\(\int \frac {(a+b \log (c x^n))^3 \log (d (\frac {1}{d}+f x^2))}{x} \, dx\) [48]

Optimal result

Integrand size = 28, antiderivative size = 101 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (\frac {1}{d}+f x^2\right )\right )}{x} \, dx=-\frac {1}{2} \left (a+b \log \left (c x^n\right )\right )^3 \operatorname {PolyLog}\left (2,-d f x^2\right )+\frac {3}{4} b n \left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}\left (3,-d f x^2\right )-\frac {3}{4} b^2 n^2 \left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (4,-d f x^2\right )+\frac {3}{8} b^3 n^3 \operatorname {PolyLog}\left (5,-d f x^2\right ) \] Output:

-1/2*(a+b*ln(c*x^n))^3*polylog(2,-d*f*x^2)+3/4*b*n*(a+b*ln(c*x^n))^2*polyl 
og(3,-d*f*x^2)-3/4*b^2*n^2*(a+b*ln(c*x^n))*polylog(4,-d*f*x^2)+3/8*b^3*n^3 
*polylog(5,-d*f*x^2)
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.39 (sec) , antiderivative size = 754, normalized size of antiderivative = 7.47 \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (\frac {1}{d}+f x^2\right )\right )}{x} \, dx =\text {Too large to display} \] Input:

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

Output:

(-(Log[x]*(b^3*n^3*Log[x]^3 - 4*b^2*n^2*Log[x]^2*(a + b*Log[c*x^n]) + 6*b* 
n*Log[x]*(a + b*Log[c*x^n])^2 - 4*(a + b*Log[c*x^n])^3)*Log[1 + d*f*x^2]) 
- 4*(a - b*n*Log[x] + b*Log[c*x^n])^3*(Log[x]*(Log[1 - I*Sqrt[d]*Sqrt[f]*x 
] + Log[1 + I*Sqrt[d]*Sqrt[f]*x]) + PolyLog[2, (-I)*Sqrt[d]*Sqrt[f]*x] + P 
olyLog[2, I*Sqrt[d]*Sqrt[f]*x]) - 6*b*n*(a - b*n*Log[x] + b*Log[c*x^n])^2* 
(Log[x]^2*Log[1 - I*Sqrt[d]*Sqrt[f]*x] + Log[x]^2*Log[1 + I*Sqrt[d]*Sqrt[f 
]*x] + 2*Log[x]*PolyLog[2, (-I)*Sqrt[d]*Sqrt[f]*x] + 2*Log[x]*PolyLog[2, I 
*Sqrt[d]*Sqrt[f]*x] - 2*PolyLog[3, (-I)*Sqrt[d]*Sqrt[f]*x] - 2*PolyLog[3, 
I*Sqrt[d]*Sqrt[f]*x]) + 4*b^2*n^2*(-a + b*n*Log[x] - b*Log[c*x^n])*(Log[x] 
^3*Log[1 - I*Sqrt[d]*Sqrt[f]*x] + Log[x]^3*Log[1 + I*Sqrt[d]*Sqrt[f]*x] + 
3*Log[x]^2*PolyLog[2, (-I)*Sqrt[d]*Sqrt[f]*x] + 3*Log[x]^2*PolyLog[2, I*Sq 
rt[d]*Sqrt[f]*x] - 6*Log[x]*PolyLog[3, (-I)*Sqrt[d]*Sqrt[f]*x] - 6*Log[x]* 
PolyLog[3, I*Sqrt[d]*Sqrt[f]*x] + 6*PolyLog[4, (-I)*Sqrt[d]*Sqrt[f]*x] + 6 
*PolyLog[4, I*Sqrt[d]*Sqrt[f]*x]) - b^3*n^3*(Log[x]^4*Log[1 - I*Sqrt[d]*Sq 
rt[f]*x] + Log[x]^4*Log[1 + I*Sqrt[d]*Sqrt[f]*x] + 4*Log[x]^3*PolyLog[2, ( 
-I)*Sqrt[d]*Sqrt[f]*x] + 4*Log[x]^3*PolyLog[2, I*Sqrt[d]*Sqrt[f]*x] - 12*L 
og[x]^2*PolyLog[3, (-I)*Sqrt[d]*Sqrt[f]*x] - 12*Log[x]^2*PolyLog[3, I*Sqrt 
[d]*Sqrt[f]*x] + 24*Log[x]*PolyLog[4, (-I)*Sqrt[d]*Sqrt[f]*x] + 24*Log[x]* 
PolyLog[4, I*Sqrt[d]*Sqrt[f]*x] - 24*PolyLog[5, (-I)*Sqrt[d]*Sqrt[f]*x] - 
24*PolyLog[5, I*Sqrt[d]*Sqrt[f]*x]))/4
 

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 101, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2821, 2830, 2830, 7143}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {\log \left (d \left (\frac {1}{d}+f x^2\right )\right ) \left (a+b \log \left (c x^n\right )\right )^3}{x} \, dx\)

\(\Big \downarrow \) 2821

\(\displaystyle \frac {3}{2} b n \int \frac {\left (a+b \log \left (c x^n\right )\right )^2 \operatorname {PolyLog}\left (2,-d f x^2\right )}{x}dx-\frac {1}{2} \operatorname {PolyLog}\left (2,-d f x^2\right ) \left (a+b \log \left (c x^n\right )\right )^3\)

\(\Big \downarrow \) 2830

\(\displaystyle \frac {3}{2} b n \left (\frac {1}{2} \operatorname {PolyLog}\left (3,-d f x^2\right ) \left (a+b \log \left (c x^n\right )\right )^2-b n \int \frac {\left (a+b \log \left (c x^n\right )\right ) \operatorname {PolyLog}\left (3,-d f x^2\right )}{x}dx\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,-d f x^2\right ) \left (a+b \log \left (c x^n\right )\right )^3\)

\(\Big \downarrow \) 2830

\(\displaystyle \frac {3}{2} b n \left (\frac {1}{2} \operatorname {PolyLog}\left (3,-d f x^2\right ) \left (a+b \log \left (c x^n\right )\right )^2-b n \left (\frac {1}{2} \operatorname {PolyLog}\left (4,-d f x^2\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{2} b n \int \frac {\operatorname {PolyLog}\left (4,-d f x^2\right )}{x}dx\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,-d f x^2\right ) \left (a+b \log \left (c x^n\right )\right )^3\)

\(\Big \downarrow \) 7143

\(\displaystyle \frac {3}{2} b n \left (\frac {1}{2} \operatorname {PolyLog}\left (3,-d f x^2\right ) \left (a+b \log \left (c x^n\right )\right )^2-b n \left (\frac {1}{2} \operatorname {PolyLog}\left (4,-d f x^2\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{4} b n \operatorname {PolyLog}\left (5,-d f x^2\right )\right )\right )-\frac {1}{2} \operatorname {PolyLog}\left (2,-d f x^2\right ) \left (a+b \log \left (c x^n\right )\right )^3\)

Input:

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

Output:

-1/2*((a + b*Log[c*x^n])^3*PolyLog[2, -(d*f*x^2)]) + (3*b*n*(((a + b*Log[c 
*x^n])^2*PolyLog[3, -(d*f*x^2)])/2 - b*n*(((a + b*Log[c*x^n])*PolyLog[4, - 
(d*f*x^2)])/2 - (b*n*PolyLog[5, -(d*f*x^2)])/4)))/2
 

Defintions of rubi rules used

rule 2821

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] + Simp[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 2830

Int[(((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))^(p_.)*PolyLog[k_, (e_.)*(x_)^(q_ 
.)])/(x_), x_Symbol] :> Simp[PolyLog[k + 1, e*x^q]*((a + b*Log[c*x^n])^p/q) 
, x] - Simp[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 7143

Int[PolyLog[n_, (c_.)*((a_.) + (b_.)*(x_))^(p_.)]/((d_.) + (e_.)*(x_)), x_S 
ymbol] :> 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]
 

Maple [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 46.62 (sec) , antiderivative size = 1296, normalized size of antiderivative = 12.83

method	result	size
risch	\(\text {Expression too large to display}\)	\(1296\)

Input:

int((a+b*ln(c*x^n))^3*ln(d*(1/d+f*x^2))/x,x,method=_RETURNVERBOSE)
 

Output:

-ln(d*f*x^2+1)*ln(x)^4*b^3*n^3+ln(x)^4*ln(1+x*(-d*f)^(1/2))*b^3*n^3+ln(x)^ 
4*ln(1-x*(-d*f)^(1/2))*b^3*n^3+ln(x)^3*dilog(1+x*(-d*f)^(1/2))*b^3*n^3+ln( 
x)^3*dilog(1-x*(-d*f)^(1/2))*b^3*n^3-1/2*ln(x)^3*polylog(2,-d*f*x^2)*b^3*n 
^3+ln(d*f*x^2+1)*ln(x)*ln(x^n)^3*b^3-ln(x)*ln(1+x*(-d*f)^(1/2))*ln(x^n)^3* 
b^3-ln(x)*ln(1-x*(-d*f)^(1/2))*ln(x^n)^3*b^3+3/4*ln(x^n)^2*polylog(3,-d*f* 
x^2)*b^3*n-3/4*ln(x^n)*polylog(4,-d*f*x^2)*b^3*n^2-3*ln(x)^2*dilog(1+x*(-d 
*f)^(1/2))*ln(x^n)*b^3*n^2-3*ln(x)^2*dilog(1-x*(-d*f)^(1/2))*ln(x^n)*b^3*n 
^2+3*ln(x)^2*ln(1+x*(-d*f)^(1/2))*ln(x^n)^2*b^3*n+3*ln(x)^2*ln(1-x*(-d*f)^ 
(1/2))*ln(x^n)^2*b^3*n+3/2*ln(x)^2*ln(x^n)*polylog(2,-d*f*x^2)*b^3*n^2+3*l 
n(x)*dilog(1+x*(-d*f)^(1/2))*ln(x^n)^2*b^3*n+3*ln(x)*dilog(1-x*(-d*f)^(1/2 
))*ln(x^n)^2*b^3*n-3/2*ln(x)*ln(x^n)^2*polylog(2,-d*f*x^2)*b^3*n+3*ln(d*f* 
x^2+1)*ln(x)^3*ln(x^n)*b^3*n^2-3*ln(x)^3*ln(1+x*(-d*f)^(1/2))*ln(x^n)*b^3* 
n^2-3*ln(x)^3*ln(1-x*(-d*f)^(1/2))*ln(x^n)*b^3*n^2-3*ln(d*f*x^2+1)*ln(x)^2 
*ln(x^n)^2*b^3*n+1/8*(I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*csgn(I*x^n 
)*csgn(I*c*x^n)*csgn(I*c)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn(I*c*x^n)^2*cs 
gn(I*c)+2*b*ln(c)+2*a)^3*(ln(x)*ln(d*f*x^2+1)-2*d*f*(1/2*ln(x)*(ln(1+x*(-d 
*f)^(1/2))+ln(1-x*(-d*f)^(1/2)))/d/f+1/2*(dilog(1+x*(-d*f)^(1/2))+dilog(1- 
x*(-d*f)^(1/2)))/d/f))-dilog(1+x*(-d*f)^(1/2))*ln(x^n)^3*b^3-dilog(1-x*(-d 
*f)^(1/2))*ln(x^n)^3*b^3+3/2*(I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-I*Pi*b*cs 
gn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-I*Pi*b*csgn(I*c*x^n)^3+I*Pi*b*csgn(I*...
 

Fricas [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (\frac {1}{d}+f x^2\right )\right )}{x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x^{2} + \frac {1}{d}\right )} d\right )}{x} \,d x } \] Input:

integrate((a+b*log(c*x^n))^3*log(d*(1/d+f*x^2))/x,x, algorithm="fricas")
 

Output:

integral((b^3*log(d*f*x^2 + 1)*log(c*x^n)^3 + 3*a*b^2*log(d*f*x^2 + 1)*log 
(c*x^n)^2 + 3*a^2*b*log(d*f*x^2 + 1)*log(c*x^n) + a^3*log(d*f*x^2 + 1))/x, 
 x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (\frac {1}{d}+f x^2\right )\right )}{x} \, dx=\text {Timed out} \] Input:

integrate((a+b*ln(c*x**n))**3*ln(d*(1/d+f*x**2))/x,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (\frac {1}{d}+f x^2\right )\right )}{x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x^{2} + \frac {1}{d}\right )} d\right )}{x} \,d x } \] Input:

integrate((a+b*log(c*x^n))^3*log(d*(1/d+f*x^2))/x,x, algorithm="maxima")
 

Output:

-1/4*(b^3*n^3*log(x)^4 - 4*b^3*log(x)*log(x^n)^3 - 4*(b^3*n^2*log(c) + a*b 
^2*n^2)*log(x)^3 + 6*(b^3*n*log(c)^2 + 2*a*b^2*n*log(c) + a^2*b*n)*log(x)^ 
2 + 6*(b^3*n*log(x)^2 - 2*(b^3*log(c) + a*b^2)*log(x))*log(x^n)^2 - 4*(b^3 
*n^2*log(x)^3 - 3*(b^3*n*log(c) + a*b^2*n)*log(x)^2 + 3*(b^3*log(c)^2 + 2* 
a*b^2*log(c) + a^2*b)*log(x))*log(x^n) - 4*(b^3*log(c)^3 + 3*a*b^2*log(c)^ 
2 + 3*a^2*b*log(c) + a^3)*log(x))*log(d*f*x^2 + 1) - integrate(-1/2*(b^3*d 
*f*n^3*x*log(x)^4 - 4*b^3*d*f*x*log(x)*log(x^n)^3 - 4*(b^3*d*f*n^2*log(c) 
+ a*b^2*d*f*n^2)*x*log(x)^3 + 6*(b^3*d*f*n*log(c)^2 + 2*a*b^2*d*f*n*log(c) 
 + a^2*b*d*f*n)*x*log(x)^2 - 4*(b^3*d*f*log(c)^3 + 3*a*b^2*d*f*log(c)^2 + 
3*a^2*b*d*f*log(c) + a^3*d*f)*x*log(x) + 6*(b^3*d*f*n*x*log(x)^2 - 2*(b^3* 
d*f*log(c) + a*b^2*d*f)*x*log(x))*log(x^n)^2 - 4*(b^3*d*f*n^2*x*log(x)^3 - 
 3*(b^3*d*f*n*log(c) + a*b^2*d*f*n)*x*log(x)^2 + 3*(b^3*d*f*log(c)^2 + 2*a 
*b^2*d*f*log(c) + a^2*b*d*f)*x*log(x))*log(x^n))/(d*f*x^2 + 1), x)
 

Giac [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (\frac {1}{d}+f x^2\right )\right )}{x} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )}^{3} \log \left ({\left (f x^{2} + \frac {1}{d}\right )} d\right )}{x} \,d x } \] Input:

integrate((a+b*log(c*x^n))^3*log(d*(1/d+f*x^2))/x,x, algorithm="giac")
 

Output:

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

Mupad [F(-1)]

Timed out. \[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (\frac {1}{d}+f x^2\right )\right )}{x} \, dx=\int \frac {\ln \left (d\,\left (f\,x^2+\frac {1}{d}\right )\right )\,{\left (a+b\,\ln \left (c\,x^n\right )\right )}^3}{x} \,d x \] Input:

int((log(d*(f*x^2 + 1/d))*(a + b*log(c*x^n))^3)/x,x)
 

Output:

int((log(d*(f*x^2 + 1/d))*(a + b*log(c*x^n))^3)/x, x)
 

Reduce [F]

\[ \int \frac {\left (a+b \log \left (c x^n\right )\right )^3 \log \left (d \left (\frac {1}{d}+f x^2\right )\right )}{x} \, dx=\left (\int \frac {\mathrm {log}\left (d f \,x^{2}+1\right )}{d f \,x^{3}+x}d x \right ) a^{3}+\left (\int \frac {\mathrm {log}\left (d f \,x^{2}+1\right ) \mathrm {log}\left (x^{n} c \right )^{3}}{x}d x \right ) b^{3}+3 \left (\int \frac {\mathrm {log}\left (d f \,x^{2}+1\right ) \mathrm {log}\left (x^{n} c \right )^{2}}{x}d x \right ) a \,b^{2}+3 \left (\int \frac {\mathrm {log}\left (d f \,x^{2}+1\right ) \mathrm {log}\left (x^{n} c \right )}{x}d x \right ) a^{2} b +\frac {\mathrm {log}\left (d f \,x^{2}+1\right )^{2} a^{3}}{4} \] Input:

int((a+b*log(c*x^n))^3*log(d*(1/d+f*x^2))/x,x)
 

Output:

(4*int(log(d*f*x**2 + 1)/(d*f*x**3 + x),x)*a**3 + 4*int((log(d*f*x**2 + 1) 
*log(x**n*c)**3)/x,x)*b**3 + 12*int((log(d*f*x**2 + 1)*log(x**n*c)**2)/x,x 
)*a*b**2 + 12*int((log(d*f*x**2 + 1)*log(x**n*c))/x,x)*a**2*b + log(d*f*x* 
*2 + 1)**2*a**3)/4