Integrand size = 19, antiderivative size = 86 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^2} \, dx=\frac {\sqrt {b^2-4 a c} n \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a}+\frac {b n \log (x)}{a}-\frac {b n \log \left (a+b x+c x^2\right )}{2 a}-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x} \] Output:
(-4*a*c+b^2)^(1/2)*n*arctanh((2*c*x+b)/(-4*a*c+b^2)^(1/2))/a+b*n*ln(x)/a-1 /2*b*n*ln(c*x^2+b*x+a)/a-ln(d*(c*x^2+b*x+a)^n)/x
Time = 0.12 (sec) , antiderivative size = 87, normalized size of antiderivative = 1.01 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^2} \, dx=\frac {2 \sqrt {-b^2+4 a c} n \arctan \left (\frac {b+2 c x}{\sqrt {-b^2+4 a c}}\right )+2 b n \log (x)-b n \log (a+x (b+c x))-\frac {2 a \log \left (d (a+x (b+c x))^n\right )}{x}}{2 a} \] Input:
Integrate[Log[d*(a + b*x + c*x^2)^n]/x^2,x]
Output:
(2*Sqrt[-b^2 + 4*a*c]*n*ArcTan[(b + 2*c*x)/Sqrt[-b^2 + 4*a*c]] + 2*b*n*Log [x] - b*n*Log[a + x*(b + c*x)] - (2*a*Log[d*(a + x*(b + c*x))^n])/x)/(2*a)
Time = 0.34 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {3005, 1200, 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.
\(\displaystyle \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^2} \, dx\) |
\(\Big \downarrow \) 3005 |
\(\displaystyle n \int \frac {b+2 c x}{x \left (c x^2+b x+a\right )}dx-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x}\) |
\(\Big \downarrow \) 1200 |
\(\displaystyle n \int \left (\frac {b}{a x}+\frac {-b^2-c x b+2 a c}{a \left (c x^2+b x+a\right )}\right )dx-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x}\) |
\(\Big \downarrow \) 2009 |
\(\displaystyle n \left (\frac {\sqrt {b^2-4 a c} \text {arctanh}\left (\frac {b+2 c x}{\sqrt {b^2-4 a c}}\right )}{a}-\frac {b \log \left (a+b x+c x^2\right )}{2 a}+\frac {b \log (x)}{a}\right )-\frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x}\) |
Input:
Int[Log[d*(a + b*x + c*x^2)^n]/x^2,x]
Output:
n*((Sqrt[b^2 - 4*a*c]*ArcTanh[(b + 2*c*x)/Sqrt[b^2 - 4*a*c]])/a + (b*Log[x ])/a - (b*Log[a + b*x + c*x^2])/(2*a)) - Log[d*(a + b*x + c*x^2)^n]/x
Int[(((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))^(n_.))/((a_.) + (b_.)* (x_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(d + e*x)^m*((f + g* x)^n/(a + b*x + c*x^2)), x], x] /; FreeQ[{a, b, c, d, e, f, g, m}, x] && In tegersQ[n]
Int[((a_.) + Log[(c_.)*(RFx_)^(p_.)]*(b_.))^(n_.)*((d_.) + (e_.)*(x_))^(m_. ), x_Symbol] :> Simp[(d + e*x)^(m + 1)*((a + b*Log[c*RFx^p])^n/(e*(m + 1))) , x] - Simp[b*n*(p/(e*(m + 1))) Int[SimplifyIntegrand[(d + e*x)^(m + 1)*( a + b*Log[c*RFx^p])^(n - 1)*(D[RFx, x]/RFx), x], x], x] /; FreeQ[{a, b, c, d, e, m, p}, x] && RationalFunctionQ[RFx, x] && IGtQ[n, 0] && (EqQ[n, 1] || IntegerQ[m]) && NeQ[m, -1]
Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.10
method | result | size |
parts | \(-\frac {\ln \left (d \left (c \,x^{2}+b x +a \right )^{n}\right )}{x}+n \left (\frac {-\frac {b \ln \left (c \,x^{2}+b x +a \right )}{2}+\frac {2 \left (2 a c -\frac {b^{2}}{2}\right ) \arctan \left (\frac {2 x c +b}{\sqrt {4 a c -b^{2}}}\right )}{\sqrt {4 a c -b^{2}}}}{a}+\frac {b \ln \left (x \right )}{a}\right )\) | \(95\) |
risch | \(-\frac {\ln \left (\left (c \,x^{2}+b x +a \right )^{n}\right )}{x}-\frac {-i \pi a \,\operatorname {csgn}\left (i d \right ) \operatorname {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) \operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )+i \pi a \,\operatorname {csgn}\left (i d \right ) {\operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )}^{2}+i \pi a \,\operatorname {csgn}\left (i \left (c \,x^{2}+b x +a \right )^{n}\right ) {\operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )}^{2}-i \pi a {\operatorname {csgn}\left (i d \left (c \,x^{2}+b x +a \right )^{n}\right )}^{3}-n \ln \left (\left (-5 a b c \sqrt {-4 a c +b^{2}}+2 \sqrt {-4 a c +b^{2}}\, b^{3}-4 a^{2} c^{2}+9 a \,b^{2} c -2 b^{4}\right ) x -2 a^{2} c \sqrt {-4 a c +b^{2}}+2 a \,b^{2} \sqrt {-4 a c +b^{2}}+8 a^{2} b c -2 a \,b^{3}\right ) \sqrt {-4 a c +b^{2}}\, x +n \ln \left (\left (-5 a b c \sqrt {-4 a c +b^{2}}+2 \sqrt {-4 a c +b^{2}}\, b^{3}-4 a^{2} c^{2}+9 a \,b^{2} c -2 b^{4}\right ) x -2 a^{2} c \sqrt {-4 a c +b^{2}}+2 a \,b^{2} \sqrt {-4 a c +b^{2}}+8 a^{2} b c -2 a \,b^{3}\right ) b x +n \ln \left (\left (5 a b c \sqrt {-4 a c +b^{2}}-2 \sqrt {-4 a c +b^{2}}\, b^{3}-4 a^{2} c^{2}+9 a \,b^{2} c -2 b^{4}\right ) x +2 a^{2} c \sqrt {-4 a c +b^{2}}-2 a \,b^{2} \sqrt {-4 a c +b^{2}}+8 a^{2} b c -2 a \,b^{3}\right ) \sqrt {-4 a c +b^{2}}\, x +n \ln \left (\left (5 a b c \sqrt {-4 a c +b^{2}}-2 \sqrt {-4 a c +b^{2}}\, b^{3}-4 a^{2} c^{2}+9 a \,b^{2} c -2 b^{4}\right ) x +2 a^{2} c \sqrt {-4 a c +b^{2}}-2 a \,b^{2} \sqrt {-4 a c +b^{2}}+8 a^{2} b c -2 a \,b^{3}\right ) b x -2 n b \ln \left (x \right ) x +2 \ln \left (d \right ) a}{2 a x}\) | \(612\) |
Input:
int(ln(d*(c*x^2+b*x+a)^n)/x^2,x,method=_RETURNVERBOSE)
Output:
-ln(d*(c*x^2+b*x+a)^n)/x+n*(1/a*(-1/2*b*ln(c*x^2+b*x+a)+2*(2*a*c-1/2*b^2)/ (4*a*c-b^2)^(1/2)*arctan((2*c*x+b)/(4*a*c-b^2)^(1/2)))+1/a*b*ln(x))
Time = 0.09 (sec) , antiderivative size = 199, normalized size of antiderivative = 2.31 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^2} \, dx=\left [\frac {2 \, b n x \log \left (x\right ) + \sqrt {b^{2} - 4 \, a c} n x \log \left (\frac {2 \, c^{2} x^{2} + 2 \, b c x + b^{2} - 2 \, a c + \sqrt {b^{2} - 4 \, a c} {\left (2 \, c x + b\right )}}{c x^{2} + b x + a}\right ) - {\left (b n x + 2 \, a n\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, a \log \left (d\right )}{2 \, a x}, \frac {2 \, b n x \log \left (x\right ) + 2 \, \sqrt {-b^{2} + 4 \, a c} n x \arctan \left (-\frac {\sqrt {-b^{2} + 4 \, a c} {\left (2 \, c x + b\right )}}{b^{2} - 4 \, a c}\right ) - {\left (b n x + 2 \, a n\right )} \log \left (c x^{2} + b x + a\right ) - 2 \, a \log \left (d\right )}{2 \, a x}\right ] \] Input:
integrate(log(d*(c*x^2+b*x+a)^n)/x^2,x, algorithm="fricas")
Output:
[1/2*(2*b*n*x*log(x) + sqrt(b^2 - 4*a*c)*n*x*log((2*c^2*x^2 + 2*b*c*x + b^ 2 - 2*a*c + sqrt(b^2 - 4*a*c)*(2*c*x + b))/(c*x^2 + b*x + a)) - (b*n*x + 2 *a*n)*log(c*x^2 + b*x + a) - 2*a*log(d))/(a*x), 1/2*(2*b*n*x*log(x) + 2*sq rt(-b^2 + 4*a*c)*n*x*arctan(-sqrt(-b^2 + 4*a*c)*(2*c*x + b)/(b^2 - 4*a*c)) - (b*n*x + 2*a*n)*log(c*x^2 + b*x + a) - 2*a*log(d))/(a*x)]
Timed out. \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^2} \, dx=\text {Timed out} \] Input:
integrate(ln(d*(c*x**2+b*x+a)**n)/x**2,x)
Output:
Timed out
Exception generated. \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^2} \, dx=\text {Exception raised: ValueError} \] Input:
integrate(log(d*(c*x^2+b*x+a)^n)/x^2,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(4*a*c-b^2>0)', see `assume?` for more deta
Time = 0.13 (sec) , antiderivative size = 99, normalized size of antiderivative = 1.15 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^2} \, dx=-\frac {b n \log \left (c x^{2} + b x + a\right )}{2 \, a} + \frac {b n \log \left (x\right )}{a} - \frac {n \log \left (c x^{2} + b x + a\right )}{x} - \frac {{\left (b^{2} n - 4 \, a c n\right )} \arctan \left (\frac {2 \, c x + b}{\sqrt {-b^{2} + 4 \, a c}}\right )}{\sqrt {-b^{2} + 4 \, a c} a} - \frac {\log \left (d\right )}{x} \] Input:
integrate(log(d*(c*x^2+b*x+a)^n)/x^2,x, algorithm="giac")
Output:
-1/2*b*n*log(c*x^2 + b*x + a)/a + b*n*log(x)/a - n*log(c*x^2 + b*x + a)/x - (b^2*n - 4*a*c*n)*arctan((2*c*x + b)/sqrt(-b^2 + 4*a*c))/(sqrt(-b^2 + 4* a*c)*a) - log(d)/x
Time = 26.36 (sec) , antiderivative size = 262, normalized size of antiderivative = 3.05 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^2} \, dx=\frac {b\,n\,\ln \left (x\right )}{a}-\frac {\ln \left (2\,b\,c^2\,n^2+4\,c^3\,n^2\,x-\frac {n\,\left (b-\sqrt {b^2-4\,a\,c}\right )\,\left (b^2\,c\,n-2\,a\,c^2\,n+b\,c^2\,n\,x+\frac {c\,n\,\left (b-\sqrt {b^2-4\,a\,c}\right )\,\left (2\,x\,b^2+a\,b-6\,a\,c\,x\right )}{2\,a}\right )}{2\,a}\right )\,\left (b\,n-n\,\sqrt {b^2-4\,a\,c}\right )}{2\,a}-\frac {\ln \left (2\,b\,c^2\,n^2+4\,c^3\,n^2\,x-\frac {n\,\left (b+\sqrt {b^2-4\,a\,c}\right )\,\left (b^2\,c\,n-2\,a\,c^2\,n+b\,c^2\,n\,x+\frac {c\,n\,\left (b+\sqrt {b^2-4\,a\,c}\right )\,\left (2\,x\,b^2+a\,b-6\,a\,c\,x\right )}{2\,a}\right )}{2\,a}\right )\,\left (b\,n+n\,\sqrt {b^2-4\,a\,c}\right )}{2\,a}-\frac {\ln \left (d\,{\left (c\,x^2+b\,x+a\right )}^n\right )}{x} \] Input:
int(log(d*(a + b*x + c*x^2)^n)/x^2,x)
Output:
(b*n*log(x))/a - (log(2*b*c^2*n^2 + 4*c^3*n^2*x - (n*(b - (b^2 - 4*a*c)^(1 /2))*(b^2*c*n - 2*a*c^2*n + b*c^2*n*x + (c*n*(b - (b^2 - 4*a*c)^(1/2))*(a* b + 2*b^2*x - 6*a*c*x))/(2*a)))/(2*a))*(b*n - n*(b^2 - 4*a*c)^(1/2)))/(2*a ) - (log(2*b*c^2*n^2 + 4*c^3*n^2*x - (n*(b + (b^2 - 4*a*c)^(1/2))*(b^2*c*n - 2*a*c^2*n + b*c^2*n*x + (c*n*(b + (b^2 - 4*a*c)^(1/2))*(a*b + 2*b^2*x - 6*a*c*x))/(2*a)))/(2*a))*(b*n + n*(b^2 - 4*a*c)^(1/2)))/(2*a) - log(d*(a + b*x + c*x^2)^n)/x
Time = 0.16 (sec) , antiderivative size = 89, normalized size of antiderivative = 1.03 \[ \int \frac {\log \left (d \left (a+b x+c x^2\right )^n\right )}{x^2} \, dx=\frac {2 \sqrt {4 a c -b^{2}}\, \mathit {atan} \left (\frac {2 c x +b}{\sqrt {4 a c -b^{2}}}\right ) n x -2 \,\mathrm {log}\left (\left (c \,x^{2}+b x +a \right )^{n} d \right ) a -\mathrm {log}\left (\left (c \,x^{2}+b x +a \right )^{n} d \right ) b x +2 \,\mathrm {log}\left (x \right ) b n x}{2 a x} \] Input:
int(log(d*(c*x^2+b*x+a)^n)/x^2,x)
Output:
(2*sqrt(4*a*c - b**2)*atan((b + 2*c*x)/sqrt(4*a*c - b**2))*n*x - 2*log((a + b*x + c*x**2)**n*d)*a - log((a + b*x + c*x**2)**n*d)*b*x + 2*log(x)*b*n* x)/(2*a*x)