Integrand size = 14, antiderivative size = 30 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x} \, dx=a \log (x)-\frac {1}{4} b \operatorname {PolyLog}\left (2,-c x^2\right )+\frac {1}{4} b \operatorname {PolyLog}\left (2,c x^2\right ) \] Output:
a*ln(x)-1/4*b*polylog(2,-c*x^2)+1/4*b*polylog(2,c*x^2)
Time = 0.02 (sec) , antiderivative size = 28, normalized size of antiderivative = 0.93 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x} \, dx=a \log (x)+\frac {1}{4} b \left (-\operatorname {PolyLog}\left (2,-c x^2\right )+\operatorname {PolyLog}\left (2,c x^2\right )\right ) \] Input:
Integrate[(a + b*ArcTanh[c*x^2])/x,x]
Output:
a*Log[x] + (b*(-PolyLog[2, -(c*x^2)] + PolyLog[2, c*x^2]))/4
Time = 0.25 (sec) , antiderivative size = 36, normalized size of antiderivative = 1.20, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {6450, 6446}
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.
| \(\displaystyle \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x} \, dx\) |
| \(\Big \downarrow \) 6450 |
| \(\displaystyle \frac {1}{2} \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^2}dx^2\) |
| \(\Big \downarrow \) 6446 |
| \(\displaystyle \frac {1}{2} \left (a \log \left (x^2\right )-\frac {1}{2} b \operatorname {PolyLog}\left (2,-c x^2\right )+\frac {1}{2} b \operatorname {PolyLog}\left (2,c x^2\right )\right )\) |
Input:
Int[(a + b*ArcTanh[c*x^2])/x,x]
Output:
(a*Log[x^2] - (b*PolyLog[2, -(c*x^2)])/2 + (b*PolyLog[2, c*x^2])/2)/2
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x ] + (-Simp[(b/2)*PolyLog[2, (-c)*x], x] + Simp[(b/2)*PolyLog[2, c*x], x]) / ; FreeQ[{a, b, c}, x]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)/(x_), x_Symbol] :> Simp[ 1/n Subst[Int[(a + b*ArcTanh[c*x])^p/x, x], x, x^n], x] /; FreeQ[{a, b, c , n}, x] && IGtQ[p, 0]
Leaf count of result is larger than twice the leaf count of optimal. \(125\) vs. \(2(26)=52\).
Time = 0.16 (sec) , antiderivative size = 126, normalized size of antiderivative = 4.20
| method | result | size |
| default | \(a \ln \left (x \right )+b \left (\ln \left (x \right ) \operatorname {arctanh}\left (c \,x^{2}\right )-2 c \left (\frac {\ln \left (x \right ) \left (\ln \left (1+\sqrt {-c}\, x \right )+\ln \left (1-\sqrt {-c}\, x \right )\right )}{4 c}+\frac {\operatorname {dilog}\left (1+\sqrt {-c}\, x \right )+\operatorname {dilog}\left (1-\sqrt {-c}\, x \right )}{4 c}-\frac {\ln \left (x \right ) \left (\ln \left (1-\sqrt {c}\, x \right )+\ln \left (1+\sqrt {c}\, x \right )\right )}{4 c}-\frac {\operatorname {dilog}\left (1-\sqrt {c}\, x \right )+\operatorname {dilog}\left (1+\sqrt {c}\, x \right )}{4 c}\right )\right )\) | \(126\) |
| parts | \(a \ln \left (x \right )+b \left (\ln \left (x \right ) \operatorname {arctanh}\left (c \,x^{2}\right )-2 c \left (\frac {\ln \left (x \right ) \left (\ln \left (1+\sqrt {-c}\, x \right )+\ln \left (1-\sqrt {-c}\, x \right )\right )}{4 c}+\frac {\operatorname {dilog}\left (1+\sqrt {-c}\, x \right )+\operatorname {dilog}\left (1-\sqrt {-c}\, x \right )}{4 c}-\frac {\ln \left (x \right ) \left (\ln \left (1-\sqrt {c}\, x \right )+\ln \left (1+\sqrt {c}\, x \right )\right )}{4 c}-\frac {\operatorname {dilog}\left (1-\sqrt {c}\, x \right )+\operatorname {dilog}\left (1+\sqrt {c}\, x \right )}{4 c}\right )\right )\) | \(126\) |
| risch | \(a \ln \left (x \right )-\frac {\ln \left (x \right ) \ln \left (-c \,x^{2}+1\right ) b}{2}+\frac {b \ln \left (x \right ) \ln \left (1+\sqrt {c}\, x \right )}{2}+\frac {b \ln \left (x \right ) \ln \left (1-\sqrt {c}\, x \right )}{2}+\frac {b \operatorname {dilog}\left (1-\sqrt {c}\, x \right )}{2}+\frac {b \operatorname {dilog}\left (1+\sqrt {c}\, x \right )}{2}-\frac {\ln \left (1+\sqrt {-c}\, x \right ) \ln \left (x \right ) b}{2}-\frac {\ln \left (1-\sqrt {-c}\, x \right ) \ln \left (x \right ) b}{2}+\frac {\ln \left (x \right ) \ln \left (c \,x^{2}+1\right ) b}{2}-\frac {\operatorname {dilog}\left (1+\sqrt {-c}\, x \right ) b}{2}-\frac {\operatorname {dilog}\left (1-\sqrt {-c}\, x \right ) b}{2}\) | \(141\) |
Input:
int((a+b*arctanh(c*x^2))/x,x,method=_RETURNVERBOSE)
Output:
a*ln(x)+b*(ln(x)*arctanh(c*x^2)-2*c*(1/4*ln(x)*(ln(1+(-c)^(1/2)*x)+ln(1-(- c)^(1/2)*x))/c+1/4*(dilog(1+(-c)^(1/2)*x)+dilog(1-(-c)^(1/2)*x))/c-1/4*ln( x)*(ln(1-c^(1/2)*x)+ln(1+c^(1/2)*x))/c-1/4*(dilog(1-c^(1/2)*x)+dilog(1+c^( 1/2)*x))/c))
\[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x} \, dx=\int { \frac {b \operatorname {artanh}\left (c x^{2}\right ) + a}{x} \,d x } \] Input:
integrate((a+b*arctanh(c*x^2))/x,x, algorithm="fricas")
Output:
integral((b*arctanh(c*x^2) + a)/x, x)
\[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x} \, dx=\int \frac {a + b \operatorname {atanh}{\left (c x^{2} \right )}}{x}\, dx \] Input:
integrate((a+b*atanh(c*x**2))/x,x)
Output:
Integral((a + b*atanh(c*x**2))/x, x)
\[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x} \, dx=\int { \frac {b \operatorname {artanh}\left (c x^{2}\right ) + a}{x} \,d x } \] Input:
integrate((a+b*arctanh(c*x^2))/x,x, algorithm="maxima")
Output:
1/2*b*integrate((log(c*x^2 + 1) - log(-c*x^2 + 1))/x, x) + a*log(x)
\[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x} \, dx=\int { \frac {b \operatorname {artanh}\left (c x^{2}\right ) + a}{x} \,d x } \] Input:
integrate((a+b*arctanh(c*x^2))/x,x, algorithm="giac")
Output:
integrate((b*arctanh(c*x^2) + a)/x, x)
Timed out. \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x} \, dx=\int \frac {a+b\,\mathrm {atanh}\left (c\,x^2\right )}{x} \,d x \] Input:
int((a + b*atanh(c*x^2))/x,x)
Output:
int((a + b*atanh(c*x^2))/x, x)
\[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x} \, dx=\left (\int \frac {\mathit {atanh} \left (c \,x^{2}\right )}{x}d x \right ) b +\mathrm {log}\left (x \right ) a \] Input:
int((a+b*atanh(c*x^2))/x,x)
Output:
int(atanh(c*x**2)/x,x)*b + log(x)*a