Integrand size = 20, antiderivative size = 105 \[ \int \frac {\text {arctanh}(a x)}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {1}{9 a c \left (c-a^2 c x^2\right )^{3/2}}-\frac {2}{3 a c^2 \sqrt {c-a^2 c x^2}}+\frac {x \text {arctanh}(a x)}{3 c \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 x \text {arctanh}(a x)}{3 c^2 \sqrt {c-a^2 c x^2}} \] Output:
-1/9/a/c/(-a^2*c*x^2+c)^(3/2)-2/3/a/c^2/(-a^2*c*x^2+c)^(1/2)+1/3*x*arctanh (a*x)/c/(-a^2*c*x^2+c)^(3/2)+2/3*x*arctanh(a*x)/c^2/(-a^2*c*x^2+c)^(1/2)
Time = 0.07 (sec) , antiderivative size = 64, normalized size of antiderivative = 0.61 \[ \int \frac {\text {arctanh}(a x)}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {c-a^2 c x^2} \left (7-6 a^2 x^2+\left (-9 a x+6 a^3 x^3\right ) \text {arctanh}(a x)\right )}{9 a c^3 \left (-1+a^2 x^2\right )^2} \] Input:
Integrate[ArcTanh[a*x]/(c - a^2*c*x^2)^(5/2),x]
Output:
-1/9*(Sqrt[c - a^2*c*x^2]*(7 - 6*a^2*x^2 + (-9*a*x + 6*a^3*x^3)*ArcTanh[a* x]))/(a*c^3*(-1 + a^2*x^2)^2)
Time = 0.35 (sec) , antiderivative size = 108, normalized size of antiderivative = 1.03, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {6522, 6520}
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.
\(\displaystyle \int \frac {\text {arctanh}(a x)}{\left (c-a^2 c x^2\right )^{5/2}} \, dx\) |
\(\Big \downarrow \) 6522 |
\(\displaystyle \frac {2 \int \frac {\text {arctanh}(a x)}{\left (c-a^2 c x^2\right )^{3/2}}dx}{3 c}+\frac {x \text {arctanh}(a x)}{3 c \left (c-a^2 c x^2\right )^{3/2}}-\frac {1}{9 a c \left (c-a^2 c x^2\right )^{3/2}}\) |
\(\Big \downarrow \) 6520 |
\(\displaystyle \frac {x \text {arctanh}(a x)}{3 c \left (c-a^2 c x^2\right )^{3/2}}+\frac {2 \left (\frac {x \text {arctanh}(a x)}{c \sqrt {c-a^2 c x^2}}-\frac {1}{a c \sqrt {c-a^2 c x^2}}\right )}{3 c}-\frac {1}{9 a c \left (c-a^2 c x^2\right )^{3/2}}\) |
Input:
Int[ArcTanh[a*x]/(c - a^2*c*x^2)^(5/2),x]
Output:
-1/9*1/(a*c*(c - a^2*c*x^2)^(3/2)) + (x*ArcTanh[a*x])/(3*c*(c - a^2*c*x^2) ^(3/2)) + (2*(-(1/(a*c*Sqrt[c - a^2*c*x^2])) + (x*ArcTanh[a*x])/(c*Sqrt[c - a^2*c*x^2])))/(3*c)
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/((d_) + (e_.)*(x_)^2)^(3/2), x_Symb ol] :> Simp[-b/(c*d*Sqrt[d + e*x^2]), x] + Simp[x*((a + b*ArcTanh[c*x])/(d* Sqrt[d + e*x^2])), x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_) + (e_.)*(x_)^2)^(q_), x_Symbo l] :> Simp[(-b)*((d + e*x^2)^(q + 1)/(4*c*d*(q + 1)^2)), x] + (-Simp[x*(d + e*x^2)^(q + 1)*((a + b*ArcTanh[c*x])/(2*d*(q + 1))), x] + Simp[(2*q + 3)/( 2*d*(q + 1)) Int[(d + e*x^2)^(q + 1)*(a + b*ArcTanh[c*x]), x], x]) /; Fre eQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && LtQ[q, -1] && NeQ[q, -3/2]
Time = 0.90 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.15
method | result | size |
orering | \(\frac {\left (4 a^{4} x^{5}-\frac {80}{9} a^{2} x^{3}+\frac {44}{9} x \right ) \operatorname {arctanh}\left (a x \right )}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {\left (6 a^{2} x^{2}-7\right ) \left (a x +1\right )^{2} \left (a x -1\right )^{2} \left (\frac {a}{\left (-a^{2} x^{2}+1\right ) \left (-a^{2} c \,x^{2}+c \right )^{\frac {5}{2}}}+\frac {5 \,\operatorname {arctanh}\left (a x \right ) a^{2} c x}{\left (-a^{2} c \,x^{2}+c \right )^{\frac {7}{2}}}\right )}{9 a^{2}}\) | \(121\) |
default | \(\frac {\left (a x +1\right ) \left (3 \,\operatorname {arctanh}\left (a x \right )-1\right ) \sqrt {-\left (a x -1\right ) c \left (a x +1\right )}}{72 a \left (a x -1\right )^{2} c^{3}}-\frac {3 \left (\operatorname {arctanh}\left (a x \right )-1\right ) \sqrt {-\left (a x -1\right ) c \left (a x +1\right )}}{8 a \,c^{3} \left (a x -1\right )}-\frac {3 \left (\operatorname {arctanh}\left (a x \right )+1\right ) \sqrt {-\left (a x -1\right ) c \left (a x +1\right )}}{8 a \left (a x +1\right ) c^{3}}+\frac {\left (a x -1\right ) \left (3 \,\operatorname {arctanh}\left (a x \right )+1\right ) \sqrt {-\left (a x -1\right ) c \left (a x +1\right )}}{72 a \left (a x +1\right )^{2} c^{3}}\) | \(160\) |
Input:
int(arctanh(a*x)/(-a^2*c*x^2+c)^(5/2),x,method=_RETURNVERBOSE)
Output:
(4*a^4*x^5-80/9*a^2*x^3+44/9*x)*arctanh(a*x)/(-a^2*c*x^2+c)^(5/2)+1/9*(6*a ^2*x^2-7)/a^2*(a*x+1)^2*(a*x-1)^2*(a/(-a^2*x^2+1)/(-a^2*c*x^2+c)^(5/2)+5*a rctanh(a*x)/(-a^2*c*x^2+c)^(7/2)*a^2*c*x)
Time = 0.09 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.80 \[ \int \frac {\text {arctanh}(a x)}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\sqrt {-a^{2} c x^{2} + c} {\left (12 \, a^{2} x^{2} - 3 \, {\left (2 \, a^{3} x^{3} - 3 \, a x\right )} \log \left (-\frac {a x + 1}{a x - 1}\right ) - 14\right )}}{18 \, {\left (a^{5} c^{3} x^{4} - 2 \, a^{3} c^{3} x^{2} + a c^{3}\right )}} \] Input:
integrate(arctanh(a*x)/(-a^2*c*x^2+c)^(5/2),x, algorithm="fricas")
Output:
1/18*sqrt(-a^2*c*x^2 + c)*(12*a^2*x^2 - 3*(2*a^3*x^3 - 3*a*x)*log(-(a*x + 1)/(a*x - 1)) - 14)/(a^5*c^3*x^4 - 2*a^3*c^3*x^2 + a*c^3)
\[ \int \frac {\text {arctanh}(a x)}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\operatorname {atanh}{\left (a x \right )}}{\left (- c \left (a x - 1\right ) \left (a x + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate(atanh(a*x)/(-a**2*c*x**2+c)**(5/2),x)
Output:
Integral(atanh(a*x)/(-c*(a*x - 1)*(a*x + 1))**(5/2), x)
Time = 0.04 (sec) , antiderivative size = 90, normalized size of antiderivative = 0.86 \[ \int \frac {\text {arctanh}(a x)}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {1}{9} \, a {\left (\frac {6}{\sqrt {-a^{2} c x^{2} + c} a^{2} c^{2}} + \frac {1}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} a^{2} c}\right )} + \frac {1}{3} \, {\left (\frac {2 \, x}{\sqrt {-a^{2} c x^{2} + c} c^{2}} + \frac {x}{{\left (-a^{2} c x^{2} + c\right )}^{\frac {3}{2}} c}\right )} \operatorname {artanh}\left (a x\right ) \] Input:
integrate(arctanh(a*x)/(-a^2*c*x^2+c)^(5/2),x, algorithm="maxima")
Output:
-1/9*a*(6/(sqrt(-a^2*c*x^2 + c)*a^2*c^2) + 1/((-a^2*c*x^2 + c)^(3/2)*a^2*c )) + 1/3*(2*x/(sqrt(-a^2*c*x^2 + c)*c^2) + x/((-a^2*c*x^2 + c)^(3/2)*c))*a rctanh(a*x)
Time = 0.16 (sec) , antiderivative size = 111, normalized size of antiderivative = 1.06 \[ \int \frac {\text {arctanh}(a x)}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=-\frac {\sqrt {-a^{2} c x^{2} + c} {\left (\frac {2 \, a^{2} x^{2}}{c} - \frac {3}{c}\right )} x \log \left (-\frac {a x + 1}{a x - 1}\right )}{6 \, {\left (a^{2} c x^{2} - c\right )}^{2}} - \frac {6 \, a^{2} c x^{2} - 7 \, c}{9 \, {\left (a^{2} c x^{2} - c\right )} \sqrt {-a^{2} c x^{2} + c} a c^{2}} \] Input:
integrate(arctanh(a*x)/(-a^2*c*x^2+c)^(5/2),x, algorithm="giac")
Output:
-1/6*sqrt(-a^2*c*x^2 + c)*(2*a^2*x^2/c - 3/c)*x*log(-(a*x + 1)/(a*x - 1))/ (a^2*c*x^2 - c)^2 - 1/9*(6*a^2*c*x^2 - 7*c)/((a^2*c*x^2 - c)*sqrt(-a^2*c*x ^2 + c)*a*c^2)
Timed out. \[ \int \frac {\text {arctanh}(a x)}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\int \frac {\mathrm {atanh}\left (a\,x\right )}{{\left (c-a^2\,c\,x^2\right )}^{5/2}} \,d x \] Input:
int(atanh(a*x)/(c - a^2*c*x^2)^(5/2),x)
Output:
int(atanh(a*x)/(c - a^2*c*x^2)^(5/2), x)
\[ \int \frac {\text {arctanh}(a x)}{\left (c-a^2 c x^2\right )^{5/2}} \, dx=\frac {\int \frac {\mathit {atanh} \left (a x \right )}{\sqrt {-a^{2} x^{2}+1}\, a^{4} x^{4}-2 \sqrt {-a^{2} x^{2}+1}\, a^{2} x^{2}+\sqrt {-a^{2} x^{2}+1}}d x}{\sqrt {c}\, c^{2}} \] Input:
int(atanh(a*x)/(-a^2*c*x^2+c)^(5/2),x)
Output:
int(atanh(a*x)/(sqrt( - a**2*x**2 + 1)*a**4*x**4 - 2*sqrt( - a**2*x**2 + 1 )*a**2*x**2 + sqrt( - a**2*x**2 + 1)),x)/(sqrt(c)*c**2)