3.1.52 \(\int x^3 (a+b \tanh ^{-1}(c x^2)) \, dx\) [52]

Optimal. Leaf size=43 \[ \frac {b x^2}{4 c}-\frac {b \tanh ^{-1}\left (c x^2\right )}{4 c^2}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \]

[Out]

1/4*b*x^2/c-1/4*b*arctanh(c*x^2)/c^2+1/4*x^4*(a+b*arctanh(c*x^2))

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Rubi [A]
time = 0.02, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6037, 281, 327, 212} \begin {gather*} \frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {b \tanh ^{-1}\left (c x^2\right )}{4 c^2}+\frac {b x^2}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^3*(a + b*ArcTanh[c*x^2]),x]

[Out]

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

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 327

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] - Dist[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 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^3 \left (a+b \tanh ^{-1}\left (c x^2\right )\right ) \, dx &=\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {1}{2} (b c) \int \frac {x^5}{1-c^2 x^4} \, dx\\ &=\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )-\frac {1}{4} (b c) \text {Subst}\left (\int \frac {x^2}{1-c^2 x^2} \, dx,x,x^2\right )\\ &=\frac {b x^2}{4 c}+\frac {1}{4} x^4 \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 )}{4 c}\\ &=\frac {b x^2}{4 c}-\frac {b \tanh ^{-1}\left (c x^2\right )}{4 c^2}+\frac {1}{4} x^4 \left (a+b \tanh ^{-1}\left (c x^2\right )\right )\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 67, normalized size = 1.56 \begin {gather*} \frac {b x^2}{4 c}+\frac {a x^4}{4}+\frac {1}{4} b x^4 \tanh ^{-1}\left (c x^2\right )+\frac {b \log \left (1-c x^2\right )}{8 c^2}-\frac {b \log \left (1+c x^2\right )}{8 c^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^3*(a + b*ArcTanh[c*x^2]),x]

[Out]

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

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Maple [A]
time = 0.04, size = 57, normalized size = 1.33

method	result	size
default	\(\frac {x^{4} a}{4}+\frac {b \,x^{4} \arctanh \left (c \,x^{2}\right )}{4}+\frac {b \,x^{2}}{4 c}+\frac {b \ln \left (c \,x^{2}-1\right )}{8 c^{2}}-\frac {b \ln \left (c \,x^{2}+1\right )}{8 c^{2}}\)	\(57\)
risch	\(\frac {x^{4} b \ln \left (c \,x^{2}+1\right )}{8}-\frac {x^{4} b \ln \left (-c \,x^{2}+1\right )}{8}+\frac {x^{4} a}{4}+\frac {b \,x^{2}}{4 c}+\frac {b \ln \left (c \,x^{2}-1\right )}{8 c^{2}}-\frac {b \ln \left (c \,x^{2}+1\right )}{8 c^{2}}+\frac {b^{2}}{16 a \,c^{2}}\)	\(85\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a+b*arctanh(c*x^2)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 58, normalized size = 1.35 \begin {gather*} \frac {1}{4} \, a x^{4} + \frac {1}{8} \, {\left (2 \, x^{4} \operatorname {artanh}\left (c x^{2}\right ) + c {\left (\frac {2 \, x^{2}}{c^{2}} - \frac {\log \left (c x^{2} + 1\right )}{c^{3}} + \frac {\log \left (c x^{2} - 1\right )}{c^{3}}\right )}\right )} b \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x^2)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 54, normalized size = 1.26 \begin {gather*} \frac {2 \, a c^{2} x^{4} + 2 \, b c x^{2} + {\left (b c^{2} x^{4} - b\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{8 \, c^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x^2)),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 3.61, size = 48, normalized size = 1.12 \begin {gather*} \begin {cases} \frac {a x^{4}}{4} + \frac {b x^{4} \operatorname {atanh}{\left (c x^{2} \right )}}{4} + \frac {b x^{2}}{4 c} - \frac {b \operatorname {atanh}{\left (c x^{2} \right )}}{4 c^{2}} & \text {for}\: c \neq 0 \\\frac {a x^{4}}{4} & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*atanh(c*x**2)),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 181 vs. \(2 (37) = 74\).
time = 0.41, size = 181, normalized size = 4.21 \begin {gather*} \frac {1}{2} \, c {\left (\frac {{\left (c x^{2} + 1\right )} b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{{\left (\frac {{\left (c x^{2} + 1\right )}^{2} c^{3}}{{\left (c x^{2} - 1\right )}^{2}} - \frac {2 \, {\left (c x^{2} + 1\right )} c^{3}}{c x^{2} - 1} + c^{3}\right )} {\left (c x^{2} - 1\right )}} + \frac {\frac {2 \, {\left (c x^{2} + 1\right )} a}{c x^{2} - 1} + \frac {{\left (c x^{2} + 1\right )} b}{c x^{2} - 1} - b}{\frac {{\left (c x^{2} + 1\right )}^{2} c^{3}}{{\left (c x^{2} - 1\right )}^{2}} - \frac {2 \, {\left (c x^{2} + 1\right )} c^{3}}{c x^{2} - 1} + c^{3}}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*arctanh(c*x^2)),x, algorithm="giac")

[Out]

1/2*c*((c*x^2 + 1)*b*log(-(c*x^2 + 1)/(c*x^2 - 1))/(((c*x^2 + 1)^2*c^3/(c*x^2 - 1)^2 - 2*(c*x^2 + 1)*c^3/(c*x^
2 - 1) + c^3)*(c*x^2 - 1)) + (2*(c*x^2 + 1)*a/(c*x^2 - 1) + (c*x^2 + 1)*b/(c*x^2 - 1) - b)/((c*x^2 + 1)^2*c^3/
(c*x^2 - 1)^2 - 2*(c*x^2 + 1)*c^3/(c*x^2 - 1) + c^3))

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Mupad [B]
time = 0.94, size = 60, normalized size = 1.40 \begin {gather*} \frac {a\,x^4}{4}+\frac {b\,x^2}{4\,c}+\frac {b\,x^4\,\ln \left (c\,x^2+1\right )}{8}-\frac {b\,x^4\,\ln \left (1-c\,x^2\right )}{8}+\frac {b\,\mathrm {atan}\left (c\,x^2\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4\,c^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(a + b*atanh(c*x^2)),x)

[Out]

(a*x^4)/4 + (b*x^2)/(4*c) + (b*atan(c*x^2*1i)*1i)/(4*c^2) + (b*x^4*log(c*x^2 + 1))/8 - (b*x^4*log(1 - c*x^2))/
8

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