3.192 \(\int (d+e x^2) \tanh ^{-1}(a x) \log (c x^n) \, dx\)

Optimal. Leaf size=180 \[ \frac{n \left (3 a^2 d+e\right ) \text{PolyLog}\left (2,a^2 x^2\right )}{12 a^3}+\frac{\left (3 a^2 d+e\right ) \log \left (1-a^2 x^2\right ) \log \left (c x^n\right )}{6 a^3}-\frac{d n \log \left (1-a^2 x^2\right )}{2 a}-\frac{e n \log \left (1-a^2 x^2\right )}{18 a^3}+d x \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac{e x^2 \log \left (c x^n\right )}{6 a}+\frac{1}{3} e x^3 \tanh ^{-1}(a x) \log \left (c x^n\right )-d n x \tanh ^{-1}(a x)-\frac{5 e n x^2}{36 a}-\frac{1}{9} e n x^3 \tanh ^{-1}(a x) \]

[Out]

(-5*e*n*x^2)/(36*a) - d*n*x*ArcTanh[a*x] - (e*n*x^3*ArcTanh[a*x])/9 + (e*x^2*Log[c*x^n])/(6*a) + d*x*ArcTanh[a
*x]*Log[c*x^n] + (e*x^3*ArcTanh[a*x]*Log[c*x^n])/3 - (d*n*Log[1 - a^2*x^2])/(2*a) - (e*n*Log[1 - a^2*x^2])/(18
*a^3) + ((3*a^2*d + e)*Log[c*x^n]*Log[1 - a^2*x^2])/(6*a^3) + ((3*a^2*d + e)*n*PolyLog[2, a^2*x^2])/(12*a^3)

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Rubi [A]  time = 0.162429, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 10, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.556, Rules used = {5976, 1593, 444, 43, 2388, 5910, 260, 5916, 266, 2391} \[ \frac{n \left (3 a^2 d+e\right ) \text{PolyLog}\left (2,a^2 x^2\right )}{12 a^3}+\frac{\left (3 a^2 d+e\right ) \log \left (1-a^2 x^2\right ) \log \left (c x^n\right )}{6 a^3}-\frac{d n \log \left (1-a^2 x^2\right )}{2 a}-\frac{e n \log \left (1-a^2 x^2\right )}{18 a^3}+d x \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac{e x^2 \log \left (c x^n\right )}{6 a}+\frac{1}{3} e x^3 \tanh ^{-1}(a x) \log \left (c x^n\right )-d n x \tanh ^{-1}(a x)-\frac{5 e n x^2}{36 a}-\frac{1}{9} e n x^3 \tanh ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

Int[(d + e*x^2)*ArcTanh[a*x]*Log[c*x^n],x]

[Out]

(-5*e*n*x^2)/(36*a) - d*n*x*ArcTanh[a*x] - (e*n*x^3*ArcTanh[a*x])/9 + (e*x^2*Log[c*x^n])/(6*a) + d*x*ArcTanh[a
*x]*Log[c*x^n] + (e*x^3*ArcTanh[a*x]*Log[c*x^n])/3 - (d*n*Log[1 - a^2*x^2])/(2*a) - (e*n*Log[1 - a^2*x^2])/(18
*a^3) + ((3*a^2*d + e)*Log[c*x^n]*Log[1 - a^2*x^2])/(6*a^3) + ((3*a^2*d + e)*n*PolyLog[2, a^2*x^2])/(12*a^3)

Rule 5976

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^
2)^q, x]}, Dist[a + b*ArcTanh[c*x], u, x] - Dist[b*c, Int[u/(1 - c^2*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x
] && (IntegerQ[q] || ILtQ[q + 1/2, 0])

Rule 1593

Int[(u_.)*((a_.)*(x_)^(p_.) + (b_.)*(x_)^(q_.))^(n_.), x_Symbol] :> Int[u*x^(n*p)*(a + b*x^(q - p))^n, x] /; F
reeQ[{a, b, p, q}, x] && IntegerQ[n] && PosQ[q - p]

Rule 444

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && EqQ[m
- n + 1, 0]

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2388

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*(Px_.)*(F_)[(d_.)*((e_.) + (f_.)*(x_))], x_Symbol] :> With[{u = IntH
ide[Px*F[d*(e + f*x)], x]}, Dist[a + b*Log[c*x^n], u, x] - Dist[b*n, Int[Dist[1/x, u, x], x], x]] /; FreeQ[{a,
b, c, d, e, f, n}, x] && PolynomialQ[Px, x] && MemberQ[{ArcTan, ArcCot, ArcTanh, ArcCoth}, F]

Rule 5910

Int[((a_.) + ArcTanh[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTanh[c*x])^p, x] - Dist[b*c*p, In
t[(x*(a + b*ArcTanh[c*x])^(p - 1))/(1 - c^2*x^2), x], x] /; FreeQ[{a, b, c}, x] && IGtQ[p, 0]

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

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 266

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[1/n, Subst[Int[x^(Simplify[(m + 1)/n] - 1)*(a
+ b*x)^p, x], x, x^n], x] /; FreeQ[{a, b, m, n, p}, x] && IntegerQ[Simplify[(m + 1)/n]]

Rule 2391

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> -Simp[PolyLog[2, -(c*e*x^n)]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rubi steps

\begin{align*} \int \left (d+e x^2\right ) \tanh ^{-1}(a x) \log \left (c x^n\right ) \, dx &=\frac{e x^2 \log \left (c x^n\right )}{6 a}+d x \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac{\left (3 a^2 d+e\right ) \log \left (c x^n\right ) \log \left (1-a^2 x^2\right )}{6 a^3}-n \int \left (\frac{e x}{6 a}+d \tanh ^{-1}(a x)+\frac{1}{3} e x^2 \tanh ^{-1}(a x)+\frac{\left (3 a^2 d+e\right ) \log \left (1-a^2 x^2\right )}{6 a^3 x}\right ) \, dx\\ &=-\frac{e n x^2}{12 a}+\frac{e x^2 \log \left (c x^n\right )}{6 a}+d x \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac{\left (3 a^2 d+e\right ) \log \left (c x^n\right ) \log \left (1-a^2 x^2\right )}{6 a^3}-(d n) \int \tanh ^{-1}(a x) \, dx-\frac{1}{3} (e n) \int x^2 \tanh ^{-1}(a x) \, dx-\frac{\left (\left (3 a^2 d+e\right ) n\right ) \int \frac{\log \left (1-a^2 x^2\right )}{x} \, dx}{6 a^3}\\ &=-\frac{e n x^2}{12 a}-d n x \tanh ^{-1}(a x)-\frac{1}{9} e n x^3 \tanh ^{-1}(a x)+\frac{e x^2 \log \left (c x^n\right )}{6 a}+d x \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac{\left (3 a^2 d+e\right ) \log \left (c x^n\right ) \log \left (1-a^2 x^2\right )}{6 a^3}+\frac{\left (3 a^2 d+e\right ) n \text{Li}_2\left (a^2 x^2\right )}{12 a^3}+(a d n) \int \frac{x}{1-a^2 x^2} \, dx+\frac{1}{9} (a e n) \int \frac{x^3}{1-a^2 x^2} \, dx\\ &=-\frac{e n x^2}{12 a}-d n x \tanh ^{-1}(a x)-\frac{1}{9} e n x^3 \tanh ^{-1}(a x)+\frac{e x^2 \log \left (c x^n\right )}{6 a}+d x \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tanh ^{-1}(a x) \log \left (c x^n\right )-\frac{d n \log \left (1-a^2 x^2\right )}{2 a}+\frac{\left (3 a^2 d+e\right ) \log \left (c x^n\right ) \log \left (1-a^2 x^2\right )}{6 a^3}+\frac{\left (3 a^2 d+e\right ) n \text{Li}_2\left (a^2 x^2\right )}{12 a^3}+\frac{1}{18} (a e n) \operatorname{Subst}\left (\int \frac{x}{1-a^2 x} \, dx,x,x^2\right )\\ &=-\frac{e n x^2}{12 a}-d n x \tanh ^{-1}(a x)-\frac{1}{9} e n x^3 \tanh ^{-1}(a x)+\frac{e x^2 \log \left (c x^n\right )}{6 a}+d x \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tanh ^{-1}(a x) \log \left (c x^n\right )-\frac{d n \log \left (1-a^2 x^2\right )}{2 a}+\frac{\left (3 a^2 d+e\right ) \log \left (c x^n\right ) \log \left (1-a^2 x^2\right )}{6 a^3}+\frac{\left (3 a^2 d+e\right ) n \text{Li}_2\left (a^2 x^2\right )}{12 a^3}+\frac{1}{18} (a e n) \operatorname{Subst}\left (\int \left (-\frac{1}{a^2}-\frac{1}{a^2 \left (-1+a^2 x\right )}\right ) \, dx,x,x^2\right )\\ &=-\frac{5 e n x^2}{36 a}-d n x \tanh ^{-1}(a x)-\frac{1}{9} e n x^3 \tanh ^{-1}(a x)+\frac{e x^2 \log \left (c x^n\right )}{6 a}+d x \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tanh ^{-1}(a x) \log \left (c x^n\right )-\frac{d n \log \left (1-a^2 x^2\right )}{2 a}-\frac{e n \log \left (1-a^2 x^2\right )}{18 a^3}+\frac{\left (3 a^2 d+e\right ) \log \left (c x^n\right ) \log \left (1-a^2 x^2\right )}{6 a^3}+\frac{\left (3 a^2 d+e\right ) n \text{Li}_2\left (a^2 x^2\right )}{12 a^3}\\ \end{align*}

Mathematica [A]  time = 0.136176, size = 167, normalized size = 0.93 \[ \frac{3 n \left (3 a^2 d+e\right ) \text{PolyLog}\left (2,a^2 x^2\right )-4 a^3 x \tanh ^{-1}(a x) \left (n \left (9 d+e x^2\right )-3 \left (3 d+e x^2\right ) \log \left (c x^n\right )\right )+18 a^2 d \log \left (1-a^2 x^2\right ) \log \left (c x^n\right )+6 a^2 e x^2 \log \left (c x^n\right )+6 e \log \left (1-a^2 x^2\right ) \log \left (c x^n\right )-18 a^2 d n \log \left (1-a^2 x^2\right )-5 a^2 e n x^2-2 e n \log \left (a^2 x^2-1\right )}{36 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^2)*ArcTanh[a*x]*Log[c*x^n],x]

[Out]

(-5*a^2*e*n*x^2 + 6*a^2*e*x^2*Log[c*x^n] - 4*a^3*x*ArcTanh[a*x]*(n*(9*d + e*x^2) - 3*(3*d + e*x^2)*Log[c*x^n])
- 18*a^2*d*n*Log[1 - a^2*x^2] + 18*a^2*d*Log[c*x^n]*Log[1 - a^2*x^2] + 6*e*Log[c*x^n]*Log[1 - a^2*x^2] - 2*e*
n*Log[-1 + a^2*x^2] + 3*(3*a^2*d + e)*n*PolyLog[2, a^2*x^2])/(36*a^3)

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Maple [C]  time = 5.25, size = 90875, normalized size = 504.9 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x^2+d)*arctanh(a*x)*ln(c*x^n),x)

[Out]

result too large to display

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Maxima [C]  time = 1.4477, size = 478, normalized size = 2.66 \begin{align*} -\frac{1}{36} \, n{\left (\frac{18 \,{\left (i \, \pi d - 2 \, d\right )} \log \left (x\right )}{a} + \frac{6 \,{\left (3 \, a^{2} d + e\right )}{\left (\log \left (a x - 1\right ) \log \left (a x\right ) +{\rm Li}_2\left (-a x + 1\right )\right )}}{a^{3}} + \frac{6 \,{\left (3 \, a^{2} d + e\right )}{\left (\log \left (a x + 1\right ) \log \left (-a x\right ) +{\rm Li}_2\left (a x + 1\right )\right )}}{a^{3}} + \frac{2 \,{\left (9 \, a^{2} d + e\right )} \log \left (a x + 1\right )}{a^{3}} + \frac{-2 i \, \pi a^{3} e x^{3} - 18 i \, \pi a^{3} d x + 5 \, a^{2} e x^{2} + 2 \,{\left (a^{3} e x^{3} + 9 \, a^{3} d x\right )} \log \left (a x + 1\right ) - 2 \,{\left (a^{3} e x^{3} + 9 \, a^{3} d x - 9 \, a^{2} d - e\right )} \log \left (a x - 1\right )}{a^{3}}\right )} + \frac{1}{36} \,{\left ({\left (6 \, x^{3} \log \left (a x + 1\right ) - a{\left (\frac{2 \, a^{2} x^{3} - 3 \, a x^{2} + 6 \, x}{a^{3}} - \frac{6 \, \log \left (a x + 1\right )}{a^{4}}\right )}\right )} e -{\left (6 \, x^{3} \log \left (-a x + 1\right ) - a{\left (\frac{2 \, a^{2} x^{3} + 3 \, a x^{2} + 6 \, x}{a^{3}} + \frac{6 \, \log \left (a x - 1\right )}{a^{4}}\right )}\right )} e - \frac{18 \,{\left (a x -{\left (a x + 1\right )} \log \left (a x + 1\right ) + 1\right )} d}{a} + \frac{18 \,{\left (a x -{\left (a x - 1\right )} \log \left (-a x + 1\right ) - 1\right )} d}{a}\right )} \log \left (c x^{n}\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*arctanh(a*x)*log(c*x^n),x, algorithm="maxima")

[Out]

-1/36*n*(18*(I*pi*d - 2*d)*log(x)/a + 6*(3*a^2*d + e)*(log(a*x - 1)*log(a*x) + dilog(-a*x + 1))/a^3 + 6*(3*a^2
*d + e)*(log(a*x + 1)*log(-a*x) + dilog(a*x + 1))/a^3 + 2*(9*a^2*d + e)*log(a*x + 1)/a^3 + (-2*I*pi*a^3*e*x^3
- 18*I*pi*a^3*d*x + 5*a^2*e*x^2 + 2*(a^3*e*x^3 + 9*a^3*d*x)*log(a*x + 1) - 2*(a^3*e*x^3 + 9*a^3*d*x - 9*a^2*d
- e)*log(a*x - 1))/a^3) + 1/36*((6*x^3*log(a*x + 1) - a*((2*a^2*x^3 - 3*a*x^2 + 6*x)/a^3 - 6*log(a*x + 1)/a^4)
)*e - (6*x^3*log(-a*x + 1) - a*((2*a^2*x^3 + 3*a*x^2 + 6*x)/a^3 + 6*log(a*x - 1)/a^4))*e - 18*(a*x - (a*x + 1)
*log(a*x + 1) + 1)*d/a + 18*(a*x - (a*x - 1)*log(-a*x + 1) - 1)*d/a)*log(c*x^n)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (e x^{2} + d\right )} \operatorname{artanh}\left (a x\right ) \log \left (c x^{n}\right ), x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*arctanh(a*x)*log(c*x^n),x, algorithm="fricas")

[Out]

integral((e*x^2 + d)*arctanh(a*x)*log(c*x^n), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x**2+d)*atanh(a*x)*ln(c*x**n),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (e x^{2} + d\right )} \operatorname{artanh}\left (a x\right ) \log \left (c x^{n}\right )\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x^2+d)*arctanh(a*x)*log(c*x^n),x, algorithm="giac")

[Out]

integrate((e*x^2 + d)*arctanh(a*x)*log(c*x^n), x)