Integrand size = 23, antiderivative size = 131 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=-\frac {\sqrt {e} \sqrt {d+e x^2}}{56 d x^7}+\frac {3 e^{3/2} \sqrt {d+e x^2}}{140 d^2 x^5}-\frac {e^{5/2} \sqrt {d+e x^2}}{35 d^3 x^3}+\frac {2 e^{7/2} \sqrt {d+e x^2}}{35 d^4 x}-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8} \]
-1/8*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^8+3/140*e^(3/2)*(e*x^2+d)^(1/2)/ d^2/x^5-1/35*e^(5/2)*(e*x^2+d)^(1/2)/d^3/x^3+2/35*e^(7/2)*(e*x^2+d)^(1/2)/ d^4/x-1/56*e^(1/2)*(e*x^2+d)^(1/2)/d/x^7
Time = 0.04 (sec) , antiderivative size = 85, normalized size of antiderivative = 0.65 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=\frac {\sqrt {e} x \sqrt {d+e x^2} \left (-5 d^3+6 d^2 e x^2-8 d e^2 x^4+16 e^3 x^6\right )-35 d^4 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{280 d^4 x^8} \]
(Sqrt[e]*x*Sqrt[d + e*x^2]*(-5*d^3 + 6*d^2*e*x^2 - 8*d*e^2*x^4 + 16*e^3*x^ 6) - 35*d^4*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(280*d^4*x^8)
Time = 0.26 (sec) , antiderivative size = 140, normalized size of antiderivative = 1.07, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {6775, 245, 245, 245, 242}
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.
\(\displaystyle \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx\) |
\(\Big \downarrow \) 6775 |
\(\displaystyle \frac {1}{8} \sqrt {e} \int \frac {1}{x^8 \sqrt {e x^2+d}}dx-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {1}{8} \sqrt {e} \left (-\frac {6 e \int \frac {1}{x^6 \sqrt {e x^2+d}}dx}{7 d}-\frac {\sqrt {d+e x^2}}{7 d x^7}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {1}{8} \sqrt {e} \left (-\frac {6 e \left (-\frac {4 e \int \frac {1}{x^4 \sqrt {e x^2+d}}dx}{5 d}-\frac {\sqrt {d+e x^2}}{5 d x^5}\right )}{7 d}-\frac {\sqrt {d+e x^2}}{7 d x^7}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}\) |
\(\Big \downarrow \) 245 |
\(\displaystyle \frac {1}{8} \sqrt {e} \left (-\frac {6 e \left (-\frac {4 e \left (-\frac {2 e \int \frac {1}{x^2 \sqrt {e x^2+d}}dx}{3 d}-\frac {\sqrt {d+e x^2}}{3 d x^3}\right )}{5 d}-\frac {\sqrt {d+e x^2}}{5 d x^5}\right )}{7 d}-\frac {\sqrt {d+e x^2}}{7 d x^7}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}\) |
\(\Big \downarrow \) 242 |
\(\displaystyle \frac {1}{8} \sqrt {e} \left (-\frac {6 e \left (-\frac {4 e \left (\frac {2 e \sqrt {d+e x^2}}{3 d^2 x}-\frac {\sqrt {d+e x^2}}{3 d x^3}\right )}{5 d}-\frac {\sqrt {d+e x^2}}{5 d x^5}\right )}{7 d}-\frac {\sqrt {d+e x^2}}{7 d x^7}\right )-\frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{8 x^8}\) |
(Sqrt[e]*(-1/7*Sqrt[d + e*x^2]/(d*x^7) - (6*e*(-1/5*Sqrt[d + e*x^2]/(d*x^5 ) - (4*e*(-1/3*Sqrt[d + e*x^2]/(d*x^3) + (2*e*Sqrt[d + e*x^2])/(3*d^2*x))) /(5*d)))/(7*d)))/8 - ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]]/(8*x^8)
3.1.8.3.1 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] /; FreeQ[{a, b, c, m, p}, x ] && EqQ[m + 2*p + 3, 0] && NeQ[m, -1]
Int[(x_)^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x^(m + 1)*((a + b*x^2)^(p + 1)/(a*(m + 1))), x] - Simp[b*((m + 2*(p + 1) + 1)/(a*(m + 1))) Int[x^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, m, p}, x] && ILtQ[Si mplify[(m + 1)/2 + p + 1], 0] && NeQ[m, -1]
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]
Time = 0.02 (sec) , antiderivative size = 158, normalized size of antiderivative = 1.21
method | result | size |
default | \(-\frac {\operatorname {arctanh}\left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{8 x^{8}}-\frac {e^{\frac {3}{2}} \left (-\frac {\sqrt {e \,x^{2}+d}}{5 d \,x^{5}}-\frac {4 e \left (-\frac {\sqrt {e \,x^{2}+d}}{3 d \,x^{3}}+\frac {2 e \sqrt {e \,x^{2}+d}}{3 d^{2} x}\right )}{5 d}\right )}{8 d}+\frac {\sqrt {e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 d \,x^{7}}-\frac {4 e \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 d \,x^{5}}+\frac {2 e \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 d^{2} x^{3}}\right )}{7 d}\right )}{8 d}\) | \(158\) |
parts | \(-\frac {\operatorname {arctanh}\left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{8 x^{8}}-\frac {e^{\frac {3}{2}} \left (-\frac {\sqrt {e \,x^{2}+d}}{5 d \,x^{5}}-\frac {4 e \left (-\frac {\sqrt {e \,x^{2}+d}}{3 d \,x^{3}}+\frac {2 e \sqrt {e \,x^{2}+d}}{3 d^{2} x}\right )}{5 d}\right )}{8 d}+\frac {\sqrt {e}\, \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{7 d \,x^{7}}-\frac {4 e \left (-\frac {\left (e \,x^{2}+d \right )^{\frac {3}{2}}}{5 d \,x^{5}}+\frac {2 e \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{15 d^{2} x^{3}}\right )}{7 d}\right )}{8 d}\) | \(158\) |
-1/8*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/x^8-1/8*e^(3/2)/d*(-1/5/d/x^5*(e*x ^2+d)^(1/2)-4/5*e/d*(-1/3/d/x^3*(e*x^2+d)^(1/2)+2/3*e/d^2/x*(e*x^2+d)^(1/2 )))+1/8*e^(1/2)/d*(-1/7/d/x^7*(e*x^2+d)^(3/2)-4/7*e/d*(-1/5/d/x^5*(e*x^2+d )^(3/2)+2/15*e/d^2/x^3*(e*x^2+d)^(3/2)))
Time = 0.28 (sec) , antiderivative size = 89, normalized size of antiderivative = 0.68 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=-\frac {35 \, d^{4} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right ) - 2 \, {\left (16 \, e^{3} x^{7} - 8 \, d e^{2} x^{5} + 6 \, d^{2} e x^{3} - 5 \, d^{3} x\right )} \sqrt {e x^{2} + d} \sqrt {e}}{560 \, d^{4} x^{8}} \]
-1/560*(35*d^4*log((2*e*x^2 + 2*sqrt(e*x^2 + d)*sqrt(e)*x + d)/d) - 2*(16* e^3*x^7 - 8*d*e^2*x^5 + 6*d^2*e*x^3 - 5*d^3*x)*sqrt(e*x^2 + d)*sqrt(e))/(d ^4*x^8)
\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=\int \frac {\operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{x^{9}}\, dx \]
Time = 0.21 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.95 \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=\frac {{\left (8 \, e^{3} x^{6} + 4 \, d e^{2} x^{4} - d^{2} e x^{2} + 3 \, d^{3}\right )} e^{\frac {3}{2}}}{120 \, \sqrt {e x^{2} + d} d^{4} x^{5}} - \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{8 \, x^{8}} - \frac {{\left (8 \, e^{3} x^{6} - 4 \, d e^{2} x^{4} + 3 \, d^{2} e x^{2} + 15 \, d^{3}\right )} \sqrt {e x^{2} + d} \sqrt {e}}{840 \, d^{4} x^{7}} \]
1/120*(8*e^3*x^6 + 4*d*e^2*x^4 - d^2*e*x^2 + 3*d^3)*e^(3/2)/(sqrt(e*x^2 + d)*d^4*x^5) - 1/8*arctanh(sqrt(e)*x/sqrt(e*x^2 + d))/x^8 - 1/840*(8*e^3*x^ 6 - 4*d*e^2*x^4 + 3*d^2*e*x^2 + 15*d^3)*sqrt(e*x^2 + d)*sqrt(e)/(d^4*x^7)
\[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=\int { \frac {\operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right )}{x^{9}} \,d x } \]
Timed out. \[ \int \frac {\text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{x^9} \, dx=\int \frac {\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right )}{x^9} \,d x \]