3.6.31 \(\int \frac {(a+b \tanh ^{-1}(c x)) (d+e \log (1-c^2 x^2))}{x^5} \, dx\) [531]

Optimal. Leaf size=244 \[ \frac {a c^2 e}{4 x^2}+\frac {5 b c^3 e}{12 x}-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)+\frac {b c^2 e \tanh ^{-1}(c x)}{4 x^2}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{12} (3 a+4 b) c^4 e \log (1-c x)+\frac {1}{12} (3 a-4 b) c^4 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 e \text {PolyLog}(2,-c x)-\frac {1}{4} b c^4 e \text {PolyLog}(2,c x) \]

[Out]

1/4*a*c^2*e/x^2+5/12*b*c^3*e/x-1/4*b*c^4*e*arctanh(c*x)+1/4*b*c^2*e*arctanh(c*x)/x^2-1/2*a*c^4*e*ln(x)+1/12*(3
*a+4*b)*c^4*e*ln(-c*x+1)+1/12*(3*a-4*b)*c^4*e*ln(c*x+1)-1/12*b*c*(d+e*ln(-c^2*x^2+1))/x^3-1/4*b*c^3*(d+e*ln(-c
^2*x^2+1))/x+1/4*b*c^4*arctanh(c*x)*(d+e*ln(-c^2*x^2+1))-1/4*(a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1))/x^4+1/4*b
*c^4*e*polylog(2,-c*x)-1/4*b*c^4*e*polylog(2,c*x)

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Rubi [A]
time = 0.18, antiderivative size = 244, normalized size of antiderivative = 1.00, number of steps used = 10, number of rules used = 7, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.259, Rules used = {6037, 331, 212, 6232, 1816, 6191, 6031} \begin {gather*} \frac {1}{12} c^4 e (3 a+4 b) \log (1-c x)+\frac {1}{12} c^4 e (3 a-4 b) \log (c x+1)-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x^4}-\frac {1}{2} a c^4 e \log (x)+\frac {a c^2 e}{4 x^2}+\frac {1}{4} b c^4 e \text {Li}_2(-c x)-\frac {1}{4} b c^4 e \text {Li}_2(c x)-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)+\frac {5 b c^3 e}{12 x}-\frac {b c \left (e \log \left (1-c^2 x^2\right )+d\right )}{12 x^3}+\frac {b c^2 e \tanh ^{-1}(c x)}{4 x^2}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (e \log \left (1-c^2 x^2\right )+d\right )-\frac {b c^3 \left (e \log \left (1-c^2 x^2\right )+d\right )}{4 x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]))/x^5,x]

[Out]

(a*c^2*e)/(4*x^2) + (5*b*c^3*e)/(12*x) - (b*c^4*e*ArcTanh[c*x])/4 + (b*c^2*e*ArcTanh[c*x])/(4*x^2) - (a*c^4*e*
Log[x])/2 + ((3*a + 4*b)*c^4*e*Log[1 - c*x])/12 + ((3*a - 4*b)*c^4*e*Log[1 + c*x])/12 - (b*c*(d + e*Log[1 - c^
2*x^2]))/(12*x^3) - (b*c^3*(d + e*Log[1 - c^2*x^2]))/(4*x) + (b*c^4*ArcTanh[c*x]*(d + e*Log[1 - c^2*x^2]))/4 -
((a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]))/(4*x^4) + (b*c^4*e*PolyLog[2, -(c*x)])/4 - (b*c^4*e*PolyLog[2
, c*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 1816

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*Pq*(a + b*x
^2)^p, x], x] /; FreeQ[{a, b, c, m}, x] && PolyQ[Pq, x] && IGtQ[p, -2]

Rule 6031

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (-Simp[(b/2)*PolyLog[2, (-c)*x]
, x] + Simp[(b/2)*PolyLog[2, c*x], x]) /; FreeQ[{a, b, c}, 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
]

Rule 6191

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[(a + b*ArcTanh[c*x])^p, (f*x)^m*(d + e*x^2)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a,
b, c, d, e, f, m}, x] && IntegerQ[q] && IGtQ[p, 0] && (GtQ[q, 0] || IntegerQ[m])

Rule 6232

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + Log[(f_.) + (g_.)*(x_)^2]*(e_.))*(x_)^(m_.), x_Symbol] :> Wit
h[{u = IntHide[x^m*(a + b*ArcTanh[c*x]), x]}, Dist[d + e*Log[f + g*x^2], u, x] - Dist[2*e*g, Int[ExpandIntegra
nd[x*(u/(f + g*x^2)), x], x], x]] /; FreeQ[{a, b, c, d, e, f, g}, x] && IntegerQ[m] && NeQ[m, -1]

Rubi steps

\begin {align*} \int \frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{x^5} \, dx &=-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\left (2 c^2 e\right ) \int \left (\frac {3 a+b c x+3 b c^3 x^3}{12 x^3 \left (-1+c^2 x^2\right )}-\frac {b \left (1+c^2 x^2\right ) \tanh ^{-1}(c x)}{4 x^3}\right ) \, dx\\ &=-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{6} \left (c^2 e\right ) \int \frac {3 a+b c x+3 b c^3 x^3}{x^3 \left (-1+c^2 x^2\right )} \, dx-\frac {1}{2} \left (b c^2 e\right ) \int \frac {\left (1+c^2 x^2\right ) \tanh ^{-1}(c x)}{x^3} \, dx\\ &=-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{6} \left (c^2 e\right ) \int \left (-\frac {3 a}{x^3}-\frac {b c}{x^2}-\frac {3 a c^2}{x}+\frac {(3 a+4 b) c^3}{2 (-1+c x)}+\frac {(3 a-4 b) c^3}{2 (1+c x)}\right ) \, dx-\frac {1}{2} \left (b c^2 e\right ) \int \left (\frac {\tanh ^{-1}(c x)}{x^3}+\frac {c^2 \tanh ^{-1}(c x)}{x}\right ) \, dx\\ &=\frac {a c^2 e}{4 x^2}+\frac {b c^3 e}{6 x}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{12} (3 a+4 b) c^4 e \log (1-c x)+\frac {1}{12} (3 a-4 b) c^4 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}-\frac {1}{2} \left (b c^2 e\right ) \int \frac {\tanh ^{-1}(c x)}{x^3} \, dx-\frac {1}{2} \left (b c^4 e\right ) \int \frac {\tanh ^{-1}(c x)}{x} \, dx\\ &=\frac {a c^2 e}{4 x^2}+\frac {b c^3 e}{6 x}+\frac {b c^2 e \tanh ^{-1}(c x)}{4 x^2}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{12} (3 a+4 b) c^4 e \log (1-c x)+\frac {1}{12} (3 a-4 b) c^4 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 e \text {Li}_2(-c x)-\frac {1}{4} b c^4 e \text {Li}_2(c x)-\frac {1}{4} \left (b c^3 e\right ) \int \frac {1}{x^2 \left (1-c^2 x^2\right )} \, dx\\ &=\frac {a c^2 e}{4 x^2}+\frac {5 b c^3 e}{12 x}+\frac {b c^2 e \tanh ^{-1}(c x)}{4 x^2}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{12} (3 a+4 b) c^4 e \log (1-c x)+\frac {1}{12} (3 a-4 b) c^4 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 e \text {Li}_2(-c x)-\frac {1}{4} b c^4 e \text {Li}_2(c x)-\frac {1}{4} \left (b c^5 e\right ) \int \frac {1}{1-c^2 x^2} \, dx\\ &=\frac {a c^2 e}{4 x^2}+\frac {5 b c^3 e}{12 x}-\frac {1}{4} b c^4 e \tanh ^{-1}(c x)+\frac {b c^2 e \tanh ^{-1}(c x)}{4 x^2}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{12} (3 a+4 b) c^4 e \log (1-c x)+\frac {1}{12} (3 a-4 b) c^4 e \log (1+c x)-\frac {b c \left (d+e \log \left (1-c^2 x^2\right )\right )}{12 x^3}-\frac {b c^3 \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x}+\frac {1}{4} b c^4 \tanh ^{-1}(c x) \left (d+e \log \left (1-c^2 x^2\right )\right )-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (d+e \log \left (1-c^2 x^2\right )\right )}{4 x^4}+\frac {1}{4} b c^4 e \text {Li}_2(-c x)-\frac {1}{4} b c^4 e \text {Li}_2(c x)\\ \end {align*}

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Mathematica [A]
time = 0.09, size = 299, normalized size = 1.23 \begin {gather*} -\frac {a d}{4 x^4}+\frac {a c^2 e}{4 x^2}+\frac {b c^3 e}{6 x}-\frac {1}{2} a c^4 e \log (x)+\frac {1}{12} \left (3 a c^4 e+4 b c^4 e\right ) \log (1-c x)-\frac {1}{2} b c^4 e \left (-\frac {\tanh ^{-1}(c x)}{2 c^2 x^2}+\frac {1}{2} \left (-\frac {1}{c x}-\frac {1}{2} \log (1-c x)+\frac {1}{2} \log (1+c x)\right )\right )+b c^4 d \left (-\frac {\tanh ^{-1}(c x)}{4 c^4 x^4}+\frac {1}{4} \left (-\frac {1}{3 c^3 x^3}-\frac {1}{c x}-\frac {1}{2} \log (1-c x)+\frac {1}{2} \log (1+c x)\right )\right )+\frac {1}{12} \left (3 a c^4 e-4 b c^4 e\right ) \log (1+c x)+\frac {e \left (-3 a-b c x-3 b c^3 x^3-3 b \tanh ^{-1}(c x)+3 b c^4 x^4 \tanh ^{-1}(c x)\right ) \log \left (1-c^2 x^2\right )}{12 x^4}-\frac {1}{4} b c^4 e (-\text {PolyLog}(2,-c x)+\text {PolyLog}(2,c x)) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*ArcTanh[c*x])*(d + e*Log[1 - c^2*x^2]))/x^5,x]

[Out]

-1/4*(a*d)/x^4 + (a*c^2*e)/(4*x^2) + (b*c^3*e)/(6*x) - (a*c^4*e*Log[x])/2 + ((3*a*c^4*e + 4*b*c^4*e)*Log[1 - c
*x])/12 - (b*c^4*e*(-1/2*ArcTanh[c*x]/(c^2*x^2) + (-(1/(c*x)) - Log[1 - c*x]/2 + Log[1 + c*x]/2)/2))/2 + b*c^4
*d*(-1/4*ArcTanh[c*x]/(c^4*x^4) + (-1/3*1/(c^3*x^3) - 1/(c*x) - Log[1 - c*x]/2 + Log[1 + c*x]/2)/4) + ((3*a*c^
4*e - 4*b*c^4*e)*Log[1 + c*x])/12 + (e*(-3*a - b*c*x - 3*b*c^3*x^3 - 3*b*ArcTanh[c*x] + 3*b*c^4*x^4*ArcTanh[c*
x])*Log[1 - c^2*x^2])/(12*x^4) - (b*c^4*e*(-PolyLog[2, -(c*x)] + PolyLog[2, c*x]))/4

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Maple [F]
time = 66.38, size = 0, normalized size = 0.00 \[\int \frac {\left (a +b \arctanh \left (c x \right )\right ) \left (d +e \ln \left (-c^{2} x^{2}+1\right )\right )}{x^{5}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1))/x^5,x)

[Out]

int((a+b*arctanh(c*x))*(d+e*ln(-c^2*x^2+1))/x^5,x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*arctanh(c*x))*(d+e*log(-c^2*x^2+1))/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*d + 1/4*((c^
2*log(c^2*x^2 - 1) - c^2*log(x^2) + 1/x^2)*c^2 - log(-c^2*x^2 + 1)/x^4)*a*e + 1/8*b*(log(-c*x + 1)^2/x^4 - 4*i
ntegrate(-1/2*(2*(c*x - 1)*log(c*x + 1)^2 - c*x*log(-c*x + 1))/(c*x^6 - x^5), x))*e - 1/4*a*d/x^4

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Fricas [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))*(d+e*log(-c^2*x^2+1))/x^5,x, algorithm="fricas")

[Out]

integral((b*d*arctanh(c*x) + a*d + (b*arctanh(c*x)*e + a*e)*log(-c^2*x^2 + 1))/x^5, x)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right ) \left (d + e \log {\left (- c^{2} x^{2} + 1 \right )}\right )}{x^{5}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*atanh(c*x))*(d+e*ln(-c**2*x**2+1))/x**5,x)

[Out]

Integral((a + b*atanh(c*x))*(d + e*log(-c**2*x**2 + 1))/x**5, x)

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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))*(d+e*log(-c^2*x^2+1))/x^5,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)*(e*log(-c^2*x^2 + 1) + d)/x^5, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (1-c^2\,x^2\right )\right )}{x^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((a + b*atanh(c*x))*(d + e*log(1 - c^2*x^2)))/x^5,x)

[Out]

int(((a + b*atanh(c*x))*(d + e*log(1 - c^2*x^2)))/x^5, x)

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