3.1.1 \(\int x^5 \tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}}) \, dx\) [1]

Optimal. Leaf size=127 \[ -\frac {5 d^2 x \sqrt {d+e x^2}}{96 e^{5/2}}+\frac {5 d x^3 \sqrt {d+e x^2}}{144 e^{3/2}}-\frac {x^5 \sqrt {d+e x^2}}{36 \sqrt {e}}+\frac {5 d^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{96 e^3}+\frac {1}{6} x^6 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]

[Out]

5/96*d^3*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/e^3+1/6*x^6*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))-5/96*d^2*x*(e*x^2+d
)^(1/2)/e^(5/2)+5/144*d*x^3*(e*x^2+d)^(1/2)/e^(3/2)-1/36*x^5*(e*x^2+d)^(1/2)/e^(1/2)

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Rubi [A]
time = 0.04, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 6, 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 {5 d^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{96 e^3}-\frac {5 d^2 x \sqrt {d+e x^2}}{96 e^{5/2}}+\frac {5 d x^3 \sqrt {d+e x^2}}{144 e^{3/2}}+\frac {1}{6} x^6 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {x^5 \sqrt {d+e x^2}}{36 \sqrt {e}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^5*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(-5*d^2*x*Sqrt[d + e*x^2])/(96*e^(5/2)) + (5*d*x^3*Sqrt[d + e*x^2])/(144*e^(3/2)) - (x^5*Sqrt[d + e*x^2])/(36*
Sqrt[e]) + (5*d^3*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(96*e^3) + (x^6*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/
6

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^5 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx &=\frac {1}{6} x^6 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{6} \sqrt {e} \int \frac {x^6}{\sqrt {d+e x^2}} \, dx\\ &=-\frac {x^5 \sqrt {d+e x^2}}{36 \sqrt {e}}+\frac {1}{6} x^6 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {(5 d) \int \frac {x^4}{\sqrt {d+e x^2}} \, dx}{36 \sqrt {e}}\\ &=\frac {5 d x^3 \sqrt {d+e x^2}}{144 e^{3/2}}-\frac {x^5 \sqrt {d+e x^2}}{36 \sqrt {e}}+\frac {1}{6} x^6 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {\left (5 d^2\right ) \int \frac {x^2}{\sqrt {d+e x^2}} \, dx}{48 e^{3/2}}\\ &=-\frac {5 d^2 x \sqrt {d+e x^2}}{96 e^{5/2}}+\frac {5 d x^3 \sqrt {d+e x^2}}{144 e^{3/2}}-\frac {x^5 \sqrt {d+e x^2}}{36 \sqrt {e}}+\frac {1}{6} x^6 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {\left (5 d^3\right ) \int \frac {1}{\sqrt {d+e x^2}} \, dx}{96 e^{5/2}}\\ &=-\frac {5 d^2 x \sqrt {d+e x^2}}{96 e^{5/2}}+\frac {5 d x^3 \sqrt {d+e x^2}}{144 e^{3/2}}-\frac {x^5 \sqrt {d+e x^2}}{36 \sqrt {e}}+\frac {1}{6} x^6 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {\left (5 d^3\right ) \text {Subst}\left (\int \frac {1}{1-e x^2} \, dx,x,\frac {x}{\sqrt {d+e x^2}}\right )}{96 e^{5/2}}\\ &=-\frac {5 d^2 x \sqrt {d+e x^2}}{96 e^{5/2}}+\frac {5 d x^3 \sqrt {d+e x^2}}{144 e^{3/2}}-\frac {x^5 \sqrt {d+e x^2}}{36 \sqrt {e}}+\frac {5 d^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{96 e^3}+\frac {1}{6} x^6 \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 = 99, normalized size = 0.78 \begin {gather*} \frac {\sqrt {e} x \sqrt {d+e x^2} \left (-15 d^2+10 d e x^2-8 e^2 x^4\right )+48 e^3 x^6 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+15 d^3 \log \left (\sqrt {e} x+\sqrt {d+e x^2}\right )}{288 e^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^5*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(Sqrt[e]*x*Sqrt[d + e*x^2]*(-15*d^2 + 10*d*e*x^2 - 8*e^2*x^4) + 48*e^3*x^6*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]
] + 15*d^3*Log[Sqrt[e]*x + Sqrt[d + e*x^2]])/(288*e^3)

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(252\) vs. \(2(97)=194\).
time = 0.04, size = 253, normalized size = 1.99

method	result	size
default	\(\frac {x^{6} \arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{6}+\frac {e^{\frac {3}{2}} \left (\frac {x^{7} \sqrt {e \,x^{2}+d}}{8 e}-\frac {7 d \left (\frac {x^{5} \sqrt {e \,x^{2}+d}}{6 e}-\frac {5 d \left (\frac {x^{3} \sqrt {e \,x^{2}+d}}{4 e}-\frac {3 d \left (\frac {x \sqrt {e \,x^{2}+d}}{2 e}-\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 e^{\frac {3}{2}}}\right )}{4 e}\right )}{6 e}\right )}{8 e}\right )}{6 d}-\frac {\sqrt {e}\, \left (\frac {x^{5} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{8 e}-\frac {5 d \left (\frac {x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{6 e}-\frac {d \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4 e}-\frac {d \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{4 e}\right )}{2 e}\right )}{8 e}\right )}{6 d}\)	\(253\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x,method=_RETURNVERBOSE)

[Out]

1/6*x^6*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/6*e^(3/2)/d*(1/8*x^7/e*(e*x^2+d)^(1/2)-7/8*d/e*(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))))))-1/6*e^(1/2)/d*(1/8*x^5*(e*x^2+d)^(3/2)/e-5/8*d/e*(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^5*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="maxima")

[Out]

1/12*x^6*log(x*e^(1/2) + sqrt(x^2*e + d)) - 1/12*x^6*log(-x*e^(1/2) + sqrt(x^2*e + d)) - 1/2*d*integrate(-1/3*
x^6*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. 380 vs. \(2 (96) = 192\).
time = 0.35, size = 380, normalized size = 2.99 \begin {gather*} \frac {3 \, {\left (16 \, x^{6} \cosh \left (\frac {1}{2}\right )^{6} + 96 \, x^{6} \cosh \left (\frac {1}{2}\right )^{5} \sinh \left (\frac {1}{2}\right ) + 240 \, x^{6} \cosh \left (\frac {1}{2}\right )^{4} \sinh \left (\frac {1}{2}\right )^{2} + 320 \, x^{6} \cosh \left (\frac {1}{2}\right )^{3} \sinh \left (\frac {1}{2}\right )^{3} + 240 \, x^{6} \cosh \left (\frac {1}{2}\right )^{2} \sinh \left (\frac {1}{2}\right )^{4} + 96 \, x^{6} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{5} + 16 \, x^{6} \sinh \left (\frac {1}{2}\right )^{6} + 5 \, d^{3}\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 (8 \, x^{5} \cosh \left (\frac {1}{2}\right )^{5} + 40 \, x^{5} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{4} + 8 \, x^{5} \sinh \left (\frac {1}{2}\right )^{5} - 10 \, d x^{3} \cosh \left (\frac {1}{2}\right )^{3} + 15 \, d^{2} x \cosh \left (\frac {1}{2}\right ) + 10 \, {\left (8 \, x^{5} \cosh \left (\frac {1}{2}\right )^{2} - d x^{3}\right )} \sinh \left (\frac {1}{2}\right )^{3} + 10 \, {\left (8 \, x^{5} \cosh \left (\frac {1}{2}\right )^{3} - 3 \, d x^{3} \cosh \left (\frac {1}{2}\right )\right )} \sinh \left (\frac {1}{2}\right )^{2} + 5 \, {\left (8 \, x^{5} \cosh \left (\frac {1}{2}\right )^{4} - 6 \, d x^{3} \cosh \left (\frac {1}{2}\right )^{2} + 3 \, d^{2} 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 )}}}{576 \, {\left (\cosh \left (\frac {1}{2}\right )^{6} + 6 \, \cosh \left (\frac {1}{2}\right )^{5} \sinh \left (\frac {1}{2}\right ) + 15 \, \cosh \left (\frac {1}{2}\right )^{4} \sinh \left (\frac {1}{2}\right )^{2} + 20 \, \cosh \left (\frac {1}{2}\right )^{3} \sinh \left (\frac {1}{2}\right )^{3} + 15 \, \cosh \left (\frac {1}{2}\right )^{2} \sinh \left (\frac {1}{2}\right )^{4} + 6 \, \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{5} + \sinh \left (\frac {1}{2}\right )^{6}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

1/576*(3*(16*x^6*cosh(1/2)^6 + 96*x^6*cosh(1/2)^5*sinh(1/2) + 240*x^6*cosh(1/2)^4*sinh(1/2)^2 + 320*x^6*cosh(1
/2)^3*sinh(1/2)^3 + 240*x^6*cosh(1/2)^2*sinh(1/2)^4 + 96*x^6*cosh(1/2)*sinh(1/2)^5 + 16*x^6*sinh(1/2)^6 + 5*d^
3)*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*(8*x^5*cosh(1/2)^5 + 40*x^5*
cosh(1/2)*sinh(1/2)^4 + 8*x^5*sinh(1/2)^5 - 10*d*x^3*cosh(1/2)^3 + 15*d^2*x*cosh(1/2) + 10*(8*x^5*cosh(1/2)^2
- d*x^3)*sinh(1/2)^3 + 10*(8*x^5*cosh(1/2)^3 - 3*d*x^3*cosh(1/2))*sinh(1/2)^2 + 5*(8*x^5*cosh(1/2)^4 - 6*d*x^3
*cosh(1/2)^2 + 3*d^2*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)^6 + 6*cosh(1/2)^5*sinh(1/2) + 15*cosh(1/2)^4*sinh(1/2)^2 + 20*cosh(1/2)^3*sinh(1/2)^3 + 15*cosh(1/2
)^2*sinh(1/2)^4 + 6*cosh(1/2)*sinh(1/2)^5 + sinh(1/2)^6)

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Sympy [A]
time = 1.64, size = 121, normalized size = 0.95 \begin {gather*} \begin {cases} \frac {5 d^{3} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{96 e^{3}} - \frac {5 d^{2} x \sqrt {d + e x^{2}}}{96 e^{\frac {5}{2}}} + \frac {5 d x^{3} \sqrt {d + e x^{2}}}{144 e^{\frac {3}{2}}} + \frac {x^{6} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{6} - \frac {x^{5} \sqrt {d + e x^{2}}}{36 \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**5*atanh(x*e**(1/2)/(e*x**2+d)**(1/2)),x)

[Out]

Piecewise((5*d**3*atanh(sqrt(e)*x/sqrt(d + e*x**2))/(96*e**3) - 5*d**2*x*sqrt(d + e*x**2)/(96*e**(5/2)) + 5*d*
x**3*sqrt(d + e*x**2)/(144*e**(3/2)) + x**6*atanh(sqrt(e)*x/sqrt(d + e*x**2))/6 - x**5*sqrt(d + e*x**2)/(36*sq
rt(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^5*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^5\,\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^5*atanh((e^(1/2)*x)/(d + e*x^2)^(1/2)),x)

[Out]

int(x^5*atanh((e^(1/2)*x)/(d + e*x^2)^(1/2)), x)

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