3.1.22 \(\int \frac {\tanh ^{-1}(\frac {\sqrt {e} x}{\sqrt {d+e x^2}})}{x^{15/2}} \, dx\) [22]

Optimal. Leaf size=201 \[ -\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}-\frac {30 e^{13/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \text {ArcTan}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{1001 d^{13/4} \sqrt {d+e x^2}} \]

[Out]

-2/13*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(13/2)+36/1001*e^(3/2)*(e*x^2+d)^(1/2)/d^2/x^(7/2)-60/1001*e^(5/2)*
(e*x^2+d)^(1/2)/d^3/x^(3/2)-4/143*e^(1/2)*(e*x^2+d)^(1/2)/d/x^(11/2)-30/1001*e^(13/4)*(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)/d^(13/4)/(e*x^2+d)^(1/2)

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Rubi [A]
time = 0.07, antiderivative size = 201, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {6356, 331, 335, 226} \begin {gather*} -\frac {30 e^{13/4} \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 )}{1001 d^{13/4} \sqrt {d+e x^2}}-\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(15/2),x]

[Out]

(-4*Sqrt[e]*Sqrt[d + e*x^2])/(143*d*x^(11/2)) + (36*e^(3/2)*Sqrt[d + e*x^2])/(1001*d^2*x^(7/2)) - (60*e^(5/2)*
Sqrt[d + e*x^2])/(1001*d^3*x^(3/2)) - (2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(13*x^(13/2)) - (30*e^(13/4)*(S
qrt[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])/(1001*d^(13/4)*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 331

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^(p + 1)/(a*c
*(m + 1))), x] - Dist[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && 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 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 \frac {\tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{15/2}} \, dx &=-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}+\frac {1}{13} \left (2 \sqrt {e}\right ) \int \frac {1}{x^{13/2} \sqrt {d+e x^2}} \, dx\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}-\frac {\left (18 e^{3/2}\right ) \int \frac {1}{x^{9/2} \sqrt {d+e x^2}} \, dx}{143 d}\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}+\frac {\left (90 e^{5/2}\right ) \int \frac {1}{x^{5/2} \sqrt {d+e x^2}} \, dx}{1001 d^2}\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}-\frac {\left (30 e^{7/2}\right ) \int \frac {1}{\sqrt {x} \sqrt {d+e x^2}} \, dx}{1001 d^3}\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}-\frac {\left (60 e^{7/2}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {d+e x^4}} \, dx,x,\sqrt {x}\right )}{1001 d^3}\\ &=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}}-\frac {30 e^{13/4} \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 )}{1001 d^{13/4} \sqrt {d+e x^2}}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 0.28, size = 163, normalized size = 0.81 \begin {gather*} \frac {2 \left (-\frac {2 \sqrt {e} x \sqrt {d+e x^2} \left (7 d^2-9 d e x^2+15 e^2 x^4\right )}{d^3}-77 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {30 \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} e^4 \sqrt {1+\frac {d}{e x^2}} x^{15/2} F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right )\right |-1\right )}{d^{7/2} \sqrt {d+e x^2}}\right )}{1001 x^{13/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/x^(15/2),x]

[Out]

(2*((-2*Sqrt[e]*x*Sqrt[d + e*x^2]*(7*d^2 - 9*d*e*x^2 + 15*e^2*x^4))/d^3 - 77*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^
2]] - (30*Sqrt[(I*Sqrt[d])/Sqrt[e]]*e^4*Sqrt[1 + d/(e*x^2)]*x^(15/2)*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt
[e]]/Sqrt[x]], -1])/(d^(7/2)*Sqrt[d + e*x^2])))/(1001*x^(13/2))

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Maple [F]
time = 0.29, size = 0, normalized size = 0.00 \[\int \frac {\arctanh \left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {15}{2}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x)

[Out]

int(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),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(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x, algorithm="maxima")

[Out]

2*d*integrate(-1/13*x*e^(1/2*log(x^2*e + d) + 1/2)/((x^4*e^2 + d*x^2*e)*x^(15/2) - (x^2*e + d)*e^(log(x^2*e +
d) + 15/2*log(x))), x) - 1/13*log(x*e^(1/2) + sqrt(x^2*e + d))/x^(13/2) + 1/13*log(-x*e^(1/2) + sqrt(x^2*e + d
))/x^(13/2)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.10, size = 355, normalized size = 1.77 \begin {gather*} -\frac {77 \, d^{3} \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 (15 \, x^{5} \cosh \left (\frac {1}{2}\right )^{5} + 75 \, x^{5} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{4} + 15 \, x^{5} \sinh \left (\frac {1}{2}\right )^{5} - 9 \, d x^{3} \cosh \left (\frac {1}{2}\right )^{3} + 7 \, d^{2} x \cosh \left (\frac {1}{2}\right ) + 3 \, {\left (50 \, x^{5} \cosh \left (\frac {1}{2}\right )^{2} - 3 \, d x^{3}\right )} \sinh \left (\frac {1}{2}\right )^{3} + 3 \, {\left (50 \, x^{5} \cosh \left (\frac {1}{2}\right )^{3} - 9 \, d x^{3} \cosh \left (\frac {1}{2}\right )\right )} \sinh \left (\frac {1}{2}\right )^{2} + {\left (75 \, x^{5} \cosh \left (\frac {1}{2}\right )^{4} - 27 \, d x^{3} \cosh \left (\frac {1}{2}\right )^{2} + 7 \, d^{2} 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 )}} + 60 \, {\left (x^{7} \cosh \left (\frac {1}{2}\right )^{6} + 6 \, x^{7} \cosh \left (\frac {1}{2}\right )^{5} \sinh \left (\frac {1}{2}\right ) + 15 \, x^{7} \cosh \left (\frac {1}{2}\right )^{4} \sinh \left (\frac {1}{2}\right )^{2} + 20 \, x^{7} \cosh \left (\frac {1}{2}\right )^{3} \sinh \left (\frac {1}{2}\right )^{3} + 15 \, x^{7} \cosh \left (\frac {1}{2}\right )^{2} \sinh \left (\frac {1}{2}\right )^{4} + 6 \, x^{7} \cosh \left (\frac {1}{2}\right ) \sinh \left (\frac {1}{2}\right )^{5} + x^{7} \sinh \left (\frac {1}{2}\right )^{6}\right )} {\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 )}{1001 \, d^{3} x^{7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x, algorithm="fricas")

[Out]

-1/1001*(77*d^3*sqrt(x)*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) + 4*(15*x^5
*cosh(1/2)^5 + 75*x^5*cosh(1/2)*sinh(1/2)^4 + 15*x^5*sinh(1/2)^5 - 9*d*x^3*cosh(1/2)^3 + 7*d^2*x*cosh(1/2) + 3
*(50*x^5*cosh(1/2)^2 - 3*d*x^3)*sinh(1/2)^3 + 3*(50*x^5*cosh(1/2)^3 - 9*d*x^3*cosh(1/2))*sinh(1/2)^2 + (75*x^5
*cosh(1/2)^4 - 27*d*x^3*cosh(1/2)^2 + 7*d^2*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))) + 60*(x^7*cosh(1/2)^6 + 6*x^7*cosh(1/2)^5*sinh(1/2) + 15*x^7*cosh(1/2)^4*sinh(1/
2)^2 + 20*x^7*cosh(1/2)^3*sinh(1/2)^3 + 15*x^7*cosh(1/2)^2*sinh(1/2)^4 + 6*x^7*cosh(1/2)*sinh(1/2)^5 + x^7*sin
h(1/2)^6)*weierstrassPInverse(-4*d/(cosh(1/2)^2 + 2*cosh(1/2)*sinh(1/2) + sinh(1/2)^2), 0, x))/(d^3*x^7)

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Sympy [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(atanh(x*e**(1/2)/(e*x**2+d)**(1/2))/x**(15/2),x)

[Out]

Timed out

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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(arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(15/2),x, algorithm="giac")

[Out]

integrate(arctanh(sqrt(e)*x/sqrt(e*x^2 + d))/x^(15/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{15/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(15/2),x)

[Out]

int(atanh((e^(1/2)*x)/(d + e*x^2)^(1/2))/x^(15/2), x)

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