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

Optimal. Leaf size=147 \[ \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{3 b n}-\frac {1}{2} m \text {Li}_2\left (-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^2+\frac {1}{2} b m n \text {Li}_3\left (-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {m \log \left (\frac {f x^2}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n}-\frac {1}{4} b^2 m n^2 \text {Li}_4\left (-\frac {f x^2}{e}\right ) \]

[Out]

1/3*(a+b*ln(c*x^n))^3*ln(d*(f*x^2+e)^m)/b/n-1/3*m*(a+b*ln(c*x^n))^3*ln(1+f*x^2/e)/b/n-1/2*m*(a+b*ln(c*x^n))^2*
polylog(2,-f*x^2/e)+1/2*b*m*n*(a+b*ln(c*x^n))*polylog(3,-f*x^2/e)-1/4*b^2*m*n^2*polylog(4,-f*x^2/e)

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Rubi [A]  time = 0.18, antiderivative size = 147, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 28, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.179, Rules used = {2375, 2337, 2374, 2383, 6589} \[ -\frac {1}{2} m \text {PolyLog}\left (2,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )^2+\frac {1}{2} b m n \text {PolyLog}\left (3,-\frac {f x^2}{e}\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b^2 m n^2 \text {PolyLog}\left (4,-\frac {f x^2}{e}\right )+\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{3 b n}-\frac {m \log \left (\frac {f x^2}{e}+1\right ) \left (a+b \log \left (c x^n\right )\right )^3}{3 b n} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*Log[c*x^n])^3*Log[d*(e + f*x^2)^m])/(3*b*n) - (m*(a + b*Log[c*x^n])^3*Log[1 + (f*x^2)/e])/(3*b*n) - (m
*(a + b*Log[c*x^n])^2*PolyLog[2, -((f*x^2)/e)])/2 + (b*m*n*(a + b*Log[c*x^n])*PolyLog[3, -((f*x^2)/e)])/2 - (b
^2*m*n^2*PolyLog[4, -((f*x^2)/e)])/4

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 2374

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> -Sim
p[(PolyLog[2, -(d*f*x^m)]*(a + b*Log[c*x^n])^p)/m, x] + Dist[(b*n*p)/m, Int[(PolyLog[2, -(d*f*x^m)]*(a + b*Log
[c*x^n])^(p - 1))/x, x], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[p, 0] && EqQ[d*e, 1]

Rule 2375

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

Rule 2383

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_.)])/(x_), x_Symbol] :> Simp[(PolyL
og[k + 1, e*x^q]*(a + b*Log[c*x^n])^p)/q, x] - Dist[(b*n*p)/q, Int[(PolyLog[k + 1, e*x^q]*(a + b*Log[c*x^n])^(
p - 1))/x, x], x] /; FreeQ[{a, b, c, e, k, n, q}, x] && GtQ[p, 0]

Rule 6589

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_Symbol] :> Simp[PolyLog[n + 1, c*(a
+ b*x)^p]/(e*p), x] /; FreeQ[{a, b, c, d, e, n, p}, x] && EqQ[b*d, a*e]

Rubi steps

\begin {align*} \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (d \left (e+f x^2\right )^m\right )}{x} \, dx &=\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{3 b n}-\frac {(2 f m) \int \frac {x \left (a+b \log \left (c x^n\right )\right )^3}{e+f x^2} \, dx}{3 b n}\\ &=\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{3 b n}-\frac {m \left (a+b \log \left (c x^n\right )\right )^3 \log \left (1+\frac {f x^2}{e}\right )}{3 b n}+m \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \log \left (1+\frac {f x^2}{e}\right )}{x} \, dx\\ &=\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{3 b n}-\frac {m \left (a+b \log \left (c x^n\right )\right )^3 \log \left (1+\frac {f x^2}{e}\right )}{3 b n}-\frac {1}{2} m \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2\left (-\frac {f x^2}{e}\right )+(b m n) \int \frac {\left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-\frac {f x^2}{e}\right )}{x} \, dx\\ &=\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{3 b n}-\frac {m \left (a+b \log \left (c x^n\right )\right )^3 \log \left (1+\frac {f x^2}{e}\right )}{3 b n}-\frac {1}{2} m \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2\left (-\frac {f x^2}{e}\right )+\frac {1}{2} b m n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3\left (-\frac {f x^2}{e}\right )-\frac {1}{2} \left (b^2 m n^2\right ) \int \frac {\text {Li}_3\left (-\frac {f x^2}{e}\right )}{x} \, dx\\ &=\frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (e+f x^2\right )^m\right )}{3 b n}-\frac {m \left (a+b \log \left (c x^n\right )\right )^3 \log \left (1+\frac {f x^2}{e}\right )}{3 b n}-\frac {1}{2} m \left (a+b \log \left (c x^n\right )\right )^2 \text {Li}_2\left (-\frac {f x^2}{e}\right )+\frac {1}{2} b m n \left (a+b \log \left (c x^n\right )\right ) \text {Li}_3\left (-\frac {f x^2}{e}\right )-\frac {1}{4} b^2 m n^2 \text {Li}_4\left (-\frac {f x^2}{e}\right )\\ \end {align*}

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Mathematica [C]  time = 0.24, size = 736, normalized size = 5.01 \[ a^2 \log (x) \log \left (d \left (e+f x^2\right )^m\right )-a^2 m \log (x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-a^2 m \log (x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 a b \log (x) \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )-m \text {Li}_2\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )^2-m \text {Li}_2\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right ) \left (a+b \log \left (c x^n\right )\right )^2-2 a b m \log (x) \log \left (c x^n\right ) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-2 a b m \log (x) \log \left (c x^n\right ) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-a b n \log ^2(x) \log \left (d \left (e+f x^2\right )^m\right )+2 a b m n \text {Li}_3\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 a b m n \text {Li}_3\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )+a b m n \log ^2(x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+a b m n \log ^2(x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-b^2 n \log ^2(x) \log \left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+b^2 \log (x) \log ^2\left (c x^n\right ) \log \left (d \left (e+f x^2\right )^m\right )+2 b^2 m n \log \left (c x^n\right ) \text {Li}_3\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+2 b^2 m n \log \left (c x^n\right ) \text {Li}_3\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )+b^2 m n \log ^2(x) \log \left (c x^n\right ) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )+b^2 m n \log ^2(x) \log \left (c x^n\right ) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )-b^2 m \log (x) \log ^2\left (c x^n\right ) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-b^2 m \log (x) \log ^2\left (c x^n\right ) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right )+\frac {1}{3} b^2 n^2 \log ^3(x) \log \left (d \left (e+f x^2\right )^m\right )-2 b^2 m n^2 \text {Li}_4\left (-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-2 b^2 m n^2 \text {Li}_4\left (\frac {i \sqrt {f} x}{\sqrt {e}}\right )-\frac {1}{3} b^2 m n^2 \log ^3(x) \log \left (1-\frac {i \sqrt {f} x}{\sqrt {e}}\right )-\frac {1}{3} b^2 m n^2 \log ^3(x) \log \left (1+\frac {i \sqrt {f} x}{\sqrt {e}}\right ) \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*m*Log[x]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]]) + a*b*m*n*Log[x]^2*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - (b^2*m*n^2*
Log[x]^3*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]])/3 - 2*a*b*m*Log[x]*Log[c*x^n]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] + b^2*m*
n*Log[x]^2*Log[c*x^n]*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]] - b^2*m*Log[x]*Log[c*x^n]^2*Log[1 - (I*Sqrt[f]*x)/Sqrt[e]
] - a^2*m*Log[x]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + a*b*m*n*Log[x]^2*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - (b^2*m*n^2
*Log[x]^3*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]])/3 - 2*a*b*m*Log[x]*Log[c*x^n]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] + b^2*m
*n*Log[x]^2*Log[c*x^n]*Log[1 + (I*Sqrt[f]*x)/Sqrt[e]] - b^2*m*Log[x]*Log[c*x^n]^2*Log[1 + (I*Sqrt[f]*x)/Sqrt[e
]] + a^2*Log[x]*Log[d*(e + f*x^2)^m] - a*b*n*Log[x]^2*Log[d*(e + f*x^2)^m] + (b^2*n^2*Log[x]^3*Log[d*(e + f*x^
2)^m])/3 + 2*a*b*Log[x]*Log[c*x^n]*Log[d*(e + f*x^2)^m] - b^2*n*Log[x]^2*Log[c*x^n]*Log[d*(e + f*x^2)^m] + b^2
*Log[x]*Log[c*x^n]^2*Log[d*(e + f*x^2)^m] - m*(a + b*Log[c*x^n])^2*PolyLog[2, ((-I)*Sqrt[f]*x)/Sqrt[e]] - m*(a
+ b*Log[c*x^n])^2*PolyLog[2, (I*Sqrt[f]*x)/Sqrt[e]] + 2*a*b*m*n*PolyLog[3, ((-I)*Sqrt[f]*x)/Sqrt[e]] + 2*b^2*
m*n*Log[c*x^n]*PolyLog[3, ((-I)*Sqrt[f]*x)/Sqrt[e]] + 2*a*b*m*n*PolyLog[3, (I*Sqrt[f]*x)/Sqrt[e]] + 2*b^2*m*n*
Log[c*x^n]*PolyLog[3, (I*Sqrt[f]*x)/Sqrt[e]] - 2*b^2*m*n^2*PolyLog[4, ((-I)*Sqrt[f]*x)/Sqrt[e]] - 2*b^2*m*n^2*
PolyLog[4, (I*Sqrt[f]*x)/Sqrt[e]]

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fricas [F]  time = 1.05, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b^{2} \log \left (c x^{n}\right )^{2} + 2 \, a b \log \left (c x^{n}\right ) + a^{2}\right )} \log \left ({\left (f x^{2} + e\right )}^{m} d\right )}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 5.16, size = 0, normalized size = 0.00 \[ \int \frac {\left (b \ln \left (c \,x^{n}\right )+a \right )^{2} \ln \left (d \left (f \,x^{2}+e \right )^{m}\right )}{x}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \frac {1}{3} \, {\left (b^{2} m n^{2} \log \relax (x)^{3} + 3 \, b^{2} m \log \relax (x) \log \left (x^{n}\right )^{2} - 3 \, {\left (b^{2} m n \log \relax (c) + a b m n\right )} \log \relax (x)^{2} - 3 \, {\left (b^{2} m n \log \relax (x)^{2} - 2 \, {\left (b^{2} m \log \relax (c) + a b m\right )} \log \relax (x)\right )} \log \left (x^{n}\right ) + 3 \, {\left (b^{2} m \log \relax (c)^{2} + 2 \, a b m \log \relax (c) + a^{2} m\right )} \log \relax (x)\right )} \log \left (f x^{2} + e\right ) - \int \frac {2 \, b^{2} f m n^{2} x^{2} \log \relax (x)^{3} - 3 \, b^{2} e \log \relax (c)^{2} \log \relax (d) - 6 \, a b e \log \relax (c) \log \relax (d) - 6 \, {\left (b^{2} f m n \log \relax (c) + a b f m n\right )} x^{2} \log \relax (x)^{2} - 3 \, a^{2} e \log \relax (d) + 6 \, {\left (b^{2} f m \log \relax (c)^{2} + 2 \, a b f m \log \relax (c) + a^{2} f m\right )} x^{2} \log \relax (x) - 3 \, {\left (b^{2} f \log \relax (c)^{2} \log \relax (d) + 2 \, a b f \log \relax (c) \log \relax (d) + a^{2} f \log \relax (d)\right )} x^{2} + 3 \, {\left (2 \, b^{2} f m x^{2} \log \relax (x) - b^{2} f x^{2} \log \relax (d) - b^{2} e \log \relax (d)\right )} \log \left (x^{n}\right )^{2} - 6 \, {\left (b^{2} f m n x^{2} \log \relax (x)^{2} + b^{2} e \log \relax (c) \log \relax (d) + a b e \log \relax (d) - 2 \, {\left (b^{2} f m \log \relax (c) + a b f m\right )} x^{2} \log \relax (x) + {\left (b^{2} f \log \relax (c) \log \relax (d) + a b f \log \relax (d)\right )} x^{2}\right )} \log \left (x^{n}\right )}{3 \, {\left (f x^{3} + e x\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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