3.2.92 \(\int (d+e x^2) \tanh ^{-1}(a x) \log (c x^n) \, dx\) [192]

Optimal. Leaf size=180 \[ -\frac {5 e n x^2}{36 a}-d n x \tanh ^{-1}(a x)-\frac {1}{9} e n x^3 \tanh ^{-1}(a x)+\frac {e x^2 \log \left (c x^n\right )}{6 a}+d x \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac {1}{3} e x^3 \tanh ^{-1}(a x) \log \left (c x^n\right )-\frac {d n \log \left (1-a^2 x^2\right )}{2 a}-\frac {e n \log \left (1-a^2 x^2\right )}{18 a^3}+\frac {\left (3 a^2 d+e\right ) \log \left (c x^n\right ) \log \left (1-a^2 x^2\right )}{6 a^3}+\frac {\left (3 a^2 d+e\right ) n \text {Li}_2\left (a^2 x^2\right )}{12 a^3} \]

[Out]

-5/36*e*n*x^2/a-d*n*x*arctanh(a*x)-1/9*e*n*x^3*arctanh(a*x)+1/6*e*x^2*ln(c*x^n)/a+d*x*arctanh(a*x)*ln(c*x^n)+1
/3*e*x^3*arctanh(a*x)*ln(c*x^n)-1/2*d*n*ln(-a^2*x^2+1)/a-1/18*e*n*ln(-a^2*x^2+1)/a^3+1/6*(3*a^2*d+e)*ln(c*x^n)
*ln(-a^2*x^2+1)/a^3+1/12*(3*a^2*d+e)*n*polylog(2,a^2*x^2)/a^3

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 180, normalized size of antiderivative = 1.00, 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 = {6123, 1607, 455, 45, 2435, 6021, 266, 6037, 272, 2438} \begin {gather*} \frac {n \left (3 a^2 d+e\right ) \text {PolyLog}\left (2,a^2 x^2\right )}{12 a^3}-\frac {d n \log \left (1-a^2 x^2\right )}{2 a}+\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 {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 {1}{3} e x^3 \tanh ^{-1}(a x) \log \left (c x^n\right )+\frac {e x^2 \log \left (c x^n\right )}{6 a}-d n x \tanh ^{-1}(a x)-\frac {1}{9} e n x^3 \tanh ^{-1}(a x)-\frac {5 e n x^2}{36 a} \end {gather*}

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 45

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 266

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 272

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 455

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 1607

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 2435

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 2438

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]

Rule 6021

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

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 6123

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])

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) \text {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) \text {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 = 0.11, size = 167, normalized size = 0.93 \begin {gather*} \frac {-5 a^2 e n x^2+6 a^2 e x^2 \log \left (c x^n\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 n \log \left (1-a^2 x^2\right )+18 a^2 d \log \left (c x^n\right ) \log \left (1-a^2 x^2\right )+6 e \log \left (c x^n\right ) \log \left (1-a^2 x^2\right )-2 e n \log \left (-1+a^2 x^2\right )+3 \left (3 a^2 d+e\right ) n \text {Li}_2\left (a^2 x^2\right )}{36 a^3} \end {gather*}

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)

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 13.38, size = 90894, normalized size = 504.97

method	result	size
risch	\(\text {Expression too large to display}\)	\(1939\)
default	\(\text {Expression too large to display}\)	\(90894\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

________________________________________________________________________________________

Maxima [C] Result contains complex when optimal does not.
time = 0.32, size = 364, normalized size = 2.02 \begin {gather*} -\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} x^{3} e - 18 i \, \pi a^{3} d x + 5 \, a^{2} x^{2} e + 2 \, {\left (a^{3} x^{3} e + 9 \, a^{3} d x\right )} \log \left (a x + 1\right ) - 2 \, {\left (a^{3} x^{3} e + 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 {gather*}

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*x^3*e
- 18*I*pi*a^3*d*x + 5*a^2*x^2*e + 2*(a^3*x^3*e + 9*a^3*d*x)*log(a*x + 1) - 2*(a^3*x^3*e + 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)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

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

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]

Integral((d + e*x**2)*log(c*x**n)*atanh(a*x), x)

________________________________________________________________________________________

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

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \ln \left (c\,x^n\right )\,\mathrm {atanh}\left (a\,x\right )\,\left (e\,x^2+d\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(log(c*x^n)*atanh(a*x)*(d + e*x^2),x)

[Out]

int(log(c*x^n)*atanh(a*x)*(d + e*x^2), x)

________________________________________________________________________________________