3.193 \(\int \frac{a+b \tanh ^{-1}(c \sqrt{x})}{x^3} \, dx\)

Optimal. Leaf size=60 \[ -\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}-\frac{b c^3}{2 \sqrt{x}}+\frac{1}{2} b c^4 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{b c}{6 x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0263707, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6097, 51, 63, 206} \[ -\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}-\frac{b c^3}{2 \sqrt{x}}+\frac{1}{2} b c^4 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{b c}{6 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])/x^3,x]

[Out]

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

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 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && 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 \frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{x^3} \, dx &=-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}+\frac{1}{4} (b c) \int \frac{1}{x^{5/2} \left (1-c^2 x\right )} \, dx\\ &=-\frac{b c}{6 x^{3/2}}-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}+\frac{1}{4} \left (b c^3\right ) \int \frac{1}{x^{3/2} \left (1-c^2 x\right )} \, dx\\ &=-\frac{b c}{6 x^{3/2}}-\frac{b c^3}{2 \sqrt{x}}-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}+\frac{1}{4} \left (b c^5\right ) \int \frac{1}{\sqrt{x} \left (1-c^2 x\right )} \, dx\\ &=-\frac{b c}{6 x^{3/2}}-\frac{b c^3}{2 \sqrt{x}}-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}+\frac{1}{2} \left (b c^5\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{b c}{6 x^{3/2}}-\frac{b c^3}{2 \sqrt{x}}+\frac{1}{2} b c^4 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2}\\ \end{align*}

Mathematica [A]  time = 0.0258193, size = 86, normalized size = 1.43 \[ -\frac{a}{2 x^2}-\frac{b c^3}{2 \sqrt{x}}-\frac{1}{4} b c^4 \log \left (1-c \sqrt{x}\right )+\frac{1}{4} b c^4 \log \left (c \sqrt{x}+1\right )-\frac{b c}{6 x^{3/2}}-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.033, size = 64, normalized size = 1.1 \begin{align*} -{\frac{a}{2\,{x}^{2}}}-{\frac{b}{2\,{x}^{2}}{\it Artanh} \left ( c\sqrt{x} \right ) }-{\frac{{c}^{4}b}{4}\ln \left ( c\sqrt{x}-1 \right ) }-{\frac{bc}{6}{x}^{-{\frac{3}{2}}}}-{\frac{b{c}^{3}}{2}{\frac{1}{\sqrt{x}}}}+{\frac{{c}^{4}b}{4}\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^3,x)

[Out]

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

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Maxima [A]  time = 0.956247, size = 86, normalized size = 1.43 \begin{align*} \frac{1}{12} \,{\left ({\left (3 \, c^{3} \log \left (c \sqrt{x} + 1\right ) - 3 \, c^{3} \log \left (c \sqrt{x} - 1\right ) - \frac{2 \,{\left (3 \, c^{2} x + 1\right )}}{x^{\frac{3}{2}}}\right )} c - \frac{6 \, \operatorname{artanh}\left (c \sqrt{x}\right )}{x^{2}}\right )} b - \frac{a}{2 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.65234, size = 149, normalized size = 2.48 \begin{align*} \frac{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 (3 \, b c^{3} x + b c\right )} \sqrt{x} - 6 \, a}{12 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^3,x, algorithm="fricas")

[Out]

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

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Sympy [A]  time = 179.21, size = 342, normalized size = 5.7 \begin{align*} \begin{cases} - \frac{a}{2 x^{2}} + \frac{b \operatorname{atanh}{\left (\sqrt{x} \sqrt{\frac{1}{x}} \right )}}{2 x^{2}} & \text{for}\: c = - \sqrt{\frac{1}{x}} \\- \frac{a}{2 x^{2}} - \frac{b \operatorname{atanh}{\left (\sqrt{x} \sqrt{\frac{1}{x}} \right )}}{2 x^{2}} & \text{for}\: c = \sqrt{\frac{1}{x}} \\- \frac{3 a c^{2} x^{\frac{3}{2}}}{6 c^{2} x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} + \frac{3 a \sqrt{x}}{6 c^{2} x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} + \frac{3 b c^{6} x^{\frac{7}{2}} \operatorname{atanh}{\left (c \sqrt{x} \right )}}{6 c^{2} x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} - \frac{3 b c^{5} x^{3}}{6 c^{2} x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} - \frac{3 b c^{4} x^{\frac{5}{2}} \operatorname{atanh}{\left (c \sqrt{x} \right )}}{6 c^{2} x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} + \frac{2 b c^{3} x^{2}}{6 c^{2} x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} - \frac{3 b c^{2} x^{\frac{3}{2}} \operatorname{atanh}{\left (c \sqrt{x} \right )}}{6 c^{2} x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} + \frac{b c x}{6 c^{2} x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} + \frac{3 b \sqrt{x} \operatorname{atanh}{\left (c \sqrt{x} \right )}}{6 c^{2} x^{\frac{7}{2}} - 6 x^{\frac{5}{2}}} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**(1/2)))/x**3,x)

[Out]

Piecewise((-a/(2*x**2) + b*atanh(sqrt(x)*sqrt(1/x))/(2*x**2), Eq(c, -sqrt(1/x))), (-a/(2*x**2) - b*atanh(sqrt(
x)*sqrt(1/x))/(2*x**2), Eq(c, sqrt(1/x))), (-3*a*c**2*x**(3/2)/(6*c**2*x**(7/2) - 6*x**(5/2)) + 3*a*sqrt(x)/(6
*c**2*x**(7/2) - 6*x**(5/2)) + 3*b*c**6*x**(7/2)*atanh(c*sqrt(x))/(6*c**2*x**(7/2) - 6*x**(5/2)) - 3*b*c**5*x*
*3/(6*c**2*x**(7/2) - 6*x**(5/2)) - 3*b*c**4*x**(5/2)*atanh(c*sqrt(x))/(6*c**2*x**(7/2) - 6*x**(5/2)) + 2*b*c*
*3*x**2/(6*c**2*x**(7/2) - 6*x**(5/2)) - 3*b*c**2*x**(3/2)*atanh(c*sqrt(x))/(6*c**2*x**(7/2) - 6*x**(5/2)) + b
*c*x/(6*c**2*x**(7/2) - 6*x**(5/2)) + 3*b*sqrt(x)*atanh(c*sqrt(x))/(6*c**2*x**(7/2) - 6*x**(5/2)), True))

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Giac [A]  time = 1.32737, size = 105, normalized size = 1.75 \begin{align*} \frac{1}{4} \, b c^{4} \log \left (c \sqrt{x} + 1\right ) - \frac{1}{4} \, b c^{4} \log \left (c \sqrt{x} - 1\right ) - \frac{b \log \left (-\frac{c \sqrt{x} + 1}{c \sqrt{x} - 1}\right )}{4 \, x^{2}} - \frac{3 \, b c^{3} x^{\frac{3}{2}} + b c \sqrt{x} + 3 \, a}{6 \, x^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^3,x, algorithm="giac")

[Out]

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