Integrand size = 16, antiderivative size = 120 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^4} \, dx=\frac {1}{3} c \left (a+b \text {arctanh}\left (c x^3\right )\right )^3-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 x^3}+b c \left (a+b \text {arctanh}\left (c x^3\right )\right )^2 \log \left (2-\frac {2}{1+c x^3}\right )-b^2 c \left (a+b \text {arctanh}\left (c x^3\right )\right ) \operatorname {PolyLog}\left (2,-1+\frac {2}{1+c x^3}\right )-\frac {1}{2} b^3 c \operatorname {PolyLog}\left (3,-1+\frac {2}{1+c x^3}\right ) \] Output:
1/3*c*(a+b*arctanh(c*x^3))^3-1/3*(a+b*arctanh(c*x^3))^3/x^3+b*c*(a+b*arcta nh(c*x^3))^2*ln(2-2/(c*x^3+1))-b^2*c*(a+b*arctanh(c*x^3))*polylog(2,-1+2/( c*x^3+1))-1/2*b^3*c*polylog(3,-1+2/(c*x^3+1))
Result contains complex when optimal does not.
Time = 0.29 (sec) , antiderivative size = 223, normalized size of antiderivative = 1.86 \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^4} \, dx=-\frac {a^3}{3 x^3}-\frac {a^2 b \text {arctanh}\left (c x^3\right )}{x^3}+3 a^2 b c \log (x)-\frac {1}{2} a^2 b c \log \left (1-c^2 x^6\right )+a b^2 c \left (\text {arctanh}\left (c x^3\right ) \left (\left (1-\frac {1}{c x^3}\right ) \text {arctanh}\left (c x^3\right )+2 \log \left (1-e^{-2 \text {arctanh}\left (c x^3\right )}\right )\right )-\operatorname {PolyLog}\left (2,e^{-2 \text {arctanh}\left (c x^3\right )}\right )\right )+\frac {1}{3} b^3 c \left (\frac {i \pi ^3}{8}-\text {arctanh}\left (c x^3\right )^3-\frac {\text {arctanh}\left (c x^3\right )^3}{c x^3}+3 \text {arctanh}\left (c x^3\right )^2 \log \left (1-e^{2 \text {arctanh}\left (c x^3\right )}\right )+3 \text {arctanh}\left (c x^3\right ) \operatorname {PolyLog}\left (2,e^{2 \text {arctanh}\left (c x^3\right )}\right )-\frac {3}{2} \operatorname {PolyLog}\left (3,e^{2 \text {arctanh}\left (c x^3\right )}\right )\right ) \] Input:
Integrate[(a + b*ArcTanh[c*x^3])^3/x^4,x]
Output:
-1/3*a^3/x^3 - (a^2*b*ArcTanh[c*x^3])/x^3 + 3*a^2*b*c*Log[x] - (a^2*b*c*Lo g[1 - c^2*x^6])/2 + a*b^2*c*(ArcTanh[c*x^3]*((1 - 1/(c*x^3))*ArcTanh[c*x^3 ] + 2*Log[1 - E^(-2*ArcTanh[c*x^3])]) - PolyLog[2, E^(-2*ArcTanh[c*x^3])]) + (b^3*c*((I/8)*Pi^3 - ArcTanh[c*x^3]^3 - ArcTanh[c*x^3]^3/(c*x^3) + 3*Ar cTanh[c*x^3]^2*Log[1 - E^(2*ArcTanh[c*x^3])] + 3*ArcTanh[c*x^3]*PolyLog[2, E^(2*ArcTanh[c*x^3])] - (3*PolyLog[3, E^(2*ArcTanh[c*x^3])])/2))/3
Time = 0.85 (sec) , antiderivative size = 133, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.375, Rules used = {6454, 6452, 6550, 6494, 6618, 7164}
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.
\(\displaystyle \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^4} \, dx\) |
\(\Big \downarrow \) 6454 |
\(\displaystyle \frac {1}{3} \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^6}dx^3\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{3} \left (3 b c \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3 \left (1-c^2 x^6\right )}dx^3-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^3}\right )\) |
\(\Big \downarrow \) 6550 |
\(\displaystyle \frac {1}{3} \left (3 b c \left (\int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^2}{x^3 \left (c x^3+1\right )}dx^3+\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b}\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^3}\right )\) |
\(\Big \downarrow \) 6494 |
\(\displaystyle \frac {1}{3} \left (3 b c \left (-2 b c \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right ) \log \left (2-\frac {2}{c x^3+1}\right )}{1-c^2 x^6}dx^3+\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b}+\log \left (2-\frac {2}{c x^3+1}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )^2\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^3}\right )\) |
\(\Big \downarrow \) 6618 |
\(\displaystyle \frac {1}{3} \left (3 b c \left (-2 b c \left (\frac {\operatorname {PolyLog}\left (2,\frac {2}{c x^3+1}-1\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{2 c}-\frac {1}{2} b \int \frac {\operatorname {PolyLog}\left (2,\frac {2}{c x^3+1}-1\right )}{1-c^2 x^6}dx^3\right )+\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b}+\log \left (2-\frac {2}{c x^3+1}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )^2\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^3}\right )\) |
\(\Big \downarrow \) 7164 |
\(\displaystyle \frac {1}{3} \left (3 b c \left (-2 b c \left (\frac {\operatorname {PolyLog}\left (2,\frac {2}{c x^3+1}-1\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )}{2 c}+\frac {b \operatorname {PolyLog}\left (3,\frac {2}{c x^3+1}-1\right )}{4 c}\right )+\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{3 b}+\log \left (2-\frac {2}{c x^3+1}\right ) \left (a+b \text {arctanh}\left (c x^3\right )\right )^2\right )-\frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^3}\right )\) |
Input:
Int[(a + b*ArcTanh[c*x^3])^3/x^4,x]
Output:
(-((a + b*ArcTanh[c*x^3])^3/x^3) + 3*b*c*((a + b*ArcTanh[c*x^3])^3/(3*b) + (a + b*ArcTanh[c*x^3])^2*Log[2 - 2/(1 + c*x^3)] - 2*b*c*(((a + b*ArcTanh[ c*x^3])*PolyLog[2, -1 + 2/(1 + c*x^3)])/(2*c) + (b*PolyLog[3, -1 + 2/(1 + c*x^3)])/(4*c))))/3
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] : > Simp[x^(m + 1)*((a + b*ArcTanh[c*x^n])^p/(m + 1)), x] - Simp[b*c*n*(p/(m + 1)) Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))), x ], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1 ] && IntegerQ[m])) && NeQ[m, -1]
Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[1/n Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x ], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[Simpl ify[(m + 1)/n]]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_))), x _Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(Log[2 - 2/(1 + e*(x/d))]/d), x] - Simp[b*c*(p/d) Int[(a + b*ArcTanh[c*x])^(p - 1)*(Log[2 - 2/(1 + e*(x/d))] /(1 - c^2*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c ^2*d^2 - e^2, 0]
Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((x_)*((d_) + (e_.)*(x_)^2)), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*d*(p + 1)), x] + Simp[1/ d Int[(a + b*ArcTanh[c*x])^p/(x*(1 + c*x)), x], x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[c^2*d + e, 0] && GtQ[p, 0]
Int[(Log[u_]*((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.))/((d_) + (e_.)*(x_)^ 2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^p*(PolyLog[2, 1 - u]/(2*c*d)), x ] - Simp[b*(p/2) Int[(a + b*ArcTanh[c*x])^(p - 1)*(PolyLog[2, 1 - u]/(d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d + e, 0] && EqQ[(1 - u)^2 - (1 - 2/(1 + c*x))^2, 0]
Int[(u_)*PolyLog[n_, v_], x_Symbol] :> With[{w = DerivativeDivides[v, u*v, x]}, Simp[w*PolyLog[n + 1, v], x] /; !FalseQ[w]] /; FreeQ[n, x]
\[\int \frac {{\left (a +b \,\operatorname {arctanh}\left (c \,x^{3}\right )\right )}^{3}}{x^{4}}d x\]
Input:
int((a+b*arctanh(c*x^3))^3/x^4,x)
Output:
int((a+b*arctanh(c*x^3))^3/x^4,x)
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{3}}{x^{4}} \,d x } \] Input:
integrate((a+b*arctanh(c*x^3))^3/x^4,x, algorithm="fricas")
Output:
integral((b^3*arctanh(c*x^3)^3 + 3*a*b^2*arctanh(c*x^3)^2 + 3*a^2*b*arctan h(c*x^3) + a^3)/x^4, x)
Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^4} \, dx=\text {Timed out} \] Input:
integrate((a+b*atanh(c*x**3))**3/x**4,x)
Output:
Timed out
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{3}}{x^{4}} \,d x } \] Input:
integrate((a+b*arctanh(c*x^3))^3/x^4,x, algorithm="maxima")
Output:
-1/2*(c*(log(c^2*x^6 - 1) - log(x^6)) + 2*arctanh(c*x^3)/x^3)*a^2*b - 1/3* a^3/x^3 - 1/24*((b^3*c*x^3 - b^3)*log(-c*x^3 + 1)^3 + 3*(2*a*b^2 + (b^3*c* x^3 + b^3)*log(c*x^3 + 1))*log(-c*x^3 + 1)^2)/x^3 - integrate(-1/8*((b^3*c *x^3 - b^3)*log(c*x^3 + 1)^3 + 6*(a*b^2*c*x^3 - a*b^2)*log(c*x^3 + 1)^2 + 3*(4*a*b^2*c*x^3 - (b^3*c*x^3 - b^3)*log(c*x^3 + 1)^2 + 2*(b^3*c^2*x^6 - ( 2*a*b^2*c - b^3*c)*x^3 + 2*a*b^2)*log(c*x^3 + 1))*log(-c*x^3 + 1))/(c*x^7 - x^4), x)
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^4} \, dx=\int { \frac {{\left (b \operatorname {artanh}\left (c x^{3}\right ) + a\right )}^{3}}{x^{4}} \,d x } \] Input:
integrate((a+b*arctanh(c*x^3))^3/x^4,x, algorithm="giac")
Output:
integrate((b*arctanh(c*x^3) + a)^3/x^4, x)
Timed out. \[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^4} \, dx=\int \frac {{\left (a+b\,\mathrm {atanh}\left (c\,x^3\right )\right )}^3}{x^4} \,d x \] Input:
int((a + b*atanh(c*x^3))^3/x^4,x)
Output:
int((a + b*atanh(c*x^3))^3/x^4, x)
\[ \int \frac {\left (a+b \text {arctanh}\left (c x^3\right )\right )^3}{x^4} \, dx=\frac {-\mathit {atanh} \left (c \,x^{3}\right )^{3} b^{3}-3 \mathit {atanh} \left (c \,x^{3}\right )^{2} a \,b^{2}+3 \mathit {atanh} \left (c \,x^{3}\right ) a^{2} b c \,x^{3}-3 \mathit {atanh} \left (c \,x^{3}\right ) a^{2} b -18 \left (\int \frac {\mathit {atanh} \left (c \,x^{3}\right )}{c^{2} x^{7}-x}d x \right ) a \,b^{2} c \,x^{3}-9 \left (\int \frac {\mathit {atanh} \left (c \,x^{3}\right )^{2}}{c^{2} x^{7}-x}d x \right ) b^{3} c \,x^{3}-3 \,\mathrm {log}\left (c^{\frac {2}{3}} x^{2}-c^{\frac {1}{3}} x +1\right ) a^{2} b c \,x^{3}-3 \,\mathrm {log}\left (c^{\frac {2}{3}} x +c^{\frac {1}{3}}\right ) a^{2} b c \,x^{3}+9 \,\mathrm {log}\left (x \right ) a^{2} b c \,x^{3}-a^{3}}{3 x^{3}} \] Input:
int((a+b*atanh(c*x^3))^3/x^4,x)
Output:
( - atanh(c*x**3)**3*b**3 - 3*atanh(c*x**3)**2*a*b**2 + 3*atanh(c*x**3)*a* *2*b*c*x**3 - 3*atanh(c*x**3)*a**2*b - 18*int(atanh(c*x**3)/(c**2*x**7 - x ),x)*a*b**2*c*x**3 - 9*int(atanh(c*x**3)**2/(c**2*x**7 - x),x)*b**3*c*x**3 - 3*log(c**(2/3)*x**2 - c**(1/3)*x + 1)*a**2*b*c*x**3 - 3*log(c**(2/3)*x + c**(1/3))*a**2*b*c*x**3 + 9*log(x)*a**2*b*c*x**3 - a**3)/(3*x**3)