Integrand size = 23, antiderivative size = 761 \[ \int \frac {a+b \text {arctanh}(c+d x)}{x^2 \left (e+f x^3\right )} \, dx=-\frac {a+b \text {arctanh}(c+d x)}{e x}+\frac {b d \log (x)}{\left (1-c^2\right ) e}-\frac {b d \log (1-c-d x)}{2 (1-c) e}-\frac {\sqrt [3]{f} (a+b \text {arctanh}(c+d x)) \log \left (\frac {2}{1+c+d x}\right )}{3 e^{4/3}}+\frac {\sqrt [3]{-1} \sqrt [3]{f} (a+b \text {arctanh}(c+d x)) \log \left (\frac {2}{1+c+d x}\right )}{3 e^{4/3}}-\frac {(-1)^{2/3} \sqrt [3]{f} (a+b \text {arctanh}(c+d x)) \log \left (\frac {2}{1+c+d x}\right )}{3 e^{4/3}}-\frac {b d \log (1+c+d x)}{2 (1+c) e}+\frac {\sqrt [3]{f} (a+b \text {arctanh}(c+d x)) \log \left (\frac {2 d \left (\sqrt [3]{e}+\sqrt [3]{f} x\right )}{\left (d \sqrt [3]{e}+(1-c) \sqrt [3]{f}\right ) (1+c+d x)}\right )}{3 e^{4/3}}-\frac {\sqrt [3]{-1} \sqrt [3]{f} (a+b \text {arctanh}(c+d x)) \log \left (\frac {2 d \left (\sqrt [3]{e}-\sqrt [3]{-1} \sqrt [3]{f} x\right )}{\left (d \sqrt [3]{e}-\sqrt [3]{-1} (1-c) \sqrt [3]{f}\right ) (1+c+d x)}\right )}{3 e^{4/3}}+\frac {(-1)^{2/3} \sqrt [3]{f} (a+b \text {arctanh}(c+d x)) \log \left (\frac {2 d \left (\sqrt [3]{e}+(-1)^{2/3} \sqrt [3]{f} x\right )}{\left (d \sqrt [3]{e}+(-1)^{2/3} (1-c) \sqrt [3]{f}\right ) (1+c+d x)}\right )}{3 e^{4/3}}+\frac {b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2}{1+c+d x}\right )}{6 e^{4/3}}-\frac {\sqrt [3]{-1} b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2}{1+c+d x}\right )}{6 e^{4/3}}+\frac {(-1)^{2/3} b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2}{1+c+d x}\right )}{6 e^{4/3}}-\frac {b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2 d \left (\sqrt [3]{e}+\sqrt [3]{f} x\right )}{\left (d \sqrt [3]{e}+(1-c) \sqrt [3]{f}\right ) (1+c+d x)}\right )}{6 e^{4/3}}+\frac {\sqrt [3]{-1} b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2 d \left (\sqrt [3]{e}-\sqrt [3]{-1} \sqrt [3]{f} x\right )}{\left (d \sqrt [3]{e}-\sqrt [3]{-1} (1-c) \sqrt [3]{f}\right ) (1+c+d x)}\right )}{6 e^{4/3}}-\frac {(-1)^{2/3} b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2 d \left (\sqrt [3]{e}+(-1)^{2/3} \sqrt [3]{f} x\right )}{\left (d \sqrt [3]{e}+(-1)^{2/3} (1-c) \sqrt [3]{f}\right ) (1+c+d x)}\right )}{6 e^{4/3}} \] Output:
-(a+b*arctanh(d*x+c))/e/x+b*d*ln(x)/(-c^2+1)/e-1/2*b*d*ln(-d*x-c+1)/(1-c)/ e-1/3*f^(1/3)*(a+b*arctanh(d*x+c))*ln(2/(d*x+c+1))/e^(4/3)+1/3*(-1)^(1/3)* f^(1/3)*(a+b*arctanh(d*x+c))*ln(2/(d*x+c+1))/e^(4/3)-1/3*(-1)^(2/3)*f^(1/3 )*(a+b*arctanh(d*x+c))*ln(2/(d*x+c+1))/e^(4/3)-1/2*b*d*ln(d*x+c+1)/(1+c)/e +1/3*f^(1/3)*(a+b*arctanh(d*x+c))*ln(2*d*(e^(1/3)+f^(1/3)*x)/(d*e^(1/3)+(1 -c)*f^(1/3))/(d*x+c+1))/e^(4/3)-1/3*(-1)^(1/3)*f^(1/3)*(a+b*arctanh(d*x+c) )*ln(2*d*(e^(1/3)-(-1)^(1/3)*f^(1/3)*x)/(d*e^(1/3)-(-1)^(1/3)*(1-c)*f^(1/3 ))/(d*x+c+1))/e^(4/3)+1/3*(-1)^(2/3)*f^(1/3)*(a+b*arctanh(d*x+c))*ln(2*d*( e^(1/3)+(-1)^(2/3)*f^(1/3)*x)/(d*e^(1/3)+(-1)^(2/3)*(1-c)*f^(1/3))/(d*x+c+ 1))/e^(4/3)+1/6*b*f^(1/3)*polylog(2,1-2/(d*x+c+1))/e^(4/3)-1/6*(-1)^(1/3)* b*f^(1/3)*polylog(2,1-2/(d*x+c+1))/e^(4/3)+1/6*(-1)^(2/3)*b*f^(1/3)*polylo g(2,1-2/(d*x+c+1))/e^(4/3)-1/6*b*f^(1/3)*polylog(2,1-2*d*(e^(1/3)+f^(1/3)* x)/(d*e^(1/3)+(1-c)*f^(1/3))/(d*x+c+1))/e^(4/3)+1/6*(-1)^(1/3)*b*f^(1/3)*p olylog(2,1-2*d*(e^(1/3)-(-1)^(1/3)*f^(1/3)*x)/(d*e^(1/3)-(-1)^(1/3)*(1-c)* f^(1/3))/(d*x+c+1))/e^(4/3)-1/6*(-1)^(2/3)*b*f^(1/3)*polylog(2,1-2*d*(e^(1 /3)+(-1)^(2/3)*f^(1/3)*x)/(d*e^(1/3)+(-1)^(2/3)*(1-c)*f^(1/3))/(d*x+c+1))/ e^(4/3)
Result contains higher order function than in optimal. Order 9 vs. order 4 in optimal.
Time = 8.74 (sec) , antiderivative size = 2550, normalized size of antiderivative = 3.35 \[ \int \frac {a+b \text {arctanh}(c+d x)}{x^2 \left (e+f x^3\right )} \, dx=\text {Result too large to show} \] Input:
Integrate[(a + b*ArcTanh[c + d*x])/(x^2*(e + f*x^3)),x]
Output:
((-6*a*e^(1/3))/x + 2*Sqrt[3]*a*f^(1/3)*ArcTan[(1 - (2*f^(1/3)*x)/e^(1/3)) /Sqrt[3]] + 2*a*f^(1/3)*Log[e^(1/3) + f^(1/3)*x] - a*f^(1/3)*Log[e^(2/3) - e^(1/3)*f^(1/3)*x + f^(2/3)*x^2] - (6*b*e^(1/3)*((-1 + c^2 + c*d*x)*ArcTa nh[c + d*x] + d*x*Log[-((d*x)/Sqrt[1 - (c + d*x)^2])]))/((-1 + c)*(1 + c)* x) + b*d*e^(1/3)*f*(-4*ArcTanh[c + d*x]*RootSum[d^3*e - f - 3*c*f - 3*c^2* f - c^3*f + 3*d^3*e*#1 + 3*f*#1 + 3*c*f*#1 - 3*c^2*f*#1 - 3*c^3*f*#1 + 3*d ^3*e*#1^2 - 3*f*#1^2 + 3*c*f*#1^2 + 3*c^2*f*#1^2 - 3*c^3*f*#1^2 + d^3*e*#1 ^3 + f*#1^3 - 3*c*f*#1^3 + 3*c^2*f*#1^3 - c^3*f*#1^3 & , (ArcTanh[c + d*x] + c*ArcTanh[c + d*x] + Log[(-1 - c - d*x + #1 - c*#1 - d*x*#1)/Sqrt[1 - ( c + d*x)^2]] + c*Log[(-1 - c - d*x + #1 - c*#1 - d*x*#1)/Sqrt[1 - (c + d*x )^2]] - ArcTanh[c + d*x]*#1 + c*ArcTanh[c + d*x]*#1 - Log[(-1 - c - d*x + #1 - c*#1 - d*x*#1)/Sqrt[1 - (c + d*x)^2]]*#1 + c*Log[(-1 - c - d*x + #1 - c*#1 - d*x*#1)/Sqrt[1 - (c + d*x)^2]]*#1)/(-(d^3*e) - f - c*f + c^2*f + c ^3*f - 2*d^3*e*#1 + 2*f*#1 - 2*c*f*#1 - 2*c^2*f*#1 + 2*c^3*f*#1 - d^3*e*#1 ^2 - f*#1^2 + 3*c*f*#1^2 - 3*c^2*f*#1^2 + c^3*f*#1^2) & ] + 2*RootSum[d^3* e - f - 3*c*f - 3*c^2*f - c^3*f + 3*d^3*e*#1 + 3*f*#1 + 3*c*f*#1 - 3*c^2*f *#1 - 3*c^3*f*#1 + 3*d^3*e*#1^2 - 3*f*#1^2 + 3*c*f*#1^2 + 3*c^2*f*#1^2 - 3 *c^3*f*#1^2 + d^3*e*#1^3 + f*#1^3 - 3*c*f*#1^3 + 3*c^2*f*#1^3 - c^3*f*#1^3 & , (I*Pi*ArcTanh[c + d*x] + 2*ArcTanh[c + d*x]*ArcTanh[(1 - #1)/(1 + #1) ] - I*Pi*Log[1 + E^(2*ArcTanh[c + d*x])] + 2*ArcTanh[c + d*x]*Log[1 - E...
Time = 1.92 (sec) , antiderivative size = 865, normalized size of antiderivative = 1.14, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {7276, 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.
| \(\displaystyle \int \frac {a+b \text {arctanh}(c+d x)}{x^2 \left (e+f x^3\right )} \, dx\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \int \left (\frac {a}{x^2 \left (e+f x^3\right )}+\frac {b \text {arctanh}(c+d x)}{x^2 \left (e+f x^3\right )}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle \frac {\sqrt [3]{f} \arctan \left (\frac {\sqrt [3]{e}-2 \sqrt [3]{f} x}{\sqrt {3} \sqrt [3]{e}}\right ) a}{\sqrt {3} e^{4/3}}+\frac {\sqrt [3]{f} \log \left (\sqrt [3]{f} x+\sqrt [3]{e}\right ) a}{3 e^{4/3}}-\frac {\sqrt [3]{f} \log \left (f^{2/3} x^2-\sqrt [3]{e} \sqrt [3]{f} x+e^{2/3}\right ) a}{6 e^{4/3}}-\frac {a}{e x}-\frac {b \text {arctanh}(c+d x)}{e x}+\frac {b d \log (x)}{\left (1-c^2\right ) e}-\frac {b d \log (-c-d x+1)}{2 (1-c) e}-\frac {(-1)^{2/3} b \sqrt [3]{f} \text {arctanh}(c+d x) \log \left (\frac {2}{c+d x+1}\right )}{3 e^{4/3}}+\frac {\sqrt [3]{-1} b \sqrt [3]{f} \text {arctanh}(c+d x) \log \left (\frac {2}{c+d x+1}\right )}{3 e^{4/3}}-\frac {b \sqrt [3]{f} \text {arctanh}(c+d x) \log \left (\frac {2}{c+d x+1}\right )}{3 e^{4/3}}-\frac {b d \log (c+d x+1)}{2 (c+1) e}+\frac {b \sqrt [3]{f} \text {arctanh}(c+d x) \log \left (\frac {2 d \left (\sqrt [3]{f} x+\sqrt [3]{e}\right )}{\left (\sqrt [3]{f} (1-c)+d \sqrt [3]{e}\right ) (c+d x+1)}\right )}{3 e^{4/3}}-\frac {\sqrt [3]{-1} b \sqrt [3]{f} \text {arctanh}(c+d x) \log \left (\frac {2 d \left (\sqrt [3]{e}-\sqrt [3]{-1} \sqrt [3]{f} x\right )}{\left (d \sqrt [3]{e}-\sqrt [3]{-1} (1-c) \sqrt [3]{f}\right ) (c+d x+1)}\right )}{3 e^{4/3}}+\frac {(-1)^{2/3} b \sqrt [3]{f} \text {arctanh}(c+d x) \log \left (\frac {2 d \left ((-1)^{2/3} \sqrt [3]{f} x+\sqrt [3]{e}\right )}{\left ((-1)^{2/3} \sqrt [3]{f} (1-c)+d \sqrt [3]{e}\right ) (c+d x+1)}\right )}{3 e^{4/3}}+\frac {(-1)^{2/3} b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{6 e^{4/3}}-\frac {\sqrt [3]{-1} b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{6 e^{4/3}}+\frac {b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{6 e^{4/3}}-\frac {b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2 d \left (\sqrt [3]{f} x+\sqrt [3]{e}\right )}{\left (\sqrt [3]{f} (1-c)+d \sqrt [3]{e}\right ) (c+d x+1)}\right )}{6 e^{4/3}}+\frac {\sqrt [3]{-1} b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2 d \left (\sqrt [3]{e}-\sqrt [3]{-1} \sqrt [3]{f} x\right )}{\left (d \sqrt [3]{e}-\sqrt [3]{-1} (1-c) \sqrt [3]{f}\right ) (c+d x+1)}\right )}{6 e^{4/3}}-\frac {(-1)^{2/3} b \sqrt [3]{f} \operatorname {PolyLog}\left (2,1-\frac {2 d \left ((-1)^{2/3} \sqrt [3]{f} x+\sqrt [3]{e}\right )}{\left ((-1)^{2/3} \sqrt [3]{f} (1-c)+d \sqrt [3]{e}\right ) (c+d x+1)}\right )}{6 e^{4/3}}\) |
Input:
Int[(a + b*ArcTanh[c + d*x])/(x^2*(e + f*x^3)),x]
Output:
-(a/(e*x)) + (a*f^(1/3)*ArcTan[(e^(1/3) - 2*f^(1/3)*x)/(Sqrt[3]*e^(1/3))]) /(Sqrt[3]*e^(4/3)) - (b*ArcTanh[c + d*x])/(e*x) + (b*d*Log[x])/((1 - c^2)* e) - (b*d*Log[1 - c - d*x])/(2*(1 - c)*e) - (b*f^(1/3)*ArcTanh[c + d*x]*Lo g[2/(1 + c + d*x)])/(3*e^(4/3)) + ((-1)^(1/3)*b*f^(1/3)*ArcTanh[c + d*x]*L og[2/(1 + c + d*x)])/(3*e^(4/3)) - ((-1)^(2/3)*b*f^(1/3)*ArcTanh[c + d*x]* Log[2/(1 + c + d*x)])/(3*e^(4/3)) - (b*d*Log[1 + c + d*x])/(2*(1 + c)*e) + (a*f^(1/3)*Log[e^(1/3) + f^(1/3)*x])/(3*e^(4/3)) + (b*f^(1/3)*ArcTanh[c + d*x]*Log[(2*d*(e^(1/3) + f^(1/3)*x))/((d*e^(1/3) + (1 - c)*f^(1/3))*(1 + c + d*x))])/(3*e^(4/3)) - ((-1)^(1/3)*b*f^(1/3)*ArcTanh[c + d*x]*Log[(2*d* (e^(1/3) - (-1)^(1/3)*f^(1/3)*x))/((d*e^(1/3) - (-1)^(1/3)*(1 - c)*f^(1/3) )*(1 + c + d*x))])/(3*e^(4/3)) + ((-1)^(2/3)*b*f^(1/3)*ArcTanh[c + d*x]*Lo g[(2*d*(e^(1/3) + (-1)^(2/3)*f^(1/3)*x))/((d*e^(1/3) + (-1)^(2/3)*(1 - c)* f^(1/3))*(1 + c + d*x))])/(3*e^(4/3)) - (a*f^(1/3)*Log[e^(2/3) - e^(1/3)*f ^(1/3)*x + f^(2/3)*x^2])/(6*e^(4/3)) + (b*f^(1/3)*PolyLog[2, 1 - 2/(1 + c + d*x)])/(6*e^(4/3)) - ((-1)^(1/3)*b*f^(1/3)*PolyLog[2, 1 - 2/(1 + c + d*x )])/(6*e^(4/3)) + ((-1)^(2/3)*b*f^(1/3)*PolyLog[2, 1 - 2/(1 + c + d*x)])/( 6*e^(4/3)) - (b*f^(1/3)*PolyLog[2, 1 - (2*d*(e^(1/3) + f^(1/3)*x))/((d*e^( 1/3) + (1 - c)*f^(1/3))*(1 + c + d*x))])/(6*e^(4/3)) + ((-1)^(1/3)*b*f^(1/ 3)*PolyLog[2, 1 - (2*d*(e^(1/3) - (-1)^(1/3)*f^(1/3)*x))/((d*e^(1/3) - (-1 )^(1/3)*(1 - c)*f^(1/3))*(1 + c + d*x))])/(6*e^(4/3)) - ((-1)^(2/3)*b*f...
Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ [n, 0]
Result contains higher order function than in optimal. Order 9 vs. order 4.
Time = 0.46 (sec) , antiderivative size = 513, normalized size of antiderivative = 0.67
| method | result | size |
| risch | \(-\frac {d b \ln \left (-d x \right )}{2 e \left (-1+c \right )}+\frac {d b \ln \left (-d x -c +1\right )}{2 e \left (-1+c \right )}+\frac {b \ln \left (-d x -c +1\right ) c}{2 e \left (-1+c \right ) x}-\frac {b \ln \left (-d x -c +1\right )}{2 e \left (-1+c \right ) x}-\frac {d b \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (f \,\textit {\_Z}^{3}+\left (3 f c -3 f \right ) \textit {\_Z}^{2}+\left (3 c^{2} f -6 f c +3 f \right ) \textit {\_Z} +c^{3} f -d^{3} e -3 c^{2} f +3 f c -f \right )}{\sum }\frac {\ln \left (-d x -c +1\right ) \ln \left (\frac {d x +\textit {\_R1} +c -1}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {d x +\textit {\_R1} +c -1}{\textit {\_R1}}\right )}{\textit {\_R1} -1+c}\right )}{6 e}-\frac {a}{e x}+\frac {d a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (f \,\textit {\_Z}^{3}+\left (3 f c -3 f \right ) \textit {\_Z}^{2}+\left (3 c^{2} f -6 f c +3 f \right ) \textit {\_Z} +c^{3} f -d^{3} e -3 c^{2} f +3 f c -f \right )}{\sum }\frac {\left (c +\textit {\_R} -1\right ) \ln \left (-d x -\textit {\_R} -c +1\right )}{\textit {\_R}^{2}+2 \textit {\_R} c +c^{2}-2 \textit {\_R} -2 c +1}\right )}{3 e}+\frac {b d \ln \left (d x \right )}{2 e \left (1+c \right )}-\frac {b d \ln \left (d x +c +1\right )}{2 \left (1+c \right ) e}-\frac {b \ln \left (d x +c +1\right ) c}{2 e \left (1+c \right ) x}-\frac {b \ln \left (d x +c +1\right )}{2 e \left (1+c \right ) x}-\frac {b d \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (f \,\textit {\_Z}^{3}+\left (-3 f c -3 f \right ) \textit {\_Z}^{2}+\left (3 c^{2} f +6 f c +3 f \right ) \textit {\_Z} -c^{3} f +d^{3} e -3 c^{2} f -3 f c -f \right )}{\sum }\frac {\ln \left (d x +c +1\right ) \ln \left (\frac {-d x +\textit {\_R1} -c -1}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c -1}{\textit {\_R1}}\right )}{\textit {\_R1} -1-c}\right )}{6 e}\) | \(513\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1347\) |
| default | \(\text {Expression too large to display}\) | \(1347\) |
| parts | \(\text {Expression too large to display}\) | \(1364\) |
Input:
int((a+b*arctanh(d*x+c))/x^2/(f*x^3+e),x,method=_RETURNVERBOSE)
Output:
-1/2*d*b/e/(-1+c)*ln(-d*x)+1/2*d*b/e*ln(-d*x-c+1)/(-1+c)+1/2*b/e*ln(-d*x-c +1)/(-1+c)/x*c-1/2*b/e*ln(-d*x-c+1)/(-1+c)/x-1/6*d*b/e*sum(1/(_R1-1+c)*(ln (-d*x-c+1)*ln((d*x+_R1+c-1)/_R1)+dilog((d*x+_R1+c-1)/_R1)),_R1=RootOf(f*_Z ^3+(3*c*f-3*f)*_Z^2+(3*c^2*f-6*c*f+3*f)*_Z+c^3*f-d^3*e-3*c^2*f+3*f*c-f))-a /e/x+1/3*d*a*sum((c+_R-1)/(_R^2+2*_R*c+c^2-2*_R-2*c+1)*ln(-d*x-_R-c+1),_R= RootOf(f*_Z^3+(3*c*f-3*f)*_Z^2+(3*c^2*f-6*c*f+3*f)*_Z+c^3*f-d^3*e-3*c^2*f+ 3*f*c-f))/e+1/2*b*d/e/(1+c)*ln(d*x)-1/2*b*d*ln(d*x+c+1)/(1+c)/e-1/2*b/e*ln (d*x+c+1)/(1+c)/x*c-1/2*b/e*ln(d*x+c+1)/(1+c)/x-1/6*b*d/e*sum(1/(_R1-1-c)* (ln(d*x+c+1)*ln((-d*x+_R1-c-1)/_R1)+dilog((-d*x+_R1-c-1)/_R1)),_R1=RootOf( f*_Z^3+(-3*c*f-3*f)*_Z^2+(3*c^2*f+6*c*f+3*f)*_Z-c^3*f+d^3*e-3*c^2*f-3*f*c- f))
\[ \int \frac {a+b \text {arctanh}(c+d x)}{x^2 \left (e+f x^3\right )} \, dx=\int { \frac {b \operatorname {artanh}\left (d x + c\right ) + a}{{\left (f x^{3} + e\right )} x^{2}} \,d x } \] Input:
integrate((a+b*arctanh(d*x+c))/x^2/(f*x^3+e),x, algorithm="fricas")
Output:
integral((b*arctanh(d*x + c) + a)/(f*x^5 + e*x^2), x)
Timed out. \[ \int \frac {a+b \text {arctanh}(c+d x)}{x^2 \left (e+f x^3\right )} \, dx=\text {Timed out} \] Input:
integrate((a+b*atanh(d*x+c))/x**2/(f*x**3+e),x)
Output:
Timed out
Exception generated. \[ \int \frac {a+b \text {arctanh}(c+d x)}{x^2 \left (e+f x^3\right )} \, dx=\text {Exception raised: ValueError} \] Input:
integrate((a+b*arctanh(d*x+c))/x^2/(f*x^3+e),x, algorithm="maxima")
Output:
Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'assume' command before evaluation *may* help (example of legal syntax is 'assume(e>0)', see `assume?` for more de tails)Is e
\[ \int \frac {a+b \text {arctanh}(c+d x)}{x^2 \left (e+f x^3\right )} \, dx=\int { \frac {b \operatorname {artanh}\left (d x + c\right ) + a}{{\left (f x^{3} + e\right )} x^{2}} \,d x } \] Input:
integrate((a+b*arctanh(d*x+c))/x^2/(f*x^3+e),x, algorithm="giac")
Output:
integrate((b*arctanh(d*x + c) + a)/((f*x^3 + e)*x^2), x)
Timed out. \[ \int \frac {a+b \text {arctanh}(c+d x)}{x^2 \left (e+f x^3\right )} \, dx=\int \frac {a+b\,\mathrm {atanh}\left (c+d\,x\right )}{x^2\,\left (f\,x^3+e\right )} \,d x \] Input:
int((a + b*atanh(c + d*x))/(x^2*(e + f*x^3)),x)
Output:
int((a + b*atanh(c + d*x))/(x^2*(e + f*x^3)), x)
\[ \int \frac {a+b \text {arctanh}(c+d x)}{x^2 \left (e+f x^3\right )} \, dx=\frac {2 \sqrt {3}\, \mathit {atan} \left (\frac {e^{\frac {1}{3}}-2 f^{\frac {1}{3}} x}{e^{\frac {1}{3}} \sqrt {3}}\right ) a f x +6 f^{\frac {2}{3}} e^{\frac {4}{3}} \left (\int \frac {\mathit {atanh} \left (d x +c \right )}{f \,x^{5}+e \,x^{2}}d x \right ) b x -6 f^{\frac {2}{3}} e^{\frac {1}{3}} a -\mathrm {log}\left (e^{\frac {2}{3}}-f^{\frac {1}{3}} e^{\frac {1}{3}} x +f^{\frac {2}{3}} x^{2}\right ) a f x +2 \,\mathrm {log}\left (e^{\frac {1}{3}}+f^{\frac {1}{3}} x \right ) a f x}{6 f^{\frac {2}{3}} e^{\frac {4}{3}} x} \] Input:
int((a+b*atanh(d*x+c))/x^2/(f*x^3+e),x)
Output:
(2*sqrt(3)*atan((e**(1/3) - 2*f**(1/3)*x)/(e**(1/3)*sqrt(3)))*a*f*x + 6*f* *(2/3)*e**(1/3)*int(atanh(c + d*x)/(e*x**2 + f*x**5),x)*b*e*x - 6*f**(2/3) *e**(1/3)*a - log(e**(2/3) - f**(1/3)*e**(1/3)*x + f**(2/3)*x**2)*a*f*x + 2*log(e**(1/3) + f**(1/3)*x)*a*f*x)/(6*f**(2/3)*e**(1/3)*e*x)