Integrand size = 14, antiderivative size = 63 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^6} \, dx=-\frac {2 b c}{15 x^3}+\frac {1}{5} b c^{5/2} \arctan \left (\sqrt {c} x\right )+\frac {1}{5} b c^{5/2} \text {arctanh}\left (\sqrt {c} x\right )-\frac {a+b \text {arctanh}\left (c x^2\right )}{5 x^5} \] Output:
-2/15*b*c/x^3+1/5*b*c^(5/2)*arctan(c^(1/2)*x)+1/5*b*c^(5/2)*arctanh(c^(1/2 )*x)-1/5*(a+b*arctanh(c*x^2))/x^5
Time = 0.03 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.44 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^6} \, dx=-\frac {a}{5 x^5}-\frac {2 b c}{15 x^3}+\frac {1}{5} b c^{5/2} \arctan \left (\sqrt {c} x\right )-\frac {b \text {arctanh}\left (c x^2\right )}{5 x^5}-\frac {1}{10} b c^{5/2} \log \left (1-\sqrt {c} x\right )+\frac {1}{10} b c^{5/2} \log \left (1+\sqrt {c} x\right ) \] Input:
Integrate[(a + b*ArcTanh[c*x^2])/x^6,x]
Output:
-1/5*a/x^5 - (2*b*c)/(15*x^3) + (b*c^(5/2)*ArcTan[Sqrt[c]*x])/5 - (b*ArcTa nh[c*x^2])/(5*x^5) - (b*c^(5/2)*Log[1 - Sqrt[c]*x])/10 + (b*c^(5/2)*Log[1 + Sqrt[c]*x])/10
Time = 0.36 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6452, 847, 756, 216, 219}
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.
\(\displaystyle \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^6} \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {2}{5} b c \int \frac {1}{x^4 \left (1-c^2 x^4\right )}dx-\frac {a+b \text {arctanh}\left (c x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 847 |
\(\displaystyle \frac {2}{5} b c \left (c^2 \int \frac {1}{1-c^2 x^4}dx-\frac {1}{3 x^3}\right )-\frac {a+b \text {arctanh}\left (c x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 756 |
\(\displaystyle \frac {2}{5} b c \left (c^2 \left (\frac {1}{2} \int \frac {1}{1-c x^2}dx+\frac {1}{2} \int \frac {1}{c x^2+1}dx\right )-\frac {1}{3 x^3}\right )-\frac {a+b \text {arctanh}\left (c x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {2}{5} b c \left (c^2 \left (\frac {1}{2} \int \frac {1}{1-c x^2}dx+\frac {\arctan \left (\sqrt {c} x\right )}{2 \sqrt {c}}\right )-\frac {1}{3 x^3}\right )-\frac {a+b \text {arctanh}\left (c x^2\right )}{5 x^5}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {2}{5} b c \left (c^2 \left (\frac {\arctan \left (\sqrt {c} x\right )}{2 \sqrt {c}}+\frac {\text {arctanh}\left (\sqrt {c} x\right )}{2 \sqrt {c}}\right )-\frac {1}{3 x^3}\right )-\frac {a+b \text {arctanh}\left (c x^2\right )}{5 x^5}\) |
Input:
Int[(a + b*ArcTanh[c*x^2])/x^6,x]
Output:
(2*b*c*(-1/3*1/x^3 + c^2*(ArcTan[Sqrt[c]*x]/(2*Sqrt[c]) + ArcTanh[Sqrt[c]* x]/(2*Sqrt[c]))))/5 - (a + b*ArcTanh[c*x^2])/(5*x^5)
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 ]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a) Int[1/(r - s*x^2), x], x] + Simp[r/(2*a) Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] && !GtQ[a /b, 0]
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x )^(m + 1)*((a + b*x^n)^(p + 1)/(a*c*(m + 1))), x] - Simp[b*((m + n*(p + 1) + 1)/(a*c^n*(m + 1))) Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a , b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p , x]
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]
Time = 0.35 (sec) , antiderivative size = 51, normalized size of antiderivative = 0.81
method | result | size |
default | \(-\frac {a}{5 x^{5}}-\frac {b \,\operatorname {arctanh}\left (c \,x^{2}\right )}{5 x^{5}}+\frac {b \,c^{\frac {5}{2}} \arctan \left (\sqrt {c}\, x \right )}{5}+\frac {b \,c^{\frac {5}{2}} \operatorname {arctanh}\left (\sqrt {c}\, x \right )}{5}-\frac {2 b c}{15 x^{3}}\) | \(51\) |
parts | \(-\frac {a}{5 x^{5}}-\frac {b \,\operatorname {arctanh}\left (c \,x^{2}\right )}{5 x^{5}}+\frac {b \,c^{\frac {5}{2}} \arctan \left (\sqrt {c}\, x \right )}{5}+\frac {b \,c^{\frac {5}{2}} \operatorname {arctanh}\left (\sqrt {c}\, x \right )}{5}-\frac {2 b c}{15 x^{3}}\) | \(51\) |
risch | \(-\frac {b \ln \left (c \,x^{2}+1\right )}{10 x^{5}}-\frac {a}{5 x^{5}}+\frac {b \ln \left (-c \,x^{2}+1\right )}{10 x^{5}}-\frac {2 b c}{15 x^{3}}+\frac {b \,c^{\frac {5}{2}} \operatorname {arctanh}\left (\sqrt {c}\, x \right )}{5}+\frac {b \,c^{\frac {5}{2}} \arctan \left (\sqrt {c}\, x \right )}{5}\) | \(68\) |
Input:
int((a+b*arctanh(c*x^2))/x^6,x,method=_RETURNVERBOSE)
Output:
-1/5*a/x^5-1/5*b/x^5*arctanh(c*x^2)+1/5*b*c^(5/2)*arctan(c^(1/2)*x)+1/5*b* c^(5/2)*arctanh(c^(1/2)*x)-2/15*b*c/x^3
Leaf count of result is larger than twice the leaf count of optimal. 100 vs. \(2 (47) = 94\).
Time = 0.09 (sec) , antiderivative size = 187, normalized size of antiderivative = 2.97 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^6} \, dx=\left [\frac {6 \, b c^{\frac {5}{2}} x^{5} \arctan \left (\sqrt {c} x\right ) + 3 \, b c^{\frac {5}{2}} x^{5} \log \left (\frac {c x^{2} + 2 \, \sqrt {c} x + 1}{c x^{2} - 1}\right ) - 4 \, b c x^{2} - 3 \, b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) - 6 \, a}{30 \, x^{5}}, -\frac {6 \, b \sqrt {-c} c^{2} x^{5} \arctan \left (\sqrt {-c} x\right ) - 3 \, b \sqrt {-c} c^{2} x^{5} \log \left (\frac {c x^{2} + 2 \, \sqrt {-c} x - 1}{c x^{2} + 1}\right ) + 4 \, b c x^{2} + 3 \, b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 6 \, a}{30 \, x^{5}}\right ] \] Input:
integrate((a+b*arctanh(c*x^2))/x^6,x, algorithm="fricas")
Output:
[1/30*(6*b*c^(5/2)*x^5*arctan(sqrt(c)*x) + 3*b*c^(5/2)*x^5*log((c*x^2 + 2* sqrt(c)*x + 1)/(c*x^2 - 1)) - 4*b*c*x^2 - 3*b*log(-(c*x^2 + 1)/(c*x^2 - 1) ) - 6*a)/x^5, -1/30*(6*b*sqrt(-c)*c^2*x^5*arctan(sqrt(-c)*x) - 3*b*sqrt(-c )*c^2*x^5*log((c*x^2 + 2*sqrt(-c)*x - 1)/(c*x^2 + 1)) + 4*b*c*x^2 + 3*b*lo g(-(c*x^2 + 1)/(c*x^2 - 1)) + 6*a)/x^5]
Leaf count of result is larger than twice the leaf count of optimal. 1948 vs. \(2 (60) = 120\).
Time = 6.92 (sec) , antiderivative size = 1948, normalized size of antiderivative = 30.92 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^6} \, dx=\text {Too large to display} \] Input:
integrate((a+b*atanh(c*x**2))/x**6,x)
Output:
Piecewise((-a/(5*x**5), Eq(c, 0)), (-(a - oo*b)/(5*x**5), Eq(c, -1/x**2)), (-(a + oo*b)/(5*x**5), Eq(c, x**(-2))), (-3*a*c*x**4*sqrt(-1/c)/(15*c*x** 9*sqrt(-1/c) + 15*c*x**9*sqrt(1/c) - 15*x**5*sqrt(-1/c)/c - 15*x**5*sqrt(1 /c)/c) - 3*a*c*x**4*sqrt(1/c)/(15*c*x**9*sqrt(-1/c) + 15*c*x**9*sqrt(1/c) - 15*x**5*sqrt(-1/c)/c - 15*x**5*sqrt(1/c)/c) + 3*a*sqrt(-1/c)/(15*c**2*x* *9*sqrt(-1/c) + 15*c**2*x**9*sqrt(1/c) - 15*x**5*sqrt(-1/c) - 15*x**5*sqrt (1/c)) + 3*a*sqrt(1/c)/(15*c**2*x**9*sqrt(-1/c) + 15*c**2*x**9*sqrt(1/c) - 15*x**5*sqrt(-1/c) - 15*x**5*sqrt(1/c)) + 3*b*c**4*x**9*sqrt(-1/c)*sqrt(1 /c)*log(x + sqrt(-1/c))/(15*c*x**9*sqrt(-1/c) + 15*c*x**9*sqrt(1/c) - 15*x **5*sqrt(-1/c)/c - 15*x**5*sqrt(1/c)/c) - 3*b*c**4*x**9*sqrt(-1/c)*sqrt(1/ c)*log(x - sqrt(1/c))/(15*c*x**9*sqrt(-1/c) + 15*c*x**9*sqrt(1/c) - 15*x** 5*sqrt(-1/c)/c - 15*x**5*sqrt(1/c)/c) - 3*b*c**4*x**9*sqrt(-1/c)*sqrt(1/c) *atanh(c*x**2)/(15*c*x**9*sqrt(-1/c) + 15*c*x**9*sqrt(1/c) - 15*x**5*sqrt( -1/c)/c - 15*x**5*sqrt(1/c)/c) + 3*b*c**3*x**9*log(x - sqrt(-1/c))/(15*c*x **9*sqrt(-1/c) + 15*c*x**9*sqrt(1/c) - 15*x**5*sqrt(-1/c)/c - 15*x**5*sqrt (1/c)/c) - 3*b*c**3*x**9*log(x - sqrt(1/c))/(15*c*x**9*sqrt(-1/c) + 15*c*x **9*sqrt(1/c) - 15*x**5*sqrt(-1/c)/c - 15*x**5*sqrt(1/c)/c) - 3*b*c**3*x** 9*atanh(c*x**2)/(15*c*x**9*sqrt(-1/c) + 15*c*x**9*sqrt(1/c) - 15*x**5*sqrt (-1/c)/c - 15*x**5*sqrt(1/c)/c) - 2*b*c**2*x**6*sqrt(-1/c)/(15*c*x**9*sqrt (-1/c) + 15*c*x**9*sqrt(1/c) - 15*x**5*sqrt(-1/c)/c - 15*x**5*sqrt(1/c)...
Time = 0.11 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.05 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^6} \, dx=\frac {1}{30} \, {\left ({\left (6 \, c^{\frac {3}{2}} \arctan \left (\sqrt {c} x\right ) - 3 \, c^{\frac {3}{2}} \log \left (\frac {c x - \sqrt {c}}{c x + \sqrt {c}}\right ) - \frac {4}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x^{2}\right )}{x^{5}}\right )} b - \frac {a}{5 \, x^{5}} \] Input:
integrate((a+b*arctanh(c*x^2))/x^6,x, algorithm="maxima")
Output:
1/30*((6*c^(3/2)*arctan(sqrt(c)*x) - 3*c^(3/2)*log((c*x - sqrt(c))/(c*x + sqrt(c))) - 4/x^3)*c - 6*arctanh(c*x^2)/x^5)*b - 1/5*a/x^5
Time = 0.17 (sec) , antiderivative size = 91, normalized size of antiderivative = 1.44 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^6} \, dx=\frac {1}{10} \, b c^{3} {\left (\frac {2 \, \arctan \left (x \sqrt {{\left | c \right |}}\right )}{\sqrt {{\left | c \right |}}} + \frac {\log \left ({\left | x + \frac {1}{\sqrt {{\left | c \right |}}} \right |}\right )}{\sqrt {{\left | c \right |}}} - \frac {\log \left ({\left | x - \frac {1}{\sqrt {{\left | c \right |}}} \right |}\right )}{\sqrt {{\left | c \right |}}}\right )} - \frac {b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{10 \, x^{5}} - \frac {2 \, b c x^{2} + 3 \, a}{15 \, x^{5}} \] Input:
integrate((a+b*arctanh(c*x^2))/x^6,x, algorithm="giac")
Output:
1/10*b*c^3*(2*arctan(x*sqrt(abs(c)))/sqrt(abs(c)) + log(abs(x + 1/sqrt(abs (c))))/sqrt(abs(c)) - log(abs(x - 1/sqrt(abs(c))))/sqrt(abs(c))) - 1/10*b* log(-(c*x^2 + 1)/(c*x^2 - 1))/x^5 - 1/15*(2*b*c*x^2 + 3*a)/x^5
Time = 3.88 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.13 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^6} \, dx=\frac {b\,c^{5/2}\,\mathrm {atan}\left (\sqrt {c}\,x\right )}{5}-\frac {\frac {2\,b\,c\,x^2}{3}+a}{5\,x^5}-\frac {b\,\ln \left (c\,x^2+1\right )}{10\,x^5}+\frac {b\,\ln \left (1-c\,x^2\right )}{10\,x^5}-\frac {b\,c^{5/2}\,\mathrm {atan}\left (\sqrt {c}\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{5} \] Input:
int((a + b*atanh(c*x^2))/x^6,x)
Output:
(b*c^(5/2)*atan(c^(1/2)*x))/5 - (a + (2*b*c*x^2)/3)/(5*x^5) - (b*c^(5/2)*a tan(c^(1/2)*x*1i)*1i)/5 - (b*log(c*x^2 + 1))/(10*x^5) + (b*log(1 - c*x^2)) /(10*x^5)
Time = 0.17 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.56 \[ \int \frac {a+b \text {arctanh}\left (c x^2\right )}{x^6} \, dx=\frac {6 \sqrt {c}\, \mathit {atan} \left (\frac {c x}{\sqrt {c}}\right ) b \,c^{2} x^{5}-6 \sqrt {c}\, \mathit {atanh} \left (c \,x^{2}\right ) b \,c^{2} x^{5}-6 \mathit {atanh} \left (c \,x^{2}\right ) b -6 \sqrt {c}\, \mathrm {log}\left (\sqrt {c}\, x -1\right ) b \,c^{2} x^{5}+3 \sqrt {c}\, \mathrm {log}\left (c \,x^{2}+1\right ) b \,c^{2} x^{5}-6 a -4 b c \,x^{2}}{30 x^{5}} \] Input:
int((a+b*atanh(c*x^2))/x^6,x)
Output:
(6*sqrt(c)*atan((c*x)/sqrt(c))*b*c**2*x**5 - 6*sqrt(c)*atanh(c*x**2)*b*c** 2*x**5 - 6*atanh(c*x**2)*b - 6*sqrt(c)*log(sqrt(c)*x - 1)*b*c**2*x**5 + 3* sqrt(c)*log(c*x**2 + 1)*b*c**2*x**5 - 6*a - 4*b*c*x**2)/(30*x**5)