\(\int \frac {\log (c (a+b x)^n)}{d+e x+f x^2} \, dx\) [349]

Optimal result

Integrand size = 23, antiderivative size = 243 \[ \int \frac {\log \left (c (a+b x)^n\right )}{d+e x+f x^2} \, dx=\frac {\log \left (c (a+b x)^n\right ) \log \left (-\frac {b \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {\log \left (c (a+b x)^n\right ) \log \left (-\frac {b \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {n \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {n \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.15 (sec) , antiderivative size = 243, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {2465, 2441, 2440, 2438} \[ \int \frac {\log \left (c (a+b x)^n\right )}{d+e x+f x^2} \, dx=\frac {\log \left (c (a+b x)^n\right ) \log \left (-\frac {b \left (-\sqrt {e^2-4 d f}+e+2 f x\right )}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {\log \left (c (a+b x)^n\right ) \log \left (-\frac {b \left (\sqrt {e^2-4 d f}+e+2 f x\right )}{2 a f-b \left (\sqrt {e^2-4 d f}+e\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {n \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {n \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}} \]

[In]

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

[Out]

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

Rule 2438

Int[Log[(c_.)*((d_) + (e_.)*(x_)^(n_.))]/(x_), x_Symbol] :> Simp[-PolyLog[2, (-c)*e*x^n]/n, x] /; FreeQ[{c, d,
e, n}, x] && EqQ[c*d, 1]

Rule 2440

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Dist[1/g, Subst[Int[(a +
b*Log[1 + c*e*(x/g)])/x, x], x, f + g*x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[g
+ c*(e*f - d*g), 0]

Rule 2441

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))/((f_.) + (g_.)*(x_)), x_Symbol] :> Simp[Log[e*((f + g
*x)/(e*f - d*g))]*((a + b*Log[c*(d + e*x)^n])/g), x] - Dist[b*e*(n/g), Int[Log[(e*(f + g*x))/(e*f - d*g)]/(d +
e*x), x], x] /; FreeQ[{a, b, c, d, e, f, g, n}, x] && NeQ[e*f - d*g, 0]

Rule 2465

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*(RFx_), x_Symbol] :> With[{u = ExpandIntegrand[
(a + b*Log[c*(d + e*x)^n])^p, RFx, x]}, Int[u, x] /; SumQ[u]] /; FreeQ[{a, b, c, d, e, n}, x] && RationalFunct
ionQ[RFx, x] && IntegerQ[p]

Rubi steps \begin{align*} \text {integral}& = \int \left (\frac {2 f \log \left (c (a+b x)^n\right )}{\sqrt {e^2-4 d f} \left (e-\sqrt {e^2-4 d f}+2 f x\right )}-\frac {2 f \log \left (c (a+b x)^n\right )}{\sqrt {e^2-4 d f} \left (e+\sqrt {e^2-4 d f}+2 f x\right )}\right ) \, dx \\ & = \frac {(2 f) \int \frac {\log \left (c (a+b x)^n\right )}{e-\sqrt {e^2-4 d f}+2 f x} \, dx}{\sqrt {e^2-4 d f}}-\frac {(2 f) \int \frac {\log \left (c (a+b x)^n\right )}{e+\sqrt {e^2-4 d f}+2 f x} \, dx}{\sqrt {e^2-4 d f}} \\ & = \frac {\log \left (c (a+b x)^n\right ) \log \left (-\frac {b \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {\log \left (c (a+b x)^n\right ) \log \left (-\frac {b \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {(b n) \int \frac {\log \left (\frac {b \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{-2 a f+b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{a+b x} \, dx}{\sqrt {e^2-4 d f}}+\frac {(b n) \int \frac {\log \left (\frac {b \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{-2 a f+b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{a+b x} \, dx}{\sqrt {e^2-4 d f}} \\ & = \frac {\log \left (c (a+b x)^n\right ) \log \left (-\frac {b \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {\log \left (c (a+b x)^n\right ) \log \left (-\frac {b \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {n \text {Subst}\left (\int \frac {\log \left (1+\frac {2 f x}{-2 a f+b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{x} \, dx,x,a+b x\right )}{\sqrt {e^2-4 d f}}+\frac {n \text {Subst}\left (\int \frac {\log \left (1+\frac {2 f x}{-2 a f+b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{x} \, dx,x,a+b x\right )}{\sqrt {e^2-4 d f}} \\ & = \frac {\log \left (c (a+b x)^n\right ) \log \left (-\frac {b \left (e-\sqrt {e^2-4 d f}+2 f x\right )}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {\log \left (c (a+b x)^n\right ) \log \left (-\frac {b \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}+\frac {n \text {Li}_2\left (\frac {2 f (a+b x)}{2 a f-b \left (e-\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}}-\frac {n \text {Li}_2\left (\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.16 (sec) , antiderivative size = 194, normalized size of antiderivative = 0.80 \[ \int \frac {\log \left (c (a+b x)^n\right )}{d+e x+f x^2} \, dx=\frac {\log \left (c (a+b x)^n\right ) \left (\log \left (\frac {b \left (-e+\sqrt {e^2-4 d f}-2 f x\right )}{-b e+2 a f+b \sqrt {e^2-4 d f}}\right )-\log \left (\frac {b \left (e+\sqrt {e^2-4 d f}+2 f x\right )}{-2 a f+b \left (e+\sqrt {e^2-4 d f}\right )}\right )\right )+n \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f+b \left (-e+\sqrt {e^2-4 d f}\right )}\right )-n \operatorname {PolyLog}\left (2,\frac {2 f (a+b x)}{2 a f-b \left (e+\sqrt {e^2-4 d f}\right )}\right )}{\sqrt {e^2-4 d f}} \]

[In]

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

[Out]

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

Maple [C] (warning: unable to verify)

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

Time = 1.29 (sec) , antiderivative size = 616, normalized size of antiderivative = 2.53

method	result	size
risch	\(-\frac {2 b \arctan \left (\frac {2 \left (b x +a \right ) f -2 a f +b e}{\sqrt {4 b^{2} d f -e^{2} b^{2}}}\right ) n \ln \left (b x +a \right )}{\sqrt {4 b^{2} d f -e^{2} b^{2}}}+\frac {2 b \arctan \left (\frac {2 \left (b x +a \right ) f -2 a f +b e}{\sqrt {4 b^{2} d f -e^{2} b^{2}}}\right ) \ln \left (\left (b x +a \right )^{n}\right )}{\sqrt {4 b^{2} d f -e^{2} b^{2}}}+\frac {b n \ln \left (b x +a \right ) \ln \left (\frac {-2 \left (b x +a \right ) f +2 a f -b e +\sqrt {-4 b^{2} d f +e^{2} b^{2}}}{2 a f -b e +\sqrt {-4 b^{2} d f +e^{2} b^{2}}}\right )}{\sqrt {-4 b^{2} d f +e^{2} b^{2}}}-\frac {b n \ln \left (b x +a \right ) \ln \left (\frac {2 \left (b x +a \right ) f -2 a f +b e +\sqrt {-4 b^{2} d f +e^{2} b^{2}}}{-2 a f +b e +\sqrt {-4 b^{2} d f +e^{2} b^{2}}}\right )}{\sqrt {-4 b^{2} d f +e^{2} b^{2}}}+\frac {b n \operatorname {dilog}\left (\frac {-2 \left (b x +a \right ) f +2 a f -b e +\sqrt {-4 b^{2} d f +e^{2} b^{2}}}{2 a f -b e +\sqrt {-4 b^{2} d f +e^{2} b^{2}}}\right )}{\sqrt {-4 b^{2} d f +e^{2} b^{2}}}-\frac {b n \operatorname {dilog}\left (\frac {2 \left (b x +a \right ) f -2 a f +b e +\sqrt {-4 b^{2} d f +e^{2} b^{2}}}{-2 a f +b e +\sqrt {-4 b^{2} d f +e^{2} b^{2}}}\right )}{\sqrt {-4 b^{2} d f +e^{2} b^{2}}}+\frac {2 \left (-\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{n}\right )^{3}}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2} \operatorname {csgn}\left (i \left (b x +a \right )^{n}\right )}{2}+\frac {i \pi \operatorname {csgn}\left (i c \left (b x +a \right )^{n}\right )^{2} \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi \,\operatorname {csgn}\left (i c \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i \left (b x +a \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \arctan \left (\frac {2 f x +e}{\sqrt {4 d f -e^{2}}}\right )}{\sqrt {4 d f -e^{2}}}\)	\(616\)

[In]

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

[Out]

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

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F(-1)]

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

[In]

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

[Out]

Timed out

Maxima [F(-2)]

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

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*d*f-e^2>0)', see `assume?` f
or more deta

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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