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

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

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 60, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6097, 51, 63, 206} \[ -\frac {a+b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 x^2}-\frac {b c^3}{2 \sqrt {x}}+\frac {1}{2} b c^4 \tanh ^{-1}\left (c \sqrt {x}\right )-\frac {b c}{6 x^{3/2}} \]

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 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^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*}

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Mathematica [A]  time = 0.03, size = 86, normalized size = 1.43 \[ -\frac {a}{2 x^2}-\frac {1}{4} b c^4 \log \left (1-c \sqrt {x}\right )+\frac {1}{4} b c^4 \log \left (c \sqrt {x}+1\right )-\frac {b c^3}{2 \sqrt {x}}-\frac {b c}{6 x^{3/2}}-\frac {b \tanh ^{-1}\left (c \sqrt {x}\right )}{2 x^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/2*a/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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fricas [A]  time = 0.56, size = 64, normalized size = 1.07 \[ \frac {3 \, {\left (b c^{4} x^{2} - b\right )} \log \left (-\frac {c^{2} x + 2 \, c \sqrt {x} + 1}{c^{2} x - 1}\right ) - 2 \, {\left (3 \, b c^{3} x + b c\right )} \sqrt {x} - 6 \, a}{12 \, x^{2}} \]

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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giac [B]  time = 0.20, size = 356, normalized size = 5.93 \[ \frac {2}{3} \, c {\left (\frac {3 \, {\left (\frac {{\left (c \sqrt {x} + 1\right )}^{3} b c^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {{\left (c \sqrt {x} + 1\right )} b c^{3}}{c \sqrt {x} - 1}\right )} \log \left (-\frac {c \sqrt {x} + 1}{c \sqrt {x} - 1}\right )}{\frac {{\left (c \sqrt {x} + 1\right )}^{4}}{{\left (c \sqrt {x} - 1\right )}^{4}} + \frac {4 \, {\left (c \sqrt {x} + 1\right )}^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )}^{2}}{{\left (c \sqrt {x} - 1\right )}^{2}} + \frac {4 \, {\left (c \sqrt {x} + 1\right )}}{c \sqrt {x} - 1} + 1} + \frac {\frac {6 \, {\left (c \sqrt {x} + 1\right )}^{3} a c^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )} a c^{3}}{c \sqrt {x} - 1} + \frac {3 \, {\left (c \sqrt {x} + 1\right )}^{3} b c^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )}^{2} b c^{3}}{{\left (c \sqrt {x} - 1\right )}^{2}} + \frac {5 \, {\left (c \sqrt {x} + 1\right )} b c^{3}}{c \sqrt {x} - 1} + 2 \, b c^{3}}{\frac {{\left (c \sqrt {x} + 1\right )}^{4}}{{\left (c \sqrt {x} - 1\right )}^{4}} + \frac {4 \, {\left (c \sqrt {x} + 1\right )}^{3}}{{\left (c \sqrt {x} - 1\right )}^{3}} + \frac {6 \, {\left (c \sqrt {x} + 1\right )}^{2}}{{\left (c \sqrt {x} - 1\right )}^{2}} + \frac {4 \, {\left (c \sqrt {x} + 1\right )}}{c \sqrt {x} - 1} + 1}\right )} \]

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]

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

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maple [A]  time = 0.04, size = 64, normalized size = 1.07 \[ -\frac {a}{2 x^{2}}-\frac {b \arctanh \left (c \sqrt {x}\right )}{2 x^{2}}-\frac {b c}{6 x^{\frac {3}{2}}}-\frac {b \,c^{3}}{2 \sqrt {x}}-\frac {c^{4} b \ln \left (c \sqrt {x}-1\right )}{4}+\frac {c^{4} b \ln \left (1+c \sqrt {x}\right )}{4} \]

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

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maxima [A]  time = 0.31, size = 64, normalized size = 1.07 \[ \frac {1}{12} \, {\left ({\left (3 \, c^{3} \log \left (c \sqrt {x} + 1\right ) - 3 \, c^{3} \log \left (c \sqrt {x} - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x + 1\right )}}{x^{\frac {3}{2}}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c \sqrt {x}\right )}{x^{2}}\right )} b - \frac {a}{2 \, x^{2}} \]

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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mupad [B]  time = 1.36, size = 61, normalized size = 1.02 \[ \frac {b\,c^4\,\mathrm {atanh}\left (c\,\sqrt {x}\right )}{2}-\frac {b\,\left (3\,\ln \left (c\,\sqrt {x}+1\right )-3\,\ln \left (1-c\,\sqrt {x}\right )+2\,c\,\sqrt {x}+6\,c^3\,x^{3/2}\right )}{12\,x^2}-\frac {a}{2\,x^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^(1/2)))/x^3,x)

[Out]

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

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sympy [A]  time = 89.35, size = 342, normalized size = 5.70 \[ \begin {cases} - \frac {a}{2 x^{2}} + \frac {b \operatorname {atanh}{\left (\sqrt {x} \sqrt {\frac {1}{x}} \right )}}{2 x^{2}} & \text {for}\: c = - \sqrt {\frac {1}{x}} \\- \frac {a}{2 x^{2}} - \frac {b \operatorname {atanh}{\left (\sqrt {x} \sqrt {\frac {1}{x}} \right )}}{2 x^{2}} & \text {for}\: c = \sqrt {\frac {1}{x}} \\- \frac {3 a c^{2} x^{\frac {3}{2}}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {3 a \sqrt {x}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {3 b c^{6} x^{\frac {7}{2}} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} - \frac {3 b c^{5} x^{3}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} - \frac {3 b c^{4} x^{\frac {5}{2}} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {2 b c^{3} x^{2}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} - \frac {3 b c^{2} x^{\frac {3}{2}} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {b c x}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} + \frac {3 b \sqrt {x} \operatorname {atanh}{\left (c \sqrt {x} \right )}}{6 c^{2} x^{\frac {7}{2}} - 6 x^{\frac {5}{2}}} & \text {otherwise} \end {cases} \]

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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