Integrand size = 16, antiderivative size = 91 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {a b x^2}{2 c}+\frac {b^2 x^2 \text {arctanh}\left (c x^2\right )}{2 c}-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{4 c^2}+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2+\frac {b^2 \log \left (1-c^2 x^4\right )}{4 c^2} \]
1/2*a*b*x^2/c+1/2*b^2*x^2*arctanh(c*x^2)/c-1/4*(a+b*arctanh(c*x^2))^2/c^2+ 1/4*x^4*(a+b*arctanh(c*x^2))^2+1/4*b^2*ln(-c^2*x^4+1)/c^2
Time = 0.04 (sec) , antiderivative size = 106, normalized size of antiderivative = 1.16 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {2 a b c x^2+a^2 c^2 x^4+2 b c x^2 \left (b+a c x^2\right ) \text {arctanh}\left (c x^2\right )+b^2 \left (-1+c^2 x^4\right ) \text {arctanh}\left (c x^2\right )^2+b (a+b) \log \left (1-c x^2\right )-a b \log \left (1+c x^2\right )+b^2 \log \left (1+c x^2\right )}{4 c^2} \]
(2*a*b*c*x^2 + a^2*c^2*x^4 + 2*b*c*x^2*(b + a*c*x^2)*ArcTanh[c*x^2] + b^2* (-1 + c^2*x^4)*ArcTanh[c*x^2]^2 + b*(a + b)*Log[1 - c*x^2] - a*b*Log[1 + c *x^2] + b^2*Log[1 + c*x^2])/(4*c^2)
Time = 0.53 (sec) , antiderivative size = 92, normalized size of antiderivative = 1.01, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6454, 6452, 6542, 2009, 6510}
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.
\(\displaystyle \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx\) |
\(\Big \downarrow \) 6454 |
\(\displaystyle \frac {1}{2} \int x^2 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2dx^2\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-b c \int \frac {x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{1-c^2 x^4}dx^2\right )\) |
\(\Big \downarrow \) 6542 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-b c \left (\frac {\int \frac {a+b \text {arctanh}\left (c x^2\right )}{1-c^2 x^4}dx^2}{c^2}-\frac {\int \left (a+b \text {arctanh}\left (c x^2\right )\right )dx^2}{c^2}\right )\right )\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-b c \left (\frac {\int \frac {a+b \text {arctanh}\left (c x^2\right )}{1-c^2 x^4}dx^2}{c^2}-\frac {a x^2+b x^2 \text {arctanh}\left (c x^2\right )+\frac {b \log \left (1-c^2 x^4\right )}{2 c}}{c^2}\right )\right )\) |
\(\Big \downarrow \) 6510 |
\(\displaystyle \frac {1}{2} \left (\frac {1}{2} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-b c \left (\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{2 b c^3}-\frac {a x^2+b x^2 \text {arctanh}\left (c x^2\right )+\frac {b \log \left (1-c^2 x^4\right )}{2 c}}{c^2}\right )\right )\) |
((x^4*(a + b*ArcTanh[c*x^2])^2)/2 - b*c*((a + b*ArcTanh[c*x^2])^2/(2*b*c^3 ) - (a*x^2 + b*x^2*ArcTanh[c*x^2] + (b*Log[1 - c^2*x^4])/(2*c))/c^2))/2
3.1.66.3.1 Defintions of rubi rules used
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_.)/((d_) + (e_.)*(x_)^2), x_Symb ol] :> Simp[(a + b*ArcTanh[c*x])^(p + 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b , c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]
Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + ( e_.)*(x_)^2), x_Symbol] :> Simp[f^2/e Int[(f*x)^(m - 2)*(a + b*ArcTanh[c* x])^p, x], x] - Simp[d*(f^2/e) Int[(f*x)^(m - 2)*((a + b*ArcTanh[c*x])^p/ (d + e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && GtQ[m, 1]
Time = 0.87 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.33
method | result | size |
parallelrisch | \(\frac {b^{2} \operatorname {arctanh}\left (c \,x^{2}\right )^{2} x^{4} c^{2}+2 x^{4} \operatorname {arctanh}\left (c \,x^{2}\right ) a b \,c^{2}+a^{2} c^{2} x^{4}+2 b^{2} \operatorname {arctanh}\left (c \,x^{2}\right ) x^{2} c +2 a b c \,x^{2}-b^{2} \operatorname {arctanh}\left (c \,x^{2}\right )^{2}+2 \ln \left (c \,x^{2}-1\right ) b^{2}-2 \,\operatorname {arctanh}\left (c \,x^{2}\right ) a b +2 \,\operatorname {arctanh}\left (c \,x^{2}\right ) b^{2}}{4 c^{2}}\) | \(121\) |
risch | \(\frac {b^{2} \left (c^{2} x^{4}-1\right ) \ln \left (c \,x^{2}+1\right )^{2}}{16 c^{2}}+\frac {b \left (-2 b \,x^{4} \ln \left (-c \,x^{2}+1\right ) a \,c^{2}+4 a^{2} c^{2} x^{4}+4 a b c \,x^{2}+2 b \ln \left (-c \,x^{2}+1\right ) a +b^{2}\right ) \ln \left (c \,x^{2}+1\right )}{16 a \,c^{2}}+\frac {b^{2} x^{4} \ln \left (-c \,x^{2}+1\right )^{2}}{16}-\frac {a b \,x^{4} \ln \left (-c \,x^{2}+1\right )}{4}+\frac {a^{2} x^{4}}{4}-\frac {b^{2} x^{2} \ln \left (-c \,x^{2}+1\right )}{4 c}+\frac {a b \,x^{2}}{2 c}-\frac {b^{2} \ln \left (-c \,x^{2}+1\right )^{2}}{16 c^{2}}+\frac {a \ln \left (-c \,x^{2}+1\right ) b}{4 c^{2}}+\frac {b^{2} \ln \left (-c \,x^{2}+1\right )}{4 c^{2}}-\frac {a \ln \left (-c \,x^{2}-1\right ) b}{4 c^{2}}+\frac {\ln \left (-c \,x^{2}-1\right ) b^{2}}{4 c^{2}}-\frac {\ln \left (-c \,x^{2}-1\right ) b^{3}}{16 a \,c^{2}}+\frac {b^{2}}{4 c^{2}}\) | \(287\) |
default | \(\text {Expression too large to display}\) | \(767\) |
parts | \(\text {Expression too large to display}\) | \(767\) |
1/4*(b^2*arctanh(c*x^2)^2*x^4*c^2+2*x^4*arctanh(c*x^2)*a*b*c^2+a^2*c^2*x^4 +2*b^2*arctanh(c*x^2)*x^2*c+2*a*b*c*x^2-b^2*arctanh(c*x^2)^2+2*ln(c*x^2-1) *b^2-2*arctanh(c*x^2)*a*b+2*arctanh(c*x^2)*b^2)/c^2
Time = 0.26 (sec) , antiderivative size = 138, normalized size of antiderivative = 1.52 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {4 \, a^{2} c^{2} x^{4} + 8 \, a b c x^{2} + {\left (b^{2} c^{2} x^{4} - b^{2}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )^{2} - 4 \, {\left (a b - b^{2}\right )} \log \left (c x^{2} + 1\right ) + 4 \, {\left (a b + b^{2}\right )} \log \left (c x^{2} - 1\right ) + 4 \, {\left (a b c^{2} x^{4} + b^{2} c x^{2}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{16 \, c^{2}} \]
1/16*(4*a^2*c^2*x^4 + 8*a*b*c*x^2 + (b^2*c^2*x^4 - b^2)*log(-(c*x^2 + 1)/( c*x^2 - 1))^2 - 4*(a*b - b^2)*log(c*x^2 + 1) + 4*(a*b + b^2)*log(c*x^2 - 1 ) + 4*(a*b*c^2*x^4 + b^2*c*x^2)*log(-(c*x^2 + 1)/(c*x^2 - 1)))/c^2
Leaf count of result is larger than twice the leaf count of optimal. 163 vs. \(2 (78) = 156\).
Time = 4.01 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.79 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\begin {cases} \frac {a^{2} x^{4}}{4} + \frac {a b x^{4} \operatorname {atanh}{\left (c x^{2} \right )}}{2} + \frac {a b x^{2}}{2 c} - \frac {a b \operatorname {atanh}{\left (c x^{2} \right )}}{2 c^{2}} + \frac {b^{2} x^{4} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{4} + \frac {b^{2} x^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{2 c} + \frac {b^{2} \log {\left (x - \sqrt {- \frac {1}{c}} \right )}}{2 c^{2}} + \frac {b^{2} \log {\left (x + \sqrt {- \frac {1}{c}} \right )}}{2 c^{2}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{4 c^{2}} - \frac {b^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{2 c^{2}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{4}}{4} & \text {otherwise} \end {cases} \]
Piecewise((a**2*x**4/4 + a*b*x**4*atanh(c*x**2)/2 + a*b*x**2/(2*c) - a*b*a tanh(c*x**2)/(2*c**2) + b**2*x**4*atanh(c*x**2)**2/4 + b**2*x**2*atanh(c*x **2)/(2*c) + b**2*log(x - sqrt(-1/c))/(2*c**2) + b**2*log(x + sqrt(-1/c))/ (2*c**2) - b**2*atanh(c*x**2)**2/(4*c**2) - b**2*atanh(c*x**2)/(2*c**2), N e(c, 0)), (a**2*x**4/4, True))
Leaf count of result is larger than twice the leaf count of optimal. 186 vs. \(2 (81) = 162\).
Time = 0.19 (sec) , antiderivative size = 186, normalized size of antiderivative = 2.04 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {1}{4} \, b^{2} x^{4} \operatorname {artanh}\left (c x^{2}\right )^{2} + \frac {1}{4} \, a^{2} x^{4} + \frac {1}{4} \, {\left (2 \, x^{4} \operatorname {artanh}\left (c x^{2}\right ) + c {\left (\frac {2 \, x^{2}}{c^{2}} - \frac {\log \left (c x^{2} + 1\right )}{c^{3}} + \frac {\log \left (c x^{2} - 1\right )}{c^{3}}\right )}\right )} a b + \frac {1}{16} \, {\left (4 \, c {\left (\frac {2 \, x^{2}}{c^{2}} - \frac {\log \left (c x^{2} + 1\right )}{c^{3}} + \frac {\log \left (c x^{2} - 1\right )}{c^{3}}\right )} \operatorname {artanh}\left (c x^{2}\right ) - \frac {2 \, {\left (\log \left (c x^{2} - 1\right ) - 2\right )} \log \left (c x^{2} + 1\right ) - \log \left (c x^{2} + 1\right )^{2} - \log \left (c x^{2} - 1\right )^{2} - 4 \, \log \left (c x^{2} - 1\right )}{c^{2}}\right )} b^{2} \]
1/4*b^2*x^4*arctanh(c*x^2)^2 + 1/4*a^2*x^4 + 1/4*(2*x^4*arctanh(c*x^2) + c *(2*x^2/c^2 - log(c*x^2 + 1)/c^3 + log(c*x^2 - 1)/c^3))*a*b + 1/16*(4*c*(2 *x^2/c^2 - log(c*x^2 + 1)/c^3 + log(c*x^2 - 1)/c^3)*arctanh(c*x^2) - (2*(l og(c*x^2 - 1) - 2)*log(c*x^2 + 1) - log(c*x^2 + 1)^2 - log(c*x^2 - 1)^2 - 4*log(c*x^2 - 1))/c^2)*b^2
Leaf count of result is larger than twice the leaf count of optimal. 361 vs. \(2 (81) = 162\).
Time = 0.31 (sec) , antiderivative size = 361, normalized size of antiderivative = 3.97 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {1}{4} \, {\left (\frac {{\left (c x^{2} + 1\right )} b^{2} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )^{2}}{{\left (\frac {{\left (c x^{2} + 1\right )}^{2} c^{3}}{{\left (c x^{2} - 1\right )}^{2}} - \frac {2 \, {\left (c x^{2} + 1\right )} c^{3}}{c x^{2} - 1} + c^{3}\right )} {\left (c x^{2} - 1\right )}} + \frac {2 \, {\left (\frac {2 \, {\left (c x^{2} + 1\right )} a b}{c x^{2} - 1} + \frac {{\left (c x^{2} + 1\right )} b^{2}}{c x^{2} - 1} - b^{2}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{\frac {{\left (c x^{2} + 1\right )}^{2} c^{3}}{{\left (c x^{2} - 1\right )}^{2}} - \frac {2 \, {\left (c x^{2} + 1\right )} c^{3}}{c x^{2} - 1} + c^{3}} + \frac {4 \, {\left (\frac {{\left (c x^{2} + 1\right )} a^{2}}{c x^{2} - 1} + \frac {{\left (c x^{2} + 1\right )} a b}{c x^{2} - 1} - a b\right )}}{\frac {{\left (c x^{2} + 1\right )}^{2} c^{3}}{{\left (c x^{2} - 1\right )}^{2}} - \frac {2 \, {\left (c x^{2} + 1\right )} c^{3}}{c x^{2} - 1} + c^{3}} - \frac {2 \, b^{2} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1} + 1\right )}{c^{3}} + \frac {2 \, b^{2} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{c^{3}}\right )} c \]
1/4*((c*x^2 + 1)*b^2*log(-(c*x^2 + 1)/(c*x^2 - 1))^2/(((c*x^2 + 1)^2*c^3/( c*x^2 - 1)^2 - 2*(c*x^2 + 1)*c^3/(c*x^2 - 1) + c^3)*(c*x^2 - 1)) + 2*(2*(c *x^2 + 1)*a*b/(c*x^2 - 1) + (c*x^2 + 1)*b^2/(c*x^2 - 1) - b^2)*log(-(c*x^2 + 1)/(c*x^2 - 1))/((c*x^2 + 1)^2*c^3/(c*x^2 - 1)^2 - 2*(c*x^2 + 1)*c^3/(c *x^2 - 1) + c^3) + 4*((c*x^2 + 1)*a^2/(c*x^2 - 1) + (c*x^2 + 1)*a*b/(c*x^2 - 1) - a*b)/((c*x^2 + 1)^2*c^3/(c*x^2 - 1)^2 - 2*(c*x^2 + 1)*c^3/(c*x^2 - 1) + c^3) - 2*b^2*log(-(c*x^2 + 1)/(c*x^2 - 1) + 1)/c^3 + 2*b^2*log(-(c*x ^2 + 1)/(c*x^2 - 1))/c^3)*c
Time = 3.85 (sec) , antiderivative size = 275, normalized size of antiderivative = 3.02 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {a^2\,x^4}{4}+\frac {b^2\,\ln \left (c\,x^2-1\right )}{4\,c^2}+\frac {b^2\,\ln \left (c\,x^2+1\right )}{4\,c^2}-\frac {b^2\,{\ln \left (c\,x^2+1\right )}^2}{16\,c^2}-\frac {b^2\,{\ln \left (1-c\,x^2\right )}^2}{16\,c^2}+\frac {b^2\,x^4\,{\ln \left (c\,x^2+1\right )}^2}{16}+\frac {b^2\,x^4\,{\ln \left (1-c\,x^2\right )}^2}{16}+\frac {b^2\,x^2\,\ln \left (c\,x^2+1\right )}{4\,c}-\frac {b^2\,x^2\,\ln \left (1-c\,x^2\right )}{4\,c}+\frac {a\,b\,\ln \left (c\,x^2-1\right )}{4\,c^2}-\frac {a\,b\,\ln \left (c\,x^2+1\right )}{4\,c^2}+\frac {a\,b\,x^4\,\ln \left (c\,x^2+1\right )}{4}-\frac {a\,b\,x^4\,\ln \left (1-c\,x^2\right )}{4}+\frac {b^2\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{8\,c^2}+\frac {a\,b\,x^2}{2\,c}-\frac {b^2\,x^4\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{8} \]
(a^2*x^4)/4 + (b^2*log(c*x^2 - 1))/(4*c^2) + (b^2*log(c*x^2 + 1))/(4*c^2) - (b^2*log(c*x^2 + 1)^2)/(16*c^2) - (b^2*log(1 - c*x^2)^2)/(16*c^2) + (b^2 *x^4*log(c*x^2 + 1)^2)/16 + (b^2*x^4*log(1 - c*x^2)^2)/16 + (b^2*x^2*log(c *x^2 + 1))/(4*c) - (b^2*x^2*log(1 - c*x^2))/(4*c) + (a*b*log(c*x^2 - 1))/( 4*c^2) - (a*b*log(c*x^2 + 1))/(4*c^2) + (a*b*x^4*log(c*x^2 + 1))/4 - (a*b* x^4*log(1 - c*x^2))/4 + (b^2*log(c*x^2 + 1)*log(1 - c*x^2))/(8*c^2) + (a*b *x^2)/(2*c) - (b^2*x^4*log(c*x^2 + 1)*log(1 - c*x^2))/8