Integrand size = 22, antiderivative size = 197 \[ \int \frac {x^4 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=-\frac {5 \sqrt {1-a^2 x^2}}{8 a^5}+\frac {\left (1-a^2 x^2\right )^{3/2}}{12 a^5}-\frac {3 x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{8 a^4}-\frac {x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{4 a^2}-\frac {3 \arctan \left (\frac {\sqrt {1-a x}}{\sqrt {1+a x}}\right ) \text {arctanh}(a x)}{4 a^5}-\frac {3 i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^5}+\frac {3 i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {1+a x}}\right )}{8 a^5} \] Output:
-5/8*(-a^2*x^2+1)^(1/2)/a^5+1/12*(-a^2*x^2+1)^(3/2)/a^5-3/8*x*(-a^2*x^2+1) ^(1/2)*arctanh(a*x)/a^4-1/4*x^3*(-a^2*x^2+1)^(1/2)*arctanh(a*x)/a^2-3/4*ar ctan((-a*x+1)^(1/2)/(a*x+1)^(1/2))*arctanh(a*x)/a^5-3/8*I*polylog(2,-I*(-a *x+1)^(1/2)/(a*x+1)^(1/2))/a^5+3/8*I*polylog(2,I*(-a*x+1)^(1/2)/(a*x+1)^(1 /2))/a^5
Time = 0.36 (sec) , antiderivative size = 160, normalized size of antiderivative = 0.81 \[ \int \frac {x^4 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\frac {\sqrt {1-a^2 x^2} \left (-13-2 a^2 x^2-15 a x \text {arctanh}(a x)-6 a x \left (-1+a^2 x^2\right ) \text {arctanh}(a x)-\frac {9 i \text {arctanh}(a x) \left (\log \left (1-i e^{-\text {arctanh}(a x)}\right )-\log \left (1+i e^{-\text {arctanh}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}-\frac {9 i \left (\operatorname {PolyLog}\left (2,-i e^{-\text {arctanh}(a x)}\right )-\operatorname {PolyLog}\left (2,i e^{-\text {arctanh}(a x)}\right )\right )}{\sqrt {1-a^2 x^2}}\right )}{24 a^5} \] Input:
Integrate[(x^4*ArcTanh[a*x])/Sqrt[1 - a^2*x^2],x]
Output:
(Sqrt[1 - a^2*x^2]*(-13 - 2*a^2*x^2 - 15*a*x*ArcTanh[a*x] - 6*a*x*(-1 + a^ 2*x^2)*ArcTanh[a*x] - ((9*I)*ArcTanh[a*x]*(Log[1 - I/E^ArcTanh[a*x]] - Log [1 + I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2] - ((9*I)*(PolyLog[2, (-I)/E^Arc Tanh[a*x]] - PolyLog[2, I/E^ArcTanh[a*x]]))/Sqrt[1 - a^2*x^2]))/(24*a^5)
Time = 0.65 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.19, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {6578, 243, 53, 2009, 6578, 241, 6512}
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.
\(\displaystyle \int \frac {x^4 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx\) |
\(\Big \downarrow \) 6578 |
\(\displaystyle \frac {3 \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{4 a^2}+\frac {\int \frac {x^3}{\sqrt {1-a^2 x^2}}dx}{4 a}-\frac {x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{4 a^2}\) |
\(\Big \downarrow \) 243 |
\(\displaystyle \frac {3 \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{4 a^2}+\frac {\int \frac {x^2}{\sqrt {1-a^2 x^2}}dx^2}{8 a}-\frac {x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{4 a^2}\) |
\(\Big \downarrow \) 53 |
\(\displaystyle \frac {3 \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{4 a^2}+\frac {\int \left (\frac {1}{a^2 \sqrt {1-a^2 x^2}}-\frac {\sqrt {1-a^2 x^2}}{a^2}\right )dx^2}{8 a}-\frac {x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{4 a^2}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {3 \int \frac {x^2 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{4 a^2}-\frac {x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{4 a^2}+\frac {\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}}{8 a}\) |
\(\Big \downarrow \) 6578 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {\text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}+\frac {\int \frac {x}{\sqrt {1-a^2 x^2}}dx}{2 a}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}\right )}{4 a^2}-\frac {x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{4 a^2}+\frac {\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}}{8 a}\) |
\(\Big \downarrow \) 241 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {\text {arctanh}(a x)}{\sqrt {1-a^2 x^2}}dx}{2 a^2}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}-\frac {\sqrt {1-a^2 x^2}}{2 a^3}\right )}{4 a^2}-\frac {x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{4 a^2}+\frac {\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}}{8 a}\) |
\(\Big \downarrow \) 6512 |
\(\displaystyle -\frac {x^3 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{4 a^2}+\frac {\frac {2 \left (1-a^2 x^2\right )^{3/2}}{3 a^4}-\frac {2 \sqrt {1-a^2 x^2}}{a^4}}{8 a}+\frac {3 \left (\frac {-\frac {2 \arctan \left (\frac {\sqrt {1-a x}}{\sqrt {a x+1}}\right ) \text {arctanh}(a x)}{a}-\frac {i \operatorname {PolyLog}\left (2,-\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}+\frac {i \operatorname {PolyLog}\left (2,\frac {i \sqrt {1-a x}}{\sqrt {a x+1}}\right )}{a}}{2 a^2}-\frac {x \sqrt {1-a^2 x^2} \text {arctanh}(a x)}{2 a^2}-\frac {\sqrt {1-a^2 x^2}}{2 a^3}\right )}{4 a^2}\) |
Input:
Int[(x^4*ArcTanh[a*x])/Sqrt[1 - a^2*x^2],x]
Output:
((-2*Sqrt[1 - a^2*x^2])/a^4 + (2*(1 - a^2*x^2)^(3/2))/(3*a^4))/(8*a) - (x^ 3*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(4*a^2) + (3*(-1/2*Sqrt[1 - a^2*x^2]/a^3 - (x*Sqrt[1 - a^2*x^2]*ArcTanh[a*x])/(2*a^2) + ((-2*ArcTan[Sqrt[1 - a*x]/ Sqrt[1 + a*x]]*ArcTanh[a*x])/a - (I*PolyLog[2, ((-I)*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a + (I*PolyLog[2, (I*Sqrt[1 - a*x])/Sqrt[1 + a*x]])/a)/(2*a^2))) /(4*a^2)
Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int [ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0] && LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])
Int[(x_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(a + b*x^2)^(p + 1)/ (2*b*(p + 1)), x] /; FreeQ[{a, b, p}, x] && NeQ[p, -1]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2 Subst[In t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I ntegerQ[(m - 1)/2]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol ] :> Simp[-2*(a + b*ArcTanh[c*x])*(ArcTan[Sqrt[1 - c*x]/Sqrt[1 + c*x]]/(c*S qrt[d])), x] + (-Simp[I*b*(PolyLog[2, (-I)*(Sqrt[1 - c*x]/Sqrt[1 + c*x])]/( c*Sqrt[d])), x] + Simp[I*b*(PolyLog[2, I*(Sqrt[1 - c*x]/Sqrt[1 + c*x])]/(c* Sqrt[d])), x]) /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[d, 0]
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/Sqrt[(d_) + (e_.)*(x_)^2], x_Symbol] :> Simp[(-f)*(f*x)^(m - 1)*Sqrt[d + e*x^2]*((a + b*ArcTanh[c*x])^p/(c^2*d*m)), x] + (Simp[b*f*(p/(c*m)) Int[(f*x)^(m - 1 )*((a + b*ArcTanh[c*x])^(p - 1)/Sqrt[d + e*x^2]), x], x] + Simp[f^2*((m - 1 )/(c^2*m)) Int[(f*x)^(m - 2)*((a + b*ArcTanh[c*x])^p/Sqrt[d + e*x^2]), x] , x]) /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0] && GtQ[m, 1]
Time = 0.77 (sec) , antiderivative size = 173, normalized size of antiderivative = 0.88
method | result | size |
default | \(-\frac {\left (6 a^{3} x^{3} \operatorname {arctanh}\left (a x \right )+2 a^{2} x^{2}+9 a x \,\operatorname {arctanh}\left (a x \right )+13\right ) \sqrt {-a^{2} x^{2}+1}}{24 a^{5}}-\frac {3 i \operatorname {arctanh}\left (a x \right ) \ln \left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{5}}+\frac {3 i \operatorname {arctanh}\left (a x \right ) \ln \left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{5}}-\frac {3 i \operatorname {dilog}\left (1+\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{5}}+\frac {3 i \operatorname {dilog}\left (1-\frac {i \left (a x +1\right )}{\sqrt {-a^{2} x^{2}+1}}\right )}{8 a^{5}}\) | \(173\) |
Input:
int(x^4*arctanh(a*x)/(-a^2*x^2+1)^(1/2),x,method=_RETURNVERBOSE)
Output:
-1/24*(6*a^3*x^3*arctanh(a*x)+2*a^2*x^2+9*a*x*arctanh(a*x)+13)*(-a^2*x^2+1 )^(1/2)/a^5-3/8*I/a^5*arctanh(a*x)*ln(1+I*(a*x+1)/(-a^2*x^2+1)^(1/2))+3/8* I/a^5*arctanh(a*x)*ln(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2))-3/8*I/a^5*dilog(1+I* (a*x+1)/(-a^2*x^2+1)^(1/2))+3/8*I/a^5*dilog(1-I*(a*x+1)/(-a^2*x^2+1)^(1/2) )
\[ \int \frac {x^4 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\int { \frac {x^{4} \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate(x^4*arctanh(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="fricas")
Output:
integral(-sqrt(-a^2*x^2 + 1)*x^4*arctanh(a*x)/(a^2*x^2 - 1), x)
\[ \int \frac {x^4 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\int \frac {x^{4} \operatorname {atanh}{\left (a x \right )}}{\sqrt {- \left (a x - 1\right ) \left (a x + 1\right )}}\, dx \] Input:
integrate(x**4*atanh(a*x)/(-a**2*x**2+1)**(1/2),x)
Output:
Integral(x**4*atanh(a*x)/sqrt(-(a*x - 1)*(a*x + 1)), x)
\[ \int \frac {x^4 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\int { \frac {x^{4} \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate(x^4*arctanh(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="maxima")
Output:
integrate(x^4*arctanh(a*x)/sqrt(-a^2*x^2 + 1), x)
\[ \int \frac {x^4 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\int { \frac {x^{4} \operatorname {artanh}\left (a x\right )}{\sqrt {-a^{2} x^{2} + 1}} \,d x } \] Input:
integrate(x^4*arctanh(a*x)/(-a^2*x^2+1)^(1/2),x, algorithm="giac")
Output:
integrate(x^4*arctanh(a*x)/sqrt(-a^2*x^2 + 1), x)
Timed out. \[ \int \frac {x^4 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\int \frac {x^4\,\mathrm {atanh}\left (a\,x\right )}{\sqrt {1-a^2\,x^2}} \,d x \] Input:
int((x^4*atanh(a*x))/(1 - a^2*x^2)^(1/2),x)
Output:
int((x^4*atanh(a*x))/(1 - a^2*x^2)^(1/2), x)
\[ \int \frac {x^4 \text {arctanh}(a x)}{\sqrt {1-a^2 x^2}} \, dx=\int \frac {\mathit {atanh} \left (a x \right ) x^{4}}{\sqrt {-a^{2} x^{2}+1}}d x \] Input:
int(x^4*atanh(a*x)/(-a^2*x^2+1)^(1/2),x)
Output:
int((atanh(a*x)*x**4)/sqrt( - a**2*x**2 + 1),x)