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

Optimal. Leaf size=182 \[ -\frac{n \left (3 a^2 d-e\right ) \text{PolyLog}\left (2,-a^2 x^2\right )}{12 a^3}-\frac{\left (3 a^2 d-e\right ) \log \left (a^2 x^2+1\right ) \log \left (c x^n\right )}{6 a^3}+\frac{d n \log \left (a^2 x^2+1\right )}{2 a}-\frac{e n \log \left (a^2 x^2+1\right )}{18 a^3}+d x \tan ^{-1}(a x) \log \left (c x^n\right )-\frac{e x^2 \log \left (c x^n\right )}{6 a}+\frac{1}{3} e x^3 \tan ^{-1}(a x) \log \left (c x^n\right )-d n x \tan ^{-1}(a x)+\frac{5 e n x^2}{36 a}-\frac{1}{9} e n x^3 \tan ^{-1}(a x) \]

[Out]

(5*e*n*x^2)/(36*a) - d*n*x*ArcTan[a*x] - (e*n*x^3*ArcTan[a*x])/9 - (e*x^2*Log[c*x^n])/(6*a) + d*x*ArcTan[a*x]*
Log[c*x^n] + (e*x^3*ArcTan[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 = 0.164447, antiderivative size = 182, 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 = {4912, 1593, 444, 43, 2388, 4846, 260, 4852, 266, 2391} \[ -\frac{n \left (3 a^2 d-e\right ) \text{PolyLog}\left (2,-a^2 x^2\right )}{12 a^3}-\frac{\left (3 a^2 d-e\right ) \log \left (a^2 x^2+1\right ) \log \left (c x^n\right )}{6 a^3}+\frac{d n \log \left (a^2 x^2+1\right )}{2 a}-\frac{e n \log \left (a^2 x^2+1\right )}{18 a^3}+d x \tan ^{-1}(a x) \log \left (c x^n\right )-\frac{e x^2 \log \left (c x^n\right )}{6 a}+\frac{1}{3} e x^3 \tan ^{-1}(a x) \log \left (c x^n\right )-d n x \tan ^{-1}(a x)+\frac{5 e n x^2}{36 a}-\frac{1}{9} e n x^3 \tan ^{-1}(a x) \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*e*n*x^2)/(36*a) - d*n*x*ArcTan[a*x] - (e*n*x^3*ArcTan[a*x])/9 - (e*x^2*Log[c*x^n])/(6*a) + d*x*ArcTan[a*x]*
Log[c*x^n] + (e*x^3*ArcTan[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 4912

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))*((d_.) + (e_.)*(x_)^2)^(q_.), x_Symbol] :> With[{u = IntHide[(d + e*x^2
)^q, x]}, Dist[a + b*ArcTan[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 4846

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x])^p, x] - Dist[b*c*p, Int[
(x*(a + b*ArcTan[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 4852

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*((d_.)*(x_))^(m_.), x_Symbol] :> Simp[((d*x)^(m + 1)*(a + b*ArcTa
n[c*x])^p)/(d*(m + 1)), x] - Dist[(b*c*p)/(d*(m + 1)), Int[((d*x)^(m + 1)*(a + b*ArcTan[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 ) \tan ^{-1}(a x) \log \left (c x^n\right ) \, dx &=-\frac{e x^2 \log \left (c x^n\right )}{6 a}+d x \tan ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tan ^{-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 \tan ^{-1}(a x)+\frac{1}{3} e x^2 \tan ^{-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 \tan ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tan ^{-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 \tan ^{-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{1}{3} (e n) \int x^2 \tan ^{-1}(a x) \, dx\\ &=\frac{e n x^2}{12 a}-d n x \tan ^{-1}(a x)-\frac{1}{9} e n x^3 \tan ^{-1}(a x)-\frac{e x^2 \log \left (c x^n\right )}{6 a}+d x \tan ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tan ^{-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 \tan ^{-1}(a x)-\frac{1}{9} e n x^3 \tan ^{-1}(a x)-\frac{e x^2 \log \left (c x^n\right )}{6 a}+d x \tan ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tan ^{-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 \tan ^{-1}(a x)-\frac{1}{9} e n x^3 \tan ^{-1}(a x)-\frac{e x^2 \log \left (c x^n\right )}{6 a}+d x \tan ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tan ^{-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 \tan ^{-1}(a x)-\frac{1}{9} e n x^3 \tan ^{-1}(a x)-\frac{e x^2 \log \left (c x^n\right )}{6 a}+d x \tan ^{-1}(a x) \log \left (c x^n\right )+\frac{1}{3} e x^3 \tan ^{-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 = 0.117193, size = 165, normalized size = 0.91 \[ \frac{3 n \left (e-3 a^2 d\right ) \text{PolyLog}\left (2,-a^2 x^2\right )-4 a^3 x \tan ^{-1}(a x) \left (n \left (9 d+e x^2\right )-3 \left (3 d+e x^2\right ) \log \left (c x^n\right )\right )-18 a^2 d \log \left (a^2 x^2+1\right ) \log \left (c x^n\right )-6 a^2 e x^2 \log \left (c x^n\right )+6 e \log \left (a^2 x^2+1\right ) \log \left (c x^n\right )+18 a^2 d n \log \left (a^2 x^2+1\right )+5 a^2 e n x^2-2 e n \log \left (a^2 x^2+1\right )}{36 a^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x^2)*ArcTan[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*ArcTan[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] - 2*e*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] + 3*(-3*a^2*d + e)*n*PolyLog[2, -(a^2*x^2)])/(36*a^3)

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Maple [F]  time = 11.802, size = 0, normalized size = 0. \begin{align*} \int \left ({x}^{2}e+d \right ) \arctan \left ( ax \right ) \ln \left ( c{x}^{n} \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} -\frac{a^{2} e x^{2} \log \left (c\right ) - 6 \, a^{3} \int{\left (e x^{2} + d\right )} \arctan \left (a x\right ) \log \left (x^{n}\right )\,{d x} - 2 \,{\left (a^{3} e x^{3} \log \left (c\right ) + 3 \, a^{3} d x \log \left (c\right )\right )} \arctan \left (a x\right ) +{\left (3 \, a^{2} d \log \left (c\right ) - e \log \left (c\right )\right )} \log \left (a^{2} x^{2} + 1\right )}{6 \, a^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(a^2*e*x^2*log(c) - 3*a^3*integrate(2*(e*x^2 + d)*arctan(a*x)*log(x^n), x) - 2*(a^3*e*x^3*log(c) + 3*a^3*
d*x*log(c))*arctan(a*x) + (3*a^2*d*log(c) - e*log(c))*log(a^2*x^2 + 1))/a^3

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

[Out]

integral((e*x^2 + d)*arctan(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)*atan(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 )} \arctan \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)*arctan(a*x)*log(c*x^n),x, algorithm="giac")

[Out]

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