3.23 \(\int x^{7/2} \tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}}) \, dx\)

Optimal. Leaf size=297 \[ -\frac {14 d^{9/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{135 e^{9/4} \sqrt {d+e x^2}}+\frac {28 d^{9/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{135 e^{9/4} \sqrt {d+e x^2}}-\frac {28 d^2 \sqrt {x} \sqrt {d+e x^2}}{135 e^2 \left (\sqrt {d}+\sqrt {e} x\right )}+\frac {28 d x^{3/2} \sqrt {d+e x^2}}{405 e^{3/2}}-\frac {4 x^{7/2} \sqrt {d+e x^2}}{81 \sqrt {e}}+\frac {2}{9} x^{9/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]

[Out]

2/9*x^(9/2)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+28/405*d*x^(3/2)*(e*x^2+d)^(1/2)/e^(3/2)-4/81*x^(7/2)*(e*x^2+d)
^(1/2)/e^(1/2)-28/135*d^2*x^(1/2)*(e*x^2+d)^(1/2)/e^2/(d^(1/2)+x*e^(1/2))+28/135*d^(9/4)*(cos(2*arctan(e^(1/4)
*x^(1/2)/d^(1/4)))^2)^(1/2)/cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))*EllipticE(sin(2*arctan(e^(1/4)*x^(1/2)/d^(1
/4))),1/2*2^(1/2))*(d^(1/2)+x*e^(1/2))*((e*x^2+d)/(d^(1/2)+x*e^(1/2))^2)^(1/2)/e^(9/4)/(e*x^2+d)^(1/2)-14/135*
d^(9/4)*(cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))^2)^(1/2)/cos(2*arctan(e^(1/4)*x^(1/2)/d^(1/4)))*EllipticF(sin(
2*arctan(e^(1/4)*x^(1/2)/d^(1/4))),1/2*2^(1/2))*(d^(1/2)+x*e^(1/2))*((e*x^2+d)/(d^(1/2)+x*e^(1/2))^2)^(1/2)/e^
(9/4)/(e*x^2+d)^(1/2)

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Rubi [A]  time = 0.19, antiderivative size = 297, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {6221, 321, 329, 305, 220, 1196} \[ -\frac {28 d^2 \sqrt {x} \sqrt {d+e x^2}}{135 e^2 \left (\sqrt {d}+\sqrt {e} x\right )}-\frac {14 d^{9/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{135 e^{9/4} \sqrt {d+e x^2}}+\frac {28 d^{9/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{135 e^{9/4} \sqrt {d+e x^2}}+\frac {28 d x^{3/2} \sqrt {d+e x^2}}{405 e^{3/2}}-\frac {4 x^{7/2} \sqrt {d+e x^2}}{81 \sqrt {e}}+\frac {2}{9} x^{9/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]

Antiderivative was successfully verified.

[In]

Int[x^(7/2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(28*d*x^(3/2)*Sqrt[d + e*x^2])/(405*e^(3/2)) - (4*x^(7/2)*Sqrt[d + e*x^2])/(81*Sqrt[e]) - (28*d^2*Sqrt[x]*Sqrt
[d + e*x^2])/(135*e^2*(Sqrt[d] + Sqrt[e]*x)) + (2*x^(9/2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/9 + (28*d^(9/4
)*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*EllipticE[2*ArcTan[(e^(1/4)*Sqrt[x])/d^(1/4)
], 1/2])/(135*e^(9/4)*Sqrt[d + e*x^2]) - (14*d^(9/4)*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]
*x)^2]*EllipticF[2*ArcTan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(135*e^(9/4)*Sqrt[d + e*x^2])

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

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^{7/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx &=\frac {2}{9} x^{9/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{9} \left (2 \sqrt {e}\right ) \int \frac {x^{9/2}}{\sqrt {d+e x^2}} \, dx\\ &=-\frac {4 x^{7/2} \sqrt {d+e x^2}}{81 \sqrt {e}}+\frac {2}{9} x^{9/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {(14 d) \int \frac {x^{5/2}}{\sqrt {d+e x^2}} \, dx}{81 \sqrt {e}}\\ &=\frac {28 d x^{3/2} \sqrt {d+e x^2}}{405 e^{3/2}}-\frac {4 x^{7/2} \sqrt {d+e x^2}}{81 \sqrt {e}}+\frac {2}{9} x^{9/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {\left (14 d^2\right ) \int \frac {\sqrt {x}}{\sqrt {d+e x^2}} \, dx}{135 e^{3/2}}\\ &=\frac {28 d x^{3/2} \sqrt {d+e x^2}}{405 e^{3/2}}-\frac {4 x^{7/2} \sqrt {d+e x^2}}{81 \sqrt {e}}+\frac {2}{9} x^{9/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {\left (28 d^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{135 e^{3/2}}\\ &=\frac {28 d x^{3/2} \sqrt {d+e x^2}}{405 e^{3/2}}-\frac {4 x^{7/2} \sqrt {d+e x^2}}{81 \sqrt {e}}+\frac {2}{9} x^{9/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {\left (28 d^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{135 e^2}+\frac {\left (28 d^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {e} x^2}{\sqrt {d}}}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{135 e^2}\\ &=\frac {28 d x^{3/2} \sqrt {d+e x^2}}{405 e^{3/2}}-\frac {4 x^{7/2} \sqrt {d+e x^2}}{81 \sqrt {e}}-\frac {28 d^2 \sqrt {x} \sqrt {d+e x^2}}{135 e^2 \left (\sqrt {d}+\sqrt {e} x\right )}+\frac {2}{9} x^{9/2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {28 d^{9/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{135 e^{9/4} \sqrt {d+e x^2}}-\frac {14 d^{9/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{135 e^{9/4} \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.15, size = 124, normalized size = 0.42 \[ \frac {2 x^{3/2} \left (-14 d^2 \sqrt {\frac {e x^2}{d}+1} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {e x^2}{d}\right )+14 d^2+45 e^{3/2} x^3 \sqrt {d+e x^2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+4 d e x^2-10 e^2 x^4\right )}{405 e^{3/2} \sqrt {d+e x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^(7/2)*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]],x]

[Out]

(2*x^(3/2)*(14*d^2 + 4*d*e*x^2 - 10*e^2*x^4 + 45*e^(3/2)*x^3*Sqrt[d + e*x^2]*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^
2]] - 14*d^2*Sqrt[1 + (e*x^2)/d]*Hypergeometric2F1[1/2, 3/4, 7/4, -((e*x^2)/d)]))/(405*e^(3/2)*Sqrt[d + e*x^2]
)

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (x^{\frac {7}{2}} \operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right ), x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="fricas")

[Out]

integral(x^(7/2)*arctanh(sqrt(e)*x/sqrt(e*x^2 + d)), x)

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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^(7/2)*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:exp(
1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=e
xp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp(1/2)^2=exp(1)exp
(1/2)^2=exp(1)exp(1/2)^2=exp(1)Unable to transpose Error: Bad Argument Value

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maple [F(-2)]  time = 180.00, size = 0, normalized size = 0.00 \[ \int x^{\frac {7}{2}} \arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x)

[Out]

int(x^(7/2)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{9} \, x^{\frac {9}{2}} \log \left (\sqrt {e} x + \sqrt {e x^{2} + d}\right ) - \frac {1}{9} \, x^{\frac {9}{2}} \log \left (-\sqrt {e} x + \sqrt {e x^{2} + d}\right ) - 2 \, d \sqrt {e} \int -\frac {x e^{\left (\frac {1}{2} \, \log \left (e x^{2} + d\right ) + \frac {7}{2} \, \log \relax (x)\right )}}{9 \, {\left (e^{2} x^{4} + d e x^{2} - {\left (e x^{2} + d\right )}^{2}\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^(7/2)*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2)),x, algorithm="maxima")

[Out]

1/9*x^(9/2)*log(sqrt(e)*x + sqrt(e*x^2 + d)) - 1/9*x^(9/2)*log(-sqrt(e)*x + sqrt(e*x^2 + d)) - 2*d*sqrt(e)*int
egrate(-1/9*x*e^(1/2*log(e*x^2 + d) + 7/2*log(x))/(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 = -0.00 \[ \int x^{7/2}\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^(7/2)*atanh((e^(1/2)*x)/(d + e*x^2)^(1/2)),x)

[Out]

int(x^(7/2)*atanh((e^(1/2)*x)/(d + e*x^2)^(1/2)), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**(7/2)*atanh(x*e**(1/2)/(e*x**2+d)**(1/2)),x)

[Out]

Timed out

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