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

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

[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 = 0.05, 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 = {6221, 321, 217, 206} \[ -\frac {5 d^2 x \sqrt {d+e x^2}}{96 e^{5/2}}+\frac {5 d^3 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{96 e^3}+\frac {5 d x^3 \sqrt {d+e x^2}}{144 e^{3/2}}-\frac {x^5 \sqrt {d+e x^2}}{36 \sqrt {e}}+\frac {1}{6} x^6 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]

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 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^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 ) \operatorname {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 = 0.08, size = 99, normalized size = 0.78 \[ \frac {15 d^3 \log \left (\sqrt {d+e x^2}+\sqrt {e} x\right )+\sqrt {e} x \sqrt {d+e x^2} \left (-15 d^2+10 d e x^2-8 e^2 x^4\right )+48 e^3 x^6 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{288 e^3} \]

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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fricas [A]  time = 0.60, size = 86, normalized size = 0.68 \[ -\frac {2 \, {\left (8 \, e^{2} x^{5} - 10 \, d e x^{3} + 15 \, d^{2} x\right )} \sqrt {e x^{2} + d} \sqrt {e} - 3 \, {\left (16 \, e^{3} x^{6} + 5 \, d^{3}\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{576 \, e^{3}} \]

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*(2*(8*e^2*x^5 - 10*d*e*x^3 + 15*d^2*x)*sqrt(e*x^2 + d)*sqrt(e) - 3*(16*e^3*x^6 + 5*d^3)*log((2*e*x^2 +
2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d))/e^3

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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^5*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 = 172, normalized size = 1.35 \[ \frac {x^{6} \arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{6}+\frac {\sqrt {e}\, x^{7} \sqrt {e \,x^{2}+d}}{48 d}-\frac {7 x^{5} \sqrt {e \,x^{2}+d}}{288 \sqrt {e}}+\frac {35 d \,x^{3} \sqrt {e \,x^{2}+d}}{1152 e^{\frac {3}{2}}}-\frac {5 d^{2} x \sqrt {e \,x^{2}+d}}{128 e^{\frac {5}{2}}}+\frac {5 d^{3} \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{96 e^{3}}-\frac {x^{5} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{48 \sqrt {e}\, d}+\frac {5 x^{3} \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{288 e^{\frac {3}{2}}}-\frac {5 d x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{384 e^{\frac {5}{2}}} \]

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)

[Out]

1/6*x^6*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/48*e^(1/2)/d*x^7*(e*x^2+d)^(1/2)-7/288*x^5*(e*x^2+d)^(1/2)/e^(1/2
)+35/1152*d*x^3*(e*x^2+d)^(1/2)/e^(3/2)-5/128*d^2*x*(e*x^2+d)^(1/2)/e^(5/2)+5/96/e^3*d^3*ln(x*e^(1/2)+(e*x^2+d
)^(1/2))-1/48/e^(1/2)/d*x^5*(e*x^2+d)^(3/2)+5/288/e^(3/2)*x^3*(e*x^2+d)^(3/2)-5/384/e^(5/2)*d*x*(e*x^2+d)^(3/2
)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{12} \, x^{6} \log \left (\sqrt {e} x + \sqrt {e x^{2} + d}\right ) - \frac {1}{12} \, x^{6} \log \left (-\sqrt {e} x + \sqrt {e x^{2} + d}\right ) - \frac {1}{2} \, d \sqrt {e} \int -\frac {\sqrt {e x^{2} + d} x^{6}}{3 \, {\left (e^{2} x^{4} + d e x^{2} - {\left (e x^{2} + d\right )}^{2}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*x^6*log(sqrt(e)*x + sqrt(e*x^2 + d)) - 1/12*x^6*log(-sqrt(e)*x + sqrt(e*x^2 + d)) - 1/2*d*sqrt(e)*integra
te(-1/3*sqrt(e*x^2 + d)*x^6/(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 \[ \int x^5\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^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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sympy [A]  time = 5.03, size = 121, normalized size = 0.95 \[ \begin {cases} \frac {5 d^{3} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{96 e^{3}} - \frac {5 d^{2} x \sqrt {d + e x^{2}}}{96 e^{\frac {5}{2}}} + \frac {5 d x^{3} \sqrt {d + e x^{2}}}{144 e^{\frac {3}{2}}} + \frac {x^{6} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{6} - \frac {x^{5} \sqrt {d + e x^{2}}}{36 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]

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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