Sqrt[d + e*x^2])/(5*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 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^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} \operatorname {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} \operatorname {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 = 0.06, size = 68, normalized size = 0.75 \[ \frac {1}{5} x^5 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {\sqrt {d+e x^2} \left (8 d^2-4 d e x^2+3 e^2 x^4\right )}{75 e^{5/2}} \]

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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fricas [A]  time = 0.40, size = 77, normalized size = 0.85 \[ \frac {15 \, e^{3} x^{5} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) - 2 \, {\left (3 \, e^{2} x^{4} - 4 \, d e x^{2} + 8 \, d^{2}\right )} \sqrt {e x^{2} + d} \sqrt {e}}{150 \, e^{3}} \]

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*e^3*x^5*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - 2*(3*e^2*x^4 - 4*d*e*x^2 + 8*d^2)*sqrt(
e*x^2 + d)*sqrt(e))/e^3

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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^4*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^5/5*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)*((25/3*sqrt(d+x^2*exp(1))*(d+x^2*exp(1))*exp
(1)^6-50/3*sqrt(d+x^2*exp(1))*(d+x^2*exp(1))*exp(1)^5*exp(1/2)^2+25/3*sqrt(d+x^2*exp(1))*(d+x^2*exp(1))*exp(1)
^4*exp(1/2)^4-50*sqrt(d+x^2*exp(1))*d*exp(1)^6+75*sqrt(d+x^2*exp(1))*d*exp(1)^5*exp(1/2)^2-25*sqrt(d+x^2*exp(1
))*d*exp(1)^4*exp(1/2)^4)/(125*exp(1)^9-375*exp(1)^8*exp(1/2)^2+375*exp(1)^7*exp(1/2)^4-125*exp(1)^6*exp(1/2)^
6)+2*d^2/2/(5*exp(1)^2-10*exp(1)*exp(1/2)^2+5*exp(1/2)^4)/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 = 176, normalized size = 1.93 \[ \frac {x^{5} \arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{5}+\frac {e^{\frac {3}{2}} \left (\frac {x^{6} \sqrt {e \,x^{2}+d}}{7 e}-\frac {6 d \left (\frac {x^{4} \sqrt {e \,x^{2}+d}}{5 e}-\frac {4 d \left (\frac {x^{2} \sqrt {e \,x^{2}+d}}{3 e}-\frac {2 d \sqrt {e \,x^{2}+d}}{3 e^{2}}\right )}{5 e}\right )}{7 e}\right )}{5 d}-\frac {\sqrt {e}\, \left (\frac {x^{4} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 e}-\frac {4 d \left (\frac {x^{2} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 e}-\frac {2 d \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 e^{2}}\right )}{7 e}\right )}{5 d} \]

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)

[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 = 0.33, size = 127, normalized size = 1.40 \[ \frac {1}{5} \, x^{5} \operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right ) - \frac {15 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} - 42 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2}}{525 \, d e^{\frac {5}{2}}} + \frac {5 \, {\left (e x^{2} + d\right )}^{\frac {7}{2}} - 21 \, {\left (e x^{2} + d\right )}^{\frac {5}{2}} d + 35 \, {\left (e x^{2} + d\right )}^{\frac {3}{2}} d^{2} - 35 \, \sqrt {e x^{2} + d} d^{3}}{175 \, d e^{\frac {5}{2}}} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int x^4\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^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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sympy [A]  time = 2.94, size = 90, normalized size = 0.99 \[ \begin {cases} - \frac {8 d^{2} \sqrt {d + e x^{2}}}{75 e^{\frac {5}{2}}} + \frac {4 d x^{2} \sqrt {d + e x^{2}}}{75 e^{\frac {3}{2}}} + \frac {x^{5} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{5} - \frac {x^{4} \sqrt {d + e x^{2}}}{25 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]

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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