Integrand size = 14, antiderivative size = 43 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {b x^2}{4 c}-\frac {b \text {arctanh}\left (c x^2\right )}{4 c^2}+\frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \] Output:
1/4*b*x^2/c-1/4*b*arctanh(c*x^2)/c^2+1/4*x^4*(a+b*arctanh(c*x^2))
Time = 0.02 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.56 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {b x^2}{4 c}+\frac {a x^4}{4}+\frac {1}{4} b x^4 \text {arctanh}\left (c x^2\right )+\frac {b \log \left (1-c x^2\right )}{8 c^2}-\frac {b \log \left (1+c x^2\right )}{8 c^2} \] Input:
Integrate[x^3*(a + b*ArcTanh[c*x^2]),x]
Output:
(b*x^2)/(4*c) + (a*x^4)/4 + (b*x^4*ArcTanh[c*x^2])/4 + (b*Log[1 - c*x^2])/ (8*c^2) - (b*Log[1 + c*x^2])/(8*c^2)
Time = 0.22 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6452, 807, 262, 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.
\(\displaystyle \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx\) |
\(\Big \downarrow \) 6452 |
\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{2} b c \int \frac {x^5}{1-c^2 x^4}dx\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{4} b c \int \frac {x^4}{1-c^2 x^4}dx^2\) |
\(\Big \downarrow \) 262 |
\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{4} b c \left (\frac {\int \frac {1}{1-c^2 x^4}dx^2}{c^2}-\frac {x^2}{c^2}\right )\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {1}{4} x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{4} b c \left (\frac {\text {arctanh}\left (c x^2\right )}{c^3}-\frac {x^2}{c^2}\right )\) |
Input:
Int[x^3*(a + b*ArcTanh[c*x^2]),x]
Output:
(x^4*(a + b*ArcTanh[c*x^2]))/4 - (b*c*(-(x^2/c^2) + ArcTanh[c*x^2]/c^3))/4
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[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[c*(c*x) ^(m - 1)*((a + b*x^2)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[a*c^2*((m - 1)/ (b*(m + 2*p + 1))) Int[(c*x)^(m - 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b , c, p}, x] && GtQ[m, 2 - 1] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c , 2, m, p, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
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 = 0.28 (sec) , antiderivative size = 46, normalized size of antiderivative = 1.07
method | result | size |
parallelrisch | \(-\frac {-\operatorname {arctanh}\left (c \,x^{2}\right ) b \,c^{2} x^{4}-a \,c^{2} x^{4}-b c \,x^{2}+b \,\operatorname {arctanh}\left (c \,x^{2}\right )}{4 c^{2}}\) | \(46\) |
default | \(\frac {x^{4} a}{4}+\frac {b \,x^{4} \operatorname {arctanh}\left (c \,x^{2}\right )}{4}+\frac {b \,x^{2}}{4 c}-\frac {b \ln \left (c \,x^{2}+1\right )}{8 c^{2}}+\frac {b \ln \left (c \,x^{2}-1\right )}{8 c^{2}}\) | \(57\) |
parts | \(\frac {x^{4} a}{4}+\frac {b \,x^{4} \operatorname {arctanh}\left (c \,x^{2}\right )}{4}+\frac {b \,x^{2}}{4 c}-\frac {b \ln \left (c \,x^{2}+1\right )}{8 c^{2}}+\frac {b \ln \left (c \,x^{2}-1\right )}{8 c^{2}}\) | \(57\) |
orering | \(\frac {5 \left (c^{2} x^{4}-1\right ) \left (a +b \,\operatorname {arctanh}\left (c \,x^{2}\right )\right )}{8 c^{2}}-\frac {\left (c \,x^{2}-1\right ) \left (c \,x^{2}+1\right ) \left (3 x^{2} \left (a +b \,\operatorname {arctanh}\left (c \,x^{2}\right )\right )+\frac {2 x^{4} b c}{-c^{2} x^{4}+1}\right )}{8 c^{2} x^{2}}\) | \(83\) |
risch | \(\frac {b \,x^{4} \ln \left (c \,x^{2}+1\right )}{8}-\frac {b \,x^{4} \ln \left (-c \,x^{2}+1\right )}{8}+\frac {x^{4} a}{4}+\frac {b \,x^{2}}{4 c}-\frac {b \ln \left (c \,x^{2}+1\right )}{8 c^{2}}+\frac {b \ln \left (c \,x^{2}-1\right )}{8 c^{2}}+\frac {b^{2}}{16 a \,c^{2}}\) | \(85\) |
Input:
int(x^3*(a+b*arctanh(c*x^2)),x,method=_RETURNVERBOSE)
Output:
-1/4*(-arctanh(c*x^2)*b*c^2*x^4-a*c^2*x^4-b*c*x^2+b*arctanh(c*x^2))/c^2
Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 1.26 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {2 \, a c^{2} x^{4} + 2 \, b c x^{2} + {\left (b c^{2} x^{4} - b\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{8 \, c^{2}} \] Input:
integrate(x^3*(a+b*arctanh(c*x^2)),x, algorithm="fricas")
Output:
1/8*(2*a*c^2*x^4 + 2*b*c*x^2 + (b*c^2*x^4 - b)*log(-(c*x^2 + 1)/(c*x^2 - 1 )))/c^2
Time = 3.07 (sec) , antiderivative size = 48, normalized size of antiderivative = 1.12 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\begin {cases} \frac {a x^{4}}{4} + \frac {b x^{4} \operatorname {atanh}{\left (c x^{2} \right )}}{4} + \frac {b x^{2}}{4 c} - \frac {b \operatorname {atanh}{\left (c x^{2} \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\\frac {a x^{4}}{4} & \text {otherwise} \end {cases} \] Input:
integrate(x**3*(a+b*atanh(c*x**2)),x)
Output:
Piecewise((a*x**4/4 + b*x**4*atanh(c*x**2)/4 + b*x**2/(4*c) - b*atanh(c*x* *2)/(4*c**2), Ne(c, 0)), (a*x**4/4, True))
Time = 0.03 (sec) , antiderivative size = 58, normalized size of antiderivative = 1.35 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {1}{4} \, a x^{4} + \frac {1}{8} \, {\left (2 \, x^{4} \operatorname {artanh}\left (c x^{2}\right ) + c {\left (\frac {2 \, x^{2}}{c^{2}} - \frac {\log \left (c x^{2} + 1\right )}{c^{3}} + \frac {\log \left (c x^{2} - 1\right )}{c^{3}}\right )}\right )} b \] Input:
integrate(x^3*(a+b*arctanh(c*x^2)),x, algorithm="maxima")
Output:
1/4*a*x^4 + 1/8*(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))*b
Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (37) = 74\).
Time = 0.13 (sec) , antiderivative size = 181, normalized size of antiderivative = 4.21 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {1}{2} \, c {\left (\frac {{\left (c x^{2} + 1\right )} b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{{\left (\frac {{\left (c x^{2} + 1\right )}^{2} c^{3}}{{\left (c x^{2} - 1\right )}^{2}} - \frac {2 \, {\left (c x^{2} + 1\right )} c^{3}}{c x^{2} - 1} + c^{3}\right )} {\left (c x^{2} - 1\right )}} + \frac {\frac {2 \, {\left (c x^{2} + 1\right )} a}{c x^{2} - 1} + \frac {{\left (c x^{2} + 1\right )} b}{c x^{2} - 1} - b}{\frac {{\left (c x^{2} + 1\right )}^{2} c^{3}}{{\left (c x^{2} - 1\right )}^{2}} - \frac {2 \, {\left (c x^{2} + 1\right )} c^{3}}{c x^{2} - 1} + c^{3}}\right )} \] Input:
integrate(x^3*(a+b*arctanh(c*x^2)),x, algorithm="giac")
Output:
1/2*c*((c*x^2 + 1)*b*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)*(c*x^2 - 1)) + (2*(c*x^2 + 1)*a/(c*x^2 - 1) + (c*x^2 + 1)*b/(c*x^2 - 1) - 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))
Time = 3.75 (sec) , antiderivative size = 60, normalized size of antiderivative = 1.40 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {a\,x^4}{4}+\frac {b\,x^2}{4\,c}+\frac {b\,x^4\,\ln \left (c\,x^2+1\right )}{8}-\frac {b\,x^4\,\ln \left (1-c\,x^2\right )}{8}+\frac {b\,\mathrm {atan}\left (c\,x^2\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,c^2} \] Input:
int(x^3*(a + b*atanh(c*x^2)),x)
Output:
(a*x^4)/4 + (b*x^2)/(4*c) + (b*atan(c*x^2*1i)*1i)/(4*c^2) + (b*x^4*log(c*x ^2 + 1))/8 - (b*x^4*log(1 - c*x^2))/8
Time = 0.16 (sec) , antiderivative size = 43, normalized size of antiderivative = 1.00 \[ \int x^3 \left (a+b \text {arctanh}\left (c x^2\right )\right ) \, dx=\frac {\mathit {atanh} \left (c \,x^{2}\right ) b \,c^{2} x^{4}-\mathit {atanh} \left (c \,x^{2}\right ) b +a \,c^{2} x^{4}+b c \,x^{2}}{4 c^{2}} \] Input:
int(x^3*(a+b*atanh(c*x^2)),x)
Output:
(atanh(c*x**2)*b*c**2*x**4 - atanh(c*x**2)*b + a*c**2*x**4 + b*c*x**2)/(4* c**2)