\(\int x^4 (a+b \text {arctanh}(\frac {c}{x^2})) \, dx\) [165]

Optimal result

Integrand size = 14, antiderivative size = 63 \[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {2}{15} b c x^3+\frac {1}{5} b c^{5/2} \arctan \left (\frac {x}{\sqrt {c}}\right )+\frac {1}{5} x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )-\frac {1}{5} b c^{5/2} \text {arctanh}\left (\frac {x}{\sqrt {c}}\right ) \] Output:

2/15*b*c*x^3+1/5*b*c^(5/2)*arctan(x/c^(1/2))+1/5*x^5*(a+b*arctanh(c/x^2))- 
1/5*b*c^(5/2)*arctanh(x/c^(1/2))
 

Mathematica [A] (verified)

Time = 0.03 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.40 \[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {2}{15} b c x^3+\frac {a x^5}{5}+\frac {1}{5} b c^{5/2} \arctan \left (\frac {x}{\sqrt {c}}\right )+\frac {1}{5} b x^5 \text {arctanh}\left (\frac {c}{x^2}\right )+\frac {1}{10} b c^{5/2} \log \left (\sqrt {c}-x\right )-\frac {1}{10} b c^{5/2} \log \left (\sqrt {c}+x\right ) \] Input:

Integrate[x^4*(a + b*ArcTanh[c/x^2]),x]
 

Output:

(2*b*c*x^3)/15 + (a*x^5)/5 + (b*c^(5/2)*ArcTan[x/Sqrt[c]])/5 + (b*x^5*ArcT 
anh[c/x^2])/5 + (b*c^(5/2)*Log[Sqrt[c] - x])/10 - (b*c^(5/2)*Log[Sqrt[c] + 
 x])/10
 

Rubi [A] (verified)

Time = 0.24 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.14, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {6452, 795, 843, 25, 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.

\(\displaystyle \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {2}{5} b c \int \frac {x^2}{1-\frac {c^2}{x^4}}dx+\frac {1}{5} x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\)

\(\Big \downarrow \) 795

\(\displaystyle \frac {2}{5} b c \int \frac {x^6}{x^4-c^2}dx+\frac {1}{5} x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\)

\(\Big \downarrow \) 843

\(\displaystyle \frac {2}{5} b c \left (c^2 \int -\frac {x^2}{c^2-x^4}dx+\frac {x^3}{3}\right )+\frac {1}{5} x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\)

\(\Big \downarrow \) 25

\(\displaystyle \frac {2}{5} b c \left (\frac {x^3}{3}-c^2 \int \frac {x^2}{c^2-x^4}dx\right )+\frac {1}{5} x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\)

\(\Big \downarrow \) 827

\(\displaystyle \frac {2}{5} b c \left (\frac {x^3}{3}-c^2 \left (\frac {1}{2} \int \frac {1}{c-x^2}dx-\frac {1}{2} \int \frac {1}{x^2+c}dx\right )\right )+\frac {1}{5} x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {2}{5} b c \left (\frac {x^3}{3}-c^2 \left (\frac {1}{2} \int \frac {1}{c-x^2}dx-\frac {\arctan \left (\frac {x}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )\right )+\frac {1}{5} x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {1}{5} x^5 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right )+\frac {2}{5} b c \left (\frac {x^3}{3}-c^2 \left (\frac {\text {arctanh}\left (\frac {x}{\sqrt {c}}\right )}{2 \sqrt {c}}-\frac {\arctan \left (\frac {x}{\sqrt {c}}\right )}{2 \sqrt {c}}\right )\right )\)

Input:

Int[x^4*(a + b*ArcTanh[c/x^2]),x]
 

Output:

(x^5*(a + b*ArcTanh[c/x^2]))/5 + (2*b*c*(x^3/3 - c^2*(-1/2*ArcTan[x/Sqrt[c 
]]/Sqrt[c] + ArcTanh[x/Sqrt[c]]/(2*Sqrt[c]))))/5
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 216

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])
 

rule 219

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])
 

rule 795

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[x^(m + n*p)* 
(b + a/x^n)^p, x] /; FreeQ[{a, b, m, n}, x] && IntegerQ[p] && NegQ[n]
 

rule 827

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]
 

rule 843

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]
 

rule 6452

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]
 

Maple [A] (verified)

Time = 0.70 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.84

method	result	size
parts	\(\frac {a \,x^{5}}{5}+\frac {b \,x^{5} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{5}+\frac {b \,c^{\frac {5}{2}} \arctan \left (\frac {x}{\sqrt {c}}\right )}{5}-\frac {b \,c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c}}{x}\right )}{5}+\frac {2 b c \,x^{3}}{15}\)	\(53\)
derivativedivides	\(\frac {a \,x^{5}}{5}+\frac {b \,x^{5} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{5}-\frac {b \,c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c}}{x}\right )}{5}-\frac {b \,c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {c}}{x}\right )}{5}+\frac {2 b c \,x^{3}}{15}\)	\(55\)
default	\(\frac {a \,x^{5}}{5}+\frac {b \,x^{5} \operatorname {arctanh}\left (\frac {c}{x^{2}}\right )}{5}-\frac {b \,c^{\frac {5}{2}} \operatorname {arctanh}\left (\frac {\sqrt {c}}{x}\right )}{5}-\frac {b \,c^{\frac {5}{2}} \arctan \left (\frac {\sqrt {c}}{x}\right )}{5}+\frac {2 b c \,x^{3}}{15}\)	\(55\)
risch	\(\frac {b \,x^{5} \ln \left (x^{2}+c \right )}{10}-\frac {b \,x^{5} \ln \left (-x^{2}+c \right )}{10}-\frac {i \pi b \,x^{5}}{10}-\frac {i \pi b \,x^{5} \operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{20}+\frac {i \pi b \,x^{5} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}}{20}-\frac {i \pi b \,x^{5} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{20}+\frac {i \pi b \,x^{5} \operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{2}}{20}+\frac {i \pi b \,x^{5} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (-x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}{20}-\frac {i \pi b \,x^{5} {\operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}^{3}}{20}-\frac {i \pi b \,x^{5} \operatorname {csgn}\left (\frac {i}{x^{2}}\right ) \operatorname {csgn}\left (i \left (x^{2}+c \right )\right ) \operatorname {csgn}\left (\frac {i \left (x^{2}+c \right )}{x^{2}}\right )}{20}+\frac {i \pi b \,x^{5} {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{2}}{10}-\frac {i \pi b \,x^{5} {\operatorname {csgn}\left (\frac {i \left (-x^{2}+c \right )}{x^{2}}\right )}^{3}}{20}+\frac {a \,x^{5}}{5}+\frac {2 b c \,x^{3}}{15}+\frac {c^{\frac {5}{2}} b \ln \left (x -\sqrt {c}\right )}{10}-\frac {c^{\frac {5}{2}} b \ln \left (\sqrt {c}+x \right )}{10}+\frac {b \,c^{2} \sqrt {-c}\, \ln \left (c^{2} \sqrt {-c}+c^{2} x \right )}{10}-\frac {b \,c^{2} \sqrt {-c}\, \ln \left (-c^{2} \sqrt {-c}+c^{2} x \right )}{10}\)	\(402\)

Input:

int(x^4*(a+b*arctanh(c/x^2)),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(x/c^(1/2))-1/5*b*c 
^(5/2)*arctanh(1/x*c^(1/2))+2/15*b*c*x^3
 

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 170, normalized size of antiderivative = 2.70 \[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\left [\frac {1}{10} \, b x^{5} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + \frac {1}{5} \, a x^{5} + \frac {2}{15} \, b c x^{3} + \frac {1}{5} \, b c^{\frac {5}{2}} \arctan \left (\frac {x}{\sqrt {c}}\right ) + \frac {1}{10} \, b c^{\frac {5}{2}} \log \left (\frac {x^{2} - 2 \, \sqrt {c} x + c}{x^{2} - c}\right ), \frac {1}{10} \, b x^{5} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + \frac {1}{5} \, a x^{5} + \frac {2}{15} \, b c x^{3} + \frac {1}{5} \, b \sqrt {-c} c^{2} \arctan \left (\frac {\sqrt {-c} x}{c}\right ) + \frac {1}{10} \, b \sqrt {-c} c^{2} \log \left (\frac {x^{2} + 2 \, \sqrt {-c} x - c}{x^{2} + c}\right )\right ] \] Input:

integrate(x^4*(a+b*arctanh(c/x^2)),x, algorithm="fricas")
 

Output:

[1/10*b*x^5*log((x^2 + c)/(x^2 - c)) + 1/5*a*x^5 + 2/15*b*c*x^3 + 1/5*b*c^ 
(5/2)*arctan(x/sqrt(c)) + 1/10*b*c^(5/2)*log((x^2 - 2*sqrt(c)*x + c)/(x^2 
- c)), 1/10*b*x^5*log((x^2 + c)/(x^2 - c)) + 1/5*a*x^5 + 2/15*b*c*x^3 + 1/ 
5*b*sqrt(-c)*c^2*arctan(sqrt(-c)*x/c) + 1/10*b*sqrt(-c)*c^2*log((x^2 + 2*s 
qrt(-c)*x - c)/(x^2 + c))]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 845 vs. \(2 (58) = 116\).

Time = 3.85 (sec) , antiderivative size = 845, normalized size of antiderivative = 13.41 \[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx =\text {Too large to display} \] Input:

integrate(x**4*(a+b*atanh(c/x**2)),x)
 

Output:

Piecewise((a*x**5/5, Eq(c, 0)), (x**5*(a - oo*b)/5, Eq(c, -x**2)), (x**5*( 
a + oo*b)/5, Eq(c, x**2)), (-6*a*c**2*x**5*sqrt(-c)/(-30*c**2*sqrt(-c) + 3 
0*x**4*sqrt(-c)) + 6*a*x**9*sqrt(-c)/(-30*c**2*sqrt(-c) + 30*x**4*sqrt(-c) 
) - 6*b*c**(9/2)*sqrt(-c)*log(-sqrt(c) + x)/(-30*c**2*sqrt(-c) + 30*x**4*s 
qrt(-c)) + 3*b*c**(9/2)*sqrt(-c)*log(x - sqrt(-c))/(-30*c**2*sqrt(-c) + 30 
*x**4*sqrt(-c)) + 3*b*c**(9/2)*sqrt(-c)*log(x + sqrt(-c))/(-30*c**2*sqrt(- 
c) + 30*x**4*sqrt(-c)) - 6*b*c**(9/2)*sqrt(-c)*atanh(c/x**2)/(-30*c**2*sqr 
t(-c) + 30*x**4*sqrt(-c)) + 6*b*c**(5/2)*x**4*sqrt(-c)*log(-sqrt(c) + x)/( 
-30*c**2*sqrt(-c) + 30*x**4*sqrt(-c)) - 3*b*c**(5/2)*x**4*sqrt(-c)*log(x - 
 sqrt(-c))/(-30*c**2*sqrt(-c) + 30*x**4*sqrt(-c)) - 3*b*c**(5/2)*x**4*sqrt 
(-c)*log(x + sqrt(-c))/(-30*c**2*sqrt(-c) + 30*x**4*sqrt(-c)) + 6*b*c**(5/ 
2)*x**4*sqrt(-c)*atanh(c/x**2)/(-30*c**2*sqrt(-c) + 30*x**4*sqrt(-c)) - 3* 
b*c**5*log(x - sqrt(-c))/(-30*c**2*sqrt(-c) + 30*x**4*sqrt(-c)) + 3*b*c**5 
*log(x + sqrt(-c))/(-30*c**2*sqrt(-c) + 30*x**4*sqrt(-c)) + 3*b*c**3*x**4* 
log(x - sqrt(-c))/(-30*c**2*sqrt(-c) + 30*x**4*sqrt(-c)) - 3*b*c**3*x**4*l 
og(x + sqrt(-c))/(-30*c**2*sqrt(-c) + 30*x**4*sqrt(-c)) - 4*b*c**3*x**3*sq 
rt(-c)/(-30*c**2*sqrt(-c) + 30*x**4*sqrt(-c)) - 6*b*c**2*x**5*sqrt(-c)*ata 
nh(c/x**2)/(-30*c**2*sqrt(-c) + 30*x**4*sqrt(-c)) + 4*b*c*x**7*sqrt(-c)/(- 
30*c**2*sqrt(-c) + 30*x**4*sqrt(-c)) + 6*b*x**9*sqrt(-c)*atanh(c/x**2)/(-3 
0*c**2*sqrt(-c) + 30*x**4*sqrt(-c)), True))
 

Maxima [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 62, normalized size of antiderivative = 0.98 \[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{5} \, a x^{5} + \frac {1}{30} \, {\left (6 \, x^{5} \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + {\left (4 \, x^{3} + 6 \, c^{\frac {3}{2}} \arctan \left (\frac {x}{\sqrt {c}}\right ) + 3 \, c^{\frac {3}{2}} \log \left (\frac {x - \sqrt {c}}{x + \sqrt {c}}\right )\right )} c\right )} b \] Input:

integrate(x^4*(a+b*arctanh(c/x^2)),x, algorithm="maxima")
 

Output:

1/5*a*x^5 + 1/30*(6*x^5*arctanh(c/x^2) + (4*x^3 + 6*c^(3/2)*arctan(x/sqrt( 
c)) + 3*c^(3/2)*log((x - sqrt(c))/(x + sqrt(c))))*c)*b
 

Giac [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06 \[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {1}{10} \, b x^{5} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + \frac {1}{5} \, a x^{5} + \frac {2}{15} \, b c x^{3} + \frac {b c^{3} \arctan \left (\frac {x}{\sqrt {-c}}\right )}{5 \, \sqrt {-c}} + \frac {1}{5} \, b c^{\frac {5}{2}} \arctan \left (\frac {x}{\sqrt {c}}\right ) \] Input:

integrate(x^4*(a+b*arctanh(c/x^2)),x, algorithm="giac")
 

Output:

1/10*b*x^5*log((x^2 + c)/(x^2 - c)) + 1/5*a*x^5 + 2/15*b*c*x^3 + 1/5*b*c^3 
*arctan(x/sqrt(-c))/sqrt(-c) + 1/5*b*c^(5/2)*arctan(x/sqrt(c))
 

Mupad [B] (verification not implemented)

Time = 3.87 (sec) , antiderivative size = 67, normalized size of antiderivative = 1.06 \[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {a\,x^5}{5}+\frac {b\,c^{5/2}\,\mathrm {atan}\left (\frac {x}{\sqrt {c}}\right )}{5}+\frac {b\,x^5\,\ln \left (x^2+c\right )}{10}+\frac {2\,b\,c\,x^3}{15}-\frac {b\,x^5\,\ln \left (x^2-c\right )}{10}+\frac {b\,c^{5/2}\,\mathrm {atan}\left (\frac {x\,1{}\mathrm {i}}{\sqrt {c}}\right )\,1{}\mathrm {i}}{5} \] Input:

int(x^4*(a + b*atanh(c/x^2)),x)
 

Output:

(a*x^5)/5 + (b*c^(5/2)*atan(x/c^(1/2)))/5 + (b*c^(5/2)*atan((x*1i)/c^(1/2) 
)*1i)/5 + (b*x^5*log(c + x^2))/10 + (2*b*c*x^3)/15 - (b*x^5*log(x^2 - c))/ 
10
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.33 \[ \int x^4 \left (a+b \text {arctanh}\left (\frac {c}{x^2}\right )\right ) \, dx=\frac {\sqrt {c}\, \mathit {atan} \left (\frac {x}{\sqrt {c}}\right ) b \,c^{2}}{5}+\frac {\sqrt {c}\, \mathit {atanh} \left (\frac {c}{x^{2}}\right ) b \,c^{2}}{5}+\frac {\mathit {atanh} \left (\frac {c}{x^{2}}\right ) b \,x^{5}}{5}+\frac {\sqrt {c}\, \mathrm {log}\left (\sqrt {c}-x \right ) b \,c^{2}}{5}-\frac {\sqrt {c}\, \mathrm {log}\left (x^{2}+c \right ) b \,c^{2}}{10}+\frac {a \,x^{5}}{5}+\frac {2 b c \,x^{3}}{15} \] Input:

int(x^4*(a+b*atanh(c/x^2)),x)
 

Output:

(6*sqrt(c)*atan(x/sqrt(c))*b*c**2 + 6*sqrt(c)*atanh(c/x**2)*b*c**2 + 6*ata 
nh(c/x**2)*b*x**5 + 6*sqrt(c)*log(sqrt(c) - x)*b*c**2 - 3*sqrt(c)*log(c + 
x**2)*b*c**2 + 6*a*x**5 + 4*b*c*x**3)/30