3.56 \(\int \frac {a+b \tanh ^{-1}(c x^2)}{x^5} \, dx\)

Optimal. Leaf size=41 \[ -\frac {a+b \tanh ^{-1}\left (c x^2\right )}{4 x^4}+\frac {1}{4} b c^2 \tanh ^{-1}\left (c x^2\right )-\frac {b c}{4 x^2} \]

[Out]

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

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Rubi [A]  time = 0.03, antiderivative size = 41, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6097, 275, 325, 206} \[ -\frac {a+b \tanh ^{-1}\left (c x^2\right )}{4 x^4}+\frac {1}{4} b c^2 \tanh ^{-1}\left (c x^2\right )-\frac {b c}{4 x^2} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x^2])/x^5,x]

[Out]

-(b*c)/(4*x^2) + (b*c^2*ArcTanh[c*x^2])/4 - (a + b*ArcTanh[c*x^2])/(4*x^4)

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

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Mathematica [A]  time = 0.01, size = 65, normalized size = 1.59 \[ -\frac {a}{4 x^4}-\frac {1}{8} b c^2 \log \left (1-c x^2\right )+\frac {1}{8} b c^2 \log \left (c x^2+1\right )-\frac {b c}{4 x^2}-\frac {b \tanh ^{-1}\left (c x^2\right )}{4 x^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x^2])/x^5,x]

[Out]

-1/4*a/x^4 - (b*c)/(4*x^2) - (b*ArcTanh[c*x^2])/(4*x^4) - (b*c^2*Log[1 - c*x^2])/8 + (b*c^2*Log[1 + c*x^2])/8

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fricas [A]  time = 1.10, size = 49, normalized size = 1.20 \[ -\frac {2 \, b c x^{2} - {\left (b c^{2} x^{4} - b\right )} \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right ) + 2 \, a}{8 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/x^5,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.14, size = 67, normalized size = 1.63 \[ \frac {1}{8} \, b c^{2} \log \left (c x^{2} + 1\right ) - \frac {1}{8} \, b c^{2} \log \left (c x^{2} - 1\right ) - \frac {b \log \left (-\frac {c x^{2} + 1}{c x^{2} - 1}\right )}{8 \, x^{4}} - \frac {b c x^{2} + a}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/x^5,x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.03, size = 55, normalized size = 1.34 \[ -\frac {a}{4 x^{4}}-\frac {b \arctanh \left (c \,x^{2}\right )}{4 x^{4}}-\frac {b c}{4 x^{2}}+\frac {b \,c^{2} \ln \left (c \,x^{2}+1\right )}{8}-\frac {b \,c^{2} \ln \left (c \,x^{2}-1\right )}{8} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x^2))/x^5,x)

[Out]

-1/4*a/x^4-1/4*b/x^4*arctanh(c*x^2)-1/4*b*c/x^2+1/8*b*c^2*ln(c*x^2+1)-1/8*b*c^2*ln(c*x^2-1)

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maxima [A]  time = 0.32, size = 51, normalized size = 1.24 \[ \frac {1}{8} \, {\left ({\left (c \log \left (c x^{2} + 1\right ) - c \log \left (c x^{2} - 1\right ) - \frac {2}{x^{2}}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x^{2}\right )}{x^{4}}\right )} b - \frac {a}{4 \, x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x^2))/x^5,x, algorithm="maxima")

[Out]

1/8*((c*log(c*x^2 + 1) - c*log(c*x^2 - 1) - 2/x^2)*c - 2*arctanh(c*x^2)/x^4)*b - 1/4*a/x^4

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mupad [B]  time = 1.00, size = 52, normalized size = 1.27 \[ \frac {b\,c^2\,\mathrm {atanh}\left (c\,x^2\right )}{4}-\frac {\frac {a}{4}+\frac {b\,\ln \left (c\,x^2+1\right )}{8}-\frac {b\,\ln \left (1-c\,x^2\right )}{8}+\frac {b\,c\,x^2}{4}}{x^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x^2))/x^5,x)

[Out]

(b*c^2*atanh(c*x^2))/4 - (a/4 + (b*log(c*x^2 + 1))/8 - (b*log(1 - c*x^2))/8 + (b*c*x^2)/4)/x^4

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sympy [A]  time = 8.07, size = 41, normalized size = 1.00 \[ - \frac {a}{4 x^{4}} + \frac {b c^{2} \operatorname {atanh}{\left (c x^{2} \right )}}{4} - \frac {b c}{4 x^{2}} - \frac {b \operatorname {atanh}{\left (c x^{2} \right )}}{4 x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x**2))/x**5,x)

[Out]

-a/(4*x**4) + b*c**2*atanh(c*x**2)/4 - b*c/(4*x**2) - b*atanh(c*x**2)/(4*x**4)

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