3.13 \(\int x^5 (a+b \tanh ^{-1}(c x))^2 \, dx\)

Optimal. Leaf size=145 \[ -\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^6}+\frac {a b x}{3 c^5}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac {b^2 x \tanh ^{-1}(c x)}{3 c^5}+\frac {4 b^2 x^2}{45 c^4}+\frac {b^2 x^4}{60 c^2}+\frac {23 b^2 \log \left (1-c^2 x^2\right )}{90 c^6} \]

[Out]

1/3*a*b*x/c^5+4/45*b^2*x^2/c^4+1/60*b^2*x^4/c^2+1/3*b^2*x*arctanh(c*x)/c^5+1/9*b*x^3*(a+b*arctanh(c*x))/c^3+1/
15*b*x^5*(a+b*arctanh(c*x))/c-1/6*(a+b*arctanh(c*x))^2/c^6+1/6*x^6*(a+b*arctanh(c*x))^2+23/90*b^2*ln(-c^2*x^2+
1)/c^6

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Rubi [A]  time = 0.33, antiderivative size = 145, normalized size of antiderivative = 1.00, number of steps used = 16, number of rules used = 7, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.500, Rules used = {5916, 5980, 266, 43, 5910, 260, 5948} \[ \frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac {a b x}{3 c^5}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^6}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac {b^2 x^4}{60 c^2}+\frac {4 b^2 x^2}{45 c^4}+\frac {23 b^2 \log \left (1-c^2 x^2\right )}{90 c^6}+\frac {b^2 x \tanh ^{-1}(c x)}{3 c^5} \]

Antiderivative was successfully verified.

[In]

Int[x^5*(a + b*ArcTanh[c*x])^2,x]

[Out]

(a*b*x)/(3*c^5) + (4*b^2*x^2)/(45*c^4) + (b^2*x^4)/(60*c^2) + (b^2*x*ArcTanh[c*x])/(3*c^5) + (b*x^3*(a + b*Arc
Tanh[c*x]))/(9*c^3) + (b*x^5*(a + b*ArcTanh[c*x]))/(15*c) - (a + b*ArcTanh[c*x])^2/(6*c^6) + (x^6*(a + b*ArcTa
nh[c*x])^2)/6 + (23*b^2*Log[1 - c^2*x^2])/(90*c^6)

Rule 43

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, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 5916

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcT
anh[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTanh[c*x])^(p - 1))/(1 -
c^2*x^2), x], x] /; FreeQ[{a, b, c, d, m}, x] && IGtQ[p, 0] && (EqQ[p, 1] || IntegerQ[m]) && NeQ[m, -1]

Rule 5948

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> 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 5980

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[f^2
/e, Int[(f*x)^(m - 2)*(a + b*ArcTanh[c*x])^p, x], x] - Dist[(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]

Rubi steps

\begin {align*} \int x^5 \left (a+b \tanh ^{-1}(c x)\right )^2 \, dx &=\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{3} (b c) \int \frac {x^6 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx\\ &=\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b \int x^4 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{3 c}-\frac {b \int \frac {x^4 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 c}\\ &=\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{15} b^2 \int \frac {x^5}{1-c^2 x^2} \, dx+\frac {b \int x^2 \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{3 c^3}-\frac {b \int \frac {x^2 \left (a+b \tanh ^{-1}(c x)\right )}{1-c^2 x^2} \, dx}{3 c^3}\\ &=\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{30} b^2 \operatorname {Subst}\left (\int \frac {x^2}{1-c^2 x} \, dx,x,x^2\right )+\frac {b \int \left (a+b \tanh ^{-1}(c x)\right ) \, dx}{3 c^5}-\frac {b \int \frac {a+b \tanh ^{-1}(c x)}{1-c^2 x^2} \, dx}{3 c^5}-\frac {b^2 \int \frac {x^3}{1-c^2 x^2} \, dx}{9 c^2}\\ &=\frac {a b x}{3 c^5}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^6}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2-\frac {1}{30} b^2 \operatorname {Subst}\left (\int \left (-\frac {1}{c^4}-\frac {x}{c^2}-\frac {1}{c^4 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )+\frac {b^2 \int \tanh ^{-1}(c x) \, dx}{3 c^5}-\frac {b^2 \operatorname {Subst}\left (\int \frac {x}{1-c^2 x} \, dx,x,x^2\right )}{18 c^2}\\ &=\frac {a b x}{3 c^5}+\frac {b^2 x^2}{30 c^4}+\frac {b^2 x^4}{60 c^2}+\frac {b^2 x \tanh ^{-1}(c x)}{3 c^5}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^6}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {b^2 \log \left (1-c^2 x^2\right )}{30 c^6}-\frac {b^2 \int \frac {x}{1-c^2 x^2} \, dx}{3 c^4}-\frac {b^2 \operatorname {Subst}\left (\int \left (-\frac {1}{c^2}-\frac {1}{c^2 \left (-1+c^2 x\right )}\right ) \, dx,x,x^2\right )}{18 c^2}\\ &=\frac {a b x}{3 c^5}+\frac {4 b^2 x^2}{45 c^4}+\frac {b^2 x^4}{60 c^2}+\frac {b^2 x \tanh ^{-1}(c x)}{3 c^5}+\frac {b x^3 \left (a+b \tanh ^{-1}(c x)\right )}{9 c^3}+\frac {b x^5 \left (a+b \tanh ^{-1}(c x)\right )}{15 c}-\frac {\left (a+b \tanh ^{-1}(c x)\right )^2}{6 c^6}+\frac {1}{6} x^6 \left (a+b \tanh ^{-1}(c x)\right )^2+\frac {23 b^2 \log \left (1-c^2 x^2\right )}{90 c^6}\\ \end {align*}

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Mathematica [A]  time = 0.07, size = 164, normalized size = 1.13 \[ \frac {30 a^2 c^6 x^6+12 a b c^5 x^5+20 a b c^3 x^3+4 b c x \tanh ^{-1}(c x) \left (15 a c^5 x^5+b \left (3 c^4 x^4+5 c^2 x^2+15\right )\right )+60 a b c x+2 b (15 a+23 b) \log (1-c x)-30 a b \log (c x+1)+30 b^2 \left (c^6 x^6-1\right ) \tanh ^{-1}(c x)^2+3 b^2 c^4 x^4+16 b^2 c^2 x^2+46 b^2 \log (c x+1)}{180 c^6} \]

Antiderivative was successfully verified.

[In]

Integrate[x^5*(a + b*ArcTanh[c*x])^2,x]

[Out]

(60*a*b*c*x + 16*b^2*c^2*x^2 + 20*a*b*c^3*x^3 + 3*b^2*c^4*x^4 + 12*a*b*c^5*x^5 + 30*a^2*c^6*x^6 + 4*b*c*x*(15*
a*c^5*x^5 + b*(15 + 5*c^2*x^2 + 3*c^4*x^4))*ArcTanh[c*x] + 30*b^2*(-1 + c^6*x^6)*ArcTanh[c*x]^2 + 2*b*(15*a +
23*b)*Log[1 - c*x] - 30*a*b*Log[1 + c*x] + 46*b^2*Log[1 + c*x])/(180*c^6)

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fricas [A]  time = 1.18, size = 193, normalized size = 1.33 \[ \frac {60 \, a^{2} c^{6} x^{6} + 24 \, a b c^{5} x^{5} + 6 \, b^{2} c^{4} x^{4} + 40 \, a b c^{3} x^{3} + 32 \, b^{2} c^{2} x^{2} + 120 \, a b c x + 15 \, {\left (b^{2} c^{6} x^{6} - b^{2}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )^{2} - 4 \, {\left (15 \, a b - 23 \, b^{2}\right )} \log \left (c x + 1\right ) + 4 \, {\left (15 \, a b + 23 \, b^{2}\right )} \log \left (c x - 1\right ) + 4 \, {\left (15 \, a b c^{6} x^{6} + 3 \, b^{2} c^{5} x^{5} + 5 \, b^{2} c^{3} x^{3} + 15 \, b^{2} c x\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{360 \, c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctanh(c*x))^2,x, algorithm="fricas")

[Out]

1/360*(60*a^2*c^6*x^6 + 24*a*b*c^5*x^5 + 6*b^2*c^4*x^4 + 40*a*b*c^3*x^3 + 32*b^2*c^2*x^2 + 120*a*b*c*x + 15*(b
^2*c^6*x^6 - b^2)*log(-(c*x + 1)/(c*x - 1))^2 - 4*(15*a*b - 23*b^2)*log(c*x + 1) + 4*(15*a*b + 23*b^2)*log(c*x
- 1) + 4*(15*a*b*c^6*x^6 + 3*b^2*c^5*x^5 + 5*b^2*c^3*x^3 + 15*b^2*c*x)*log(-(c*x + 1)/(c*x - 1)))/c^6

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giac [B]  time = 0.18, size = 889, normalized size = 6.13 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctanh(c*x))^2,x, algorithm="giac")

[Out]

1/90*(15*(3*(c*x + 1)^5*b^2/(c*x - 1)^5 + 10*(c*x + 1)^3*b^2/(c*x - 1)^3 + 3*(c*x + 1)*b^2/(c*x - 1))*log(-(c*
x + 1)/(c*x - 1))^2/((c*x + 1)^6*c^7/(c*x - 1)^6 - 6*(c*x + 1)^5*c^7/(c*x - 1)^5 + 15*(c*x + 1)^4*c^7/(c*x - 1
)^4 - 20*(c*x + 1)^3*c^7/(c*x - 1)^3 + 15*(c*x + 1)^2*c^7/(c*x - 1)^2 - 6*(c*x + 1)*c^7/(c*x - 1) + c^7) + 2*(
90*(c*x + 1)^5*a*b/(c*x - 1)^5 + 300*(c*x + 1)^3*a*b/(c*x - 1)^3 + 90*(c*x + 1)*a*b/(c*x - 1) + 45*(c*x + 1)^5
*b^2/(c*x - 1)^5 - 135*(c*x + 1)^4*b^2/(c*x - 1)^4 + 230*(c*x + 1)^3*b^2/(c*x - 1)^3 - 210*(c*x + 1)^2*b^2/(c*
x - 1)^2 + 93*(c*x + 1)*b^2/(c*x - 1) - 23*b^2)*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^6*c^7/(c*x - 1)^6 - 6*(c*
x + 1)^5*c^7/(c*x - 1)^5 + 15*(c*x + 1)^4*c^7/(c*x - 1)^4 - 20*(c*x + 1)^3*c^7/(c*x - 1)^3 + 15*(c*x + 1)^2*c^
7/(c*x - 1)^2 - 6*(c*x + 1)*c^7/(c*x - 1) + c^7) + 4*(45*(c*x + 1)^5*a^2/(c*x - 1)^5 + 150*(c*x + 1)^3*a^2/(c*
x - 1)^3 + 45*(c*x + 1)*a^2/(c*x - 1) + 45*(c*x + 1)^5*a*b/(c*x - 1)^5 - 135*(c*x + 1)^4*a*b/(c*x - 1)^4 + 230
*(c*x + 1)^3*a*b/(c*x - 1)^3 - 210*(c*x + 1)^2*a*b/(c*x - 1)^2 + 93*(c*x + 1)*a*b/(c*x - 1) - 23*a*b + 11*(c*x
+ 1)^5*b^2/(c*x - 1)^5 - 38*(c*x + 1)^4*b^2/(c*x - 1)^4 + 54*(c*x + 1)^3*b^2/(c*x - 1)^3 - 38*(c*x + 1)^2*b^2
/(c*x - 1)^2 + 11*(c*x + 1)*b^2/(c*x - 1))/((c*x + 1)^6*c^7/(c*x - 1)^6 - 6*(c*x + 1)^5*c^7/(c*x - 1)^5 + 15*(
c*x + 1)^4*c^7/(c*x - 1)^4 - 20*(c*x + 1)^3*c^7/(c*x - 1)^3 + 15*(c*x + 1)^2*c^7/(c*x - 1)^2 - 6*(c*x + 1)*c^7
/(c*x - 1) + c^7) - 46*b^2*log(-(c*x + 1)/(c*x - 1) + 1)/c^7 + 46*b^2*log(-(c*x + 1)/(c*x - 1))/c^7)*c

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maple [B]  time = 0.03, size = 314, normalized size = 2.17 \[ \frac {x^{6} a^{2}}{6}+\frac {b^{2} x^{6} \arctanh \left (c x \right )^{2}}{6}+\frac {b^{2} \arctanh \left (c x \right ) x^{5}}{15 c}+\frac {b^{2} \arctanh \left (c x \right ) x^{3}}{9 c^{3}}+\frac {b^{2} x \arctanh \left (c x \right )}{3 c^{5}}+\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x -1\right )}{6 c^{6}}-\frac {b^{2} \arctanh \left (c x \right ) \ln \left (c x +1\right )}{6 c^{6}}+\frac {b^{2} \ln \left (c x -1\right )^{2}}{24 c^{6}}-\frac {b^{2} \ln \left (c x -1\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{12 c^{6}}+\frac {b^{2} \ln \left (c x +1\right )^{2}}{24 c^{6}}-\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (c x +1\right )}{12 c^{6}}+\frac {b^{2} \ln \left (-\frac {c x}{2}+\frac {1}{2}\right ) \ln \left (\frac {1}{2}+\frac {c x}{2}\right )}{12 c^{6}}+\frac {b^{2} x^{4}}{60 c^{2}}+\frac {4 b^{2} x^{2}}{45 c^{4}}+\frac {23 b^{2} \ln \left (c x -1\right )}{90 c^{6}}+\frac {23 b^{2} \ln \left (c x +1\right )}{90 c^{6}}+\frac {a b \,x^{6} \arctanh \left (c x \right )}{3}+\frac {x^{5} a b}{15 c}+\frac {a b \,x^{3}}{9 c^{3}}+\frac {a b x}{3 c^{5}}+\frac {a b \ln \left (c x -1\right )}{6 c^{6}}-\frac {a b \ln \left (c x +1\right )}{6 c^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a+b*arctanh(c*x))^2,x)

[Out]

1/6*x^6*a^2+1/6*b^2*x^6*arctanh(c*x)^2+1/15/c*b^2*arctanh(c*x)*x^5+1/9/c^3*b^2*arctanh(c*x)*x^3+1/3*b^2*x*arct
anh(c*x)/c^5+1/6/c^6*b^2*arctanh(c*x)*ln(c*x-1)-1/6/c^6*b^2*arctanh(c*x)*ln(c*x+1)+1/24/c^6*b^2*ln(c*x-1)^2-1/
12/c^6*b^2*ln(c*x-1)*ln(1/2+1/2*c*x)+1/24/c^6*b^2*ln(c*x+1)^2-1/12/c^6*b^2*ln(-1/2*c*x+1/2)*ln(c*x+1)+1/12/c^6
*b^2*ln(-1/2*c*x+1/2)*ln(1/2+1/2*c*x)+1/60*b^2*x^4/c^2+4/45*b^2*x^2/c^4+23/90/c^6*b^2*ln(c*x-1)+23/90/c^6*b^2*
ln(c*x+1)+1/3*a*b*x^6*arctanh(c*x)+1/15/c*x^5*a*b+1/9*a*b*x^3/c^3+1/3*a*b*x/c^5+1/6/c^6*a*b*ln(c*x-1)-1/6/c^6*
a*b*ln(c*x+1)

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maxima [A]  time = 0.37, size = 215, normalized size = 1.48 \[ \frac {1}{6} \, b^{2} x^{6} \operatorname {artanh}\left (c x\right )^{2} + \frac {1}{6} \, a^{2} x^{6} + \frac {1}{90} \, {\left (30 \, x^{6} \operatorname {artanh}\left (c x\right ) + c {\left (\frac {2 \, {\left (3 \, c^{4} x^{5} + 5 \, c^{2} x^{3} + 15 \, x\right )}}{c^{6}} - \frac {15 \, \log \left (c x + 1\right )}{c^{7}} + \frac {15 \, \log \left (c x - 1\right )}{c^{7}}\right )}\right )} a b + \frac {1}{360} \, {\left (4 \, c {\left (\frac {2 \, {\left (3 \, c^{4} x^{5} + 5 \, c^{2} x^{3} + 15 \, x\right )}}{c^{6}} - \frac {15 \, \log \left (c x + 1\right )}{c^{7}} + \frac {15 \, \log \left (c x - 1\right )}{c^{7}}\right )} \operatorname {artanh}\left (c x\right ) + \frac {6 \, c^{4} x^{4} + 32 \, c^{2} x^{2} - 2 \, {\left (15 \, \log \left (c x - 1\right ) - 46\right )} \log \left (c x + 1\right ) + 15 \, \log \left (c x + 1\right )^{2} + 15 \, \log \left (c x - 1\right )^{2} + 92 \, \log \left (c x - 1\right )}{c^{6}}\right )} b^{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^5*(a+b*arctanh(c*x))^2,x, algorithm="maxima")

[Out]

1/6*b^2*x^6*arctanh(c*x)^2 + 1/6*a^2*x^6 + 1/90*(30*x^6*arctanh(c*x) + c*(2*(3*c^4*x^5 + 5*c^2*x^3 + 15*x)/c^6
- 15*log(c*x + 1)/c^7 + 15*log(c*x - 1)/c^7))*a*b + 1/360*(4*c*(2*(3*c^4*x^5 + 5*c^2*x^3 + 15*x)/c^6 - 15*log
(c*x + 1)/c^7 + 15*log(c*x - 1)/c^7)*arctanh(c*x) + (6*c^4*x^4 + 32*c^2*x^2 - 2*(15*log(c*x - 1) - 46)*log(c*x
+ 1) + 15*log(c*x + 1)^2 + 15*log(c*x - 1)^2 + 92*log(c*x - 1))/c^6)*b^2

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mupad [B]  time = 1.04, size = 171, normalized size = 1.18 \[ \frac {46\,b^2\,\ln \left (c^2\,x^2-1\right )-30\,b^2\,{\mathrm {atanh}\left (c\,x\right )}^2+30\,a^2\,c^6\,x^6+16\,b^2\,c^2\,x^2+3\,b^2\,c^4\,x^4-60\,a\,b\,\mathrm {atanh}\left (c\,x\right )+20\,b^2\,c^3\,x^3\,\mathrm {atanh}\left (c\,x\right )+12\,b^2\,c^5\,x^5\,\mathrm {atanh}\left (c\,x\right )+60\,b^2\,c\,x\,\mathrm {atanh}\left (c\,x\right )+30\,b^2\,c^6\,x^6\,{\mathrm {atanh}\left (c\,x\right )}^2+20\,a\,b\,c^3\,x^3+12\,a\,b\,c^5\,x^5+60\,a\,b\,c\,x+60\,a\,b\,c^6\,x^6\,\mathrm {atanh}\left (c\,x\right )}{180\,c^6} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^5*(a + b*atanh(c*x))^2,x)

[Out]

(46*b^2*log(c^2*x^2 - 1) - 30*b^2*atanh(c*x)^2 + 30*a^2*c^6*x^6 + 16*b^2*c^2*x^2 + 3*b^2*c^4*x^4 - 60*a*b*atan
h(c*x) + 20*b^2*c^3*x^3*atanh(c*x) + 12*b^2*c^5*x^5*atanh(c*x) + 60*b^2*c*x*atanh(c*x) + 30*b^2*c^6*x^6*atanh(
c*x)^2 + 20*a*b*c^3*x^3 + 12*a*b*c^5*x^5 + 60*a*b*c*x + 60*a*b*c^6*x^6*atanh(c*x))/(180*c^6)

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sympy [A]  time = 2.74, size = 211, normalized size = 1.46 \[ \begin {cases} \frac {a^{2} x^{6}}{6} + \frac {a b x^{6} \operatorname {atanh}{\left (c x \right )}}{3} + \frac {a b x^{5}}{15 c} + \frac {a b x^{3}}{9 c^{3}} + \frac {a b x}{3 c^{5}} - \frac {a b \operatorname {atanh}{\left (c x \right )}}{3 c^{6}} + \frac {b^{2} x^{6} \operatorname {atanh}^{2}{\left (c x \right )}}{6} + \frac {b^{2} x^{5} \operatorname {atanh}{\left (c x \right )}}{15 c} + \frac {b^{2} x^{4}}{60 c^{2}} + \frac {b^{2} x^{3} \operatorname {atanh}{\left (c x \right )}}{9 c^{3}} + \frac {4 b^{2} x^{2}}{45 c^{4}} + \frac {b^{2} x \operatorname {atanh}{\left (c x \right )}}{3 c^{5}} + \frac {23 b^{2} \log {\left (x - \frac {1}{c} \right )}}{45 c^{6}} - \frac {b^{2} \operatorname {atanh}^{2}{\left (c x \right )}}{6 c^{6}} + \frac {23 b^{2} \operatorname {atanh}{\left (c x \right )}}{45 c^{6}} & \text {for}\: c \neq 0 \\\frac {a^{2} x^{6}}{6} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**5*(a+b*atanh(c*x))**2,x)

[Out]

Piecewise((a**2*x**6/6 + a*b*x**6*atanh(c*x)/3 + a*b*x**5/(15*c) + a*b*x**3/(9*c**3) + a*b*x/(3*c**5) - a*b*at
anh(c*x)/(3*c**6) + b**2*x**6*atanh(c*x)**2/6 + b**2*x**5*atanh(c*x)/(15*c) + b**2*x**4/(60*c**2) + b**2*x**3*
atanh(c*x)/(9*c**3) + 4*b**2*x**2/(45*c**4) + b**2*x*atanh(c*x)/(3*c**5) + 23*b**2*log(x - 1/c)/(45*c**6) - b*
*2*atanh(c*x)**2/(6*c**6) + 23*b**2*atanh(c*x)/(45*c**6), Ne(c, 0)), (a**2*x**6/6, True))

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