Integrand size = 21, antiderivative size = 679 \[ \int \frac {x (a+b \text {arctanh}(c+d x))}{e+f x^3} \, dx=\frac {(a+b \text {arctanh}(c+d x)) \log \left (\frac {2}{1+c+d x}\right )}{3 \sqrt [3]{e} f^{2/3}}-\frac {\sqrt [3]{-1} (a+b \text {arctanh}(c+d x)) \log \left (\frac {2}{1+c+d x}\right )}{3 \sqrt [3]{e} f^{2/3}}+\frac {(-1)^{2/3} (a+b \text {arctanh}(c+d x)) \log \left (\frac {2}{1+c+d x}\right )}{3 \sqrt [3]{e} f^{2/3}}-\frac {(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 \sqrt [3]{e} f^{2/3}}+\frac {\sqrt [3]{-1} (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 \sqrt [3]{e} f^{2/3}}-\frac {(-1)^{2/3} (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 \sqrt [3]{e} f^{2/3}}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c+d x}\right )}{6 \sqrt [3]{e} f^{2/3}}+\frac {\sqrt [3]{-1} b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c+d x}\right )}{6 \sqrt [3]{e} f^{2/3}}-\frac {(-1)^{2/3} b \operatorname {PolyLog}\left (2,1-\frac {2}{1+c+d x}\right )}{6 \sqrt [3]{e} f^{2/3}}+\frac {b \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 \sqrt [3]{e} f^{2/3}}-\frac {\sqrt [3]{-1} b \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 \sqrt [3]{e} f^{2/3}}+\frac {(-1)^{2/3} b \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 \sqrt [3]{e} f^{2/3}} \] Output:
1/3*(a+b*arctanh(d*x+c))*ln(2/(d*x+c+1))/e^(1/3)/f^(2/3)-1/3*(-1)^(1/3)*(a +b*arctanh(d*x+c))*ln(2/(d*x+c+1))/e^(1/3)/f^(2/3)+1/3*(-1)^(2/3)*(a+b*arc tanh(d*x+c))*ln(2/(d*x+c+1))/e^(1/3)/f^(2/3)-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^(1/3)/f^(2/3 )+1/3*(-1)^(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^(1/3)/f^(2/3)-1/3*(-1) ^(2/3)*(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^(1/3)/f^(2/3)-1/6*b*polylog(2,1- 2/(d*x+c+1))/e^(1/3)/f^(2/3)+1/6*(-1)^(1/3)*b*polylog(2,1-2/(d*x+c+1))/e^( 1/3)/f^(2/3)-1/6*(-1)^(2/3)*b*polylog(2,1-2/(d*x+c+1))/e^(1/3)/f^(2/3)+1/6 *b*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^(1/3)/f^(2/3)-1/6*(-1)^(1/3)*b*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))/(d*x+c+1))/e^(1/3)/f^(2/3)+1/ 6*(-1)^(2/3)*b*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^(1/3)/f^(2/3)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 0.74 (sec) , antiderivative size = 652, normalized size of antiderivative = 0.96 \[ \int \frac {x (a+b \text {arctanh}(c+d x))}{e+f x^3} \, dx=\frac {3 a x^2 \operatorname {Hypergeometric2F1}\left (\frac {2}{3},1,\frac {5}{3},-\frac {f x^3}{e}\right )+\frac {b e^{2/3} \left (\log (1-c-d x) \log \left (\frac {d \left (\sqrt [3]{e}+\sqrt [3]{f} x\right )}{d \sqrt [3]{e}-(-1+c) \sqrt [3]{f}}\right )-\log (1+c+d x) \log \left (\frac {d \left (\sqrt [3]{e}+\sqrt [3]{f} x\right )}{d \sqrt [3]{e}-(1+c) \sqrt [3]{f}}\right )-\sqrt [3]{-1} \log (1-c-d x) \log \left (\frac {d \left (\sqrt [3]{e}-\sqrt [3]{-1} \sqrt [3]{f} x\right )}{d \sqrt [3]{e}+\sqrt [3]{-1} (-1+c) \sqrt [3]{f}}\right )+\sqrt [3]{-1} \log (1+c+d x) \log \left (\frac {d \left (\sqrt [3]{e}-\sqrt [3]{-1} \sqrt [3]{f} x\right )}{d \sqrt [3]{e}+\sqrt [3]{-1} (1+c) \sqrt [3]{f}}\right )+(-1)^{2/3} \log (1-c-d x) \log \left (\frac {d \left (\sqrt [3]{e}+(-1)^{2/3} \sqrt [3]{f} x\right )}{d \sqrt [3]{e}-(-1)^{2/3} (-1+c) \sqrt [3]{f}}\right )-(-1)^{2/3} \log (1+c+d x) \log \left (\frac {d \left (\sqrt [3]{e}+(-1)^{2/3} \sqrt [3]{f} x\right )}{d \sqrt [3]{e}-(-1)^{2/3} (1+c) \sqrt [3]{f}}\right )+\operatorname {PolyLog}\left (2,-\frac {\sqrt [3]{f} (-1+c+d x)}{d \sqrt [3]{e}-(-1+c) \sqrt [3]{f}}\right )-\sqrt [3]{-1} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{f} (-1+c+d x)}{d \sqrt [3]{e}+\sqrt [3]{-1} (-1+c) \sqrt [3]{f}}\right )+(-1)^{2/3} \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{f} (-1+c+d x)}{-d \sqrt [3]{e}+(-1)^{2/3} (-1+c) \sqrt [3]{f}}\right )-\operatorname {PolyLog}\left (2,-\frac {\sqrt [3]{f} (1+c+d x)}{d \sqrt [3]{e}-(1+c) \sqrt [3]{f}}\right )+\sqrt [3]{-1} \operatorname {PolyLog}\left (2,\frac {\sqrt [3]{-1} \sqrt [3]{f} (1+c+d x)}{d \sqrt [3]{e}+\sqrt [3]{-1} (1+c) \sqrt [3]{f}}\right )-(-1)^{2/3} \operatorname {PolyLog}\left (2,\frac {(-1)^{2/3} \sqrt [3]{f} (1+c+d x)}{-d \sqrt [3]{e}+(-1)^{2/3} (1+c) \sqrt [3]{f}}\right )\right )}{f^{2/3}}}{6 e} \] Input:
Integrate[(x*(a + b*ArcTanh[c + d*x]))/(e + f*x^3),x]
Output:
(3*a*x^2*Hypergeometric2F1[2/3, 1, 5/3, -((f*x^3)/e)] + (b*e^(2/3)*(Log[1 - c - d*x]*Log[(d*(e^(1/3) + f^(1/3)*x))/(d*e^(1/3) - (-1 + c)*f^(1/3))] - Log[1 + c + d*x]*Log[(d*(e^(1/3) + f^(1/3)*x))/(d*e^(1/3) - (1 + c)*f^(1/ 3))] - (-1)^(1/3)*Log[1 - c - d*x]*Log[(d*(e^(1/3) - (-1)^(1/3)*f^(1/3)*x) )/(d*e^(1/3) + (-1)^(1/3)*(-1 + c)*f^(1/3))] + (-1)^(1/3)*Log[1 + c + d*x] *Log[(d*(e^(1/3) - (-1)^(1/3)*f^(1/3)*x))/(d*e^(1/3) + (-1)^(1/3)*(1 + c)* f^(1/3))] + (-1)^(2/3)*Log[1 - c - d*x]*Log[(d*(e^(1/3) + (-1)^(2/3)*f^(1/ 3)*x))/(d*e^(1/3) - (-1)^(2/3)*(-1 + c)*f^(1/3))] - (-1)^(2/3)*Log[1 + c + d*x]*Log[(d*(e^(1/3) + (-1)^(2/3)*f^(1/3)*x))/(d*e^(1/3) - (-1)^(2/3)*(1 + c)*f^(1/3))] + PolyLog[2, -((f^(1/3)*(-1 + c + d*x))/(d*e^(1/3) - (-1 + c)*f^(1/3)))] - (-1)^(1/3)*PolyLog[2, ((-1)^(1/3)*f^(1/3)*(-1 + c + d*x))/ (d*e^(1/3) + (-1)^(1/3)*(-1 + c)*f^(1/3))] + (-1)^(2/3)*PolyLog[2, ((-1)^( 2/3)*f^(1/3)*(-1 + c + d*x))/(-(d*e^(1/3)) + (-1)^(2/3)*(-1 + c)*f^(1/3))] - PolyLog[2, -((f^(1/3)*(1 + c + d*x))/(d*e^(1/3) - (1 + c)*f^(1/3)))] + (-1)^(1/3)*PolyLog[2, ((-1)^(1/3)*f^(1/3)*(1 + c + d*x))/(d*e^(1/3) + (-1) ^(1/3)*(1 + c)*f^(1/3))] - (-1)^(2/3)*PolyLog[2, ((-1)^(2/3)*f^(1/3)*(1 + c + d*x))/(-(d*e^(1/3)) + (-1)^(2/3)*(1 + c)*f^(1/3))]))/f^(2/3))/(6*e)
Time = 1.51 (sec) , antiderivative size = 778, normalized size of antiderivative = 1.15, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, 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.
| \(\displaystyle \int \frac {x (a+b \text {arctanh}(c+d x))}{e+f x^3} \, dx\) |
| \(\Big \downarrow \) 7276 |
| \(\displaystyle \int \left (\frac {a x}{e+f x^3}+\frac {b x \text {arctanh}(c+d x)}{e+f x^3}\right )dx\) |
| \(\Big \downarrow \) 2009 |
| \(\displaystyle -\frac {a \arctan \left (\frac {\sqrt [3]{e}-2 \sqrt [3]{f} x}{\sqrt {3} \sqrt [3]{e}}\right )}{\sqrt {3} \sqrt [3]{e} f^{2/3}}+\frac {a \log \left (e^{2/3}-\sqrt [3]{e} \sqrt [3]{f} x+f^{2/3} x^2\right )}{6 \sqrt [3]{e} f^{2/3}}-\frac {a \log \left (\sqrt [3]{e}+\sqrt [3]{f} x\right )}{3 \sqrt [3]{e} f^{2/3}}+\frac {(-1)^{2/3} b \text {arctanh}(c+d x) \log \left (\frac {2}{c+d x+1}\right )}{3 \sqrt [3]{e} f^{2/3}}-\frac {\sqrt [3]{-1} b \text {arctanh}(c+d x) \log \left (\frac {2}{c+d x+1}\right )}{3 \sqrt [3]{e} f^{2/3}}+\frac {b \text {arctanh}(c+d x) \log \left (\frac {2}{c+d x+1}\right )}{3 \sqrt [3]{e} f^{2/3}}-\frac {b \text {arctanh}(c+d x) \log \left (\frac {2 d \left (\sqrt [3]{e}+\sqrt [3]{f} x\right )}{(c+d x+1) \left ((1-c) \sqrt [3]{f}+d \sqrt [3]{e}\right )}\right )}{3 \sqrt [3]{e} f^{2/3}}+\frac {\sqrt [3]{-1} b \text {arctanh}(c+d x) \log \left (\frac {2 d \left (\sqrt [3]{e}-\sqrt [3]{-1} \sqrt [3]{f} x\right )}{(c+d x+1) \left (d \sqrt [3]{e}-\sqrt [3]{-1} (1-c) \sqrt [3]{f}\right )}\right )}{3 \sqrt [3]{e} f^{2/3}}-\frac {(-1)^{2/3} b \text {arctanh}(c+d x) \log \left (\frac {2 d \left (\sqrt [3]{e}+(-1)^{2/3} \sqrt [3]{f} x\right )}{(c+d x+1) \left ((-1)^{2/3} (1-c) \sqrt [3]{f}+d \sqrt [3]{e}\right )}\right )}{3 \sqrt [3]{e} f^{2/3}}-\frac {(-1)^{2/3} b \operatorname {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{6 \sqrt [3]{e} f^{2/3}}+\frac {\sqrt [3]{-1} b \operatorname {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{6 \sqrt [3]{e} f^{2/3}}-\frac {b \operatorname {PolyLog}\left (2,1-\frac {2}{c+d x+1}\right )}{6 \sqrt [3]{e} f^{2/3}}+\frac {b \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 \sqrt [3]{e} f^{2/3}}-\frac {\sqrt [3]{-1} b \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 \sqrt [3]{e} f^{2/3}}+\frac {(-1)^{2/3} b \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 \sqrt [3]{e} f^{2/3}}\) |
Input:
Int[(x*(a + b*ArcTanh[c + d*x]))/(e + f*x^3),x]
Output:
-((a*ArcTan[(e^(1/3) - 2*f^(1/3)*x)/(Sqrt[3]*e^(1/3))])/(Sqrt[3]*e^(1/3)*f ^(2/3))) + (b*ArcTanh[c + d*x]*Log[2/(1 + c + d*x)])/(3*e^(1/3)*f^(2/3)) - ((-1)^(1/3)*b*ArcTanh[c + d*x]*Log[2/(1 + c + d*x)])/(3*e^(1/3)*f^(2/3)) + ((-1)^(2/3)*b*ArcTanh[c + d*x]*Log[2/(1 + c + d*x)])/(3*e^(1/3)*f^(2/3)) - (a*Log[e^(1/3) + f^(1/3)*x])/(3*e^(1/3)*f^(2/3)) - (b*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^(1/3)*f^(2/3)) + ((-1)^(1/3)*b*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^(1/3)*f^(2/3)) - ((-1)^(2/3)*b*ArcTanh[c + d*x]*Log[(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^(1/3)*f^(2/3)) + (a*Log[e^(2/3) - e^(1/3)*f^(1/3) *x + f^(2/3)*x^2])/(6*e^(1/3)*f^(2/3)) - (b*PolyLog[2, 1 - 2/(1 + c + d*x) ])/(6*e^(1/3)*f^(2/3)) + ((-1)^(1/3)*b*PolyLog[2, 1 - 2/(1 + c + d*x)])/(6 *e^(1/3)*f^(2/3)) - ((-1)^(2/3)*b*PolyLog[2, 1 - 2/(1 + c + d*x)])/(6*e^(1 /3)*f^(2/3)) + (b*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^(1/3)*f^(2/3)) - ((-1)^(1/3)*b*Pol yLog[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^(1/3)*f^(2/3)) + ((-1)^(2/3)*b*Po lyLog[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))*(1 + c + d*x))])/(6*e^(1/3)*f^(2/3))
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 = 0.72 (sec) , antiderivative size = 338, normalized size of antiderivative = 0.50
| method | result | size |
| risch | \(\frac {d b \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (f \,\textit {\_Z}^{3}+\left (3 f c -3 f \right ) \textit {\_Z}^{2}+\left (3 c^{2} f -6 f c +3 f \right ) \textit {\_Z} +c^{3} f -d^{3} e -3 c^{2} f +3 f c -f \right )}{\sum }\frac {\ln \left (-d x -c +1\right ) \ln \left (\frac {d x +\textit {\_R1} +c -1}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {d x +\textit {\_R1} +c -1}{\textit {\_R1}}\right )}{\textit {\_R1} -1+c}\right )}{6 f}-\frac {d a \left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (f \,\textit {\_Z}^{3}+\left (3 f c -3 f \right ) \textit {\_Z}^{2}+\left (3 c^{2} f -6 f c +3 f \right ) \textit {\_Z} +c^{3} f -d^{3} e -3 c^{2} f +3 f c -f \right )}{\sum }\frac {\left (c +\textit {\_R} -1\right ) \ln \left (-d x -\textit {\_R} -c +1\right )}{\textit {\_R}^{2}+2 \textit {\_R} c +c^{2}-2 \textit {\_R} -2 c +1}\right )}{3 f}+\frac {b d \left (\munderset {\textit {\_R1} =\operatorname {RootOf}\left (f \,\textit {\_Z}^{3}+\left (-3 f c -3 f \right ) \textit {\_Z}^{2}+\left (3 c^{2} f +6 f c +3 f \right ) \textit {\_Z} -c^{3} f +d^{3} e -3 c^{2} f -3 f c -f \right )}{\sum }\frac {\ln \left (d x +c +1\right ) \ln \left (\frac {-d x +\textit {\_R1} -c -1}{\textit {\_R1}}\right )+\operatorname {dilog}\left (\frac {-d x +\textit {\_R1} -c -1}{\textit {\_R1}}\right )}{\textit {\_R1} -1-c}\right )}{6 f}\) | \(338\) |
| derivativedivides | \(\text {Expression too large to display}\) | \(1258\) |
| default | \(\text {Expression too large to display}\) | \(1258\) |
| parts | \(\text {Expression too large to display}\) | \(1262\) |
Input:
int(x*(a+b*arctanh(d*x+c))/(f*x^3+e),x,method=_RETURNVERBOSE)
Output:
1/6*d*b/f*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))-1/3*d*a/f*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))+1/6*b*d/f*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))
\[ \int \frac {x (a+b \text {arctanh}(c+d x))}{e+f x^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )} x}{f x^{3} + e} \,d x } \] Input:
integrate(x*(a+b*arctanh(d*x+c))/(f*x^3+e),x, algorithm="fricas")
Output:
integral((b*x*arctanh(d*x + c) + a*x)/(f*x^3 + e), x)
Timed out. \[ \int \frac {x (a+b \text {arctanh}(c+d x))}{e+f x^3} \, dx=\text {Timed out} \] Input:
integrate(x*(a+b*atanh(d*x+c))/(f*x**3+e),x)
Output:
Timed out
Exception generated. \[ \int \frac {x (a+b \text {arctanh}(c+d x))}{e+f x^3} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(x*(a+b*arctanh(d*x+c))/(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
\[ \int \frac {x (a+b \text {arctanh}(c+d x))}{e+f x^3} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (d x + c\right ) + a\right )} x}{f x^{3} + e} \,d x } \] Input:
integrate(x*(a+b*arctanh(d*x+c))/(f*x^3+e),x, algorithm="giac")
Output:
integrate((b*arctanh(d*x + c) + a)*x/(f*x^3 + e), x)
Timed out. \[ \int \frac {x (a+b \text {arctanh}(c+d x))}{e+f x^3} \, dx=\int \frac {x\,\left (a+b\,\mathrm {atanh}\left (c+d\,x\right )\right )}{f\,x^3+e} \,d x \] Input:
int((x*(a + b*atanh(c + d*x)))/(e + f*x^3),x)
Output:
int((x*(a + b*atanh(c + d*x)))/(e + f*x^3), x)
\[ \int \frac {x (a+b \text {arctanh}(c+d x))}{e+f x^3} \, dx=\frac {-2 \sqrt {3}\, \mathit {atan} \left (\frac {e^{\frac {1}{3}}-2 f^{\frac {1}{3}} x}{e^{\frac {1}{3}} \sqrt {3}}\right ) a +6 f^{\frac {2}{3}} e^{\frac {1}{3}} \left (\int \frac {\mathit {atanh} \left (d x +c \right ) x}{f \,x^{3}+e}d x \right ) b +\mathrm {log}\left (e^{\frac {2}{3}}-f^{\frac {1}{3}} e^{\frac {1}{3}} x +f^{\frac {2}{3}} x^{2}\right ) a -2 \,\mathrm {log}\left (e^{\frac {1}{3}}+f^{\frac {1}{3}} x \right ) a}{6 f^{\frac {2}{3}} e^{\frac {1}{3}}} \] Input:
int(x*(a+b*atanh(d*x+c))/(f*x^3+e),x)
Output:
( - 2*sqrt(3)*atan((e**(1/3) - 2*f**(1/3)*x)/(e**(1/3)*sqrt(3)))*a + 6*f** (2/3)*e**(1/3)*int((atanh(c + d*x)*x)/(e + f*x**3),x)*b + log(e**(2/3) - f **(1/3)*e**(1/3)*x + f**(2/3)*x**2)*a - 2*log(e**(1/3) + f**(1/3)*x)*a)/(6 *f**(2/3)*e**(1/3))