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

Optimal. Leaf size=75 \[ \frac{1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )+\frac{b x^{3/2}}{9 c^3}+\frac{b \sqrt{x}}{3 c^5}-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 c^6}+\frac{b x^{5/2}}{15 c} \]

[Out]

(b*Sqrt[x])/(3*c^5) + (b*x^(3/2))/(9*c^3) + (b*x^(5/2))/(15*c) - (b*ArcTanh[c*Sqrt[x]])/(3*c^6) + (x^3*(a + b*
ArcTanh[c*Sqrt[x]]))/3

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Rubi [A]  time = 0.0345661, antiderivative size = 75, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6097, 50, 63, 206} \[ \frac{1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )+\frac{b x^{3/2}}{9 c^3}+\frac{b \sqrt{x}}{3 c^5}-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 c^6}+\frac{b x^{5/2}}{15 c} \]

Antiderivative was successfully verified.

[In]

Int[x^2*(a + b*ArcTanh[c*Sqrt[x]]),x]

[Out]

(b*Sqrt[x])/(3*c^5) + (b*x^(3/2))/(9*c^3) + (b*x^(5/2))/(15*c) - (b*ArcTanh[c*Sqrt[x]])/(3*c^6) + (x^3*(a + b*
ArcTanh[c*Sqrt[x]]))/3

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]

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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])

Rubi steps

\begin{align*} \int x^2 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right ) \, dx &=\frac{1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{1}{6} (b c) \int \frac{x^{5/2}}{1-c^2 x} \, dx\\ &=\frac{b x^{5/2}}{15 c}+\frac{1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{b \int \frac{x^{3/2}}{1-c^2 x} \, dx}{6 c}\\ &=\frac{b x^{3/2}}{9 c^3}+\frac{b x^{5/2}}{15 c}+\frac{1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{b \int \frac{\sqrt{x}}{1-c^2 x} \, dx}{6 c^3}\\ &=\frac{b \sqrt{x}}{3 c^5}+\frac{b x^{3/2}}{9 c^3}+\frac{b x^{5/2}}{15 c}+\frac{1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{b \int \frac{1}{\sqrt{x} \left (1-c^2 x\right )} \, dx}{6 c^5}\\ &=\frac{b \sqrt{x}}{3 c^5}+\frac{b x^{3/2}}{9 c^3}+\frac{b x^{5/2}}{15 c}+\frac{1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )-\frac{b \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )}{3 c^5}\\ &=\frac{b \sqrt{x}}{3 c^5}+\frac{b x^{3/2}}{9 c^3}+\frac{b x^{5/2}}{15 c}-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 c^6}+\frac{1}{3} x^3 \left (a+b \tanh ^{-1}\left (c \sqrt{x}\right )\right )\\ \end{align*}

Mathematica [A]  time = 0.0236325, size = 101, normalized size = 1.35 \[ \frac{a x^3}{3}+\frac{b x^{3/2}}{9 c^3}+\frac{b \sqrt{x}}{3 c^5}+\frac{b \log \left (1-c \sqrt{x}\right )}{6 c^6}-\frac{b \log \left (c \sqrt{x}+1\right )}{6 c^6}+\frac{b x^{5/2}}{15 c}+\frac{1}{3} b x^3 \tanh ^{-1}\left (c \sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(b*Sqrt[x])/(3*c^5) + (b*x^(3/2))/(9*c^3) + (b*x^(5/2))/(15*c) + (a*x^3)/3 + (b*x^3*ArcTanh[c*Sqrt[x]])/3 + (b
*Log[1 - c*Sqrt[x]])/(6*c^6) - (b*Log[1 + c*Sqrt[x]])/(6*c^6)

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Maple [A]  time = 0.026, size = 75, normalized size = 1. \begin{align*}{\frac{{x}^{3}a}{3}}+{\frac{b{x}^{3}}{3}{\it Artanh} \left ( c\sqrt{x} \right ) }+{\frac{b}{15\,c}{x}^{{\frac{5}{2}}}}+{\frac{b}{9\,{c}^{3}}{x}^{{\frac{3}{2}}}}+{\frac{b}{3\,{c}^{5}}\sqrt{x}}+{\frac{b}{6\,{c}^{6}}\ln \left ( c\sqrt{x}-1 \right ) }-{\frac{b}{6\,{c}^{6}}\ln \left ( 1+c\sqrt{x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2*(a+b*arctanh(c*x^(1/2))),x)

[Out]

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

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Maxima [A]  time = 0.958278, size = 105, normalized size = 1.4 \begin{align*} \frac{1}{3} \, a x^{3} + \frac{1}{90} \,{\left (30 \, x^{3} \operatorname{artanh}\left (c \sqrt{x}\right ) + c{\left (\frac{2 \,{\left (3 \, c^{4} x^{\frac{5}{2}} + 5 \, c^{2} x^{\frac{3}{2}} + 15 \, \sqrt{x}\right )}}{c^{6}} - \frac{15 \, \log \left (c \sqrt{x} + 1\right )}{c^{7}} + \frac{15 \, \log \left (c \sqrt{x} - 1\right )}{c^{7}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctanh(c*x^(1/2))),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.77937, size = 185, normalized size = 2.47 \begin{align*} \frac{30 \, a c^{6} x^{3} + 15 \,{\left (b c^{6} x^{3} - b\right )} \log \left (-\frac{c^{2} x + 2 \, c \sqrt{x} + 1}{c^{2} x - 1}\right ) + 2 \,{\left (3 \, b c^{5} x^{2} + 5 \, b c^{3} x + 15 \, b c\right )} \sqrt{x}}{90 \, c^{6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctanh(c*x^(1/2))),x, algorithm="fricas")

[Out]

1/90*(30*a*c^6*x^3 + 15*(b*c^6*x^3 - b)*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1)) + 2*(3*b*c^5*x^2 + 5*b*c^3
*x + 15*b*c)*sqrt(x))/c^6

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int x^{2} \left (a + b \operatorname{atanh}{\left (c \sqrt{x} \right )}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2*(a+b*atanh(c*x**(1/2))),x)

[Out]

Integral(x**2*(a + b*atanh(c*sqrt(x))), x)

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Giac [A]  time = 1.17049, size = 131, normalized size = 1.75 \begin{align*} \frac{1}{3} \, a x^{3} + \frac{1}{90} \,{\left (15 \, x^{3} \log \left (-\frac{c \sqrt{x} + 1}{c \sqrt{x} - 1}\right ) - c{\left (\frac{15 \, \log \left ({\left | c \sqrt{x} + 1 \right |}\right )}{c^{7}} - \frac{15 \, \log \left ({\left | c \sqrt{x} - 1 \right |}\right )}{c^{7}} - \frac{2 \,{\left (3 \, c^{8} x^{\frac{5}{2}} + 5 \, c^{6} x^{\frac{3}{2}} + 15 \, c^{4} \sqrt{x}\right )}}{c^{10}}\right )}\right )} b \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2*(a+b*arctanh(c*x^(1/2))),x, algorithm="giac")

[Out]

1/3*a*x^3 + 1/90*(15*x^3*log(-(c*sqrt(x) + 1)/(c*sqrt(x) - 1)) - c*(15*log(abs(c*sqrt(x) + 1))/c^7 - 15*log(ab
s(c*sqrt(x) - 1))/c^7 - 2*(3*c^8*x^(5/2) + 5*c^6*x^(3/2) + 15*c^4*sqrt(x))/c^10))*b