\(\int x^3 (a+b \log (c x^n)) \log (d (\frac {1}{d}+f x^2)) \, dx\) [30]

Optimal result

Integrand size = 26, antiderivative size = 180 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (\frac {1}{d}+f x^2\right )\right ) \, dx=-\frac {3 b n x^2}{16 d f}+\frac {1}{16} b n x^4+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{4 d f}-\frac {1}{8} x^4 \left (a+b \log \left (c x^n\right )\right )+\frac {b n \log \left (1+d f x^2\right )}{16 d^2 f^2}-\frac {1}{16} b n x^4 \log \left (1+d f x^2\right )-\frac {\left (a+b \log \left (c x^n\right )\right ) \log \left (1+d f x^2\right )}{4 d^2 f^2}+\frac {1}{4} x^4 \left (a+b \log \left (c x^n\right )\right ) \log \left (1+d f x^2\right )-\frac {b n \operatorname {PolyLog}\left (2,-d f x^2\right )}{8 d^2 f^2} \] Output:

-3/16*b*n*x^2/d/f+1/16*b*n*x^4+1/4*x^2*(a+b*ln(c*x^n))/d/f-1/8*x^4*(a+b*ln 
(c*x^n))+1/16*b*n*ln(d*f*x^2+1)/d^2/f^2-1/16*b*n*x^4*ln(d*f*x^2+1)-1/4*(a+ 
b*ln(c*x^n))*ln(d*f*x^2+1)/d^2/f^2+1/4*x^4*(a+b*ln(c*x^n))*ln(d*f*x^2+1)-1 
/8*b*n*polylog(2,-d*f*x^2)/d^2/f^2
 

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.15 (sec) , antiderivative size = 348, normalized size of antiderivative = 1.93 \[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (\frac {1}{d}+f x^2\right )\right ) \, dx=\frac {a x^2}{4 d f}-\frac {a x^4}{8}+\frac {1}{32} b x^4 \left (n-4 \left (-n \log (x)+\log \left (c x^n\right )\right )\right )+\frac {b x^2 \left (-n+4 \left (-n \log (x)+\log \left (c x^n\right )\right )\right )}{16 d f}-\frac {a \log \left (1+d f x^2\right )}{4 d^2 f^2}+\frac {1}{4} a x^4 \log \left (1+d f x^2\right )+\frac {b \left (n-4 \left (-n \log (x)+\log \left (c x^n\right )\right )\right ) \log \left (1+d f x^2\right )}{16 d^2 f^2}+\frac {1}{16} b x^4 \left (-n+4 n \log (x)+4 \left (-n \log (x)+\log \left (c x^n\right )\right )\right ) \log \left (1+d f x^2\right )-\frac {1}{2} b d f n \left (-\frac {-\frac {x^2}{4}+\frac {1}{2} x^2 \log (x)}{d^2 f^2}+\frac {-\frac {x^4}{16}+\frac {1}{4} x^4 \log (x)}{d f}+\frac {\log (x) \log \left (1+i \sqrt {d} \sqrt {f} x\right )+\operatorname {PolyLog}\left (2,-i \sqrt {d} \sqrt {f} x\right )}{2 d^3 f^3}+\frac {\log (x) \log \left (1-i \sqrt {d} \sqrt {f} x\right )+\operatorname {PolyLog}\left (2,i \sqrt {d} \sqrt {f} x\right )}{2 d^3 f^3}\right ) \] Input:

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

Output:

(a*x^2)/(4*d*f) - (a*x^4)/8 + (b*x^4*(n - 4*(-(n*Log[x]) + Log[c*x^n])))/3 
2 + (b*x^2*(-n + 4*(-(n*Log[x]) + Log[c*x^n])))/(16*d*f) - (a*Log[1 + d*f* 
x^2])/(4*d^2*f^2) + (a*x^4*Log[1 + d*f*x^2])/4 + (b*(n - 4*(-(n*Log[x]) + 
Log[c*x^n]))*Log[1 + d*f*x^2])/(16*d^2*f^2) + (b*x^4*(-n + 4*n*Log[x] + 4* 
(-(n*Log[x]) + Log[c*x^n]))*Log[1 + d*f*x^2])/16 - (b*d*f*n*(-((-1/4*x^2 + 
 (x^2*Log[x])/2)/(d^2*f^2)) + (-1/16*x^4 + (x^4*Log[x])/4)/(d*f) + (Log[x] 
*Log[1 + I*Sqrt[d]*Sqrt[f]*x] + PolyLog[2, (-I)*Sqrt[d]*Sqrt[f]*x])/(2*d^3 
*f^3) + (Log[x]*Log[1 - I*Sqrt[d]*Sqrt[f]*x] + PolyLog[2, I*Sqrt[d]*Sqrt[f 
]*x])/(2*d^3*f^3)))/2
 

Rubi [A] (verified)

Time = 0.44 (sec) , antiderivative size = 175, normalized size of antiderivative = 0.97, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {2823, 2009}

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.

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

\(\Big \downarrow \) 2823

\(\displaystyle -b n \int \left (\frac {1}{4} \log \left (d f x^2+1\right ) x^3-\frac {x^3}{8}+\frac {x}{4 d f}-\frac {\log \left (d f x^2+1\right )}{4 d^2 f^2 x}\right )dx-\frac {\log \left (d f x^2+1\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^2 f^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{4 d f}+\frac {1}{4} x^4 \log \left (d f x^2+1\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} x^4 \left (a+b \log \left (c x^n\right )\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {\log \left (d f x^2+1\right ) \left (a+b \log \left (c x^n\right )\right )}{4 d^2 f^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{4 d f}+\frac {1}{4} x^4 \log \left (d f x^2+1\right ) \left (a+b \log \left (c x^n\right )\right )-\frac {1}{8} x^4 \left (a+b \log \left (c x^n\right )\right )-b n \left (\frac {\operatorname {PolyLog}\left (2,-d f x^2\right )}{8 d^2 f^2}-\frac {\log \left (d f x^2+1\right )}{16 d^2 f^2}+\frac {3 x^2}{16 d f}+\frac {1}{16} x^4 \log \left (d f x^2+1\right )-\frac {x^4}{16}\right )\)

Input:

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

Output:

(x^2*(a + b*Log[c*x^n]))/(4*d*f) - (x^4*(a + b*Log[c*x^n]))/8 - ((a + b*Lo 
g[c*x^n])*Log[1 + d*f*x^2])/(4*d^2*f^2) + (x^4*(a + b*Log[c*x^n])*Log[1 + 
d*f*x^2])/4 - b*n*((3*x^2)/(16*d*f) - x^4/16 - Log[1 + d*f*x^2]/(16*d^2*f^ 
2) + (x^4*Log[1 + d*f*x^2])/16 + PolyLog[2, -(d*f*x^2)]/(8*d^2*f^2))
 

Defintions of rubi rules used

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2823

Int[Log[(d_.)*((e_) + (f_.)*(x_)^(m_.))^(r_.)]*((a_.) + Log[(c_.)*(x_)^(n_. 
)]*(b_.))*((g_.)*(x_))^(q_.), x_Symbol] :> With[{u = IntHide[(g*x)^q*Log[d* 
(e + f*x^m)^r], x]}, Simp[(a + b*Log[c*x^n])   u, x] - Simp[b*n   Int[1/x 
 u, x], x]] /; FreeQ[{a, b, c, d, e, f, g, r, m, n, q}, x] && (IntegerQ[(q 
+ 1)/m] || (RationalQ[m] && RationalQ[q])) && NeQ[q, -1]
 

Maple [C] (warning: unable to verify)

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

Time = 54.37 (sec) , antiderivative size = 393, normalized size of antiderivative = 2.18

method	result	size
risch	\(\left (\frac {b \,x^{4} \ln \left (d \left (\frac {1}{d}+f \,x^{2}\right )\right )}{4}-\frac {b \left (d^{2} f^{2} x^{4}-2 d f \,x^{2}+2 \ln \left (d \left (\frac {1}{d}+f \,x^{2}\right )\right )+1\right )}{8 d^{2} f^{2}}\right ) \ln \left (x^{n}\right )+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {x^{4} \ln \left (d f \,x^{2}+1\right )}{4}-\frac {d f \left (\frac {\frac {1}{2} x^{4} d f -x^{2}}{2 d^{2} f^{2}}+\frac {\ln \left (d f \,x^{2}+1\right )}{2 d^{3} f^{3}}\right )}{2}\right )+\frac {b n \,x^{4}}{16}-\frac {3 b n \,x^{2}}{16 d f}+\frac {n b \ln \left (x \right )}{8 d^{2} f^{2}}-\frac {b n \,x^{4} \ln \left (d f \,x^{2}+1\right )}{16}+\frac {b n \ln \left (d f \,x^{2}+1\right )}{16 d^{2} f^{2}}+\frac {n b \ln \left (x \right ) \ln \left (d f \,x^{2}+1\right )}{4 d^{2} f^{2}}-\frac {n b \ln \left (x \right ) \ln \left (1+x \sqrt {-d f}\right )}{4 d^{2} f^{2}}-\frac {n b \ln \left (x \right ) \ln \left (1-x \sqrt {-d f}\right )}{4 d^{2} f^{2}}-\frac {n b \operatorname {dilog}\left (1+x \sqrt {-d f}\right )}{4 d^{2} f^{2}}-\frac {n b \operatorname {dilog}\left (1-x \sqrt {-d f}\right )}{4 d^{2} f^{2}}\)	\(393\)

Input:

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

Output:

(1/4*b*x^4*ln(d*(1/d+f*x^2))-1/8*b*(d^2*f^2*x^4-2*d*f*x^2+2*ln(d*(1/d+f*x^ 
2))+1)/d^2/f^2)*ln(x^n)+(1/2*I*Pi*b*csgn(I*x^n)*csgn(I*c*x^n)^2-1/2*I*Pi*b 
*csgn(I*x^n)*csgn(I*c*x^n)*csgn(I*c)-1/2*I*Pi*b*csgn(I*c*x^n)^3+1/2*I*Pi*b 
*csgn(I*c*x^n)^2*csgn(I*c)+b*ln(c)+a)*(1/4*x^4*ln(d*f*x^2+1)-1/2*d*f*(1/2/ 
d^2/f^2*(1/2*x^4*d*f-x^2)+1/2/d^3/f^3*ln(d*f*x^2+1)))+1/16*b*n*x^4-3/16*b* 
n*x^2/d/f+1/8*n*b/d^2/f^2*ln(x)-1/16*b*n*x^4*ln(d*f*x^2+1)+1/16*b*n*ln(d*f 
*x^2+1)/d^2/f^2+1/4*n*b/d^2/f^2*ln(x)*ln(d*f*x^2+1)-1/4*n*b/d^2/f^2*ln(x)* 
ln(1+x*(-d*f)^(1/2))-1/4*n*b/d^2/f^2*ln(x)*ln(1-x*(-d*f)^(1/2))-1/4*n*b/d^ 
2/f^2*dilog(1+x*(-d*f)^(1/2))-1/4*n*b/d^2/f^2*dilog(1-x*(-d*f)^(1/2))
 

Fricas [F]

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

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

Output:

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

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

1/16*(4*b*x^4*log(x^n) - (b*(n - 4*log(c)) - 4*a)*x^4)*log(d*f*x^2 + 1) - 
integrate(1/8*(4*b*d*f*x^5*log(x^n) + (4*a*d*f - (d*f*n - 4*d*f*log(c))*b) 
*x^5)/(d*f*x^2 + 1), x)
 

Giac [F(-2)]

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

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

Output:

Exception raised: TypeError >> an error occurred running a Giac command:IN 
PUT:sage2:=int(sage0,sageVARx):;OUTPUT:sym2poly/r2sym(const gen & e,const 
index_m & i,const vecteur & l) Error: Bad Argument Value
                                                                                    
                                                                                    
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

\[ \int x^3 \left (a+b \log \left (c x^n\right )\right ) \log \left (d \left (\frac {1}{d}+f x^2\right )\right ) \, dx=\frac {8 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{d f \,x^{3}+x}d x \right ) b n +4 \,\mathrm {log}\left (d f \,x^{2}+1\right ) \mathrm {log}\left (x^{n} c \right ) b \,d^{2} f^{2} n \,x^{4}+4 \,\mathrm {log}\left (d f \,x^{2}+1\right ) a \,d^{2} f^{2} n \,x^{4}-4 \,\mathrm {log}\left (d f \,x^{2}+1\right ) a n -\mathrm {log}\left (d f \,x^{2}+1\right ) b \,d^{2} f^{2} n^{2} x^{4}+\mathrm {log}\left (d f \,x^{2}+1\right ) b \,n^{2}-4 \mathrm {log}\left (x^{n} c \right )^{2} b -2 \,\mathrm {log}\left (x^{n} c \right ) b \,d^{2} f^{2} n \,x^{4}+4 \,\mathrm {log}\left (x^{n} c \right ) b d f n \,x^{2}-2 a \,d^{2} f^{2} n \,x^{4}+4 a d f n \,x^{2}+b \,d^{2} f^{2} n^{2} x^{4}-3 b d f \,n^{2} x^{2}}{16 d^{2} f^{2} n} \] Input:

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

Output:

(8*int(log(x**n*c)/(d*f*x**3 + x),x)*b*n + 4*log(d*f*x**2 + 1)*log(x**n*c) 
*b*d**2*f**2*n*x**4 + 4*log(d*f*x**2 + 1)*a*d**2*f**2*n*x**4 - 4*log(d*f*x 
**2 + 1)*a*n - log(d*f*x**2 + 1)*b*d**2*f**2*n**2*x**4 + log(d*f*x**2 + 1) 
*b*n**2 - 4*log(x**n*c)**2*b - 2*log(x**n*c)*b*d**2*f**2*n*x**4 + 4*log(x* 
*n*c)*b*d*f*n*x**2 - 2*a*d**2*f**2*n*x**4 + 4*a*d*f*n*x**2 + b*d**2*f**2*n 
**2*x**4 - 3*b*d*f*n**2*x**2)/(16*d**2*f**2*n)