3.1.11 \(\int \frac {a+b \tanh ^{-1}(c x)}{x^5} \, dx\) [11]

Optimal. Leaf size=48 \[ -\frac {b c}{12 x^3}-\frac {b c^3}{4 x}+\frac {1}{4} b c^4 \tanh ^{-1}(c x)-\frac {a+b \tanh ^{-1}(c x)}{4 x^4} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 48, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {6037, 331, 212} \begin {gather*} -\frac {a+b \tanh ^{-1}(c x)}{4 x^4}+\frac {1}{4} b c^4 \tanh ^{-1}(c x)-\frac {b c^3}{4 x}-\frac {b c}{12 x^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*ArcTanh[c*x])/x^5,x]

[Out]

-1/12*(b*c)/x^3 - (b*c^3)/(4*x) + (b*c^4*ArcTanh[c*x])/4 - (a + b*ArcTanh[c*x])/(4*x^4)

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 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 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}(c x)}{x^5} \, dx &=-\frac {a+b \tanh ^{-1}(c x)}{4 x^4}+\frac {1}{4} (b c) \int \frac {1}{x^4 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {b c}{12 x^3}-\frac {a+b \tanh ^{-1}(c x)}{4 x^4}+\frac {1}{4} \left (b c^3\right ) \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx\\ &=-\frac {b c}{12 x^3}-\frac {b c^3}{4 x}-\frac {a+b \tanh ^{-1}(c x)}{4 x^4}+\frac {1}{4} \left (b c^5\right ) \int \frac {1}{1-c^2 x^2} \, dx\\ &=-\frac {b c}{12 x^3}-\frac {b c^3}{4 x}+\frac {1}{4} b c^4 \tanh ^{-1}(c x)-\frac {a+b \tanh ^{-1}(c x)}{4 x^4}\\ \end {align*}

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Mathematica [A]
time = 0.01, size = 70, normalized size = 1.46 \begin {gather*} -\frac {a}{4 x^4}-\frac {b c}{12 x^3}-\frac {b c^3}{4 x}-\frac {b \tanh ^{-1}(c x)}{4 x^4}-\frac {1}{8} b c^4 \log (1-c x)+\frac {1}{8} b c^4 \log (1+c x) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*ArcTanh[c*x])/x^5,x]

[Out]

-1/4*a/x^4 - (b*c)/(12*x^3) - (b*c^3)/(4*x) - (b*ArcTanh[c*x])/(4*x^4) - (b*c^4*Log[1 - c*x])/8 + (b*c^4*Log[1
+ c*x])/8

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Maple [A]
time = 0.02, size = 64, normalized size = 1.33

method	result	size
derivativedivides	\(c^{4} \left (-\frac {a}{4 c^{4} x^{4}}-\frac {b \arctanh \left (c x \right )}{4 c^{4} x^{4}}-\frac {b \ln \left (c x -1\right )}{8}-\frac {b}{12 c^{3} x^{3}}-\frac {b}{4 c x}+\frac {b \ln \left (c x +1\right )}{8}\right )\)	\(64\)
default	\(c^{4} \left (-\frac {a}{4 c^{4} x^{4}}-\frac {b \arctanh \left (c x \right )}{4 c^{4} x^{4}}-\frac {b \ln \left (c x -1\right )}{8}-\frac {b}{12 c^{3} x^{3}}-\frac {b}{4 c x}+\frac {b \ln \left (c x +1\right )}{8}\right )\)	\(64\)
risch	\(-\frac {b \ln \left (c x +1\right )}{8 x^{4}}-\frac {3 x^{4} b \ln \left (-c x +1\right ) c^{4}-3 b \,c^{4} \ln \left (-c x -1\right ) x^{4}+6 b \,c^{3} x^{3}+2 b c x -3 b \ln \left (-c x +1\right )+6 a}{24 x^{4}}\)	\(79\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))/x^5,x,method=_RETURNVERBOSE)

[Out]

c^4*(-1/4*a/c^4/x^4-1/4*b/c^4/x^4*arctanh(c*x)-1/8*b*ln(c*x-1)-1/12*b/c^3/x^3-1/4*b/c/x+1/8*b*ln(c*x+1))

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Maxima [A]
time = 0.28, size = 60, normalized size = 1.25 \begin {gather*} \frac {1}{24} \, {\left ({\left (3 \, c^{3} \log \left (c x + 1\right ) - 3 \, c^{3} \log \left (c x - 1\right ) - \frac {2 \, {\left (3 \, c^{2} x^{2} + 1\right )}}{x^{3}}\right )} c - \frac {6 \, \operatorname {artanh}\left (c x\right )}{x^{4}}\right )} b - \frac {a}{4 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*((3*c^3*log(c*x + 1) - 3*c^3*log(c*x - 1) - 2*(3*c^2*x^2 + 1)/x^3)*c - 6*arctanh(c*x)/x^4)*b - 1/4*a/x^4

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Fricas [A]
time = 0.35, size = 52, normalized size = 1.08 \begin {gather*} -\frac {6 \, b c^{3} x^{3} + 2 \, b c x - 3 \, {\left (b c^{4} x^{4} - b\right )} \log \left (-\frac {c x + 1}{c x - 1}\right ) + 6 \, a}{24 \, x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(6*b*c^3*x^3 + 2*b*c*x - 3*(b*c^4*x^4 - b)*log(-(c*x + 1)/(c*x - 1)) + 6*a)/x^4

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Sympy [A]
time = 0.31, size = 46, normalized size = 0.96 \begin {gather*} - \frac {a}{4 x^{4}} + \frac {b c^{4} \operatorname {atanh}{\left (c x \right )}}{4} - \frac {b c^{3}}{4 x} - \frac {b c}{12 x^{3}} - \frac {b \operatorname {atanh}{\left (c x \right )}}{4 x^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a/(4*x**4) + b*c**4*atanh(c*x)/4 - b*c**3/(4*x) - b*c/(12*x**3) - b*atanh(c*x)/(4*x**4)

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (40) = 80\).
time = 0.42, size = 292, normalized size = 6.08 \begin {gather*} \frac {1}{3} \, c {\left (\frac {3 \, {\left (\frac {{\left (c x + 1\right )}^{3} b c^{3}}{{\left (c x - 1\right )}^{3}} + \frac {{\left (c x + 1\right )} b c^{3}}{c x - 1}\right )} \log \left (-\frac {c x + 1}{c x - 1}\right )}{\frac {{\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )}}{c x - 1} + 1} + \frac {\frac {6 \, {\left (c x + 1\right )}^{3} a c^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )} a c^{3}}{c x - 1} + \frac {3 \, {\left (c x + 1\right )}^{3} b c^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2} b c^{3}}{{\left (c x - 1\right )}^{2}} + \frac {5 \, {\left (c x + 1\right )} b c^{3}}{c x - 1} + 2 \, b c^{3}}{\frac {{\left (c x + 1\right )}^{4}}{{\left (c x - 1\right )}^{4}} + \frac {4 \, {\left (c x + 1\right )}^{3}}{{\left (c x - 1\right )}^{3}} + \frac {6 \, {\left (c x + 1\right )}^{2}}{{\left (c x - 1\right )}^{2}} + \frac {4 \, {\left (c x + 1\right )}}{c x - 1} + 1}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*c*(3*((c*x + 1)^3*b*c^3/(c*x - 1)^3 + (c*x + 1)*b*c^3/(c*x - 1))*log(-(c*x + 1)/(c*x - 1))/((c*x + 1)^4/(c
*x - 1)^4 + 4*(c*x + 1)^3/(c*x - 1)^3 + 6*(c*x + 1)^2/(c*x - 1)^2 + 4*(c*x + 1)/(c*x - 1) + 1) + (6*(c*x + 1)^
3*a*c^3/(c*x - 1)^3 + 6*(c*x + 1)*a*c^3/(c*x - 1) + 3*(c*x + 1)^3*b*c^3/(c*x - 1)^3 + 6*(c*x + 1)^2*b*c^3/(c*x
- 1)^2 + 5*(c*x + 1)*b*c^3/(c*x - 1) + 2*b*c^3)/((c*x + 1)^4/(c*x - 1)^4 + 4*(c*x + 1)^3/(c*x - 1)^3 + 6*(c*x
+ 1)^2/(c*x - 1)^2 + 4*(c*x + 1)/(c*x - 1) + 1))

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Mupad [B]
time = 1.05, size = 59, normalized size = 1.23 \begin {gather*} \frac {b\,\ln \left (1-c\,x\right )}{8\,x^4}-\frac {b\,\ln \left (c\,x+1\right )}{8\,x^4}-\frac {b\,c^3\,x^3+\frac {b\,c\,x}{3}+a}{4\,x^4}-\frac {b\,c^4\,\mathrm {atan}\left (c\,x\,1{}\mathrm {i}\right )\,1{}\mathrm {i}}{4} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*atanh(c*x))/x^5,x)

[Out]

(b*log(1 - c*x))/(8*x^4) - (b*c^4*atan(c*x*1i)*1i)/4 - (b*log(c*x + 1))/(8*x^4) - (a + b*c^3*x^3 + (b*c*x)/3)/
(4*x^4)

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