3.529 \(\int \frac {(a+b \tanh ^{-1}(c x)) (d+e \log (1-c^2 x^2))}{x^3} \, dx\)

Optimal. Leaf size=157 \[ -\frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{2 x^2}+\frac {1}{2} c^2 e (a+b) \log (1-c x)+\frac {1}{2} c^2 e (a-b) \log (c x+1)-a c^2 e \log (x)-\frac {b c \left (e \log \left (1-c^2 x^2\right )+d\right )}{2 x}+\frac {1}{2} b c^2 \tanh ^{-1}(c x) \left (e \log \left (1-c^2 x^2\right )+d\right )+\frac {1}{2} b c^2 e \text {Li}_2(-c x)-\frac {1}{2} b c^2 e \text {Li}_2(c x) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.14, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 6, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {5916, 325, 206, 6085, 801, 5912} \[ \frac {1}{2} b c^2 e \text {PolyLog}(2,-c x)-\frac {1}{2} b c^2 e \text {PolyLog}(2,c x)-\frac {\left (a+b \tanh ^{-1}(c x)\right ) \left (e \log \left (1-c^2 x^2\right )+d\right )}{2 x^2}+\frac {1}{2} c^2 e (a+b) \log (1-c x)+\frac {1}{2} c^2 e (a-b) \log (c x+1)-a c^2 e \log (x)-\frac {b c \left (e \log \left (1-c^2 x^2\right )+d\right )}{2 x}+\frac {1}{2} b c^2 \tanh ^{-1}(c x) \left (e \log \left (1-c^2 x^2\right )+d\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rule 5912

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

Rule 5916

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

Rule 6085

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

________________________________________________________________________________________

Mathematica [A]  time = 0.17, size = 152, normalized size = 0.97 \[ \frac {1}{2} \left (-\frac {e \log \left (1-c^2 x^2\right ) \left (a+\left (b-b c^2 x^2\right ) \tanh ^{-1}(c x)+b c x\right )}{x^2}+c^2 e (a+b) \log (1-c x)+c^2 e (a-b) \log (c x+1)-2 a c^2 e \log (x)-\frac {a d}{x^2}+b c^2 e (\text {Li}_2(-c x)-\text {Li}_2(c x))-\frac {b d \left (c x (c x \log (1-c x)-c x \log (c x+1)+2)+2 \tanh ^{-1}(c x)\right )}{2 x^2}\right ) \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

________________________________________________________________________________________

fricas [F]  time = 0.43, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b d \operatorname {artanh}\left (c x\right ) + a d + {\left (b e \operatorname {artanh}\left (c x\right ) + a e\right )} \log \left (-c^{2} x^{2} + 1\right )}{x^{3}}, x\right ) \]

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^3,x, algorithm="fricas")

[Out]

integral((b*d*arctanh(c*x) + a*d + (b*e*arctanh(c*x) + a*e)*log(-c^2*x^2 + 1))/x^3, x)

________________________________________________________________________________________

giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \operatorname {artanh}\left (c x\right ) + a\right )} {\left (e \log \left (-c^{2} x^{2} + 1\right ) + d\right )}}{x^{3}}\,{d x} \]

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^3,x, algorithm="giac")

[Out]

integrate((b*arctanh(c*x) + a)*(e*log(-c^2*x^2 + 1) + d)/x^3, x)

________________________________________________________________________________________

maple [F]  time = 26.70, size = 0, normalized size = 0.00 \[ \int \frac {\left (a +b \arctanh \left (c x \right )\right ) \left (d +e \ln \left (-c^{2} x^{2}+1\right )\right )}{x^{3}}\, dx \]

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^3,x)

[Out]

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

________________________________________________________________________________________

maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{4} \, {\left ({\left (c \log \left (c x + 1\right ) - c \log \left (c x - 1\right ) - \frac {2}{x}\right )} c - \frac {2 \, \operatorname {artanh}\left (c x\right )}{x^{2}}\right )} b d + \frac {1}{2} \, {\left (c^{2} {\left (\log \left (c^{2} x^{2} - 1\right ) - \log \left (x^{2}\right )\right )} - \frac {\log \left (-c^{2} x^{2} + 1\right )}{x^{2}}\right )} a e + \frac {1}{4} \, b e {\left (\frac {\log \left (-c x + 1\right )^{2}}{x^{2}} - 2 \, \int -\frac {{\left (c x - 1\right )} \log \left (c x + 1\right )^{2} - c x \log \left (-c x + 1\right )}{c x^{4} - x^{3}}\,{d x}\right )} - \frac {a d}{2 \, x^{2}} \]

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^3,x, algorithm="maxima")

[Out]

1/4*((c*log(c*x + 1) - c*log(c*x - 1) - 2/x)*c - 2*arctanh(c*x)/x^2)*b*d + 1/2*(c^2*(log(c^2*x^2 - 1) - log(x^
2)) - log(-c^2*x^2 + 1)/x^2)*a*e + 1/4*b*e*(log(-c*x + 1)^2/x^2 - 2*integrate(-((c*x - 1)*log(c*x + 1)^2 - c*x
*log(-c*x + 1))/(c*x^4 - x^3), x)) - 1/2*a*d/x^2

________________________________________________________________________________________

mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (a+b\,\mathrm {atanh}\left (c\,x\right )\right )\,\left (d+e\,\ln \left (1-c^2\,x^2\right )\right )}{x^3} \,d x \]

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^3,x)

[Out]

int(((a + b*atanh(c*x))*(d + e*log(1 - c^2*x^2)))/x^3, x)

________________________________________________________________________________________

sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (a + b \operatorname {atanh}{\left (c x \right )}\right ) \left (d + e \log {\left (- c^{2} x^{2} + 1 \right )}\right )}{x^{3}}\, dx \]

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**3,x)

[Out]

Integral((a + b*atanh(c*x))*(d + e*log(-c**2*x**2 + 1))/x**3, x)

________________________________________________________________________________________