Integrand size = 25, antiderivative size = 272 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{5/2}} \, dx=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{3 d \sqrt {x}}+\frac {4 e \sqrt {x} \sqrt {d+e x^2}}{3 d \left (\sqrt {d}+\sqrt {e} x\right )}-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 x^{3/2}}-\frac {4 e^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{3 d^{3/4} \sqrt {d+e x^2}}+\frac {2 e^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{3 d^{3/4} \sqrt {d+e x^2}} \]
-2/3*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(3/2)-4/3*e^(1/2)*(e*x^2+d)^(1/2 )/d/x^(1/2)+4/3*e*x^(1/2)*(e*x^2+d)^(1/2)/d/(d^(1/2)+x*e^(1/2))-4/3*e^(3/4 )*(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)/d^(3/4)/(e *x^2+d)^(1/2)+2/3*e^(3/4)*(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^(3/4)/(e*x^2+d)^(1/2)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.08 (sec) , antiderivative size = 118, normalized size of antiderivative = 0.43 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{5/2}} \, dx=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{3 d \sqrt {x}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 x^{3/2}}+\frac {4 e^{3/2} x^{3/2} \sqrt {1+\frac {e x^2}{d}} \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {3}{4},\frac {7}{4},-\frac {e x^2}{d}\right )}{9 d \sqrt {d+e x^2}} \]
(-4*Sqrt[e]*Sqrt[d + e*x^2])/(3*d*Sqrt[x]) - (2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(3*x^(3/2)) + (4*e^(3/2)*x^(3/2)*Sqrt[1 + (e*x^2)/d]*Hypergeom etric2F1[1/2, 3/4, 7/4, -((e*x^2)/d)])/(9*d*Sqrt[d + e*x^2])
Time = 0.40 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {6775, 264, 266, 834, 27, 761, 1510}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
| \(\displaystyle \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{5/2}} \, dx\) |
| \(\Big \downarrow \) 6775 |
| \(\displaystyle \frac {2}{3} \sqrt {e} \int \frac {1}{x^{3/2} \sqrt {e x^2+d}}dx-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 264 |
| \(\displaystyle \frac {2}{3} \sqrt {e} \left (\frac {e \int \frac {\sqrt {x}}{\sqrt {e x^2+d}}dx}{d}-\frac {2 \sqrt {d+e x^2}}{d \sqrt {x}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 266 |
| \(\displaystyle \frac {2}{3} \sqrt {e} \left (\frac {2 e \int \frac {x}{\sqrt {e x^2+d}}d\sqrt {x}}{d}-\frac {2 \sqrt {d+e x^2}}{d \sqrt {x}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 834 |
| \(\displaystyle \frac {2}{3} \sqrt {e} \left (\frac {2 e \left (\frac {\sqrt {d} \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}-\frac {\sqrt {d} \int \frac {\sqrt {d}-\sqrt {e} x}{\sqrt {d} \sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}\right )}{d}-\frac {2 \sqrt {d+e x^2}}{d \sqrt {x}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 27 |
| \(\displaystyle \frac {2}{3} \sqrt {e} \left (\frac {2 e \left (\frac {\sqrt {d} \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}-\frac {\int \frac {\sqrt {d}-\sqrt {e} x}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}\right )}{d}-\frac {2 \sqrt {d+e x^2}}{d \sqrt {x}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 761 |
| \(\displaystyle \frac {2}{3} \sqrt {e} \left (\frac {2 e \left (\frac {\sqrt [4]{d} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{2 e^{3/4} \sqrt {d+e x^2}}-\frac {\int \frac {\sqrt {d}-\sqrt {e} x}{\sqrt {e x^2+d}}d\sqrt {x}}{\sqrt {e}}\right )}{d}-\frac {2 \sqrt {d+e x^2}}{d \sqrt {x}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 x^{3/2}}\) |
| \(\Big \downarrow \) 1510 |
| \(\displaystyle \frac {2}{3} \sqrt {e} \left (\frac {2 e \left (\frac {\sqrt [4]{d} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{2 e^{3/4} \sqrt {d+e x^2}}-\frac {\frac {\sqrt [4]{d} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} E\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right )|\frac {1}{2}\right )}{\sqrt [4]{e} \sqrt {d+e x^2}}-\frac {\sqrt {x} \sqrt {d+e x^2}}{\sqrt {d}+\sqrt {e} x}}{\sqrt {e}}\right )}{d}-\frac {2 \sqrt {d+e x^2}}{d \sqrt {x}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{3 x^{3/2}}\) |
(-2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(3*x^(3/2)) + (2*Sqrt[e]*((-2*Sq rt[d + e*x^2])/(d*Sqrt[x]) + (2*e*(-((-((Sqrt[x]*Sqrt[d + e*x^2])/(Sqrt[d] + Sqrt[e]*x)) + (d^(1/4)*(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])/(e^(1 /4)*Sqrt[d + e*x^2]))/Sqrt[e]) + (d^(1/4)*(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])/(2*e^(3/4)*Sqrt[d + e*x^2])))/d))/3
3.1.26.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^( m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + 2*p + 3)/(a*c ^2*(m + 1))) Int[(c*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, p }, x] && LtQ[m, -1] && IntBinomialQ[a, b, c, 2, m, p, x]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De nominator[m]}, Simp[k/c Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) ^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I ntBinomialQ[a, b, c, 2, m, p, x]
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]
Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, S imp[1/q Int[1/Sqrt[a + b*x^4], x], x] - Simp[1/q Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
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]))*E llipticE[2*ArcTan[q*x], 1/2], x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e }, x] && PosQ[c/a]
Int[ArcTanh[((c_.)*(x_))/Sqrt[(a_.) + (b_.)*(x_)^2]]*((d_.)*(x_))^(m_.), x_ Symbol] :> Simp[(d*x)^(m + 1)*(ArcTanh[(c*x)/Sqrt[a + b*x^2]]/(d*(m + 1))), x] - Simp[c/(d*(m + 1)) Int[(d*x)^(m + 1)/Sqrt[a + b*x^2], x], x] /; Fre eQ[{a, b, c, d, m}, x] && EqQ[b, c^2] && NeQ[m, -1]
\[\int \frac {\operatorname {arctanh}\left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {5}{2}}}d x\]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 82, normalized size of antiderivative = 0.30 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{5/2}} \, dx=-\frac {4 \, e x^{2} {\rm weierstrassZeta}\left (-\frac {4 \, d}{e}, 0, {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right )\right ) + 4 \, \sqrt {e x^{2} + d} \sqrt {e} x^{\frac {3}{2}} + d \sqrt {x} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{3 \, d x^{2}} \]
-1/3*(4*e*x^2*weierstrassZeta(-4*d/e, 0, weierstrassPInverse(-4*d/e, 0, x) ) + 4*sqrt(e*x^2 + d)*sqrt(e)*x^(3/2) + d*sqrt(x)*log((2*e*x^2 + 2*sqrt(e* x^2 + d)*sqrt(e)*x + d)/d))/(d*x^2)
\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{5/2}} \, dx=\int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{\frac {5}{2}}}\, dx \]
\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{5/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {5}{2}}} \,d x } \]
2*d*sqrt(e)*integrate(-1/3*sqrt(e*x^2 + d)*x/((e^2*x^4 + d*e*x^2)*x^(5/2) - (e*x^2 + d)*e^(log(e*x^2 + d) + 5/2*log(x))), x) - 1/3*log(sqrt(e)*x + s qrt(e*x^2 + d))/x^(3/2) + 1/3*log(-sqrt(e)*x + sqrt(e*x^2 + d))/x^(3/2)
\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{5/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{5/2}} \, dx=\int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{5/2}} \,d x \]