\(\int x^7 (a+b \text {arctanh}(c x^2))^2 \, dx\) [64]

Optimal result

Integrand size = 16, antiderivative size = 125 \[ \int x^7 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {a b x^2}{4 c^3}+\frac {b^2 x^4}{24 c^2}+\frac {b^2 x^2 \text {arctanh}\left (c x^2\right )}{4 c^3}+\frac {b x^6 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{12 c}-\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{8 c^4}+\frac {1}{8} x^8 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2+\frac {b^2 \log \left (1-c^2 x^4\right )}{6 c^4} \] Output:

1/4*a*b*x^2/c^3+1/24*b^2*x^4/c^2+1/4*b^2*x^2*arctanh(c*x^2)/c^3+1/12*b*x^6 
*(a+b*arctanh(c*x^2))/c-1/8*(a+b*arctanh(c*x^2))^2/c^4+1/8*x^8*(a+b*arctan 
h(c*x^2))^2+1/6*b^2*ln(-c^2*x^4+1)/c^4
 

Mathematica [A] (verified)

Time = 0.05 (sec) , antiderivative size = 146, normalized size of antiderivative = 1.17 \[ \int x^7 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {6 a b c x^2+b^2 c^2 x^4+2 a b c^3 x^6+3 a^2 c^4 x^8+2 b c x^2 \left (3 a c^3 x^6+b \left (3+c^2 x^4\right )\right ) \text {arctanh}\left (c x^2\right )+3 b^2 \left (-1+c^4 x^8\right ) \text {arctanh}\left (c x^2\right )^2+b (3 a+4 b) \log \left (1-c x^2\right )-3 a b \log \left (1+c x^2\right )+4 b^2 \log \left (1+c x^2\right )}{24 c^4} \] Input:

Integrate[x^7*(a + b*ArcTanh[c*x^2])^2,x]
 

Output:

(6*a*b*c*x^2 + b^2*c^2*x^4 + 2*a*b*c^3*x^6 + 3*a^2*c^4*x^8 + 2*b*c*x^2*(3* 
a*c^3*x^6 + b*(3 + c^2*x^4))*ArcTanh[c*x^2] + 3*b^2*(-1 + c^4*x^8)*ArcTanh 
[c*x^2]^2 + b*(3*a + 4*b)*Log[1 - c*x^2] - 3*a*b*Log[1 + c*x^2] + 4*b^2*Lo 
g[1 + c*x^2])/(24*c^4)
 

Rubi [A] (verified)

Time = 1.03 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.22, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.625, Rules used = {6454, 6452, 6542, 6452, 243, 49, 2009, 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.

\(\displaystyle \int x^7 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx\)

\(\Big \downarrow \) 6454

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

\(\Big \downarrow \) 6452

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} x^8 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {1}{2} b c \int \frac {x^8 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{1-c^2 x^4}dx^2\right )\)

\(\Big \downarrow \) 6542

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} x^8 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {1}{2} b c \left (\frac {\int \frac {x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{1-c^2 x^4}dx^2}{c^2}-\frac {\int x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )dx^2}{c^2}\right )\right )\)

\(\Big \downarrow \) 6452

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} x^8 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {1}{2} b c \left (\frac {\int \frac {x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{1-c^2 x^4}dx^2}{c^2}-\frac {\frac {1}{3} x^6 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{3} b c \int \frac {x^6}{1-c^2 x^4}dx^2}{c^2}\right )\right )\)

\(\Big \downarrow \) 243

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} x^8 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {1}{2} b c \left (\frac {\int \frac {x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{1-c^2 x^4}dx^2}{c^2}-\frac {\frac {1}{3} x^6 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{6} b c \int \frac {x^4}{1-c^2 x^4}dx^4}{c^2}\right )\right )\)

\(\Big \downarrow \) 49

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} x^8 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {1}{2} b c \left (\frac {\int \frac {x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{1-c^2 x^4}dx^2}{c^2}-\frac {\frac {1}{3} x^6 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{6} b c \int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (c^2 x^4-1\right )}\right )dx^4}{c^2}\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} x^8 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {1}{2} b c \left (\frac {\int \frac {x^4 \left (a+b \text {arctanh}\left (c x^2\right )\right )}{1-c^2 x^4}dx^2}{c^2}-\frac {\frac {1}{3} x^6 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{6} b c \left (-\frac {x^4}{c^2}-\frac {\log \left (1-c^2 x^4\right )}{c^4}\right )}{c^2}\right )\right )\)

\(\Big \downarrow \) 6542

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} x^8 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {1}{2} b c \left (\frac {\frac {\int \frac {a+b \text {arctanh}\left (c x^2\right )}{1-c^2 x^4}dx^2}{c^2}-\frac {\int \left (a+b \text {arctanh}\left (c x^2\right )\right )dx^2}{c^2}}{c^2}-\frac {\frac {1}{3} x^6 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{6} b c \left (-\frac {x^4}{c^2}-\frac {\log \left (1-c^2 x^4\right )}{c^4}\right )}{c^2}\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} x^8 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {1}{2} b c \left (\frac {\frac {\int \frac {a+b \text {arctanh}\left (c x^2\right )}{1-c^2 x^4}dx^2}{c^2}-\frac {a x^2+b x^2 \text {arctanh}\left (c x^2\right )+\frac {b \log \left (1-c^2 x^4\right )}{2 c}}{c^2}}{c^2}-\frac {\frac {1}{3} x^6 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{6} b c \left (-\frac {x^4}{c^2}-\frac {\log \left (1-c^2 x^4\right )}{c^4}\right )}{c^2}\right )\right )\)

\(\Big \downarrow \) 6510

\(\displaystyle \frac {1}{2} \left (\frac {1}{4} x^8 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2-\frac {1}{2} b c \left (\frac {\frac {\left (a+b \text {arctanh}\left (c x^2\right )\right )^2}{2 b c^3}-\frac {a x^2+b x^2 \text {arctanh}\left (c x^2\right )+\frac {b \log \left (1-c^2 x^4\right )}{2 c}}{c^2}}{c^2}-\frac {\frac {1}{3} x^6 \left (a+b \text {arctanh}\left (c x^2\right )\right )-\frac {1}{6} b c \left (-\frac {x^4}{c^2}-\frac {\log \left (1-c^2 x^4\right )}{c^4}\right )}{c^2}\right )\right )\)

Input:

Int[x^7*(a + b*ArcTanh[c*x^2])^2,x]
 

Output:

((x^8*(a + b*ArcTanh[c*x^2])^2)/4 - (b*c*(-(((x^6*(a + b*ArcTanh[c*x^2]))/ 
3 - (b*c*(-(x^4/c^2) - Log[1 - c^2*x^4]/c^4))/6)/c^2) + ((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)/c^2))/2)/2
 

Defintions of rubi rules used

rule 49

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int 
[ExpandIntegrand[(a + b*x)^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d}, x] 
&& IGtQ[m, 0] && IGtQ[m + n + 2, 0]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

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]
 

rule 6454

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]]
 

rule 6510

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]
 

rule 6542

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]
 

Maple [A] (verified)

Time = 0.73 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.30

method	result	size
parallelrisch	\(\frac {3 b^{2} \operatorname {arctanh}\left (c \,x^{2}\right )^{2} x^{8} c^{4}+6 a b \,\operatorname {arctanh}\left (c \,x^{2}\right ) x^{8} c^{4}+3 a^{2} c^{4} x^{8}+2 b^{2} \operatorname {arctanh}\left (c \,x^{2}\right ) x^{6} c^{3}+2 a b \,c^{3} x^{6}+b^{2} c^{2} x^{4}+6 b^{2} \operatorname {arctanh}\left (c \,x^{2}\right ) x^{2} c +6 a b c \,x^{2}-3 b^{2} \operatorname {arctanh}\left (c \,x^{2}\right )^{2}+8 \ln \left (c \,x^{2}-1\right ) b^{2}-6 \,\operatorname {arctanh}\left (c \,x^{2}\right ) a b +8 \,\operatorname {arctanh}\left (c \,x^{2}\right ) b^{2}+b^{2}}{24 c^{4}}\)	\(163\)
risch	\(\frac {b^{2} \left (c^{4} x^{8}-1\right ) \ln \left (c \,x^{2}+1\right )^{2}}{32 c^{4}}+\frac {b \left (-3 x^{8} b \ln \left (-c \,x^{2}+1\right ) c^{4}+6 a \,c^{4} x^{8}+2 b \,c^{3} x^{6}+6 b c \,x^{2}+3 b \ln \left (-c \,x^{2}+1\right )\right ) \ln \left (c \,x^{2}+1\right )}{48 c^{4}}+\frac {b^{2} x^{8} \ln \left (-c \,x^{2}+1\right )^{2}}{32}-\frac {a b \,x^{8} \ln \left (-c \,x^{2}+1\right )}{8}+\frac {a^{2} x^{8}}{8}-\frac {b^{2} x^{6} \ln \left (-c \,x^{2}+1\right )}{24 c}+\frac {a b \,x^{6}}{12 c}+\frac {b^{2} x^{4}}{24 c^{2}}-\frac {b^{2} x^{2} \ln \left (-c \,x^{2}+1\right )}{8 c^{3}}+\frac {a b \,x^{2}}{4 c^{3}}-\frac {b^{2} \ln \left (-c \,x^{2}+1\right )^{2}}{32 c^{4}}-\frac {b \ln \left (-c \,x^{2}-1\right ) a}{8 c^{4}}+\frac {b^{2} \ln \left (-c \,x^{2}-1\right )}{6 c^{4}}+\frac {b \ln \left (-c \,x^{2}+1\right ) a}{8 c^{4}}+\frac {b^{2} \ln \left (-c \,x^{2}+1\right )}{6 c^{4}}\)	\(298\)
default	\(\text {Expression too large to display}\)	\(799\)
parts	\(\text {Expression too large to display}\)	\(799\)

Input:

int(x^7*(a+b*arctanh(c*x^2))^2,x,method=_RETURNVERBOSE)
 

Output:

1/24*(3*b^2*arctanh(c*x^2)^2*x^8*c^4+6*a*b*arctanh(c*x^2)*x^8*c^4+3*a^2*c^ 
4*x^8+2*b^2*arctanh(c*x^2)*x^6*c^3+2*a*b*c^3*x^6+b^2*c^2*x^4+6*b^2*arctanh 
(c*x^2)*x^2*c+6*a*b*c*x^2-3*b^2*arctanh(c*x^2)^2+8*ln(c*x^2-1)*b^2-6*arcta 
nh(c*x^2)*a*b+8*arctanh(c*x^2)*b^2+b^2)/c^4
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.41 \[ \int x^7 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {12 \, a^{2} c^{4} x^{8} + 8 \, a b c^{3} x^{6} + 4 \, b^{2} c^{2} x^{4} + 24 \, a b c x^{2} + 3 \, {\left (b^{2} c^{4} x^{8} - b^{2}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )^{2} - 4 \, {\left (3 \, a b - 4 \, b^{2}\right )} \log \left (c x^{2} + 1\right ) + 4 \, {\left (3 \, a b + 4 \, b^{2}\right )} \log \left (c x^{2} - 1\right ) + 4 \, {\left (3 \, a b c^{4} x^{8} + b^{2} c^{3} x^{6} + 3 \, b^{2} c x^{2}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{96 \, c^{4}} \] Input:

integrate(x^7*(a+b*arctanh(c*x^2))^2,x, algorithm="fricas")
 

Output:

1/96*(12*a^2*c^4*x^8 + 8*a*b*c^3*x^6 + 4*b^2*c^2*x^4 + 24*a*b*c*x^2 + 3*(b 
^2*c^4*x^8 - b^2)*log(-(c*x^2 + 1)/(c*x^2 - 1))^2 - 4*(3*a*b - 4*b^2)*log( 
c*x^2 + 1) + 4*(3*a*b + 4*b^2)*log(c*x^2 - 1) + 4*(3*a*b*c^4*x^8 + b^2*c^3 
*x^6 + 3*b^2*c*x^2)*log(-(c*x^2 + 1)/(c*x^2 - 1)))/c^4
 

Sympy [A] (verification not implemented)

Time = 7.68 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.65 \[ \int x^7 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\begin {cases} \frac {a^{2} x^{8}}{8} + \frac {a b x^{8} \operatorname {atanh}{\left (c x^{2} \right )}}{4} + \frac {a b x^{6}}{12 c} + \frac {a b x^{2}}{4 c^{3}} - \frac {a b \operatorname {atanh}{\left (c x^{2} \right )}}{4 c^{4}} + \frac {b^{2} x^{8} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{8} + \frac {b^{2} x^{6} \operatorname {atanh}{\left (c x^{2} \right )}}{12 c} + \frac {b^{2} x^{4}}{24 c^{2}} + \frac {b^{2} x^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{4 c^{3}} + \frac {b^{2} \log {\left (x - \sqrt {- \frac {1}{c}} \right )}}{3 c^{4}} + \frac {b^{2} \log {\left (x + \sqrt {- \frac {1}{c}} \right )}}{3 c^{4}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c x^{2} \right )}}{8 c^{4}} - \frac {b^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{3 c^{4}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{8}}{8} & \text {otherwise} \end {cases} \] Input:

integrate(x**7*(a+b*atanh(c*x**2))**2,x)
 

Output:

Piecewise((a**2*x**8/8 + a*b*x**8*atanh(c*x**2)/4 + a*b*x**6/(12*c) + a*b* 
x**2/(4*c**3) - a*b*atanh(c*x**2)/(4*c**4) + b**2*x**8*atanh(c*x**2)**2/8 
+ b**2*x**6*atanh(c*x**2)/(12*c) + b**2*x**4/(24*c**2) + b**2*x**2*atanh(c 
*x**2)/(4*c**3) + b**2*log(x - sqrt(-1/c))/(3*c**4) + b**2*log(x + sqrt(-1 
/c))/(3*c**4) - b**2*atanh(c*x**2)**2/(8*c**4) - b**2*atanh(c*x**2)/(3*c** 
4), Ne(c, 0)), (a**2*x**8/8, True))
 

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 217, normalized size of antiderivative = 1.74 \[ \int x^7 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {1}{8} \, b^{2} x^{8} \operatorname {artanh}\left (c x^{2}\right )^{2} + \frac {1}{8} \, a^{2} x^{8} + \frac {1}{24} \, {\left (6 \, x^{8} \operatorname {artanh}\left (c x^{2}\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{6} + 3 \, x^{2}\right )}}{c^{4}} - \frac {3 \, \log \left (c x^{2} + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x^{2} - 1\right )}{c^{5}}\right )}\right )} a b + \frac {1}{96} \, {\left (4 \, c {\left (\frac {2 \, {\left (c^{2} x^{6} + 3 \, x^{2}\right )}}{c^{4}} - \frac {3 \, \log \left (c x^{2} + 1\right )}{c^{5}} + \frac {3 \, \log \left (c x^{2} - 1\right )}{c^{5}}\right )} \operatorname {artanh}\left (c x^{2}\right ) + \frac {4 \, c^{2} x^{4} - 2 \, {\left (3 \, \log \left (c x^{2} - 1\right ) - 8\right )} \log \left (c x^{2} + 1\right ) + 3 \, \log \left (c x^{2} + 1\right )^{2} + 3 \, \log \left (c x^{2} - 1\right )^{2} + 16 \, \log \left (c x^{2} - 1\right )}{c^{4}}\right )} b^{2} \] Input:

integrate(x^7*(a+b*arctanh(c*x^2))^2,x, algorithm="maxima")
 

Output:

1/8*b^2*x^8*arctanh(c*x^2)^2 + 1/8*a^2*x^8 + 1/24*(6*x^8*arctanh(c*x^2) + 
c*(2*(c^2*x^6 + 3*x^2)/c^4 - 3*log(c*x^2 + 1)/c^5 + 3*log(c*x^2 - 1)/c^5)) 
*a*b + 1/96*(4*c*(2*(c^2*x^6 + 3*x^2)/c^4 - 3*log(c*x^2 + 1)/c^5 + 3*log(c 
*x^2 - 1)/c^5)*arctanh(c*x^2) + (4*c^2*x^4 - 2*(3*log(c*x^2 - 1) - 8)*log( 
c*x^2 + 1) + 3*log(c*x^2 + 1)^2 + 3*log(c*x^2 - 1)^2 + 16*log(c*x^2 - 1))/ 
c^4)*b^2
 

Giac [A] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.40 \[ \int x^7 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {1}{8} \, a^{2} x^{8} + \frac {a b x^{6}}{12 \, c} + \frac {b^{2} x^{4}}{24 \, c^{2}} + \frac {1}{32} \, {\left (b^{2} x^{8} - \frac {b^{2}}{c^{4}}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )^{2} + \frac {1}{24} \, {\left (3 \, a b x^{8} + \frac {b^{2} x^{6}}{c} + \frac {3 \, b^{2} x^{2}}{c^{3}}\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + \frac {a b x^{2}}{4 \, c^{3}} - \frac {{\left (3 \, a b - 4 \, b^{2}\right )} \log \left (c x^{2} + 1\right )}{24 \, c^{4}} + \frac {{\left (3 \, a b + 4 \, b^{2}\right )} \log \left (c x^{2} - 1\right )}{24 \, c^{4}} \] Input:

integrate(x^7*(a+b*arctanh(c*x^2))^2,x, algorithm="giac")
 

Output:

1/8*a^2*x^8 + 1/12*a*b*x^6/c + 1/24*b^2*x^4/c^2 + 1/32*(b^2*x^8 - b^2/c^4) 
*log(-(c*x^2 + 1)/(c*x^2 - 1))^2 + 1/24*(3*a*b*x^8 + b^2*x^6/c + 3*b^2*x^2 
/c^3)*log(-(c*x^2 + 1)/(c*x^2 - 1)) + 1/4*a*b*x^2/c^3 - 1/24*(3*a*b - 4*b^ 
2)*log(c*x^2 + 1)/c^4 + 1/24*(3*a*b + 4*b^2)*log(c*x^2 - 1)/c^4
 

Mupad [B] (verification not implemented)

Time = 4.41 (sec) , antiderivative size = 335, normalized size of antiderivative = 2.68 \[ \int x^7 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {a^2\,x^8}{8}+\frac {b^2\,\ln \left (c\,x^2-1\right )}{6\,c^4}+\frac {b^2\,\ln \left (c\,x^2+1\right )}{6\,c^4}-\frac {b^2\,{\ln \left (c\,x^2+1\right )}^2}{32\,c^4}-\frac {b^2\,{\ln \left (1-c\,x^2\right )}^2}{32\,c^4}+\frac {b^2\,x^4}{24\,c^2}+\frac {b^2\,x^8\,{\ln \left (c\,x^2+1\right )}^2}{32}+\frac {b^2\,x^8\,{\ln \left (1-c\,x^2\right )}^2}{32}+\frac {b^2\,x^2\,\ln \left (c\,x^2+1\right )}{8\,c^3}-\frac {b^2\,x^2\,\ln \left (1-c\,x^2\right )}{8\,c^3}+\frac {b^2\,x^6\,\ln \left (c\,x^2+1\right )}{24\,c}-\frac {b^2\,x^6\,\ln \left (1-c\,x^2\right )}{24\,c}+\frac {a\,b\,\ln \left (c\,x^2-1\right )}{8\,c^4}-\frac {a\,b\,\ln \left (c\,x^2+1\right )}{8\,c^4}+\frac {a\,b\,x^8\,\ln \left (c\,x^2+1\right )}{8}-\frac {a\,b\,x^8\,\ln \left (1-c\,x^2\right )}{8}+\frac {b^2\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{16\,c^4}+\frac {a\,b\,x^2}{4\,c^3}+\frac {a\,b\,x^6}{12\,c}-\frac {b^2\,x^8\,\ln \left (c\,x^2+1\right )\,\ln \left (1-c\,x^2\right )}{16} \] Input:

int(x^7*(a + b*atanh(c*x^2))^2,x)
 

Output:

(a^2*x^8)/8 + (b^2*log(c*x^2 - 1))/(6*c^4) + (b^2*log(c*x^2 + 1))/(6*c^4) 
- (b^2*log(c*x^2 + 1)^2)/(32*c^4) - (b^2*log(1 - c*x^2)^2)/(32*c^4) + (b^2 
*x^4)/(24*c^2) + (b^2*x^8*log(c*x^2 + 1)^2)/32 + (b^2*x^8*log(1 - c*x^2)^2 
)/32 + (b^2*x^2*log(c*x^2 + 1))/(8*c^3) - (b^2*x^2*log(1 - c*x^2))/(8*c^3) 
 + (b^2*x^6*log(c*x^2 + 1))/(24*c) - (b^2*x^6*log(1 - c*x^2))/(24*c) + (a* 
b*log(c*x^2 - 1))/(8*c^4) - (a*b*log(c*x^2 + 1))/(8*c^4) + (a*b*x^8*log(c* 
x^2 + 1))/8 - (a*b*x^8*log(1 - c*x^2))/8 + (b^2*log(c*x^2 + 1)*log(1 - c*x 
^2))/(16*c^4) + (a*b*x^2)/(4*c^3) + (a*b*x^6)/(12*c) - (b^2*x^8*log(c*x^2 
+ 1)*log(1 - c*x^2))/16
 

Reduce [B] (verification not implemented)

Time = 0.17 (sec) , antiderivative size = 159, normalized size of antiderivative = 1.27 \[ \int x^7 \left (a+b \text {arctanh}\left (c x^2\right )\right )^2 \, dx=\frac {3 \mathit {atanh} \left (c \,x^{2}\right )^{2} b^{2} c^{4} x^{8}-3 \mathit {atanh} \left (c \,x^{2}\right )^{2} b^{2}+6 \mathit {atanh} \left (c \,x^{2}\right ) a b \,c^{4} x^{8}-6 \mathit {atanh} \left (c \,x^{2}\right ) a b +2 \mathit {atanh} \left (c \,x^{2}\right ) b^{2} c^{3} x^{6}+6 \mathit {atanh} \left (c \,x^{2}\right ) b^{2} c \,x^{2}-8 \mathit {atanh} \left (c \,x^{2}\right ) b^{2}+8 \,\mathrm {log}\left (c \,x^{2}+1\right ) b^{2}+3 a^{2} c^{4} x^{8}+2 a b \,c^{3} x^{6}+6 a b c \,x^{2}+b^{2} c^{2} x^{4}}{24 c^{4}} \] Input:

int(x^7*(a+b*atanh(c*x^2))^2,x)
 

Output:

(3*atanh(c*x**2)**2*b**2*c**4*x**8 - 3*atanh(c*x**2)**2*b**2 + 6*atanh(c*x 
**2)*a*b*c**4*x**8 - 6*atanh(c*x**2)*a*b + 2*atanh(c*x**2)*b**2*c**3*x**6 
+ 6*atanh(c*x**2)*b**2*c*x**2 - 8*atanh(c*x**2)*b**2 + 8*log(c*x**2 + 1)*b 
**2 + 3*a**2*c**4*x**8 + 2*a*b*c**3*x**6 + 6*a*b*c*x**2 + b**2*c**2*x**4)/ 
(24*c**4)