\(\int \frac {x^5 (a+b \log (c x^n))}{(d+e x^2)^2} \, dx\) [221]

Optimal result
Mathematica [C] (verified)
Rubi [A] (verified)
Maple [C] (warning: unable to verify)
Fricas [F]
Sympy [A] (verification not implemented)
Maxima [F]
Giac [F]
Mupad [F(-1)]
Reduce [F]

Optimal result

Integrand size = 23, antiderivative size = 129 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=-\frac {b n x^2}{4 e^2}+\frac {x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2}+\frac {d x^2 \left (a+b \log \left (c x^n\right )\right )}{2 e^2 \left (d+e x^2\right )}-\frac {b d n \log \left (d+e x^2\right )}{4 e^3}-\frac {d \left (a+b \log \left (c x^n\right )\right ) \log \left (1+\frac {e x^2}{d}\right )}{e^3}-\frac {b d n \operatorname {PolyLog}\left (2,-\frac {e x^2}{d}\right )}{2 e^3} \] Output:

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

Mathematica [C] (verified)

Result contains complex when optimal does not.

Time = 0.61 (sec) , antiderivative size = 287, normalized size of antiderivative = 2.22 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\frac {2 e x^2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )-\frac {2 d^2 \left (a-b n \log (x)+b \log \left (c x^n\right )\right )}{d+e x^2}-4 d \left (a-b n \log (x)+b \log \left (c x^n\right )\right ) \log \left (d+e x^2\right )+b n \left (\frac {d \sqrt {e} x \log (x)}{-i \sqrt {d}+\sqrt {e} x}+\frac {d \sqrt {e} x \log (x)}{i \sqrt {d}+\sqrt {e} x}+e x^2 (-1+2 \log (x))-d \log \left (i \sqrt {d}-\sqrt {e} x\right )-d \log \left (i \sqrt {d}+\sqrt {e} x\right )-4 d \left (\log (x) \log \left (1+\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\operatorname {PolyLog}\left (2,-\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )-4 d \left (\log (x) \log \left (1-\frac {i \sqrt {e} x}{\sqrt {d}}\right )+\operatorname {PolyLog}\left (2,\frac {i \sqrt {e} x}{\sqrt {d}}\right )\right )\right )}{4 e^3} \] Input:

Integrate[(x^5*(a + b*Log[c*x^n]))/(d + e*x^2)^2,x]
 

Output:

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

Rubi [A] (verified)

Time = 0.45 (sec) , antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.087, Rules used = {2793, 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.

\(\displaystyle \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx\)

\(\Big \downarrow \) 2793

\(\displaystyle \int \left (\frac {d^2 x \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )^2}-\frac {2 d x \left (a+b \log \left (c x^n\right )\right )}{e^2 \left (d+e x^2\right )}+\frac {x \left (a+b \log \left (c x^n\right )\right )}{e^2}\right )dx\)

\(\Big \downarrow \) 2009

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

Input:

Int[(x^5*(a + b*Log[c*x^n]))/(d + e*x^2)^2,x]
 

Output:

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

Defintions of rubi rules used

rule 2009
Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2793
Int[((a_.) + Log[(c_.)*(x_)^(n_.)]*(b_.))*((f_.)*(x_))^(m_.)*((d_) + (e_.)* 
(x_)^(r_.))^(q_.), x_Symbol] :> With[{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] && Integer 
Q[r]))
 
Maple [C] (warning: unable to verify)

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

Time = 0.82 (sec) , antiderivative size = 351, normalized size of antiderivative = 2.72

method result size
risch \(\frac {b \ln \left (x^{n}\right ) x^{2}}{2 e^{2}}-\frac {b \ln \left (x^{n}\right ) d^{2}}{2 e^{3} \left (e \,x^{2}+d \right )}-\frac {b \ln \left (x^{n}\right ) d \ln \left (e \,x^{2}+d \right )}{e^{3}}-\frac {b n \,x^{2}}{4 e^{2}}-\frac {b d n \ln \left (e \,x^{2}+d \right )}{4 e^{3}}+\frac {b n d \ln \left (x \right )}{2 e^{3}}+\frac {b n d \ln \left (x \right ) \ln \left (e \,x^{2}+d \right )}{e^{3}}-\frac {b n d \ln \left (x \right ) \ln \left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{e^{3}}-\frac {b n d \ln \left (x \right ) \ln \left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{e^{3}}-\frac {b n d \operatorname {dilog}\left (\frac {-e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{e^{3}}-\frac {b n d \operatorname {dilog}\left (\frac {e x +\sqrt {-d e}}{\sqrt {-d e}}\right )}{e^{3}}+\left (\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right )^{2}}{2}-\frac {i b \pi \,\operatorname {csgn}\left (i x^{n}\right ) \operatorname {csgn}\left (i c \,x^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{3}}{2}+\frac {i b \pi \operatorname {csgn}\left (i c \,x^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+b \ln \left (c \right )+a \right ) \left (\frac {x^{2}}{2 e^{2}}-\frac {d \left (\frac {d}{e \left (e \,x^{2}+d \right )}+\frac {2 \ln \left (e \,x^{2}+d \right )}{e}\right )}{2 e^{2}}\right )\) \(351\)

Input:

int(x^5*(a+b*ln(c*x^n))/(e*x^2+d)^2,x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [F]

\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \] Input:

integrate(x^5*(a+b*log(c*x^n))/(e*x^2+d)^2,x, algorithm="fricas")
 

Output:

integral((b*x^5*log(c*x^n) + a*x^5)/(e^2*x^4 + 2*d*e*x^2 + d^2), x)
 

Sympy [A] (verification not implemented)

Time = 45.28 (sec) , antiderivative size = 316, normalized size of antiderivative = 2.45 \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\frac {a d^{2} \left (\begin {cases} \frac {x^{2}}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x^{2}} & \text {otherwise} \end {cases}\right )}{2 e^{2}} - \frac {a d \left (\begin {cases} \frac {x^{2}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{2} \right )}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}} + \frac {a x^{2}}{2 e^{2}} - \frac {b d^{2} n \left (\begin {cases} \frac {x^{2}}{2 d^{2}} & \text {for}\: e = 0 \\- \frac {\log {\left (x \right )}}{d e} + \frac {\log {\left (\frac {d}{e} + x^{2} \right )}}{2 d e} & \text {otherwise} \end {cases}\right )}{2 e^{2}} + \frac {b d^{2} \left (\begin {cases} \frac {x^{2}}{d^{2}} & \text {for}\: e = 0 \\- \frac {1}{d e + e^{2} x^{2}} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{2 e^{2}} + \frac {b d n \left (\begin {cases} \frac {x^{2}}{2 d} & \text {for}\: e = 0 \\\frac {\begin {cases} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \wedge \left |{x}\right | < 1 \\\log {\left (d \right )} \log {\left (x \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \left |{x}\right | < 1 \\- \log {\left (d \right )} \log {\left (\frac {1}{x} \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {for}\: \frac {1}{\left |{x}\right |} < 1 \\- {G_{2, 2}^{2, 0}\left (\begin {matrix} & 1, 1 \\0, 0 & \end {matrix} \middle | {x} \right )} \log {\left (d \right )} + {G_{2, 2}^{0, 2}\left (\begin {matrix} 1, 1 & \\ & 0, 0 \end {matrix} \middle | {x} \right )} \log {\left (d \right )} - \frac {\operatorname {Li}_{2}\left (\frac {e x^{2} e^{i \pi }}{d}\right )}{2} & \text {otherwise} \end {cases}}{e} & \text {otherwise} \end {cases}\right )}{e^{2}} - \frac {b d \left (\begin {cases} \frac {x^{2}}{d} & \text {for}\: e = 0 \\\frac {\log {\left (d + e x^{2} \right )}}{e} & \text {otherwise} \end {cases}\right ) \log {\left (c x^{n} \right )}}{e^{2}} - \frac {b n x^{2}}{4 e^{2}} + \frac {b x^{2} \log {\left (c x^{n} \right )}}{2 e^{2}} \] Input:

integrate(x**5*(a+b*ln(c*x**n))/(e*x**2+d)**2,x)
 

Output:

a*d**2*Piecewise((x**2/d**2, Eq(e, 0)), (-1/(d*e + e**2*x**2), True))/(2*e 
**2) - a*d*Piecewise((x**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))/e**2 + 
 a*x**2/(2*e**2) - b*d**2*n*Piecewise((x**2/(2*d**2), Eq(e, 0)), (-log(x)/ 
(d*e) + log(d/e + x**2)/(2*d*e), True))/(2*e**2) + b*d**2*Piecewise((x**2/ 
d**2, Eq(e, 0)), (-1/(d*e + e**2*x**2), True))*log(c*x**n)/(2*e**2) + b*d* 
n*Piecewise((x**2/(2*d), Eq(e, 0)), (Piecewise((-polylog(2, e*x**2*exp_pol 
ar(I*pi)/d)/2, (Abs(x) < 1) & (1/Abs(x) < 1)), (log(d)*log(x) - polylog(2, 
 e*x**2*exp_polar(I*pi)/d)/2, Abs(x) < 1), (-log(d)*log(1/x) - polylog(2, 
e*x**2*exp_polar(I*pi)/d)/2, 1/Abs(x) < 1), (-meijerg(((), (1, 1)), ((0, 0 
), ()), x)*log(d) + meijerg(((1, 1), ()), ((), (0, 0)), x)*log(d) - polylo 
g(2, e*x**2*exp_polar(I*pi)/d)/2, True))/e, True))/e**2 - b*d*Piecewise((x 
**2/d, Eq(e, 0)), (log(d + e*x**2)/e, True))*log(c*x**n)/e**2 - b*n*x**2/( 
4*e**2) + b*x**2*log(c*x**n)/(2*e**2)
 

Maxima [F]

\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \] Input:

integrate(x^5*(a+b*log(c*x^n))/(e*x^2+d)^2,x, algorithm="maxima")
 

Output:

-1/2*a*(d^2/(e^4*x^2 + d*e^3) - x^2/e^2 + 2*d*log(e*x^2 + d)/e^3) + b*inte 
grate((x^5*log(c) + x^5*log(x^n))/(e^2*x^4 + 2*d*e*x^2 + d^2), x)
 

Giac [F]

\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int { \frac {{\left (b \log \left (c x^{n}\right ) + a\right )} x^{5}}{{\left (e x^{2} + d\right )}^{2}} \,d x } \] Input:

integrate(x^5*(a+b*log(c*x^n))/(e*x^2+d)^2,x, algorithm="giac")
 

Output:

integrate((b*log(c*x^n) + a)*x^5/(e*x^2 + d)^2, x)
                                                                                    
                                                                                    
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\int \frac {x^5\,\left (a+b\,\ln \left (c\,x^n\right )\right )}{{\left (e\,x^2+d\right )}^2} \,d x \] Input:

int((x^5*(a + b*log(c*x^n)))/(d + e*x^2)^2,x)
 

Output:

int((x^5*(a + b*log(c*x^n)))/(d + e*x^2)^2, x)
 

Reduce [F]

\[ \int \frac {x^5 \left (a+b \log \left (c x^n\right )\right )}{\left (d+e x^2\right )^2} \, dx=\frac {8 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{2} x^{5}+2 d e \,x^{3}+d^{2} x}d x \right ) b \,d^{4} n +8 \left (\int \frac {\mathrm {log}\left (x^{n} c \right )}{e^{2} x^{5}+2 d e \,x^{3}+d^{2} x}d x \right ) b \,d^{3} e n \,x^{2}-4 \,\mathrm {log}\left (e \,x^{2}+d \right ) a \,d^{2} n -4 \,\mathrm {log}\left (e \,x^{2}+d \right ) a d e n \,x^{2}-3 \,\mathrm {log}\left (e \,x^{2}+d \right ) b \,d^{2} n^{2}-3 \,\mathrm {log}\left (e \,x^{2}+d \right ) b d e \,n^{2} x^{2}-4 \mathrm {log}\left (x^{n} c \right )^{2} b \,d^{2}-4 \mathrm {log}\left (x^{n} c \right )^{2} b d e \,x^{2}+8 \,\mathrm {log}\left (x^{n} c \right ) b d e n \,x^{2}+2 \,\mathrm {log}\left (x^{n} c \right ) b \,e^{2} n \,x^{4}+4 a d e n \,x^{2}+2 a \,e^{2} n \,x^{4}-b d e \,n^{2} x^{2}-b \,e^{2} n^{2} x^{4}}{4 e^{3} n \left (e \,x^{2}+d \right )} \] Input:

int(x^5*(a+b*log(c*x^n))/(e*x^2+d)^2,x)
 

Output:

(8*int(log(x**n*c)/(d**2*x + 2*d*e*x**3 + e**2*x**5),x)*b*d**4*n + 8*int(l 
og(x**n*c)/(d**2*x + 2*d*e*x**3 + e**2*x**5),x)*b*d**3*e*n*x**2 - 4*log(d 
+ e*x**2)*a*d**2*n - 4*log(d + e*x**2)*a*d*e*n*x**2 - 3*log(d + e*x**2)*b* 
d**2*n**2 - 3*log(d + e*x**2)*b*d*e*n**2*x**2 - 4*log(x**n*c)**2*b*d**2 - 
4*log(x**n*c)**2*b*d*e*x**2 + 8*log(x**n*c)*b*d*e*n*x**2 + 2*log(x**n*c)*b 
*e**2*n*x**4 + 4*a*d*e*n*x**2 + 2*a*e**2*n*x**4 - b*d*e*n**2*x**2 - b*e**2 
*n**2*x**4)/(4*e**3*n*(d + e*x**2))