3.157 \(\int x^7 (a+b \tanh ^{-1}(\frac {c}{x^2})) \, dx\)

Optimal. Leaf size=54 \[ \frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )-\frac {1}{8} b c^4 \tanh ^{-1}\left (\frac {x^2}{c}\right )+\frac {1}{8} b c^3 x^2+\frac {1}{24} b c x^6 \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.357, Rules used = {6097, 263, 275, 302, 207} \[ \frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{8} b c^3 x^2-\frac {1}{8} b c^4 \tanh ^{-1}\left (\frac {x^2}{c}\right )+\frac {1}{24} b c x^6 \]

Antiderivative was successfully verified.

[In]

Int[x^7*(a + b*ArcTanh[c/x^2]),x]

[Out]

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

Rule 207

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTanh[(Rt[b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && NegQ[a/b] && (LtQ[a, 0] || GtQ[b, 0])

Rule 263

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 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[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]

Rule 302

Int[(x_)^(m_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Int[PolynomialDivide[x^m, a + b*x^n, x], x] /; FreeQ[{a,
b}, x] && IGtQ[m, 0] && IGtQ[n, 0] && GtQ[m, 2*n - 1]

Rule 6097

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
nh[c*x^n]))/(d*(m + 1)), x] - Dist[(b*c*n)/(d*(m + 1)), Int[(x^(n - 1)*(d*x)^(m + 1))/(1 - c^2*x^(2*n)), x], x
] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[m, -1]

Rubi steps

\begin {align*} \int x^7 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right ) \, dx &=\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{4} (b c) \int \frac {x^5}{1-\frac {c^2}{x^4}} \, dx\\ &=\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{4} (b c) \int \frac {x^9}{-c^2+x^4} \, dx\\ &=\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{8} (b c) \operatorname {Subst}\left (\int \frac {x^4}{-c^2+x^2} \, dx,x,x^2\right )\\ &=\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{8} (b c) \operatorname {Subst}\left (\int \left (c^2+x^2+\frac {c^4}{-c^2+x^2}\right ) \, dx,x,x^2\right )\\ &=\frac {1}{8} b c^3 x^2+\frac {1}{24} b c x^6+\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )+\frac {1}{8} \left (b c^5\right ) \operatorname {Subst}\left (\int \frac {1}{-c^2+x^2} \, dx,x,x^2\right )\\ &=\frac {1}{8} b c^3 x^2+\frac {1}{24} b c x^6+\frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (\frac {c}{x^2}\right )\right )-\frac {1}{8} b c^4 \tanh ^{-1}\left (\frac {x^2}{c}\right )\\ \end {align*}

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Mathematica [A]  time = 0.01, size = 73, normalized size = 1.35 \[ \frac {a x^8}{8}+\frac {1}{16} b c^4 \log \left (x^2-c\right )-\frac {1}{16} b c^4 \log \left (c+x^2\right )+\frac {1}{8} b c^3 x^2+\frac {1}{24} b c x^6+\frac {1}{8} b x^8 \tanh ^{-1}\left (\frac {c}{x^2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*c^3*x^2)/8 + (b*c*x^6)/24 + (a*x^8)/8 + (b*x^8*ArcTanh[c/x^2])/8 + (b*c^4*Log[-c + x^2])/16 - (b*c^4*Log[c
+ x^2])/16

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fricas [A]  time = 0.53, size = 53, normalized size = 0.98 \[ \frac {1}{8} \, a x^{8} + \frac {1}{24} \, b c x^{6} + \frac {1}{8} \, b c^{3} x^{2} + \frac {1}{16} \, {\left (b x^{8} - b c^{4}\right )} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.15, size = 71, normalized size = 1.31 \[ \frac {1}{16} \, b x^{8} \log \left (\frac {x^{2} + c}{x^{2} - c}\right ) + \frac {1}{8} \, a x^{8} + \frac {1}{24} \, b c x^{6} + \frac {1}{8} \, b c^{3} x^{2} - \frac {1}{16} \, b c^{4} \log \left (x^{2} + c\right ) + \frac {1}{16} \, b c^{4} \log \left (-x^{2} + c\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.04, size = 64, normalized size = 1.19 \[ \frac {x^{8} a}{8}+\frac {b \,x^{8} \arctanh \left (\frac {c}{x^{2}}\right )}{8}+\frac {b c \,x^{6}}{24}+\frac {b \,c^{3} x^{2}}{8}-\frac {b \,c^{4} \ln \left (1+\frac {c}{x^{2}}\right )}{16}+\frac {b \,c^{4} \ln \left (\frac {c}{x^{2}}-1\right )}{16} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 62, normalized size = 1.15 \[ \frac {1}{8} \, a x^{8} + \frac {1}{48} \, {\left (6 \, x^{8} \operatorname {artanh}\left (\frac {c}{x^{2}}\right ) + {\left (2 \, x^{6} + 6 \, c^{2} x^{2} - 3 \, c^{3} \log \left (x^{2} + c\right ) + 3 \, c^{3} \log \left (x^{2} - c\right )\right )} c\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*a*x^8 + 1/48*(6*x^8*arctanh(c/x^2) + (2*x^6 + 6*c^2*x^2 - 3*c^3*log(x^2 + c) + 3*c^3*log(x^2 - c))*c)*b

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mupad [B]  time = 1.01, size = 66, normalized size = 1.22 \[ \frac {a\,x^8}{8}+\frac {b\,c^3\,x^2}{8}+\frac {b\,x^8\,\ln \left (x^2+c\right )}{16}+\frac {b\,c\,x^6}{24}-\frac {b\,x^8\,\ln \left (x^2-c\right )}{16}+\frac {b\,c^4\,\mathrm {atan}\left (\frac {x^2\,1{}\mathrm {i}}{c}\right )\,1{}\mathrm {i}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7*(a + b*atanh(c/x^2)),x)

[Out]

(a*x^8)/8 + (b*c^3*x^2)/8 + (b*x^8*log(c + x^2))/16 + (b*c^4*atan((x^2*1i)/c)*1i)/8 + (b*c*x^6)/24 - (b*x^8*lo
g(x^2 - c))/16

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sympy [A]  time = 9.89, size = 51, normalized size = 0.94 \[ \frac {a x^{8}}{8} - \frac {b c^{4} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{8} + \frac {b c^{3} x^{2}}{8} + \frac {b c x^{6}}{24} + \frac {b x^{8} \operatorname {atanh}{\left (\frac {c}{x^{2}} \right )}}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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