3.2.88 \(\int (d+e x^2) \tan ^{-1}(a x) \log (c x^n) \, dx\) [188]

Optimal. Leaf size=182 \[ \frac {5 e n x^2}{36 a}-d n x \tan ^{-1}(a x)-\frac {1}{9} e n x^3 \tan ^{-1}(a x)-\frac {e x^2 \log \left (c x^n\right )}{6 a}+d x \tan ^{-1}(a x) \log \left (c x^n\right )+\frac {1}{3} e x^3 \tan ^{-1}(a x) \log \left (c x^n\right )+\frac {d n \log \left (1+a^2 x^2\right )}{2 a}-\frac {e n \log \left (1+a^2 x^2\right )}{18 a^3}-\frac {\left (3 a^2 d-e\right ) \log \left (c x^n\right ) \log \left (1+a^2 x^2\right )}{6 a^3}-\frac {\left (3 a^2 d-e\right ) n \text {Li}_2\left (-a^2 x^2\right )}{12 a^3} \]

[Out]

5/36*e*n*x^2/a-d*n*x*arctan(a*x)-1/9*e*n*x^3*arctan(a*x)-1/6*e*x^2*ln(c*x^n)/a+d*x*arctan(a*x)*ln(c*x^n)+1/3*e
*x^3*arctan(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

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Rubi [A]
time = 0.11, antiderivative size = 182, 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 = {5032, 1607, 455, 45, 2435, 4930, 266, 4946, 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 (a^2 x^2+1\right )}{2 a}-\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 {e n \log \left (a^2 x^2+1\right )}{18 a^3}+d x \text {ArcTan}(a x) \log \left (c x^n\right )+\frac {1}{3} e x^3 \text {ArcTan}(a x) \log \left (c x^n\right )-d n x \text {ArcTan}(a x)-\frac {1}{9} e n x^3 \text {ArcTan}(a x)-\frac {e x^2 \log \left (c x^n\right )}{6 a}+\frac {5 e n x^2}{36 a} \end {gather*}

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 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 4930

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.), x_Symbol] :> Simp[x*(a + b*ArcTan[c*x^n])^p, x] - Dist[b*c
*n*p, Int[x^n*((a + b*ArcTan[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 4946

Int[((a_.) + ArcTan[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)*((a + b*ArcTan[c*x^
n])^p/(m + 1)), x] - Dist[b*c*n*(p/(m + 1)), Int[x^(m + n)*((a + b*ArcTan[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 5032

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])

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) \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 \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) \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 \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*}

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Mathematica [A]
time = 0.09, size = 165, normalized size = 0.91 \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 \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 n \log \left (1+a^2 x^2\right )-2 e 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 )+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)*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 [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 13.61, size = 78943, normalized size = 433.75

method	result	size
risch	\(\text {Expression too large to display}\)	\(2700\)
default	\(\text {Expression too large to display}\)	\(78943\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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

[Out]

-1/6*(a^2*x^2*e*log(c) - 3*a^3*integrate(2*(x^2*e + d)*arctan(a*x)*log(x^n), x) - 2*(a^3*x^3*e*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 = 0.00, 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)*arctan(a*x)*log(c*x^n),x, algorithm="fricas")

[Out]

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

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Sympy [A]
time = 54.39, size = 221, normalized size = 1.21 \begin {gather*} - d n \left (\begin {cases} 0 & \text {for}\: a = 0 \\\begin {cases} x \operatorname {atan}{\left (a x \right )} - \frac {\log {\left (a^{2} x^{2} + 1 \right )}}{2 a} & \text {for}\: a \neq 0 \\0 & \text {otherwise} \end {cases} + \frac {\operatorname {Li}_{2}\left (a^{2} x^{2} e^{i \pi }\right )}{4 a} & \text {otherwise} \end {cases}\right ) + d \left (\begin {cases} 0 & \text {for}\: a = 0 \\x \operatorname {atan}{\left (a x \right )} - \frac {\log {\left (a^{2} x^{2} + 1 \right )}}{2 a} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )} - \frac {e n x^{3} \operatorname {atan}{\left (a x \right )}}{9} + \frac {e x^{3} \log {\left (c x^{n} \right )} \operatorname {atan}{\left (a x \right )}}{3} + \frac {5 e n x^{2}}{36 a} - \frac {e n \left (\begin {cases} \frac {x^{2}}{2} & \text {for}\: a = 0 \\- \frac {\operatorname {Li}_{2}\left (a^{2} x^{2} e^{i \pi }\right )}{2 a^{2}} & \text {otherwise} \end {cases}\right )}{6 a} - \frac {e n \left (\begin {cases} x^{2} & \text {for}\: a^{2} = 0 \\\frac {\log {\left (a^{2} x^{2} + 1 \right )}}{a^{2}} & \text {otherwise} \end {cases}\right )}{18 a} - \frac {e x^{2} \log {\left (c x^{n} \right )}}{6 a} + \frac {e \left (\begin {cases} x^{2} & \text {for}\: a^{2} = 0 \\\frac {\log {\left (a^{2} x^{2} + 1 \right )}}{a^{2}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{6 a} \end {gather*}

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]

-d*n*Piecewise((0, Eq(a, 0)), (Piecewise((x*atan(a*x) - log(a**2*x**2 + 1)/(2*a), Ne(a, 0)), (0, True)) + poly
log(2, a**2*x**2*exp_polar(I*pi))/(4*a), True)) + d*Piecewise((0, Eq(a, 0)), (x*atan(a*x) - log(a**2*x**2 + 1)
/(2*a), True))*log(c*x**n) - e*n*x**3*atan(a*x)/9 + e*x**3*log(c*x**n)*atan(a*x)/3 + 5*e*n*x**2/(36*a) - e*n*P
iecewise((x**2/2, Eq(a, 0)), (-polylog(2, a**2*x**2*exp_polar(I*pi))/(2*a**2), True))/(6*a) - e*n*Piecewise((x
**2, Eq(a**2, 0)), (log(a**2*x**2 + 1)/a**2, True))/(18*a) - e*x**2*log(c*x**n)/(6*a) + e*Piecewise((x**2, Eq(
a**2, 0)), (log(a**2*x**2 + 1)/a**2, True))*log(c*x**n)/(6*a)

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

[Out]

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

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

[Out]

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

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