3.1.50 \(\int x^7 (a+b \tanh ^{-1}(c x^2)) \, dx\) [50]

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

[Out]

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

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Rubi [A]
time = 0.03, antiderivative size = 54, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6037, 281, 308, 212} \begin {gather*} \frac {1}{8} x^8 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {b \tanh ^{-1}\left (c x^2\right )}{8 c^4}+\frac {b x^2}{8 c^3}+\frac {b x^6}{24 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^7*(a + b*ArcTanh[c*x^2]),x]

[Out]

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

Rule 212

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] && (GtQ[a, 0] || LtQ[b, 0])

Rule 281

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 308

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 6037

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] - Dist[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
]

Rubi steps

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

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Mathematica [A]
time = 0.02, size = 78, normalized size = 1.44 \begin {gather*} \frac {b x^2}{8 c^3}+\frac {b x^6}{24 c}+\frac {a x^8}{8}+\frac {1}{8} b x^8 \tanh ^{-1}\left (c x^2\right )+\frac {b \log \left (1-c x^2\right )}{16 c^4}-\frac {b \log \left (1+c x^2\right )}{16 c^4} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(b*x^2)/(8*c^3) + (b*x^6)/(24*c) + (a*x^8)/8 + (b*x^8*ArcTanh[c*x^2])/8 + (b*Log[1 - c*x^2])/(16*c^4) - (b*Log
[1 + c*x^2])/(16*c^4)

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Maple [A]
time = 0.03, size = 66, normalized size = 1.22

method	result	size
default	\(\frac {x^{8} a}{8}+\frac {b \,x^{8} \arctanh \left (c \,x^{2}\right )}{8}+\frac {b \,x^{6}}{24 c}+\frac {b \,x^{2}}{8 c^{3}}+\frac {b \ln \left (c \,x^{2}-1\right )}{16 c^{4}}-\frac {b \ln \left (c \,x^{2}+1\right )}{16 c^{4}}\)	\(66\)
risch	\(\frac {x^{8} b \ln \left (c \,x^{2}+1\right )}{16}-\frac {x^{8} b \ln \left (-c \,x^{2}+1\right )}{16}+\frac {x^{8} a}{8}+\frac {b \,x^{6}}{24 c}+\frac {b \,x^{2}}{8 c^{3}}-\frac {b \ln \left (c \,x^{2}+1\right )}{16 c^{4}}+\frac {b \ln \left (c \,x^{2}-1\right )}{16 c^{4}}\)	\(83\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.27, size = 69, normalized size = 1.28 \begin {gather*} \frac {1}{8} \, a x^{8} + \frac {1}{48} \, {\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 )} b \end {gather*}

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) + 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))*b

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Fricas [A]
time = 0.34, size = 64, normalized size = 1.19 \begin {gather*} \frac {6 \, a c^{4} x^{8} + 2 \, b c^{3} x^{6} + 6 \, b c x^{2} + 3 \, {\left (b c^{4} x^{8} - b\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{48 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(6*a*c^4*x^8 + 2*b*c^3*x^6 + 6*b*c*x^2 + 3*(b*c^4*x^8 - b)*log(-(c*x^2 + 1)/(c*x^2 - 1)))/c^4

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Sympy [A]
time = 7.60, size = 58, normalized size = 1.07 \begin {gather*} \begin {cases} \frac {a x^{8}}{8} + \frac {b x^{8} \operatorname {atanh}{\left (c x^{2} \right )}}{8} + \frac {b x^{6}}{24 c} + \frac {b x^{2}}{8 c^{3}} - \frac {b \operatorname {atanh}{\left (c x^{2} \right )}}{8 c^{4}} & \text {for}\: c \neq 0 \\\frac {a x^{8}}{8} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.43, size = 78, normalized size = 1.44 \begin {gather*} \frac {1}{16} \, b x^{8} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + \frac {1}{8} \, a x^{8} + \frac {b x^{6}}{24 \, c} + \frac {b x^{2}}{8 \, c^{3}} - \frac {b \log \left (c x^{2} + 1\right )}{16 \, c^{4}} + \frac {b \log \left (c x^{2} - 1\right )}{16 \, c^{4}} \end {gather*}

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

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Mupad [B]
time = 1.06, size = 69, normalized size = 1.28 \begin {gather*} \frac {a\,x^8}{8}+\frac {b\,x^2}{8\,c^3}+\frac {b\,x^6}{24\,c}+\frac {b\,x^8\,\ln \left (c\,x^2+1\right )}{16}-\frac {b\,x^8\,\ln \left (1-c\,x^2\right )}{16}+\frac {b\,\mathrm {atan}\left (c\,x^2\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,c^4} \end {gather*}

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

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