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

Optimal result

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

[Out]

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

Rubi [A] (verified)

Time = 0.13 (sec) , antiderivative size = 229, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2456, 2441, 2440, 2438} \[ \int \frac {\log \left (c (a+b x)^n\right )}{d+e x^2} \, dx=\frac {\log \left (c (a+b x)^n\right ) \log \left (\frac {b \left (\sqrt {-d}-\sqrt {e} x\right )}{a \sqrt {e}+b \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {\log \left (c (a+b x)^n\right ) \log \left (\frac {b \left (\sqrt {-d}+\sqrt {e} x\right )}{b \sqrt {-d}-a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}-\frac {n \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} (a+b x)}{b \sqrt {-d}-a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}}+\frac {n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (a+b x)}{\sqrt {e} a+b \sqrt {-d}}\right )}{2 \sqrt {-d} \sqrt {e}} \]

[In]

Int[Log[c*(a + b*x)^n]/(d + e*x^2),x]

[Out]

(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, -((S
qrt[e]*(a + b*x))/(b*Sqrt[-d] - a*Sqrt[e]))])/(2*Sqrt[-d]*Sqrt[e]) + (n*PolyLog[2, (Sqrt[e]*(a + b*x))/(b*Sqrt
[-d] + a*Sqrt[e])])/(2*Sqrt[-d]*Sqrt[e])

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 2456

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_))^(n_.)]*(b_.))^(p_.)*((f_) + (g_.)*(x_)^(r_))^(q_.), x_Symbol] :> In
t[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]
&& IGtQ[p, 0] && IntegerQ[q] && (GtQ[q, 0] || (IntegerQ[r] && NeQ[r, 1]))

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

Mathematica [A] (verified)

Time = 0.09 (sec) , antiderivative size = 178, normalized size of antiderivative = 0.78 \[ \int \frac {\log \left (c (a+b x)^n\right )}{d+e x^2} \, dx=\frac {\log \left (c (a+b x)^n\right ) \left (\log \left (\frac {b \left (\sqrt {-d}-\sqrt {e} x\right )}{b \sqrt {-d}+a \sqrt {e}}\right )-\log \left (\frac {b \left (\sqrt {-d}+\sqrt {e} x\right )}{b \sqrt {-d}-a \sqrt {e}}\right )\right )-n \operatorname {PolyLog}\left (2,-\frac {\sqrt {e} (a+b x)}{b \sqrt {-d}-a \sqrt {e}}\right )+n \operatorname {PolyLog}\left (2,\frac {\sqrt {e} (a+b x)}{b \sqrt {-d}+a \sqrt {e}}\right )}{2 \sqrt {-d} \sqrt {e}} \]

[In]

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

[Out]

(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*PolyLog[2, -((Sqrt[e]*(a + b*x))/(b*Sqrt[-d] - a*Sqrt[e]))] + n*PolyLog[2, (Sq
rt[e]*(a + b*x))/(b*Sqrt[-d] + a*Sqrt[e])])/(2*Sqrt[-d]*Sqrt[e])

Maple [C] (warning: unable to verify)

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

Time = 0.75 (sec) , antiderivative size = 392, normalized size of antiderivative = 1.71

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

[In]

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

[Out]

-1/(d*e)^(1/2)*arctan(1/2*(2*e*(b*x+a)-2*a*e)/(d*e)^(1/2)/b)*n*ln(b*x+a)+1/(d*e)^(1/2)*arctan(1/2*(2*e*(b*x+a)
-2*a*e)/(d*e)^(1/2)/b)*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)+a*e)/(b*(-d*e)^
(1/2)+a*e))-1/2*n*ln(b*x+a)/(-d*e)^(1/2)*ln((b*(-d*e)^(1/2)+e*(b*x+a)-a*e)/(b*(-d*e)^(1/2)-a*e))+1/2*n/(-d*e)^
(1/2)*dilog((b*(-d*e)^(1/2)-e*(b*x+a)+a*e)/(b*(-d*e)^(1/2)+a*e))-1/2*n/(-d*e)^(1/2)*dilog((b*(-d*e)^(1/2)+e*(b
*x+a)-a*e)/(b*(-d*e)^(1/2)-a*e))+(-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))/
(d*e)^(1/2)*arctan(x*e/(d*e)^(1/2))

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-2)]

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

[In]

integrate(log(c*(b*x+a)^n)/(e*x^2+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(e>0)', see `assume?` for more
details)Is e

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

int(log(c*(a + b*x)^n)/(d + e*x^2),x)

[Out]

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