3.1.23 \(\int x^{7/2} \tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}}) \, dx\) [23]

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

[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 = 0.14, 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 = {6356, 327, 335, 311, 226, 1210} \begin {gather*} -\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 \text {ArcTan}\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 \text {ArcTan}\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 ) \end {gather*}

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 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 311

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 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 335

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 1210

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)]/(q*Sqrt[a + c*x^4
]))*EllipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

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^{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 ) \text {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 ) \text {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 ) \text {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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.09, size = 124, normalized size = 0.42 \begin {gather*} \frac {2 x^{3/2} \left (14 d^2+4 d e x^2-10 e^2 x^4+45 e^{3/2} x^3 \sqrt {d+e x^2} \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-14 d^2 \sqrt {1+\frac {e x^2}{d}} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\frac {e x^2}{d}\right )\right )}{405 e^{3/2} \sqrt {d+e x^2}} \end {gather*}

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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Maple [F]
time = 0.31, 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 \begin {gather*} \text {Failed to integrate} \end {gather*}

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(x*e^(1/2) + sqrt(x^2*e + d)) - 1/9*x^(9/2)*log(-x*e^(1/2) + sqrt(x^2*e + d)) - 2*d*integrate(-
1/9*x*e^(1/2*log(x^2*e + d) + 7/2*log(x) + 1/2)/(x^4*e^2 + d*x^2*e - (x^2*e + d)^2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.11, size = 312, normalized size = 1.05 \begin {gather*} \frac {84 \, d^{2} {\rm weierstrassZeta}\left (-\frac {4 \, d}{\cosh \left (\frac {1}{2}\right )^{2} + 2 \, \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right ) + \sinh \left (\frac {1}{2}\right )^{2}}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, d}{\cosh \left (\frac {1}{2}\right )^{2} + 2 \, \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right ) + \sinh \left (\frac {1}{2}\right )^{2}}, 0, x\right )\right ) + 45 \, {\left (x^{4} \cosh \left (\frac {1}{2}\right )^{4} + 4 \, x^{4} \cosh \left (\frac {1}{2}\right )^{3} \sinh \left (\frac {1}{2}\right ) + 6 \, x^{4} \cosh \left (\frac {1}{2}\right )^{2} \sinh \left (\frac {1}{2}\right )^{2} + 4 \, x^{4} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{3} + x^{4} \sinh \left (\frac {1}{2}\right )^{4}\right )} \sqrt {x} \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 ) - 4 \, {\left (5 \, x^{3} \cosh \left (\frac {1}{2}\right )^{3} + 15 \, x^{3} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{2} + 5 \, x^{3} \sinh \left (\frac {1}{2}\right )^{3} - 7 \, d x \cosh \left (\frac {1}{2}\right ) + {\left (15 \, x^{3} \cosh \left (\frac {1}{2}\right )^{2} - 7 \, d x\right )} \sinh \left (\frac {1}{2}\right )\right )} \sqrt {x} \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 )}}}{405 \, {\left (\cosh \left (\frac {1}{2}\right )^{4} + 4 \, \cosh \left (\frac {1}{2}\right )^{3} \sinh \left (\frac {1}{2}\right ) + 6 \, \cosh \left (\frac {1}{2}\right )^{2} \sinh \left (\frac {1}{2}\right )^{2} + 4 \, \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{3} + \sinh \left (\frac {1}{2}\right )^{4}\right )}} \end {gather*}

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]

1/405*(84*d^2*weierstrassZeta(-4*d/(cosh(1/2)^2 + 2*cosh(1/2)*sinh(1/2) + sinh(1/2)^2), 0, weierstrassPInverse
(-4*d/(cosh(1/2)^2 + 2*cosh(1/2)*sinh(1/2) + sinh(1/2)^2), 0, x)) + 45*(x^4*cosh(1/2)^4 + 4*x^4*cosh(1/2)^3*si
nh(1/2) + 6*x^4*cosh(1/2)^2*sinh(1/2)^2 + 4*x^4*cosh(1/2)*sinh(1/2)^3 + x^4*sinh(1/2)^4)*sqrt(x)*log((2*x^2*co
sh(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) - 4*(5*x^3*cosh(1/2)^3 + 15*x^3*cosh(1/2)*sinh(1
/2)^2 + 5*x^3*sinh(1/2)^3 - 7*d*x*cosh(1/2) + (15*x^3*cosh(1/2)^2 - 7*d*x)*sinh(1/2))*sqrt(x)*sqrt(((x^2 + d)*
cosh(1/2) + (x^2 - d)*sinh(1/2))/(cosh(1/2) - sinh(1/2))))/(cosh(1/2)^4 + 4*cosh(1/2)^3*sinh(1/2) + 6*cosh(1/2
)^2*sinh(1/2)^2 + 4*cosh(1/2)*sinh(1/2)^3 + sinh(1/2)^4)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

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]

Exception raised: SystemError >> excessive stack use: stack is 3876 deep

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int x^{7/2}\,\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^(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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