3.220 \(\int \frac{a+b \tanh ^{-1}(c x^{3/2})}{x^4} \, dx\)

Optimal. Leaf size=47 \[ -\frac{a+b \tanh ^{-1}\left (c x^{3/2}\right )}{3 x^3}+\frac{1}{3} b c^2 \tanh ^{-1}\left (c x^{3/2}\right )-\frac{b c}{3 x^{3/2}} \]

[Out]

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

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Rubi [A]  time = 0.0322121, antiderivative size = 47, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.312, Rules used = {6097, 325, 329, 275, 206} \[ -\frac{a+b \tanh ^{-1}\left (c x^{3/2}\right )}{3 x^3}+\frac{1}{3} b c^2 \tanh ^{-1}\left (c x^{3/2}\right )-\frac{b c}{3 x^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^(3/2)])/x^4,x]

[Out]

-(b*c)/(3*x^(3/2)) + (b*c^2*ArcTanh[c*x^(3/2)])/3 - (a + b*ArcTanh[c*x^(3/2)])/(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 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 275

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
+ 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

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

Mathematica [A]  time = 0.0289697, size = 73, normalized size = 1.55 \[ -\frac{a}{3 x^3}-\frac{1}{6} b c^2 \log \left (1-c x^{3/2}\right )+\frac{1}{6} b c^2 \log \left (c x^{3/2}+1\right )-\frac{b c}{3 x^{3/2}}-\frac{b \tanh ^{-1}\left (c x^{3/2}\right )}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^(3/2)])/x^4,x]

[Out]

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

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Maple [A]  time = 0.036, size = 55, normalized size = 1.2 \begin{align*} -{\frac{a}{3\,{x}^{3}}}-{\frac{b}{3\,{x}^{3}}{\it Artanh} \left ( c{x}^{{\frac{3}{2}}} \right ) }-{\frac{b{c}^{2}}{6}\ln \left ( c{x}^{{\frac{3}{2}}}-1 \right ) }+{\frac{b{c}^{2}}{6}\ln \left ( c{x}^{{\frac{3}{2}}}+1 \right ) }-{\frac{bc}{3}{x}^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^(3/2)))/x^4,x)

[Out]

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

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Maxima [A]  time = 0.984514, size = 69, normalized size = 1.47 \begin{align*} \frac{1}{6} \,{\left ({\left (c \log \left (c x^{\frac{3}{2}} + 1\right ) - c \log \left (c x^{\frac{3}{2}} - 1\right ) - \frac{2}{x^{\frac{3}{2}}}\right )} c - \frac{2 \, \operatorname{artanh}\left (c x^{\frac{3}{2}}\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^(3/2)))/x^4,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.87849, size = 132, normalized size = 2.81 \begin{align*} -\frac{2 \, b c x^{\frac{3}{2}} -{\left (b c^{2} x^{3} - b\right )} \log \left (-\frac{c^{2} x^{3} + 2 \, c x^{\frac{3}{2}} + 1}{c^{2} x^{3} - 1}\right ) + 2 \, a}{6 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(3/2)))/x^4,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.2122, size = 90, normalized size = 1.91 \begin{align*} \frac{1}{6} \, b c^{2} \log \left (c x^{\frac{3}{2}} + 1\right ) - \frac{1}{6} \, b c^{2} \log \left (c x^{\frac{3}{2}} - 1\right ) - \frac{b \log \left (-\frac{c x^{\frac{3}{2}} + 1}{c x^{\frac{3}{2}} - 1}\right )}{6 \, x^{3}} - \frac{b c x^{\frac{3}{2}} + a}{3 \, x^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^(3/2)))/x^4,x, algorithm="giac")

[Out]

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