Integrand size = 25, antiderivative size = 173 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=-\frac {4 \sqrt {e} \sqrt {d+e x^2}}{63 d x^{7/2}}+\frac {20 e^{3/2} \sqrt {d+e x^2}}{189 d^2 x^{3/2}}-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}+\frac {10 e^{9/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{189 d^{9/4} \sqrt {d+e x^2}} \]
-2/9*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^(9/2)+20/189*e^(3/2)*(e*x^2+d)^( 1/2)/d^2/x^(3/2)-4/63*e^(1/2)*(e*x^2+d)^(1/2)/d/x^(7/2)+10/189*e^(9/4)*(co s(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^(9/4)/(e*x^2+ d)^(1/2)
Result contains complex when optimal does not.
Time = 0.24 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.89 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\frac {4 \sqrt {e} x \sqrt {d+e x^2} \left (-3 d+5 e x^2\right )-42 d^2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{189 d^2 x^{9/2}}+\frac {20 \sqrt {\frac {i \sqrt {d}}{\sqrt {e}}} e^3 \sqrt {1+\frac {d}{e x^2}} x \operatorname {EllipticF}\left (i \text {arcsinh}\left (\frac {\sqrt {\frac {i \sqrt {d}}{\sqrt {e}}}}{\sqrt {x}}\right ),-1\right )}{189 d^{5/2} \sqrt {d+e x^2}} \]
(4*Sqrt[e]*x*Sqrt[d + e*x^2]*(-3*d + 5*e*x^2) - 42*d^2*ArcTanh[(Sqrt[e]*x) /Sqrt[d + e*x^2]])/(189*d^2*x^(9/2)) + (20*Sqrt[(I*Sqrt[d])/Sqrt[e]]*e^3*S qrt[1 + d/(e*x^2)]*x*EllipticF[I*ArcSinh[Sqrt[(I*Sqrt[d])/Sqrt[e]]/Sqrt[x] ], -1])/(189*d^(5/2)*Sqrt[d + e*x^2])
Time = 0.30 (sec) , antiderivative size = 182, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {6775, 264, 264, 266, 761}
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.
\(\displaystyle \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx\) |
\(\Big \downarrow \) 6775 |
\(\displaystyle \frac {2}{9} \sqrt {e} \int \frac {1}{x^{9/2} \sqrt {e x^2+d}}dx-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2}{9} \sqrt {e} \left (-\frac {5 e \int \frac {1}{x^{5/2} \sqrt {e x^2+d}}dx}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}\) |
\(\Big \downarrow \) 264 |
\(\displaystyle \frac {2}{9} \sqrt {e} \left (-\frac {5 e \left (-\frac {e \int \frac {1}{\sqrt {x} \sqrt {e x^2+d}}dx}{3 d}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2}{9} \sqrt {e} \left (-\frac {5 e \left (-\frac {2 e \int \frac {1}{\sqrt {e x^2+d}}d\sqrt {x}}{3 d}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}\) |
\(\Big \downarrow \) 761 |
\(\displaystyle \frac {2}{9} \sqrt {e} \left (-\frac {5 e \left (-\frac {e^{3/4} \left (\sqrt {d}+\sqrt {e} x\right ) \sqrt {\frac {d+e x^2}{\left (\sqrt {d}+\sqrt {e} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{e} \sqrt {x}}{\sqrt [4]{d}}\right ),\frac {1}{2}\right )}{3 d^{5/4} \sqrt {d+e x^2}}-\frac {2 \sqrt {d+e x^2}}{3 d x^{3/2}}\right )}{7 d}-\frac {2 \sqrt {d+e x^2}}{7 d x^{7/2}}\right )-\frac {2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{9 x^{9/2}}\) |
(-2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(9*x^(9/2)) + (2*Sqrt[e]*((-2*Sq rt[d + e*x^2])/(7*d*x^(7/2)) - (5*e*((-2*Sqrt[d + e*x^2])/(3*d*x^(3/2)) - (e^(3/4)*(Sqrt[d] + Sqrt[e]*x)*Sqrt[(d + e*x^2)/(Sqrt[d] + Sqrt[e]*x)^2]*E llipticF[2*ArcTan[(e^(1/4)*Sqrt[x])/d^(1/4)], 1/2])/(3*d^(5/4)*Sqrt[d + e* x^2])))/(7*d)))/9
3.1.21.3.1 Defintions of rubi rules used
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[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]
\[\int \frac {\operatorname {arctanh}\left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{x^{\frac {11}{2}}}d x\]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.09 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.52 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\frac {20 \, e^{2} x^{5} {\rm weierstrassPInverse}\left (-\frac {4 \, d}{e}, 0, x\right ) - 21 \, d^{2} \sqrt {x} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) + 4 \, {\left (5 \, e x^{3} - 3 \, d x\right )} \sqrt {e x^{2} + d} \sqrt {e} \sqrt {x}}{189 \, d^{2} x^{5}} \]
1/189*(20*e^2*x^5*weierstrassPInverse(-4*d/e, 0, x) - 21*d^2*sqrt(x)*log(( 2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) + 4*(5*e*x^3 - 3*d*x)*sqrt(e *x^2 + d)*sqrt(e)*sqrt(x))/(d^2*x^5)
Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\text {Timed out} \]
\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {11}{2}}} \,d x } \]
2*d*sqrt(e)*integrate(-1/9*sqrt(e*x^2 + d)*x/((e^2*x^4 + d*e*x^2)*x^(11/2) - (e*x^2 + d)*e^(log(e*x^2 + d) + 11/2*log(x))), x) - 1/9*log(sqrt(e)*x + sqrt(e*x^2 + d))/x^(9/2) + 1/9*log(-sqrt(e)*x + sqrt(e*x^2 + d))/x^(9/2)
\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{\frac {11}{2}}} \,d x } \]
Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^{11/2}} \, dx=\int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^{11/2}} \,d x \]