3.3.20 \(\int \frac {a+b \tanh ^{-1}(c x^{3/2})}{x^4} \, dx\) [220]

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

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 47, normalized size of antiderivative = 1.00, 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 = {6037, 331, 335, 281, 212} \begin {gather*} -\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}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^(3/2)])/x^4,x]

[Out]

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

Rule 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 281

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 331

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 335

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 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
]

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 ) \text {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 ) \text {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*}

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Mathematica [A]
time = 0.02, size = 73, normalized size = 1.55 \begin {gather*} -\frac {a}{3 x^3}-\frac {b c}{3 x^{3/2}}-\frac {b \tanh ^{-1}\left (c x^{3/2}\right )}{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 (1+c x^{3/2}\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*a/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 = 0.06, size = 55, normalized size = 1.17

method	result	size
derivativedivides	\(-\frac {a}{3 x^{3}}-\frac {b \arctanh \left (c \,x^{\frac {3}{2}}\right )}{3 x^{3}}-\frac {b \,c^{2} \ln \left (c \,x^{\frac {3}{2}}-1\right )}{6}-\frac {b c}{3 x^{\frac {3}{2}}}+\frac {b \,c^{2} \ln \left (c \,x^{\frac {3}{2}}+1\right )}{6}\)	\(55\)
default	\(-\frac {a}{3 x^{3}}-\frac {b \arctanh \left (c \,x^{\frac {3}{2}}\right )}{3 x^{3}}-\frac {b \,c^{2} \ln \left (c \,x^{\frac {3}{2}}-1\right )}{6}-\frac {b c}{3 x^{\frac {3}{2}}}+\frac {b \,c^{2} \ln \left (c \,x^{\frac {3}{2}}+1\right )}{6}\)	\(55\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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

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Maxima [A]
time = 0.27, size = 51, normalized size = 1.09 \begin {gather*} \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 {gather*}

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 = 0.35, size = 59, normalized size = 1.26 \begin {gather*} -\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 {gather*}

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)] 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)))/x**4,x)

[Out]

Timed out

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Giac [A]
time = 0.44, size = 67, normalized size = 1.43 \begin {gather*} \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 {gather*}

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

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Mupad [B]
time = 1.36, size = 114, normalized size = 2.43 \begin {gather*} \frac {b\,c^2\,\ln \left (\frac {c\,x^{3/2}+1}{c\,x^{3/2}-1}\right )}{6}-\frac {a}{3\,x^3}-\frac {b\,c}{3\,x^{3/2}}-\frac {b\,\ln \left (c\,x^{3/2}+1\right )}{6\,x^3}+\frac {b\,x\,\ln \left (1-c\,x^{3/2}\right )}{3\,\left (2\,x^4-2\,c^2\,x^7\right )}-\frac {b\,c^2\,x^4\,\ln \left (1-c\,x^{3/2}\right )}{3\,\left (2\,x^4-2\,c^2\,x^7\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^(3/2)))/x^4,x)

[Out]

(b*c^2*log((c*x^(3/2) + 1)/(c*x^(3/2) - 1)))/6 - a/(3*x^3) - (b*c)/(3*x^(3/2)) - (b*log(c*x^(3/2) + 1))/(6*x^3
) + (b*x*log(1 - c*x^(3/2)))/(3*(2*x^4 - 2*c^2*x^7)) - (b*c^2*x^4*log(1 - c*x^(3/2)))/(3*(2*x^4 - 2*c^2*x^7))

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