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

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

[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 = 0.10, 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 = {6221, 325, 329, 220} \[ -\frac {60 e^{5/2} \sqrt {d+e x^2}}{1001 d^3 x^{3/2}}+\frac {36 e^{3/2} \sqrt {d+e x^2}}{1001 d^2 x^{7/2}}-\frac {30 e^{13/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{1001 d^{13/4} \sqrt {d+e x^2}}-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{143 d x^{11/2}}-\frac {2 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{13 x^{13/2}} \]

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 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 325

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 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 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 \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 ) \operatorname {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]  time = 0.47, size = 163, normalized size = 0.81 \[ \frac {2 \left (-\frac {30 e^4 x^{15/2} \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} \sqrt {\frac {d}{e x^2}+1} F\left (\left .i \sinh ^{-1}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right )\right |-1\right )}{d^{7/2} \sqrt {d+e x^2}}-\frac {2 \sqrt {e} x \sqrt {d+e x^2} \left (7 d^2-9 d e x^2+15 e^2 x^4\right )}{d^3}-77 \tanh ^{-1}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )\right )}{1001 x^{13/2}} \]

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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fricas [F]  time = 0.52, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {15}{2}}}, x\right ) \]

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]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {15}{2}}}\,{d x} \]

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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maple [F(-2)]  time = 180.00, 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 \[ 2 \, d \sqrt {e} \int -\frac {\sqrt {e x^{2} + d} x}{13 \, {\left ({\left (e^{2} x^{4} + d e x^{2}\right )} x^{\frac {15}{2}} - {\left (e x^{2} + d\right )} e^{\left (\log \left (e x^{2} + d\right ) + \frac {15}{2} \, \log \relax (x)\right )}\right )}}\,{d x} - \frac {\log \left (\sqrt {e} x + \sqrt {e x^{2} + d}\right )}{13 \, x^{\frac {13}{2}}} + \frac {\log \left (-\sqrt {e} x + \sqrt {e x^{2} + d}\right )}{13 \, x^{\frac {13}{2}}} \]

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

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

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

[Out]

Timed out

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