Integrand size = 14, antiderivative size = 65 \[ \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {2 b x^3}{15 c}+\frac {b \arctan \left (\sqrt {c} x\right )}{5 c^{5/2}}-\frac {b \text {arctanh}\left (\sqrt {c} x\right )}{5 c^{5/2}}+\frac {1}{5} x^5 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \]
2/15*b*x^3/c+1/5*b*arctan(x*c^(1/2))/c^(5/2)+1/5*x^5*(a+b*arctanh(c*x^2))- 1/5*b*arctanh(x*c^(1/2))/c^(5/2)
Time = 0.03 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.43 \[ \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {2 b x^3}{15 c}+\frac {a x^5}{5}+\frac {b \arctan \left (\sqrt {c} x\right )}{5 c^{5/2}}+\frac {1}{5} b x^5 \text {arctanh}\left (c x^2\right )+\frac {b \log \left (1-\sqrt {c} x\right )}{10 c^{5/2}}-\frac {b \log \left (1+\sqrt {c} x\right )}{10 c^{5/2}} \]
(2*b*x^3)/(15*c) + (a*x^5)/5 + (b*ArcTan[Sqrt[c]*x])/(5*c^(5/2)) + (b*x^5* ArcTanh[c*x^2])/5 + (b*Log[1 - Sqrt[c]*x])/(10*c^(5/2)) - (b*Log[1 + Sqrt[ c]*x])/(10*c^(5/2))
Time = 0.25 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.14, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6452, 843, 827, 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.
\(\displaystyle \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{5} x^5 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {2}{5} b c \int \frac {x^6}{1-c^2 x^4}dx\) |
\(\Big \downarrow \) 843 |
\(\displaystyle \frac {1}{5} x^5 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {2}{5} b c \left (\frac {\int \frac {x^2}{1-c^2 x^4}dx}{c^2}-\frac {x^3}{3 c^2}\right )\) |
\(\Big \downarrow \) 827 |
\(\displaystyle \frac {1}{5} x^5 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {2}{5} b c \left (\frac {\frac {\int \frac {1}{1-c x^2}dx}{2 c}-\frac {\int \frac {1}{c x^2+1}dx}{2 c}}{c^2}-\frac {x^3}{3 c^2}\right )\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {1}{5} x^5 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {2}{5} b c \left (\frac {\frac {\int \frac {1}{1-c x^2}dx}{2 c}-\frac {\arctan \left (\sqrt {c} x\right )}{2 c^{3/2}}}{c^2}-\frac {x^3}{3 c^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{5} x^5 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {2}{5} b c \left (\frac {\frac {\text {arctanh}\left (\sqrt {c} x\right )}{2 c^{3/2}}-\frac {\arctan \left (\sqrt {c} x\right )}{2 c^{3/2}}}{c^2}-\frac {x^3}{3 c^2}\right )\) |
(-2*b*c*(-1/3*x^3/c^2 + (-1/2*ArcTan[Sqrt[c]*x]/c^(3/2) + ArcTanh[Sqrt[c]* x]/(2*c^(3/2)))/c^2))/5 + (x^5*(a + b*ArcTanh[c*x^2]))/5
3.1.58.3.1 Defintions of rubi rules used
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[(x_)^2/((a_) + (b_.)*(x_)^4), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2]], s = Denominator[Rt[-a/b, 2]]}, Simp[s/(2*b) Int[1/(r + s*x^2), x], x] - Simp[s/(2*b) 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^(n - 1)*(c*x)^(m - n + 1)*((a + b*x^n)^(p + 1)/(b*(m + n*p + 1))), x] - Simp[ a*c^n*((m - n + 1)/(b*(m + n*p + 1))) Int[(c*x)^(m - n)*(a + b*x^n)^p, x] , x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n* p + 1, 0] && 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 = 0.65 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.82
method | result | size |
default | \(\frac {a \,x^{5}}{5}+\frac {b \,x^{5} \operatorname {arctanh}\left (c \,x^{2}\right )}{5}+\frac {2 b \,x^{3}}{15 c}-\frac {b \,\operatorname {arctanh}\left (x \sqrt {c}\right )}{5 c^{\frac {5}{2}}}+\frac {b \arctan \left (x \sqrt {c}\right )}{5 c^{\frac {5}{2}}}\) | \(53\) |
parts | \(\frac {a \,x^{5}}{5}+\frac {b \,x^{5} \operatorname {arctanh}\left (c \,x^{2}\right )}{5}+\frac {2 b \,x^{3}}{15 c}-\frac {b \,\operatorname {arctanh}\left (x \sqrt {c}\right )}{5 c^{\frac {5}{2}}}+\frac {b \arctan \left (x \sqrt {c}\right )}{5 c^{\frac {5}{2}}}\) | \(53\) |
risch | \(\frac {b \,x^{5} \ln \left (c \,x^{2}+1\right )}{10}-\frac {b \,x^{5} \ln \left (-c \,x^{2}+1\right )}{10}+\frac {a \,x^{5}}{5}+\frac {2 b \,x^{3}}{15 c}+\frac {b \ln \left (1-x \sqrt {c}\right )}{10 c^{\frac {5}{2}}}-\frac {b \ln \left (1+x \sqrt {c}\right )}{10 c^{\frac {5}{2}}}+\frac {\sqrt {-c}\, \ln \left (-\sqrt {-c}\, c -c^{2} x \right ) b}{10 c^{3}}-\frac {\sqrt {-c}\, \ln \left (-\sqrt {-c}\, c +c^{2} x \right ) b}{10 c^{3}}\) | \(128\) |
1/5*a*x^5+1/5*b*x^5*arctanh(c*x^2)+2/15*b*x^3/c-1/5*b*arctanh(x*c^(1/2))/c ^(5/2)+1/5*b*arctan(x*c^(1/2))/c^(5/2)
Leaf count of result is larger than twice the leaf count of optimal. 102 vs. \(2 (49) = 98\).
Time = 0.28 (sec) , antiderivative size = 197, normalized size of antiderivative = 3.03 \[ \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\left [\frac {3 \, b c^{3} x^{5} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 6 \, a c^{3} x^{5} + 4 \, b c^{2} x^{3} + 6 \, b \sqrt {c} \arctan \left (\sqrt {c} x\right ) + 3 \, b \sqrt {c} \log \left (\frac {c x^{2} - 2 \, \sqrt {c} x + 1}{c x^{2} - 1}\right )}{30 \, c^{3}}, \frac {3 \, b c^{3} x^{5} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 6 \, a c^{3} x^{5} + 4 \, b c^{2} x^{3} + 6 \, b \sqrt {-c} \arctan \left (\sqrt {-c} x\right ) - 3 \, b \sqrt {-c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-c} x - 1}{c x^{2} + 1}\right )}{30 \, c^{3}}\right ] \]
[1/30*(3*b*c^3*x^5*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 6*a*c^3*x^5 + 4*b*c^2*x ^3 + 6*b*sqrt(c)*arctan(sqrt(c)*x) + 3*b*sqrt(c)*log((c*x^2 - 2*sqrt(c)*x + 1)/(c*x^2 - 1)))/c^3, 1/30*(3*b*c^3*x^5*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 6*a*c^3*x^5 + 4*b*c^2*x^3 + 6*b*sqrt(-c)*arctan(sqrt(-c)*x) - 3*b*sqrt(-c) *log((c*x^2 - 2*sqrt(-c)*x - 1)/(c*x^2 + 1)))/c^3]
Leaf count of result is larger than twice the leaf count of optimal. 944 vs. \(2 (58) = 116\).
Time = 3.67 (sec) , antiderivative size = 944, normalized size of antiderivative = 14.52 \[ \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\text {Too large to display} \]
Piecewise((12*a*c**3*x**5*sqrt(-1/c)*sqrt(1/c)/(15*c**4*(-1/c)**(3/2)*sqrt (1/c) - 15*c**4*sqrt(-1/c)*(1/c)**(3/2) + 90*c**3*sqrt(-1/c)*sqrt(1/c)) + 12*b*c**3*x**5*sqrt(-1/c)*sqrt(1/c)*atanh(c*x**2)/(15*c**4*(-1/c)**(3/2)*s qrt(1/c) - 15*c**4*sqrt(-1/c)*(1/c)**(3/2) + 90*c**3*sqrt(-1/c)*sqrt(1/c)) + 8*b*c**2*x**3*sqrt(-1/c)*sqrt(1/c)/(15*c**4*(-1/c)**(3/2)*sqrt(1/c) - 1 5*c**4*sqrt(-1/c)*(1/c)**(3/2) + 90*c**3*sqrt(-1/c)*sqrt(1/c)) - 3*b*c*(-1 /c)**(3/2)*log(x + sqrt(-1/c))/(15*c**4*(-1/c)**(3/2)*sqrt(1/c) - 15*c**4* sqrt(-1/c)*(1/c)**(3/2) + 90*c**3*sqrt(-1/c)*sqrt(1/c)) + 3*b*c*(1/c)**(3/ 2)*log(x + sqrt(-1/c))/(15*c**4*(-1/c)**(3/2)*sqrt(1/c) - 15*c**4*sqrt(-1/ c)*(1/c)**(3/2) + 90*c**3*sqrt(-1/c)*sqrt(1/c)) - 6*b*sqrt(-1/c)*log(x - s qrt(-1/c))/(15*c**4*(-1/c)**(3/2)*sqrt(1/c) - 15*c**4*sqrt(-1/c)*(1/c)**(3 /2) + 90*c**3*sqrt(-1/c)*sqrt(1/c)) - 9*b*sqrt(-1/c)*log(x + sqrt(-1/c))/( 15*c**4*(-1/c)**(3/2)*sqrt(1/c) - 15*c**4*sqrt(-1/c)*(1/c)**(3/2) + 90*c** 3*sqrt(-1/c)*sqrt(1/c)) + 12*b*sqrt(-1/c)*log(x - sqrt(1/c))/(15*c**4*(-1/ c)**(3/2)*sqrt(1/c) - 15*c**4*sqrt(-1/c)*(1/c)**(3/2) + 90*c**3*sqrt(-1/c) *sqrt(1/c)) + 12*b*sqrt(-1/c)*atanh(c*x**2)/(15*c**4*(-1/c)**(3/2)*sqrt(1/ c) - 15*c**4*sqrt(-1/c)*(1/c)**(3/2) + 90*c**3*sqrt(-1/c)*sqrt(1/c)) + 6*b *sqrt(1/c)*log(x - sqrt(-1/c))/(15*c**4*(-1/c)**(3/2)*sqrt(1/c) - 15*c**4* sqrt(-1/c)*(1/c)**(3/2) + 90*c**3*sqrt(-1/c)*sqrt(1/c)) - 9*b*sqrt(1/c)*lo g(x + sqrt(-1/c))/(15*c**4*(-1/c)**(3/2)*sqrt(1/c) - 15*c**4*sqrt(-1/c)...
Time = 0.29 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.06 \[ \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {1}{5} \, a x^{5} + \frac {1}{30} \, {\left (6 \, x^{5} \operatorname {artanh}\left (c x^{2}\right ) + c {\left (\frac {4 \, x^{3}}{c^{2}} + \frac {6 \, \arctan \left (\sqrt {c} x\right )}{c^{\frac {7}{2}}} + \frac {3 \, \log \left (\frac {c x - \sqrt {c}}{c x + \sqrt {c}}\right )}{c^{\frac {7}{2}}}\right )}\right )} b \]
1/5*a*x^5 + 1/30*(6*x^5*arctanh(c*x^2) + c*(4*x^3/c^2 + 6*arctan(sqrt(c)*x )/c^(7/2) + 3*log((c*x - sqrt(c))/(c*x + sqrt(c)))/c^(7/2)))*b
Time = 0.37 (sec) , antiderivative size = 73, normalized size of antiderivative = 1.12 \[ \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {1}{10} \, b x^{5} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + \frac {1}{5} \, a x^{5} + \frac {2 \, b x^{3}}{15 \, c} + \frac {b \arctan \left (\sqrt {c} x\right )}{5 \, c^{\frac {5}{2}}} + \frac {b \arctan \left (\frac {c x}{\sqrt {-c}}\right )}{5 \, \sqrt {-c} c^{2}} \]
1/10*b*x^5*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 1/5*a*x^5 + 2/15*b*x^3/c + 1/5* b*arctan(sqrt(c)*x)/c^(5/2) + 1/5*b*arctan(c*x/sqrt(-c))/(sqrt(-c)*c^2)
Time = 3.46 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.11 \[ \int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {a\,x^5}{5}+\frac {2\,b\,x^3}{15\,c}+\frac {b\,\mathrm {atan}\left (\sqrt {c}\,x\right )}{5\,c^{5/2}}+\frac {b\,x^5\,\ln \left (c\,x^2+1\right )}{10}-\frac {b\,x^5\,\ln \left (1-c\,x^2\right )}{10}+\frac {b\,\mathrm {atan}\left (\sqrt {c}\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{5\,c^{5/2}} \]