3.214 \(\int x^2 (a+b \tanh ^{-1}(c x^{3/2})) \, dx\)

Optimal. Leaf size=49 \[ \frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )-\frac {b \tanh ^{-1}\left (c x^{3/2}\right )}{3 c^2}+\frac {b x^{3/2}}{3 c} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 49, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {6097, 321, 329, 275, 206} \[ \frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )-\frac {b \tanh ^{-1}\left (c x^{3/2}\right )}{3 c^2}+\frac {b x^{3/2}}{3 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTanh[c*x^(3/2)]),x]

[Out]

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

Rule 206

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

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 321

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 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

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^2 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right ) \, dx &=\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )-\frac {1}{2} (b c) \int \frac {x^{7/2}}{1-c^2 x^3} \, dx\\ &=\frac {b x^{3/2}}{3 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )-\frac {b \int \frac {\sqrt {x}}{1-c^2 x^3} \, dx}{2 c}\\ &=\frac {b x^{3/2}}{3 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )-\frac {b \operatorname {Subst}\left (\int \frac {x^2}{1-c^2 x^6} \, dx,x,\sqrt {x}\right )}{c}\\ &=\frac {b x^{3/2}}{3 c}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )-\frac {b \operatorname {Subst}\left (\int \frac {1}{1-c^2 x^2} \, dx,x,x^{3/2}\right )}{3 c}\\ &=\frac {b x^{3/2}}{3 c}-\frac {b \tanh ^{-1}\left (c x^{3/2}\right )}{3 c^2}+\frac {1}{3} x^3 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.03, size = 75, normalized size = 1.53 \[ \frac {a x^3}{3}+\frac {b \log \left (1-c x^{3/2}\right )}{6 c^2}-\frac {b \log \left (c x^{3/2}+1\right )}{6 c^2}+\frac {b x^{3/2}}{3 c}+\frac {1}{3} b x^3 \tanh ^{-1}\left (c x^{3/2}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.67, size = 64, normalized size = 1.31 \[ \frac {2 \, a c^{2} x^{3} + 2 \, b c x^{\frac {3}{2}} + {\left (b c^{2} x^{3} - b\right )} \log \left (-\frac {c^{2} x^{3} + 2 \, c x^{\frac {3}{2}} + 1}{c^{2} x^{3} - 1}\right )}{6 \, c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.17, size = 97, normalized size = 1.98 \[ \frac {1}{3} \, a x^{3} + \frac {2}{3} \, b c {\left (\frac {1}{c^{3} {\left (\frac {c x^{\frac {3}{2}} + 1}{c x^{\frac {3}{2}} - 1} - 1\right )}} + \frac {{\left (c x^{\frac {3}{2}} + 1\right )} \log \left (-\frac {c x^{\frac {3}{2}} + 1}{c x^{\frac {3}{2}} - 1}\right )}{{\left (c x^{\frac {3}{2}} - 1\right )} c^{3} {\left (\frac {c x^{\frac {3}{2}} + 1}{c x^{\frac {3}{2}} - 1} - 1\right )}^{2}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 57, normalized size = 1.16 \[ \frac {x^{3} a}{3}+\frac {b \,x^{3} \arctanh \left (c \,x^{\frac {3}{2}}\right )}{3}+\frac {b \,x^{\frac {3}{2}}}{3 c}+\frac {b \ln \left (c \,x^{\frac {3}{2}}-1\right )}{6 c^{2}}-\frac {b \ln \left (c \,x^{\frac {3}{2}}+1\right )}{6 c^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arctanh(c*x^(3/2))),x)

[Out]

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

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maxima [A]  time = 0.32, size = 58, normalized size = 1.18 \[ \frac {1}{3} \, a x^{3} + \frac {1}{6} \, {\left (2 \, x^{3} \operatorname {artanh}\left (c x^{\frac {3}{2}}\right ) + c {\left (\frac {2 \, x^{\frac {3}{2}}}{c^{2}} - \frac {\log \left (c x^{\frac {3}{2}} + 1\right )}{c^{3}} + \frac {\log \left (c x^{\frac {3}{2}} - 1\right )}{c^{3}}\right )}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.76, size = 110, normalized size = 2.24 \[ \frac {a\,x^3}{3}+\frac {b\,x^{3/2}}{3\,c}+\frac {b\,\ln \left (\frac {c\,x^{3/2}-1}{c\,x^{3/2}+1}\right )}{6\,c^2}+\frac {b\,x^3\,\ln \left (c\,x^{3/2}+1\right )}{6}+\frac {b\,x^3\,\ln \left (1-c\,x^{3/2}\right )}{3\,\left (2\,c^2\,x^3-2\right )}-\frac {b\,c^2\,x^6\,\ln \left (1-c\,x^{3/2}\right )}{3\,\left (2\,c^2\,x^3-2\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a + b*atanh(c*x^(3/2))),x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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