3.222 \(\int \frac {x^3 (a+b \log (c x^n))}{(d+e x^2)^2} \, dx\)

Optimal. Leaf size=95 \[ \frac {\log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e \left (d+e x^2\right )}+\frac {b n \text {Li}_2\left (-\frac {e x^2}{d}\right )}{4 e^2}+\frac {b n \log \left (d+e x^2\right )}{4 e^2} \]

[Out]

-1/2*x^2*(a+b*ln(c*x^n))/e/(e*x^2+d)+1/4*b*n*ln(e*x^2+d)/e^2+1/2*(a+b*ln(c*x^n))*ln(1+e*x^2/d)/e^2+1/4*b*n*pol
ylog(2,-e*x^2/d)/e^2

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Rubi [A]  time = 0.19, antiderivative size = 95, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.304, Rules used = {266, 43, 2351, 2335, 260, 2337, 2391} \[ \frac {b n \text {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{4 e^2}+\frac {\log \left (\frac {e x^2}{d}+1\right ) \left (a+b \log \left (c x^n\right )\right )}{2 e^2}-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e \left (d+e x^2\right )}+\frac {b n \log \left (d+e x^2\right )}{4 e^2} \]

Antiderivative was successfully verified.

[In]

Int[(x^3*(a + b*Log[c*x^n]))/(d + e*x^2)^2,x]

[Out]

-(x^2*(a + b*Log[c*x^n]))/(2*e*(d + e*x^2)) + (b*n*Log[d + e*x^2])/(4*e^2) + ((a + b*Log[c*x^n])*Log[1 + (e*x^
2)/d])/(2*e^2) + (b*n*PolyLog[2, -((e*x^2)/d)])/(4*e^2)

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 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 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 2335

Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)*(x_)^(r_.))^(q_), x_Symbol] :> Simp
[((f*x)^(m + 1)*(d + e*x^r)^(q + 1)*(a + b*Log[c*x^n]))/(d*f*(m + 1)), x] - Dist[(b*n)/(d*(m + 1)), Int[(f*x)^
m*(d + e*x^r)^(q + 1), x], x] /; FreeQ[{a, b, c, d, e, f, m, n, q, r}, x] && EqQ[m + r*(q + 1) + 1, 0] && NeQ[
m, -1]

Rule 2337

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*((f_.)*(x_))^(m_.))/((d_) + (e_.)*(x_)^(r_)), x_Symbol] :> Si
mp[(f^m*Log[1 + (e*x^r)/d]*(a + b*Log[c*x^n])^p)/(e*r), x] - Dist[(b*f^m*n*p)/(e*r), Int[(Log[1 + (e*x^r)/d]*(
a + b*Log[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n, r}, x] && EqQ[m, r - 1] && IGtQ[p, 0] &
& (IntegerQ[m] || GtQ[f, 0]) && NeQ[r, n]

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 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^3 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx &=\int \left (-\frac {d x \left (a+b \log \left (c x^n\right )\right )}{e \left (d+e x^2\right )^2}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{e \left (d+e x^2\right )}\right ) \, dx\\ &=\frac {\int \frac {x \left (a+b \log \left (c x^n\right )\right )}{d+e x^2} \, dx}{e}-\frac {d \int \frac {x \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx}{e}\\ &=-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e \left (d+e x^2\right )}+\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e^2}-\frac {(b n) \int \frac {\log \left (1+\frac {e x^2}{d}\right )}{x} \, dx}{2 e^2}+\frac {(b n) \int \frac {x}{d+e x^2} \, dx}{2 e}\\ &=-\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e \left (d+e x^2\right )}+\frac {b n \log \left (d+e x^2\right )}{4 e^2}+\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{2 e^2}+\frac {b n \text {Li}_2\left (-\frac {e x^2}{d}\right )}{4 e^2}\\ \end {align*}

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Mathematica [C]  time = 0.24, size = 321, normalized size = 3.38 \[ \frac {2 \log \left (d+e x^2\right ) \left (a+b \log \left (c x^n\right )-b n \log (x)\right )+\frac {2 d \left (a+b \log \left (c x^n\right )-b n \log (x)\right )}{d+e x^2}+\frac {b n \left (2 \left (d+e x^2\right ) \text {Li}_2\left (-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+2 \left (d+e x^2\right ) \text {Li}_2\left (\frac {i \sqrt {e} x}{\sqrt {d}}\right )+e x^2 \log \left (-\sqrt {e} x+i \sqrt {d}\right )+e x^2 \log \left (\sqrt {e} x+i \sqrt {d}\right )+2 e x^2 \log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+2 e x^2 \log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+d \log \left (-\sqrt {e} x+i \sqrt {d}\right )+d \log \left (\sqrt {e} x+i \sqrt {d}\right )+2 d \log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+2 d \log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )-2 e x^2 \log (x)\right )}{d+e x^2}}{4 e^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^3*(a + b*Log[c*x^n]))/(d + e*x^2)^2,x]

[Out]

((2*d*(a - b*n*Log[x] + b*Log[c*x^n]))/(d + e*x^2) + 2*(a - b*n*Log[x] + b*Log[c*x^n])*Log[d + e*x^2] + (b*n*(
-2*e*x^2*Log[x] + d*Log[I*Sqrt[d] - Sqrt[e]*x] + e*x^2*Log[I*Sqrt[d] - Sqrt[e]*x] + d*Log[I*Sqrt[d] + Sqrt[e]*
x] + e*x^2*Log[I*Sqrt[d] + Sqrt[e]*x] + 2*d*Log[x]*Log[1 - (I*Sqrt[e]*x)/Sqrt[d]] + 2*e*x^2*Log[x]*Log[1 - (I*
Sqrt[e]*x)/Sqrt[d]] + 2*d*Log[x]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]] + 2*e*x^2*Log[x]*Log[1 + (I*Sqrt[e]*x)/Sqrt[d]
] + 2*(d + e*x^2)*PolyLog[2, ((-I)*Sqrt[e]*x)/Sqrt[d]] + 2*(d + e*x^2)*PolyLog[2, (I*Sqrt[e]*x)/Sqrt[d]]))/(d
+ e*x^2))/(4*e^2)

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fricas [F]  time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {b x^{3} \log \left (c x^{n}\right ) + a x^{3}}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^2,x, algorithm="fricas")

[Out]

integral((b*x^3*log(c*x^n) + a*x^3)/(e^2*x^4 + 2*d*e*x^2 + d^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{3}}{{\left (e x^{2} + d\right )}^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^2,x, algorithm="giac")

[Out]

integrate((b*log(c*x^n) + a)*x^3/(e*x^2 + d)^2, x)

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maple [C]  time = 0.19, size = 511, normalized size = 5.38 \[ -\frac {i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )}{4 \left (e \,x^{2}+d \right ) e^{2}}+\frac {i \pi b d \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 \left (e \,x^{2}+d \right ) e^{2}}+\frac {i \pi b d \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2}}{4 \left (e \,x^{2}+d \right ) e^{2}}-\frac {i \pi b d \mathrm {csgn}\left (i c \,x^{n}\right )^{3}}{4 \left (e \,x^{2}+d \right ) e^{2}}-\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{4 e^{2}}+\frac {i \pi b \,\mathrm {csgn}\left (i c \right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e \,x^{2}+d \right )}{4 e^{2}}+\frac {i \pi b \,\mathrm {csgn}\left (i x^{n}\right ) \mathrm {csgn}\left (i c \,x^{n}\right )^{2} \ln \left (e \,x^{2}+d \right )}{4 e^{2}}-\frac {i \pi b \mathrm {csgn}\left (i c \,x^{n}\right )^{3} \ln \left (e \,x^{2}+d \right )}{4 e^{2}}+\frac {b n \ln \relax (x ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{2}}+\frac {b n \ln \relax (x ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{2}}-\frac {b n \ln \relax (x ) \ln \left (e \,x^{2}+d \right )}{2 e^{2}}+\frac {b d \ln \relax (c )}{2 \left (e \,x^{2}+d \right ) e^{2}}+\frac {b d \ln \left (x^{n}\right )}{2 \left (e \,x^{2}+d \right ) e^{2}}+\frac {b n \dilog \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{2}}+\frac {b n \dilog \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{2 e^{2}}-\frac {b n \ln \relax (x )}{2 e^{2}}+\frac {b n \ln \left (e \,x^{2}+d \right )}{4 e^{2}}+\frac {b \ln \relax (c ) \ln \left (e \,x^{2}+d \right )}{2 e^{2}}+\frac {b \ln \left (x^{n}\right ) \ln \left (e \,x^{2}+d \right )}{2 e^{2}}+\frac {a d}{2 \left (e \,x^{2}+d \right ) e^{2}}+\frac {a \ln \left (e \,x^{2}+d \right )}{2 e^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^3*(b*ln(c*x^n)+a)/(e*x^2+d)^2,x)

[Out]

1/2*b*ln(x^n)*d/e^2/(e*x^2+d)+1/2*b*ln(x^n)/e^2*ln(e*x^2+d)-1/2*b*n/e^2*ln(x)*ln(e*x^2+d)+1/2*b*n/e^2*ln(x)*ln
((-e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n/e^2*ln(x)*ln((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n/e^2*dilog((-e
*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/2*b*n/e^2*dilog((e*x+(-d*e)^(1/2))/(-d*e)^(1/2))+1/4*b*n*ln(e*x^2+d)/e^2-1/2*
b*n/e^2*ln(x)+1/4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)*d/e^2/(e*x^2+d)-1/4*I*b*Pi*csgn(I*c*x^n)^3*d/e^2/(e*x^2+d)-
1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)*d/e^2/(e*x^2+d)+1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2*d/e^2/(e
*x^2+d)+1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)^2/e^2*ln(e*x^2+d)+1/4*I*b*Pi*csgn(I*c*x^n)^2*csgn(I*c)/e^2*ln(e*x
^2+d)-1/4*I*b*Pi*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)/e^2*ln(e*x^2+d)-1/4*I*b*Pi*csgn(I*c*x^n)^3/e^2*ln(e*x^2+d
)+1/2*b*ln(c)*d/e^2/(e*x^2+d)+1/2*b*ln(c)/e^2*ln(e*x^2+d)+1/2*a*d/e^2/(e*x^2+d)+1/2*a/e^2*ln(e*x^2+d)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{2} \, a {\left (\frac {d}{e^{3} x^{2} + d e^{2}} + \frac {\log \left (e x^{2} + d\right )}{e^{2}}\right )} + b \int \frac {x^{3} \log \relax (c) + x^{3} \log \left (x^{n}\right )}{e^{2} x^{4} + 2 \, d e x^{2} + d^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^3*(a+b*log(c*x^n))/(e*x^2+d)^2,x, algorithm="maxima")

[Out]

1/2*a*(d/(e^3*x^2 + d*e^2) + log(e*x^2 + d)/e^2) + b*integrate((x^3*log(c) + x^3*log(x^n))/(e^2*x^4 + 2*d*e*x^
2 + d^2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {x^3\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^3*(a + b*log(c*x^n)))/(d + e*x^2)^2,x)

[Out]

int((x^3*(a + b*log(c*x^n)))/(d + e*x^2)^2, x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**3*(a+b*ln(c*x**n))/(e*x**2+d)**2,x)

[Out]

Timed out

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