Integrand size = 21, antiderivative size = 27 \[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)} \, dx=\frac {3 \text {Chi}(\text {arctanh}(a x))}{4 a}+\frac {\text {Chi}(3 \text {arctanh}(a x))}{4 a} \] Output:
3/4*Chi(arctanh(a*x))/a+1/4*Chi(3*arctanh(a*x))/a
Time = 0.05 (sec) , antiderivative size = 24, normalized size of antiderivative = 0.89 \[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)} \, dx=-\frac {-3 \text {Chi}(\text {arctanh}(a x))-\text {Chi}(3 \text {arctanh}(a x))}{4 a} \] Input:
Integrate[1/((1 - a^2*x^2)^(5/2)*ArcTanh[a*x]),x]
Output:
-1/4*(-3*CoshIntegral[ArcTanh[a*x]] - CoshIntegral[3*ArcTanh[a*x]])/a
Time = 0.32 (sec) , antiderivative size = 25, normalized size of antiderivative = 0.93, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {6530, 3042, 3793, 2009}
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.
\(\displaystyle \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)} \, dx\) |
\(\Big \downarrow \) 6530 |
\(\displaystyle \frac {\int \frac {1}{\left (1-a^2 x^2\right )^{3/2} \text {arctanh}(a x)}d\text {arctanh}(a x)}{a}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\sin \left (i \text {arctanh}(a x)+\frac {\pi }{2}\right )^3}{\text {arctanh}(a x)}d\text {arctanh}(a x)}{a}\) |
\(\Big \downarrow \) 3793 |
\(\displaystyle \frac {\int \left (\frac {\cosh (3 \text {arctanh}(a x))}{4 \text {arctanh}(a x)}+\frac {3}{4 \sqrt {1-a^2 x^2} \text {arctanh}(a x)}\right )d\text {arctanh}(a x)}{a}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {\frac {3}{4} \text {Chi}(\text {arctanh}(a x))+\frac {1}{4} \text {Chi}(3 \text {arctanh}(a x))}{a}\) |
Input:
Int[1/((1 - a^2*x^2)^(5/2)*ArcTanh[a*x]),x]
Output:
((3*CoshIntegral[ArcTanh[a*x]])/4 + CoshIntegral[3*ArcTanh[a*x]]/4)/a
Int[((c_.) + (d_.)*(x_))^(m_)*sin[(e_.) + (f_.)*(x_)]^(n_), x_Symbol] :> In t[ExpandTrigReduce[(c + d*x)^m, Sin[e + f*x]^n, x], x] /; FreeQ[{c, d, e, f , m}, x] && IGtQ[n, 1] && ( !RationalQ[m] || (GeQ[m, -1] && LtQ[m, 1]))
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_) + (e_.)*(x_)^2)^(q_), x _Symbol] :> Simp[d^q/c Subst[Int[(a + b*x)^p/Cosh[x]^(2*(q + 1)), x], x, ArcTanh[c*x]], x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && I LtQ[2*(q + 1), 0] && (IntegerQ[q] || GtQ[d, 0])
Time = 0.49 (sec) , antiderivative size = 21, normalized size of antiderivative = 0.78
method | result | size |
default | \(\frac {3 \,\operatorname {Chi}\left (\operatorname {arctanh}\left (a x \right )\right )+\operatorname {Chi}\left (3 \,\operatorname {arctanh}\left (a x \right )\right )}{4 a}\) | \(21\) |
Input:
int(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x),x,method=_RETURNVERBOSE)
Output:
1/4*(3*Chi(arctanh(a*x))+Chi(3*arctanh(a*x)))/a
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} \operatorname {artanh}\left (a x\right )} \,d x } \] Input:
integrate(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x),x, algorithm="fricas")
Output:
integral(-sqrt(-a^2*x^2 + 1)/((a^6*x^6 - 3*a^4*x^4 + 3*a^2*x^2 - 1)*arctan h(a*x)), x)
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)} \, dx=\int \frac {1}{\left (- \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}} \operatorname {atanh}{\left (a x \right )}}\, dx \] Input:
integrate(1/(-a**2*x**2+1)**(5/2)/atanh(a*x),x)
Output:
Integral(1/((-(a*x - 1)*(a*x + 1))**(5/2)*atanh(a*x)), x)
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} \operatorname {artanh}\left (a x\right )} \,d x } \] Input:
integrate(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x),x, algorithm="maxima")
Output:
integrate(1/((-a^2*x^2 + 1)^(5/2)*arctanh(a*x)), x)
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)} \, dx=\int { \frac {1}{{\left (-a^{2} x^{2} + 1\right )}^{\frac {5}{2}} \operatorname {artanh}\left (a x\right )} \,d x } \] Input:
integrate(1/(-a^2*x^2+1)^(5/2)/arctanh(a*x),x, algorithm="giac")
Output:
integrate(1/((-a^2*x^2 + 1)^(5/2)*arctanh(a*x)), x)
Timed out. \[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)} \, dx=\int \frac {1}{\mathrm {atanh}\left (a\,x\right )\,{\left (1-a^2\,x^2\right )}^{5/2}} \,d x \] Input:
int(1/(atanh(a*x)*(1 - a^2*x^2)^(5/2)),x)
Output:
int(1/(atanh(a*x)*(1 - a^2*x^2)^(5/2)), x)
\[ \int \frac {1}{\left (1-a^2 x^2\right )^{5/2} \text {arctanh}(a x)} \, dx=\int \frac {1}{\sqrt {-a^{2} x^{2}+1}\, \mathit {atanh} \left (a x \right ) a^{4} x^{4}-2 \sqrt {-a^{2} x^{2}+1}\, \mathit {atanh} \left (a x \right ) a^{2} x^{2}+\sqrt {-a^{2} x^{2}+1}\, \mathit {atanh} \left (a x \right )}d x \] Input:
int(1/(-a^2*x^2+1)^(5/2)/atanh(a*x),x)
Output:
int(1/(sqrt( - a**2*x**2 + 1)*atanh(a*x)*a**4*x**4 - 2*sqrt( - a**2*x**2 + 1)*atanh(a*x)*a**2*x**2 + sqrt( - a**2*x**2 + 1)*atanh(a*x)),x)