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

Optimal. Leaf size=40 \[ -\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x}+b c^2 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {b c}{\sqrt {x}} \]

[Out]

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

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Rubi [A]  time = 0.02, antiderivative size = 40, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6097, 51, 63, 206} \[ -\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{x}+b c^2 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {b c}{\sqrt {x}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])/x^2,x]

[Out]

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

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])

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

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Mathematica [A]  time = 0.03, size = 67, normalized size = 1.68 \[ -\frac {a}{x}-\frac {1}{2} b c^2 \log \left (1-c \sqrt {x}\right )+\frac {1}{2} b c^2 \log \left (c \sqrt {x}+1\right )-\frac {b c}{\sqrt {x}}-\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{x} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.80, size = 53, normalized size = 1.32 \[ -\frac {2 \, b c \sqrt {x} - {\left (b c^{2} x - b\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) + 2 \, a}{2 \, x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [B]  time = 0.43, size = 168, normalized size = 4.20 \[ 2 \, {\left (\frac {{\left (c \sqrt {x} + 1\right )} b c \log \left (-\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1}\right )}{{\left (c \sqrt {x} - 1\right )} {\left (\frac {{\left (c \sqrt {x} + 1\right )}^{2}}{{\left (c \sqrt {x} - 1\right )}^{2}} + \frac {2 \, {\left (c \sqrt {x} + 1\right )}}{c \sqrt {x} - 1} + 1\right )}} + \frac {\frac {2 \, {\left (c \sqrt {x} + 1\right )} a c}{c \sqrt {x} - 1} + \frac {{\left (c \sqrt {x} + 1\right )} b c}{c \sqrt {x} - 1} + b c}{\frac {{\left (c \sqrt {x} + 1\right )}^{2}}{{\left (c \sqrt {x} - 1\right )}^{2}} + \frac {2 \, {\left (c \sqrt {x} + 1\right )}}{c \sqrt {x} - 1} + 1}\right )} c \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 55, normalized size = 1.38 \[ -\frac {a}{x}-\frac {b \arctanh \left (c \sqrt {x}\right )}{x}-\frac {b c}{\sqrt {x}}-\frac {c^{2} b \ln \left (c \sqrt {x}-1\right )}{2}+\frac {c^{2} b \ln \left (1+c \sqrt {x}\right )}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.32, size = 51, normalized size = 1.28 \[ \frac {1}{2} \, {\left ({\left (c \log \left (c \sqrt {x} + 1\right ) - c \log \left (c \sqrt {x} - 1\right ) - \frac {2}{\sqrt {x}}\right )} c - \frac {2 \, \operatorname {artanh}\left (c \sqrt {x}\right )}{x}\right )} b - \frac {a}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.12, size = 52, normalized size = 1.30 \[ b\,c\,\mathrm {atan}\left (\frac {c^2\,\sqrt {x}}{\sqrt {-c^2}}\right )\,\sqrt {-c^2}-\frac {a}{x}-\frac {b\,\mathrm {atanh}\left (c\,\sqrt {x}\right )+b\,c\,\sqrt {x}}{x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^(1/2)))/x^2,x)

[Out]

b*c*atan((c^2*x^(1/2))/(-c^2)^(1/2))*(-c^2)^(1/2) - a/x - (b*atanh(c*x^(1/2)) + b*c*x^(1/2))/x

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sympy [A]  time = 21.87, size = 231, normalized size = 5.78 \[ \begin {cases} - \frac {a}{x} + \frac {b \operatorname {atanh}{\left (\sqrt {x} \sqrt {\frac {1}{x}} \right )}}{x} & \text {for}\: c = - \sqrt {\frac {1}{x}} \\- \frac {a}{x} - \frac {b \operatorname {atanh}{\left (\sqrt {x} \sqrt {\frac {1}{x}} \right )}}{x} & \text {for}\: c = \sqrt {\frac {1}{x}} \\- \frac {a c^{2} x^{\frac {3}{2}}}{c^{2} x^{\frac {5}{2}} - x^{\frac {3}{2}}} + \frac {a \sqrt {x}}{c^{2} x^{\frac {5}{2}} - x^{\frac {3}{2}}} + \frac {b c^{4} x^{\frac {5}{2}} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{c^{2} x^{\frac {5}{2}} - x^{\frac {3}{2}}} - \frac {b c^{3} x^{2}}{c^{2} x^{\frac {5}{2}} - x^{\frac {3}{2}}} - \frac {2 b c^{2} x^{\frac {3}{2}} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{c^{2} x^{\frac {5}{2}} - x^{\frac {3}{2}}} + \frac {b c x}{c^{2} x^{\frac {5}{2}} - x^{\frac {3}{2}}} + \frac {b \sqrt {x} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{c^{2} x^{\frac {5}{2}} - x^{\frac {3}{2}}} & \text {otherwise} \end {cases} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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