3.1.26 \(\int \frac {(a+b \log (c x^n)) \log (d (\frac {1}{d}+f x^2))}{x} \, dx\) [26]

Optimal. Leaf size=39 \[ -\frac {1}{2} \left (a+b \log \left (c x^n\right )\right ) \text {Li}_2\left (-d f x^2\right )+\frac {1}{4} b n \text {Li}_3\left (-d f x^2\right ) \]

[Out]

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

________________________________________________________________________________________

Rubi [A]
time = 0.02, antiderivative size = 39, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2421, 6724} \begin {gather*} \frac {1}{4} b n \text {PolyLog}\left (3,-d f x^2\right )-\frac {1}{2} \text {PolyLog}\left (2,-d f x^2\right ) \left (a+b \log \left (c x^n\right )\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((a + b*Log[c*x^n])*Log[d*(d^(-1) + f*x^2)])/x,x]

[Out]

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

Rule 2421

Int[(Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))]*((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.))/(x_), x_Symbol] :> Simp
[(-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*L
og[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 6724

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

________________________________________________________________________________________

Mathematica [A]
time = 0.01, size = 50, normalized size = 1.28 \begin {gather*} -\frac {1}{2} a \text {Li}_2\left (-d f x^2\right )-\frac {1}{2} b \log \left (c x^n\right ) \text {Li}_2\left (-d f x^2\right )+\frac {1}{4} b n \text {Li}_3\left (-d f x^2\right ) \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((a + b*Log[c*x^n])*Log[d*(d^(-1) + f*x^2)])/x,x]

[Out]

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

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.13, size = 1026, normalized size = 26.31

method	result	size
risch	\(\text {Expression too large to display}\)	\(1026\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*ln(c*x^n))*ln(d*(1/d+f*x^2))/x,x,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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((a+b*log(c*x^n))*log(d*(1/d+f*x^2))/x,x, algorithm="maxima")

[Out]

-1/2*(b*n*log(x)^2 - 2*b*log(x)*log(x^n) - 2*(b*log(c) + a)*log(x))*log(d*f*x^2 + 1) - integrate(-(b*d*f*n*x*l
og(x)^2 - 2*b*d*f*x*log(x)*log(x^n) - 2*(b*d*f*log(c) + a*d*f)*x*log(x))/(d*f*x^2 + 1), x)

________________________________________________________________________________________

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((a+b*log(c*x^n))*log(d*(1/d+f*x^2))/x,x, algorithm="fricas")

[Out]

integral((b*log(d*f*x^2 + 1)*log(c*x^n) + a*log(d*f*x^2 + 1))/x, x)

________________________________________________________________________________________

Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*ln(c*x**n))*ln(d*(1/d+f*x**2))/x,x)

[Out]

Timed out

________________________________________________________________________________________

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((a+b*log(c*x^n))*log(d*(1/d+f*x^2))/x,x, algorithm="giac")

[Out]

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

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.03 \begin {gather*} \int \frac {\ln \left (d\,\left (f\,x^2+\frac {1}{d}\right )\right )\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((log(d*(f*x^2 + 1/d))*(a + b*log(c*x^n)))/x,x)

[Out]

int((log(d*(f*x^2 + 1/d))*(a + b*log(c*x^n)))/x, x)

________________________________________________________________________________________