3.3.23 \(\int \frac {(a+b \tanh ^{-1}(c x^{3/2}))^2}{x^4} \, dx\) [223]

Optimal. Leaf size=96 \[ -\frac {2 b c \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )}{3 x^{3/2}}+\frac {1}{3} c^2 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )^2-\frac {\left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )^2}{3 x^3}+b^2 c^2 \log (x)-\frac {1}{3} b^2 c^2 \log \left (1-c^2 x^3\right ) \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 96, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 8, integrand size = 18, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.444, Rules used = {6039, 6037, 6129, 272, 36, 29, 31, 6095} \begin {gather*} \frac {1}{3} c^2 \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )^2-\frac {2 b c \left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )}{3 x^{3/2}}-\frac {\left (a+b \tanh ^{-1}\left (c x^{3/2}\right )\right )^2}{3 x^3}-\frac {1}{3} b^2 c^2 \log \left (1-c^2 x^3\right )+b^2 c^2 \log (x) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 29

Int[(x_)^(-1), x_Symbol] :> Simp[Log[x], x]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 36

Int[1/(((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))), x_Symbol] :> Dist[b/(b*c - a*d), Int[1/(a + b*x), x], x] -
Dist[d/(b*c - a*d), Int[1/(c + d*x), x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0]

Rule 272

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

Rule 6037

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTanh[c*
x^n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTanh[c*x^n])^(p - 1)/(1 - c^2*x^(2*n))
), x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0] && (EqQ[p, 1] || (EqQ[n, 1] && IntegerQ[m])) && NeQ[m, -1
]

Rule 6039

Int[((a_.) + ArcTanh[(c_.)*(x_)^(n_)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m
+ 1)/n] - 1)*(a + b*ArcTanh[c*x])^p, x], x, x^n], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 1] && IntegerQ[S
implify[(m + 1)/n]]

Rule 6095

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(a + b*ArcTanh[c*x])^(p
+ 1)/(b*c*d*(p + 1)), x] /; FreeQ[{a, b, c, d, e, p}, x] && EqQ[c^2*d + e, 0] && NeQ[p, -1]

Rule 6129

Int[(((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Dist[1/d
, Int[(f*x)^m*(a + b*ArcTanh[c*x])^p, x], x] - Dist[e/(d*f^2), Int[(f*x)^(m + 2)*((a + b*ArcTanh[c*x])^p/(d +
e*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && GtQ[p, 0] && LtQ[m, -1]

Rubi steps

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

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Mathematica [A]
time = 0.09, size = 123, normalized size = 1.28 \begin {gather*} \frac {1}{3} \left (-\frac {a^2}{x^3}-\frac {2 a b c}{x^{3/2}}-\frac {2 b \left (a+b c x^{3/2}\right ) \tanh ^{-1}\left (c x^{3/2}\right )}{x^3}+\frac {b^2 \left (-1+c^2 x^3\right ) \tanh ^{-1}\left (c x^{3/2}\right )^2}{x^3}+3 b^2 c^2 \log (x)-b (a+b) c^2 \log \left (1-c x^{3/2}\right )+(a-b) b c^2 \log \left (1+c x^{3/2}\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.11, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arctanh \left (c \,x^{\frac {3}{2}}\right )\right )^{2}}{x^{4}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 175 vs. \(2 (80) = 160\).
time = 0.28, size = 175, normalized size = 1.82 \begin {gather*} \frac {1}{3} \, {\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 )} a b + \frac {1}{12} \, {\left ({\left (2 \, {\left (\log \left (c x^{\frac {3}{2}} - 1\right ) - 2\right )} \log \left (c x^{\frac {3}{2}} + 1\right ) - \log \left (c x^{\frac {3}{2}} + 1\right )^{2} - \log \left (c x^{\frac {3}{2}} - 1\right )^{2} - 4 \, \log \left (c x^{\frac {3}{2}} - 1\right ) + 12 \, \log \left (x\right )\right )} c^{2} + 4 \, {\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 \operatorname {artanh}\left (c x^{\frac {3}{2}}\right )\right )} b^{2} - \frac {b^{2} \operatorname {artanh}\left (c x^{\frac {3}{2}}\right )^{2}}{3 \, x^{3}} - \frac {a^{2}}{3 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((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)*a*b + 1/12*((2*(l
og(c*x^(3/2) - 1) - 2)*log(c*x^(3/2) + 1) - log(c*x^(3/2) + 1)^2 - log(c*x^(3/2) - 1)^2 - 4*log(c*x^(3/2) - 1)
+ 12*log(x))*c^2 + 4*(c*log(c*x^(3/2) + 1) - c*log(c*x^(3/2) - 1) - 2/x^(3/2))*c*arctanh(c*x^(3/2)))*b^2 - 1/
3*b^2*arctanh(c*x^(3/2))^2/x^3 - 1/3*a^2/x^3

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 173 vs. \(2 (80) = 160\).
time = 0.37, size = 173, normalized size = 1.80 \begin {gather*} \frac {24 \, b^{2} c^{2} x^{3} \log \left (\sqrt {x}\right ) + 4 \, {\left (a b - b^{2}\right )} c^{2} x^{3} \log \left (c x^{\frac {3}{2}} + 1\right ) - 4 \, {\left (a b + b^{2}\right )} c^{2} x^{3} \log \left (c x^{\frac {3}{2}} - 1\right ) - 8 \, a b c x^{\frac {3}{2}} + {\left (b^{2} c^{2} x^{3} - b^{2}\right )} \log \left (-\frac {c^{2} x^{3} + 2 \, c x^{\frac {3}{2}} + 1}{c^{2} x^{3} - 1}\right )^{2} - 4 \, a^{2} - 4 \, {\left (b^{2} c x^{\frac {3}{2}} + a b\right )} \log \left (-\frac {c^{2} x^{3} + 2 \, c x^{\frac {3}{2}} + 1}{c^{2} x^{3} - 1}\right )}{12 \, x^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(24*b^2*c^2*x^3*log(sqrt(x)) + 4*(a*b - b^2)*c^2*x^3*log(c*x^(3/2) + 1) - 4*(a*b + b^2)*c^2*x^3*log(c*x^(
3/2) - 1) - 8*a*b*c*x^(3/2) + (b^2*c^2*x^3 - b^2)*log(-(c^2*x^3 + 2*c*x^(3/2) + 1)/(c^2*x^3 - 1))^2 - 4*a^2 -
4*(b^2*c*x^(3/2) + a*b)*log(-(c^2*x^3 + 2*c*x^(3/2) + 1)/(c^2*x^3 - 1)))/x^3

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**(3/2)))**2/x**4,x)

[Out]

Timed out

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [B]
time = 2.04, size = 281, normalized size = 2.93 \begin {gather*} \frac {2\,b^2\,c^2\,\ln \left (x^{3/2}\right )}{3}-\frac {a^2}{3\,x^3}-\frac {b^2\,c^2\,\ln \left (c\,x^{3/2}-1\right )}{3}-\frac {b^2\,c^2\,\ln \left (c\,x^{3/2}+1\right )}{3}+\frac {b^2\,c^2\,{\ln \left (c\,x^{3/2}+1\right )}^2}{12}+\frac {b^2\,c^2\,{\ln \left (1-c\,x^{3/2}\right )}^2}{12}-\frac {b^2\,{\ln \left (c\,x^{3/2}+1\right )}^2}{12\,x^3}-\frac {b^2\,{\ln \left (1-c\,x^{3/2}\right )}^2}{12\,x^3}-\frac {a\,b\,c^2\,\ln \left (c\,x^{3/2}-1\right )}{3}+\frac {a\,b\,c^2\,\ln \left (c\,x^{3/2}+1\right )}{3}-\frac {2\,a\,b\,c}{3\,x^{3/2}}-\frac {a\,b\,\ln \left (c\,x^{3/2}+1\right )}{3\,x^3}+\frac {a\,b\,\ln \left (1-c\,x^{3/2}\right )}{3\,x^3}-\frac {b^2\,c^2\,\ln \left (c\,x^{3/2}+1\right )\,\ln \left (1-c\,x^{3/2}\right )}{6}-\frac {b^2\,c\,\ln \left (c\,x^{3/2}+1\right )}{3\,x^{3/2}}+\frac {b^2\,c\,\ln \left (1-c\,x^{3/2}\right )}{3\,x^{3/2}}+\frac {b^2\,\ln \left (c\,x^{3/2}+1\right )\,\ln \left (1-c\,x^{3/2}\right )}{6\,x^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^(3/2)))^2/x^4,x)

[Out]

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

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