Integrand size = 21, antiderivative size = 75 \[ \int x \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=-\frac {x \sqrt {d+e x^2}}{4 \sqrt {e}}+\frac {d \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{4 e}+\frac {1}{2} x^2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \]
1/4*d*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))/e+1/2*x^2*arctanh(x*e^(1/2)/(e*x^ 2+d)^(1/2))-1/4*x*(e*x^2+d)^(1/2)/e^(1/2)
Time = 0.03 (sec) , antiderivative size = 76, normalized size of antiderivative = 1.01 \[ \int x \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=-\frac {x \sqrt {d+e x^2}}{4 \sqrt {e}}+\frac {1}{2} x^2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )+\frac {d \log \left (\sqrt {e} x+\sqrt {d+e x^2}\right )}{4 e} \]
-1/4*(x*Sqrt[d + e*x^2])/Sqrt[e] + (x^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2 ]])/2 + (d*Log[Sqrt[e]*x + Sqrt[d + e*x^2]])/(4*e)
Time = 0.22 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6775, 262, 224, 219}
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.
\(\displaystyle \int x \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx\) |
\(\Big \downarrow \) 6775 |
\(\displaystyle \frac {1}{2} x^2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{2} \sqrt {e} \int \frac {x^2}{\sqrt {e x^2+d}}dx\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{2} x^2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{2} \sqrt {e} \left (\frac {x \sqrt {d+e x^2}}{2 e}-\frac {d \int \frac {1}{\sqrt {e x^2+d}}dx}{2 e}\right )\) |
\(\Big \downarrow \) 224 |
\(\displaystyle \frac {1}{2} x^2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{2} \sqrt {e} \left (\frac {x \sqrt {d+e x^2}}{2 e}-\frac {d \int \frac {1}{1-\frac {e x^2}{e x^2+d}}d\frac {x}{\sqrt {e x^2+d}}}{2 e}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{2} x^2 \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )-\frac {1}{2} \sqrt {e} \left (\frac {x \sqrt {d+e x^2}}{2 e}-\frac {d \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right )}{2 e^{3/2}}\right )\) |
(x^2*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/2 - (Sqrt[e]*((x*Sqrt[d + e*x^2 ])/(2*e) - (d*ArcTanh[(Sqrt[e]*x)/Sqrt[d + e*x^2]])/(2*e^(3/2))))/2
3.1.3.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] && !GtQ[a, 0]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
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]
Leaf count of result is larger than twice the leaf count of optimal. \(156\) vs. \(2(57)=114\).
Time = 0.01 (sec) , antiderivative size = 157, normalized size of antiderivative = 2.09
method | result | size |
default | \(\frac {x^{2} \operatorname {arctanh}\left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{2}+\frac {e^{\frac {3}{2}} \left (\frac {x^{3} \sqrt {e \,x^{2}+d}}{4 e}-\frac {3 d \left (\frac {x \sqrt {e \,x^{2}+d}}{2 e}-\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 e^{\frac {3}{2}}}\right )}{4 e}\right )}{2 d}-\frac {\sqrt {e}\, \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4 e}-\frac {d \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{4 e}\right )}{2 d}\) | \(157\) |
parts | \(\frac {x^{2} \operatorname {arctanh}\left (\frac {x \sqrt {e}}{\sqrt {e \,x^{2}+d}}\right )}{2}+\frac {e^{\frac {3}{2}} \left (\frac {x^{3} \sqrt {e \,x^{2}+d}}{4 e}-\frac {3 d \left (\frac {x \sqrt {e \,x^{2}+d}}{2 e}-\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 e^{\frac {3}{2}}}\right )}{4 e}\right )}{2 d}-\frac {\sqrt {e}\, \left (\frac {x \left (e \,x^{2}+d \right )^{\frac {3}{2}}}{4 e}-\frac {d \left (\frac {x \sqrt {e \,x^{2}+d}}{2}+\frac {d \ln \left (x \sqrt {e}+\sqrt {e \,x^{2}+d}\right )}{2 \sqrt {e}}\right )}{4 e}\right )}{2 d}\) | \(157\) |
1/2*x^2*arctanh(x*e^(1/2)/(e*x^2+d)^(1/2))+1/2*e^(3/2)/d*(1/4*x^3/e*(e*x^2 +d)^(1/2)-3/4*d/e*(1/2*x/e*(e*x^2+d)^(1/2)-1/2*d/e^(3/2)*ln(x*e^(1/2)+(e*x ^2+d)^(1/2))))-1/2*e^(1/2)/d*(1/4*x*(e*x^2+d)^(3/2)/e-1/4*d/e*(1/2*x*(e*x^ 2+d)^(1/2)+1/2*d/e^(1/2)*ln(x*e^(1/2)+(e*x^2+d)^(1/2))))
Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.79 \[ \int x \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=-\frac {2 \, \sqrt {e x^{2} + d} \sqrt {e} x - {\left (2 \, e x^{2} + d\right )} \log \left (\frac {2 \, e x^{2} + 2 \, \sqrt {e x^{2} + d} \sqrt {e} x + d}{d}\right )}{8 \, e} \]
-1/8*(2*sqrt(e*x^2 + d)*sqrt(e)*x - (2*e*x^2 + d)*log((2*e*x^2 + 2*sqrt(e* x^2 + d)*sqrt(e)*x + d)/d))/e
Time = 0.22 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.88 \[ \int x \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\begin {cases} \frac {d \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{4 e} + \frac {x^{2} \operatorname {atanh}{\left (\frac {\sqrt {e} x}{\sqrt {d + e x^{2}}} \right )}}{2} - \frac {x \sqrt {d + e x^{2}}}{4 \sqrt {e}} & \text {for}\: e \neq 0 \\0 & \text {otherwise} \end {cases} \]
Piecewise((d*atanh(sqrt(e)*x/sqrt(d + e*x**2))/(4*e) + x**2*atanh(sqrt(e)* x/sqrt(d + e*x**2))/2 - x*sqrt(d + e*x**2)/(4*sqrt(e)), Ne(e, 0)), (0, Tru e))
\[ \int x \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\int { x \operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right ) \,d x } \]
1/4*x^2*log(sqrt(e)*x + sqrt(e*x^2 + d)) - 1/4*x^2*log(-sqrt(e)*x + sqrt(e *x^2 + d)) - 1/2*d*sqrt(e)*integrate(-sqrt(e*x^2 + d)*x^2/(e^2*x^4 + d*e*x ^2 - (e*x^2 + d)^2), x)
\[ \int x \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\int { x \operatorname {artanh}\left (\frac {\sqrt {e} x}{\sqrt {e x^{2} + d}}\right ) \,d x } \]
Timed out. \[ \int x \text {arctanh}\left (\frac {\sqrt {e} x}{\sqrt {d+e x^2}}\right ) \, dx=\int x\,\mathrm {atanh}\left (\frac {\sqrt {e}\,x}{\sqrt {e\,x^2+d}}\right ) \,d x \]