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

Optimal. Leaf size=62 \[ \frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 c^4}+\frac {b \sqrt {x}}{2 c^3}+\frac {b x^{3/2}}{6 c} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 62, 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 = {6097, 50, 63, 206} \[ \frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )+\frac {b \sqrt {x}}{2 c^3}-\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 c^4}+\frac {b x^{3/2}}{6 c} \]

Antiderivative was successfully verified.

[In]

Int[x*(a + b*ArcTanh[c*Sqrt[x]]),x]

[Out]

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

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])

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 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right ) \, dx &=\frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-\frac {1}{4} (b c) \int \frac {x^{3/2}}{1-c^2 x} \, dx\\ &=\frac {b x^{3/2}}{6 c}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )-\frac {b \int \frac {\sqrt {x}}{1-c^2 x} \, dx}{4 c}\\ &=\frac {b \sqrt {x}}{2 c^3}+\frac {b x^{3/2}}{6 c}+\frac {1}{2} x^2 \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}{4 c^3}\\ &=\frac {b \sqrt {x}}{2 c^3}+\frac {b x^{3/2}}{6 c}+\frac {1}{2} x^2 \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 )}{2 c^3}\\ &=\frac {b \sqrt {x}}{2 c^3}+\frac {b x^{3/2}}{6 c}-\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 c^4}+\frac {1}{2} x^2 \left (a+b \tanh ^{-1}\left (c \sqrt {x}\right )\right )\\ \end {align*}

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Mathematica [A]  time = 0.02, size = 88, normalized size = 1.42 \[ \frac {a x^2}{2}+\frac {b \log \left (1-c \sqrt {x}\right )}{4 c^4}-\frac {b \log \left (c \sqrt {x}+1\right )}{4 c^4}+\frac {b \sqrt {x}}{2 c^3}+\frac {b x^{3/2}}{6 c}+\frac {1}{2} b x^2 \tanh ^{-1}\left (c \sqrt {x}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.61, size = 70, normalized size = 1.13 \[ \frac {6 \, a c^{4} x^{2} + 3 \, {\left (b c^{4} x^{2} - b\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) + 2 \, {\left (b c^{3} x + 3 \, b c\right )} \sqrt {x}}{12 \, c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(6*a*c^4*x^2 + 3*(b*c^4*x^2 - b)*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1)) + 2*(b*c^3*x + 3*b*c)*sqrt(x
))/c^4

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giac [B]  time = 0.21, size = 239, normalized size = 3.85 \[ \frac {1}{2} \, a x^{2} + \frac {2}{3} \, b c {\left (\frac {\frac {3 \, {\left (c \sqrt {x} + 1\right )}^{2}}{{\left (c \sqrt {x} - 1\right )}^{2}} - \frac {3 \, {\left (c \sqrt {x} + 1\right )}}{c \sqrt {x} - 1} + 2}{c^{5} {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} - 1\right )}^{3}} + \frac {3 \, {\left (\frac {{\left (c \sqrt {x} + 1\right )}^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {c \sqrt {x} + 1}{c \sqrt {x} - 1}\right )} \log \left (-\frac {\frac {c {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} + 1\right )}}{\frac {{\left (c \sqrt {x} + 1\right )} c}{c \sqrt {x} - 1} - c} + 1}{\frac {c {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} + 1\right )}}{\frac {{\left (c \sqrt {x} + 1\right )} c}{c \sqrt {x} - 1} - c} - 1}\right )}{c^{5} {\left (\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1} - 1\right )}^{4}}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2 + 2/3*b*c*((3*(c*sqrt(x) + 1)^2/(c*sqrt(x) - 1)^2 - 3*(c*sqrt(x) + 1)/(c*sqrt(x) - 1) + 2)/(c^5*((c*
sqrt(x) + 1)/(c*sqrt(x) - 1) - 1)^3) + 3*((c*sqrt(x) + 1)^3/(c*sqrt(x) - 1)^3 + (c*sqrt(x) + 1)/(c*sqrt(x) - 1
))*log(-(c*((c*sqrt(x) + 1)/(c*sqrt(x) - 1) + 1)/((c*sqrt(x) + 1)*c/(c*sqrt(x) - 1) - c) + 1)/(c*((c*sqrt(x) +
1)/(c*sqrt(x) - 1) + 1)/((c*sqrt(x) + 1)*c/(c*sqrt(x) - 1) - c) - 1))/(c^5*((c*sqrt(x) + 1)/(c*sqrt(x) - 1) -
1)^4))

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maple [A]  time = 0.03, size = 66, normalized size = 1.06 \[ \frac {a \,x^{2}}{2}+\frac {b \,x^{2} \arctanh \left (c \sqrt {x}\right )}{2}+\frac {b \,x^{\frac {3}{2}}}{6 c}+\frac {b \sqrt {x}}{2 c^{3}}+\frac {b \ln \left (c \sqrt {x}-1\right )}{4 c^{4}}-\frac {b \ln \left (1+c \sqrt {x}\right )}{4 c^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2+1/2*b*x^2*arctanh(c*x^(1/2))+1/6*b*x^(3/2)/c+1/2*b*x^(1/2)/c^3+1/4/c^4*b*ln(c*x^(1/2)-1)-1/4/c^4*b*l
n(1+c*x^(1/2))

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maxima [A]  time = 0.31, size = 69, normalized size = 1.11 \[ \frac {1}{2} \, a x^{2} + \frac {1}{12} \, {\left (6 \, x^{2} \operatorname {artanh}\left (c \sqrt {x}\right ) + c {\left (\frac {2 \, {\left (c^{2} x^{\frac {3}{2}} + 3 \, \sqrt {x}\right )}}{c^{4}} - \frac {3 \, \log \left (c \sqrt {x} + 1\right )}{c^{5}} + \frac {3 \, \log \left (c \sqrt {x} - 1\right )}{c^{5}}\right )}\right )} b \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*a*x^2 + 1/12*(6*x^2*arctanh(c*sqrt(x)) + c*(2*(c^2*x^(3/2) + 3*sqrt(x))/c^4 - 3*log(c*sqrt(x) + 1)/c^5 + 3
*log(c*sqrt(x) - 1)/c^5))*b

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mupad [B]  time = 1.17, size = 49, normalized size = 0.79 \[ \frac {\frac {b\,c^3\,x^{3/2}}{6}-\frac {b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{2}+\frac {b\,c\,\sqrt {x}}{2}}{c^4}+\frac {a\,x^2}{2}+\frac {b\,x^2\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a + b*atanh(c*x^(1/2))),x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int x \left (a + b \operatorname {atanh}{\left (c \sqrt {x} \right )}\right )\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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