3.194 \(\int \frac{a+b \tanh ^{-1}(c \sqrt{x})}{x^4} \, dx\)

Optimal. Leaf size=73 \[ -\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 x^3}-\frac{b c^3}{9 x^{3/2}}-\frac{b c^5}{3 \sqrt{x}}+\frac{1}{3} b c^6 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{b c}{15 x^{5/2}} \]

[Out]

-(b*c)/(15*x^(5/2)) - (b*c^3)/(9*x^(3/2)) - (b*c^5)/(3*Sqrt[x]) + (b*c^6*ArcTanh[c*Sqrt[x]])/3 - (a + b*ArcTan
h[c*Sqrt[x]])/(3*x^3)

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Rubi [A]  time = 0.0335061, antiderivative size = 73, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.25, Rules used = {6097, 51, 63, 206} \[ -\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 x^3}-\frac{b c^3}{9 x^{3/2}}-\frac{b c^5}{3 \sqrt{x}}+\frac{1}{3} b c^6 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{b c}{15 x^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*Sqrt[x]])/x^4,x]

[Out]

-(b*c)/(15*x^(5/2)) - (b*c^3)/(9*x^(3/2)) - (b*c^5)/(3*Sqrt[x]) + (b*c^6*ArcTanh[c*Sqrt[x]])/3 - (a + b*ArcTan
h[c*Sqrt[x]])/(3*x^3)

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^4} \, dx &=-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 x^3}+\frac{1}{6} (b c) \int \frac{1}{x^{7/2} \left (1-c^2 x\right )} \, dx\\ &=-\frac{b c}{15 x^{5/2}}-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 x^3}+\frac{1}{6} \left (b c^3\right ) \int \frac{1}{x^{5/2} \left (1-c^2 x\right )} \, dx\\ &=-\frac{b c}{15 x^{5/2}}-\frac{b c^3}{9 x^{3/2}}-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 x^3}+\frac{1}{6} \left (b c^5\right ) \int \frac{1}{x^{3/2} \left (1-c^2 x\right )} \, dx\\ &=-\frac{b c}{15 x^{5/2}}-\frac{b c^3}{9 x^{3/2}}-\frac{b c^5}{3 \sqrt{x}}-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 x^3}+\frac{1}{6} \left (b c^7\right ) \int \frac{1}{\sqrt{x} \left (1-c^2 x\right )} \, dx\\ &=-\frac{b c}{15 x^{5/2}}-\frac{b c^3}{9 x^{3/2}}-\frac{b c^5}{3 \sqrt{x}}-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 x^3}+\frac{1}{3} \left (b c^7\right ) \operatorname{Subst}\left (\int \frac{1}{1-c^2 x^2} \, dx,x,\sqrt{x}\right )\\ &=-\frac{b c}{15 x^{5/2}}-\frac{b c^3}{9 x^{3/2}}-\frac{b c^5}{3 \sqrt{x}}+\frac{1}{3} b c^6 \tanh ^{-1}\left (c \sqrt{x}\right )-\frac{a+b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 x^3}\\ \end{align*}

Mathematica [A]  time = 0.0302242, size = 99, normalized size = 1.36 \[ -\frac{a}{3 x^3}-\frac{b c^3}{9 x^{3/2}}-\frac{b c^5}{3 \sqrt{x}}-\frac{1}{6} b c^6 \log \left (1-c \sqrt{x}\right )+\frac{1}{6} b c^6 \log \left (c \sqrt{x}+1\right )-\frac{b c}{15 x^{5/2}}-\frac{b \tanh ^{-1}\left (c \sqrt{x}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*Sqrt[x]])/x^4,x]

[Out]

-a/(3*x^3) - (b*c)/(15*x^(5/2)) - (b*c^3)/(9*x^(3/2)) - (b*c^5)/(3*Sqrt[x]) - (b*ArcTanh[c*Sqrt[x]])/(3*x^3) -
(b*c^6*Log[1 - c*Sqrt[x]])/6 + (b*c^6*Log[1 + c*Sqrt[x]])/6

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Maple [A]  time = 0.037, size = 73, normalized size = 1. \begin{align*} -{\frac{a}{3\,{x}^{3}}}-{\frac{b}{3\,{x}^{3}}{\it Artanh} \left ( c\sqrt{x} \right ) }-{\frac{{c}^{6}b}{6}\ln \left ( c\sqrt{x}-1 \right ) }-{\frac{bc}{15}{x}^{-{\frac{5}{2}}}}-{\frac{b{c}^{3}}{9}{x}^{-{\frac{3}{2}}}}-{\frac{b{c}^{5}}{3}{\frac{1}{\sqrt{x}}}}+{\frac{{c}^{6}b}{6}\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^4,x)

[Out]

-1/3*a/x^3-1/3*b/x^3*arctanh(c*x^(1/2))-1/6*c^6*b*ln(c*x^(1/2)-1)-1/15*b*c/x^(5/2)-1/9*b*c^3/x^(3/2)-1/3*b*c^5
/x^(1/2)+1/6*c^6*b*ln(1+c*x^(1/2))

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Maxima [A]  time = 0.957319, size = 97, normalized size = 1.33 \begin{align*} \frac{1}{90} \,{\left ({\left (15 \, c^{5} \log \left (c \sqrt{x} + 1\right ) - 15 \, c^{5} \log \left (c \sqrt{x} - 1\right ) - \frac{2 \,{\left (15 \, c^{4} x^{2} + 5 \, c^{2} x + 3\right )}}{x^{\frac{5}{2}}}\right )} c - \frac{30 \, \operatorname{artanh}\left (c \sqrt{x}\right )}{x^{3}}\right )} b - \frac{a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^4,x, algorithm="maxima")

[Out]

1/90*((15*c^5*log(c*sqrt(x) + 1) - 15*c^5*log(c*sqrt(x) - 1) - 2*(15*c^4*x^2 + 5*c^2*x + 3)/x^(5/2))*c - 30*ar
ctanh(c*sqrt(x))/x^3)*b - 1/3*a/x^3

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Fricas [A]  time = 1.62725, size = 174, normalized size = 2.38 \begin{align*} \frac{15 \,{\left (b c^{6} x^{3} - b\right )} \log \left (-\frac{c^{2} x + 2 \, c \sqrt{x} + 1}{c^{2} x - 1}\right ) - 2 \,{\left (15 \, b c^{5} x^{2} + 5 \, b c^{3} x + 3 \, b c\right )} \sqrt{x} - 30 \, a}{90 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^4,x, algorithm="fricas")

[Out]

1/90*(15*(b*c^6*x^3 - b)*log(-(c^2*x + 2*c*sqrt(x) + 1)/(c^2*x - 1)) - 2*(15*b*c^5*x^2 + 5*b*c^3*x + 3*b*c)*sq
rt(x) - 30*a)/x^3

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**(1/2)))/x**4,x)

[Out]

Timed out

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Giac [A]  time = 1.39889, size = 119, normalized size = 1.63 \begin{align*} \frac{1}{6} \, b c^{6} \log \left (c \sqrt{x} + 1\right ) - \frac{1}{6} \, b c^{6} \log \left (c \sqrt{x} - 1\right ) - \frac{b \log \left (-\frac{c \sqrt{x} + 1}{c \sqrt{x} - 1}\right )}{6 \, x^{3}} - \frac{15 \, b c^{5} x^{\frac{5}{2}} + 5 \, b c^{3} x^{\frac{3}{2}} + 3 \, b c \sqrt{x} + 15 \, a}{45 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(1/2)))/x^4,x, algorithm="giac")

[Out]

1/6*b*c^6*log(c*sqrt(x) + 1) - 1/6*b*c^6*log(c*sqrt(x) - 1) - 1/6*b*log(-(c*sqrt(x) + 1)/(c*sqrt(x) - 1))/x^3
- 1/45*(15*b*c^5*x^(5/2) + 5*b*c^3*x^(3/2) + 3*b*c*sqrt(x) + 15*a)/x^3