3.217 \(\int \frac{x^2 (a+b \log (c x^n))}{d+e x^2} \, dx\)

Optimal. Leaf size=132 \[ \frac{i b \sqrt{d} n \text{PolyLog}\left (2,-\frac{i \sqrt{e} x}{\sqrt{d}}\right )}{2 e^{3/2}}-\frac{i b \sqrt{d} n \text{PolyLog}\left (2,\frac{i \sqrt{e} x}{\sqrt{d}}\right )}{2 e^{3/2}}-\frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^{3/2}}+\frac{a x}{e}+\frac{b x \log \left (c x^n\right )}{e}-\frac{b n x}{e} \]

[Out]

(a*x)/e - (b*n*x)/e + (b*x*Log[c*x^n])/e - (Sqrt[d]*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(a + b*Log[c*x^n]))/e^(3/2) +
((I/2)*b*Sqrt[d]*n*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]])/e^(3/2) - ((I/2)*b*Sqrt[d]*n*PolyLog[2, (I*Sqrt[e]*x)
/Sqrt[d]])/e^(3/2)

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Rubi [A]  time = 0.164603, antiderivative size = 132, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 8, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.348, Rules used = {321, 205, 2351, 2295, 2324, 12, 4848, 2391} \[ \frac{i b \sqrt{d} n \text{PolyLog}\left (2,-\frac{i \sqrt{e} x}{\sqrt{d}}\right )}{2 e^{3/2}}-\frac{i b \sqrt{d} n \text{PolyLog}\left (2,\frac{i \sqrt{e} x}{\sqrt{d}}\right )}{2 e^{3/2}}-\frac{\sqrt{d} \tan ^{-1}\left (\frac{\sqrt{e} x}{\sqrt{d}}\right ) \left (a+b \log \left (c x^n\right )\right )}{e^{3/2}}+\frac{a x}{e}+\frac{b x \log \left (c x^n\right )}{e}-\frac{b n x}{e} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*x)/e - (b*n*x)/e + (b*x*Log[c*x^n])/e - (Sqrt[d]*ArcTan[(Sqrt[e]*x)/Sqrt[d]]*(a + b*Log[c*x^n]))/e^(3/2) +
((I/2)*b*Sqrt[d]*n*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]])/e^(3/2) - ((I/2)*b*Sqrt[d]*n*PolyLog[2, (I*Sqrt[e]*x)
/Sqrt[d]])/e^(3/2)

Rule 321

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

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2351

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_.), x_Symbol] :> Wit
h[{u = ExpandIntegrand[a + b*Log[c*x^n], (f*x)^m*(d + e*x^r)^q, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c,
d, e, f, m, n, q, r}, x] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[m] && IntegerQ[r]))

Rule 2295

Int[Log[(c_.)*(x_)^(n_.)], x_Symbol] :> Simp[x*Log[c*x^n], x] - Simp[n*x, x] /; FreeQ[{c, n}, x]

Rule 2324

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))/((d_) + (e_.)*(x_)^2), x_Symbol] :> With[{u = IntHide[1/(d + e*x^2),
x]}, Simp[u*(a + b*Log[c*x^n]), x] - Dist[b*n, Int[u/x, x], x]] /; FreeQ[{a, b, c, d, e, n}, x]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 4848

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))/(x_), x_Symbol] :> Simp[a*Log[x], x] + (Dist[(I*b)/2, Int[Log[1 - I*c*x
]/x, x], x] - Dist[(I*b)/2, Int[Log[1 + I*c*x]/x, x], x]) /; FreeQ[{a, b, c}, x]

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

Mathematica [A]  time = 0.109913, size = 170, normalized size = 1.29 \[ \frac{b \sqrt{-d} n \text{PolyLog}\left (2,\frac{\sqrt{e} x}{\sqrt{-d}}\right )-b \sqrt{-d} n \text{PolyLog}\left (2,\frac{d \sqrt{e} x}{(-d)^{3/2}}\right )-\sqrt{-d} \log \left (\frac{\sqrt{e} x}{\sqrt{-d}}+1\right ) \left (a+b \log \left (c x^n\right )\right )+\sqrt{-d} \log \left (\frac{d \sqrt{e} x}{(-d)^{3/2}}+1\right ) \left (a+b \log \left (c x^n\right )\right )+2 a \sqrt{e} x+2 b \sqrt{e} x \log \left (c x^n\right )-2 b \sqrt{e} n x}{2 e^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [C]  time = 0.174, size = 512, normalized size = 3.9 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

b*ln(x^n)/e*x+b*d/e/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*n*ln(x)-b*d/e/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))*ln(x
^n)-b*n*x/e-1/2*b*n*d/e/(-d*e)^(1/2)*ln(x)*ln((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n*d/e/(-d*e)^(1/2)*ln(x)
*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*b*n*d/e/(-d*e)^(1/2)*dilog((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n*
d/e/(-d*e)^(1/2)*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))-1/2*I*b*Pi*csgn(I*c*x^n)^3/e*x-1/2*I*b*Pi*csgn(I*x^n)*
csgn(I*c*x^n)^2*d/e/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))-1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/e*x+1/2
*I*b*Pi*csgn(I*c*x^n)^3*d/e/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))+1/2*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e*x+1/2
*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*d/e/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))-1/2*I*b*Pi*csgn(I*c*x^n)^2
*csgn(I*c)*d/e/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))+1/2*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/e*x+b*ln(c)/e*x-b*ln(c
)*d/e/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))+a*x/e-a*d/e/(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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