\(\int \frac {\log (c (a+b x)^n)}{d+e x^2} \, dx\) [331]

Optimal result

Integrand size = 20, antiderivative size = 229 \[ \int \frac {\log \left (c (a+b x)^n\right )}{d+e x^2} \, dx=\frac {\log \left (c (a+b x)^n\right ) \log \left (\frac {b \left (\sqrt {-d}-\sqrt {e} x\right )}{b \sqrt {-d}+a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\log \left (c (a+b x)^n\right ) \log \left (\frac {b \left (\sqrt {-d}+\sqrt {e} x\right )}{b \sqrt {-d}-a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {n \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} (a+b x)}{b \sqrt {-d}-a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (a+b x)}{b \sqrt {-d}+a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}} \] Output:

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

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.78 \[ \int \frac {\log \left (c (a+b x)^n\right )}{d+e x^2} \, dx=\frac {\log \left (c (a+b x)^n\right ) \left (\log \left (\frac {b \left (\sqrt {-d}-\sqrt {e} x\right )}{b \sqrt {-d}+a \sqrt {e}}\right )-\log \left (\frac {b \left (\sqrt {-d}+\sqrt {e} x\right )}{b \sqrt {-d}-a \sqrt {e}}\right )\right )-n \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} (a+b x)}{b \sqrt {-d}-a \sqrt {e}}\right )+n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (a+b x)}{b \sqrt {-d}+a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}} \] Input:

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

Output:

(Log[c*(a + b*x)^n]*(Log[(b*(Sqrt[-d] - Sqrt[e]*x))/(b*Sqrt[-d] + a*Sqrt[e 
])] - Log[(b*(Sqrt[-d] + Sqrt[e]*x))/(b*Sqrt[-d] - a*Sqrt[e])]) - n*PolyLo 
g[2, -((Sqrt[e]*(a + b*x))/(b*Sqrt[-d] - a*Sqrt[e]))] + n*PolyLog[2, (Sqrt 
[e]*(a + b*x))/(b*Sqrt[-d] + a*Sqrt[e])])/(2*Sqrt[-d]*Sqrt[e])
 

Rubi [A] (verified)

Time = 0.69 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {2856, 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.

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

\(\Big \downarrow \) 2856

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

\(\Big \downarrow \) 2009

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

Input:

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

Output:

(Log[c*(a + b*x)^n]*Log[(b*(Sqrt[-d] - Sqrt[e]*x))/(b*Sqrt[-d] + a*Sqrt[e] 
)])/(2*Sqrt[-d]*Sqrt[e]) - (Log[c*(a + b*x)^n]*Log[(b*(Sqrt[-d] + Sqrt[e]* 
x))/(b*Sqrt[-d] - a*Sqrt[e])])/(2*Sqrt[-d]*Sqrt[e]) - (n*PolyLog[2, -((Sqr 
t[e]*(a + b*x))/(b*Sqrt[-d] - a*Sqrt[e]))])/(2*Sqrt[-d]*Sqrt[e]) + (n*Poly 
Log[2, (Sqrt[e]*(a + b*x))/(b*Sqrt[-d] + a*Sqrt[e])])/(2*Sqrt[-d]*Sqrt[e])
 

Defintions of rubi rules used

rule 2009

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

rule 2856

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_. 
)*(x_)^(r_))^(q_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*Log[c*(d + e*x) 
^n])^p, (f + g*x^r)^q, x], x] /; FreeQ[{a, b, c, d, e, f, g, n, r}, x] && I 
GtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))
 

Maple [C] (warning: unable to verify)

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

Time = 1.47 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.71

method	result	size
risch	\(-\frac {\arctan \left (\frac {2 e \left (b x +a \right )-2 e a}{2 b \sqrt {d e}}\right ) \ln \left (b x +a \right ) n}{\sqrt {d e}}+\frac {\arctan \left (\frac {2 e \left (b x +a \right )-2 e a}{2 b \sqrt {d e}}\right ) \ln \left (\left (b x +a \right )^{n}\right )}{\sqrt {d e}}+\frac {n \ln \left (b x +a \right ) \ln \left (\frac {b \sqrt {-d e}-e \left (b x +a \right )+e a}{b \sqrt {-d e}+e a}\right )}{2 \sqrt {-d e}}-\frac {n \ln \left (b x +a \right ) \ln \left (\frac {b \sqrt {-d e}+e \left (b x +a \right )-e a}{b \sqrt {-d e}-e a}\right )}{2 \sqrt {-d e}}+\frac {n \operatorname {dilog}\left (\frac {b \sqrt {-d e}-e \left (b x +a \right )+e a}{b \sqrt {-d e}+e a}\right )}{2 \sqrt {-d e}}-\frac {n \operatorname {dilog}\left (\frac {b \sqrt {-d e}+e \left (b x +a \right )-e a}{b \sqrt {-d e}-e a}\right )}{2 \sqrt {-d e}}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i c \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{n}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \arctan \left (\frac {x e}{\sqrt {d e}}\right )}{\sqrt {d e}}\)	\(392\)

Input:

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

Output:

-1/(d*e)^(1/2)*arctan(1/2*(2*e*(b*x+a)-2*e*a)/b/(d*e)^(1/2))*ln(b*x+a)*n+1 
/(d*e)^(1/2)*arctan(1/2*(2*e*(b*x+a)-2*e*a)/b/(d*e)^(1/2))*ln((b*x+a)^n)+1 
/2*n*ln(b*x+a)/(-d*e)^(1/2)*ln((b*(-d*e)^(1/2)-e*(b*x+a)+e*a)/(b*(-d*e)^(1 
/2)+e*a))-1/2*n*ln(b*x+a)/(-d*e)^(1/2)*ln((b*(-d*e)^(1/2)+e*(b*x+a)-e*a)/( 
b*(-d*e)^(1/2)-e*a))+1/2*n/(-d*e)^(1/2)*dilog((b*(-d*e)^(1/2)-e*(b*x+a)+e* 
a)/(b*(-d*e)^(1/2)+e*a))-1/2*n/(-d*e)^(1/2)*dilog((b*(-d*e)^(1/2)+e*(b*x+a 
)-e*a)/(b*(-d*e)^(1/2)-e*a))+(1/2*I*Pi*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^ 
n)^2-1/2*I*Pi*csgn(I*(b*x+a)^n)*csgn(I*c*(b*x+a)^n)*csgn(I*c)-1/2*I*Pi*csg 
n(I*c*(b*x+a)^n)^3+1/2*I*Pi*csgn(I*c*(b*x+a)^n)^2*csgn(I*c)+ln(c))/(d*e)^( 
1/2)*arctan(x*e/(d*e)^(1/2))
 

Fricas [F]

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

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

Output:

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

Sympy [F]

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

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

Output:

Integral(log(c*(a + b*x)**n)/(d + e*x**2), x)
 

Maxima [F(-2)]

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

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

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(e>0)', see `assume?` for more de 
tails)Is e
 

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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