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

Optimal. Leaf size=39 \[ a x-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{c^2}+\frac{b \sqrt{x}}{c}+b x \tanh ^{-1}\left (c \sqrt{x}\right ) \]

[Out]

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

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Rubi [A]  time = 0.0204042, antiderivative size = 39, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 12, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {6091, 50, 63, 206} \[ a x-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{c^2}+\frac{b \sqrt{x}}{c}+b x \tanh ^{-1}\left (c \sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

Int[a + b*ArcTanh[c*Sqrt[x]],x]

[Out]

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

Rule 6091

Int[ArcTanh[(c_.)*(x_)^(n_)], x_Symbol] :> Simp[x*ArcTanh[c*x^n], x] - Dist[c*n, Int[x^n/(1 - c^2*x^(2*n)), x]
, x] /; FreeQ[{c, n}, x]

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

Mathematica [A]  time = 0.0244588, size = 42, normalized size = 1.08 \[ a x-b c \left (\frac{\tanh ^{-1}\left (c \sqrt{x}\right )}{c^3}-\frac{\sqrt{x}}{c^2}\right )+b x \tanh ^{-1}\left (c \sqrt{x}\right ) \]

Antiderivative was successfully verified.

[In]

Integrate[a + b*ArcTanh[c*Sqrt[x]],x]

[Out]

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

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Maple [A]  time = 0.024, size = 50, normalized size = 1.3 \begin{align*} ax+bx{\it Artanh} \left ( c\sqrt{x} \right ) +{\frac{b}{c}\sqrt{x}}+{\frac{b}{2\,{c}^{2}}\ln \left ( c\sqrt{x}-1 \right ) }-{\frac{b}{2\,{c}^{2}}\ln \left ( 1+c\sqrt{x} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.971143, size = 72, normalized size = 1.85 \begin{align*} \frac{1}{2} \,{\left (c{\left (\frac{2 \, \sqrt{x}}{c^{2}} - \frac{\log \left (c \sqrt{x} + 1\right )}{c^{3}} + \frac{\log \left (c \sqrt{x} - 1\right )}{c^{3}}\right )} + 2 \, x \operatorname{artanh}\left (c \sqrt{x}\right )\right )} b + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.71946, size = 131, normalized size = 3.36 \begin{align*} \frac{2 \, a c^{2} x + 2 \, b c \sqrt{x} +{\left (b c^{2} x - b\right )} \log \left (-\frac{c^{2} x + 2 \, c \sqrt{x} + 1}{c^{2} x - 1}\right )}{2 \, c^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \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(a+b*atanh(c*x**(1/2)),x)

[Out]

Integral(a + b*atanh(c*sqrt(x)), x)

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Giac [A]  time = 1.22042, size = 90, normalized size = 2.31 \begin{align*} \frac{1}{2} \,{\left (c{\left (\frac{2 \, \sqrt{x}}{c^{2}} - \frac{\log \left ({\left | c \sqrt{x} + 1 \right |}\right )}{c^{3}} + \frac{\log \left ({\left | c \sqrt{x} - 1 \right |}\right )}{c^{3}}\right )} + x \log \left (-\frac{c \sqrt{x} + 1}{c \sqrt{x} - 1}\right )\right )} b + a x \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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