\(\int \frac {\log (c (a+b x^2)^n)}{a+b x^2} \, dx\) [357]

Optimal result

Integrand size = 22, antiderivative size = 163 \[ \int \frac {\log \left (c \left (a+b x^2\right )^n\right )}{a+b x^2} \, dx=\frac {i n \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{\sqrt {a} \sqrt {b}}+\frac {2 n \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{\sqrt {a} \sqrt {b}}+\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^n\right )}{\sqrt {a} \sqrt {b}}+\frac {i n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{\sqrt {a} \sqrt {b}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {211, 2520, 12, 5040, 4964, 2449, 2352} \[ \int \frac {\log \left (c \left (a+b x^2\right )^n\right )}{a+b x^2} \, dx=\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^n\right )}{\sqrt {a} \sqrt {b}}+\frac {i n \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{\sqrt {a} \sqrt {b}}+\frac {2 n \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{\sqrt {a} \sqrt {b}}+\frac {i n \operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \]

[In]

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

[Out]

(I*n*ArcTan[(Sqrt[b]*x)/Sqrt[a]]^2)/(Sqrt[a]*Sqrt[b]) + (2*n*ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[(2*Sqrt[a])/(Sqrt
[a] + I*Sqrt[b]*x)])/(Sqrt[a]*Sqrt[b]) + (ArcTan[(Sqrt[b]*x)/Sqrt[a]]*Log[c*(a + b*x^2)^n])/(Sqrt[a]*Sqrt[b])
+ (I*n*PolyLog[2, 1 - (2*Sqrt[a])/(Sqrt[a] + I*Sqrt[b]*x)])/(Sqrt[a]*Sqrt[b])

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 211

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]/a)*ArcTan[x/Rt[a/b, 2]], x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 2352

Int[Log[(c_.)*(x_)]/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-e^(-1))*PolyLog[2, 1 - c*x], x] /; FreeQ[{c, d, e
}, x] && EqQ[e + c*d, 0]

Rule 2449

Int[Log[(c_.)/((d_) + (e_.)*(x_))]/((f_) + (g_.)*(x_)^2), x_Symbol] :> Dist[-e/g, Subst[Int[Log[2*d*x]/(1 - 2*
d*x), x], x, 1/(d + e*x)], x] /; FreeQ[{c, d, e, f, g}, x] && EqQ[c, 2*d] && EqQ[e^2*f + d^2*g, 0]

Rule 2520

Int[((a_.) + Log[(c_.)*((d_) + (e_.)*(x_)^(n_))^(p_.)]*(b_.))/((f_) + (g_.)*(x_)^2), x_Symbol] :> With[{u = In
tHide[1/(f + g*x^2), x]}, Simp[u*(a + b*Log[c*(d + e*x^n)^p]), x] - Dist[b*e*n*p, Int[u*(x^(n - 1)/(d + e*x^n)
), x], x]] /; FreeQ[{a, b, c, d, e, f, g, n, p}, x] && IntegerQ[n]

Rule 4964

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> Simp[(-(a + b*ArcTan[c*x])^p)*(
Log[2/(1 + e*(x/d))]/e), x] + Dist[b*c*(p/e), Int[(a + b*ArcTan[c*x])^(p - 1)*(Log[2/(1 + e*(x/d))]/(1 + c^2*x
^2)), x], x] /; FreeQ[{a, b, c, d, e}, x] && IGtQ[p, 0] && EqQ[c^2*d^2 + e^2, 0]

Rule 5040

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> Simp[(-I)*((a + b*ArcT
an[c*x])^(p + 1)/(b*e*(p + 1))), x] - Dist[1/(c*d), Int[(a + b*ArcTan[c*x])^p/(I - c*x), x], x] /; FreeQ[{a, b
, c, d, e}, x] && EqQ[e, c^2*d] && IGtQ[p, 0]

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

Mathematica [A] (verified)

Time = 0.04 (sec) , antiderivative size = 128, normalized size of antiderivative = 0.79 \[ \int \frac {\log \left (c \left (a+b x^2\right )^n\right )}{a+b x^2} \, dx=\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (i n \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )+2 n \log \left (\frac {2 i}{i-\frac {\sqrt {b} x}{\sqrt {a}}}\right )+\log \left (c \left (a+b x^2\right )^n\right )\right )+i n \operatorname {PolyLog}\left (2,\frac {i \sqrt {a}+\sqrt {b} x}{-i \sqrt {a}+\sqrt {b} x}\right )}{\sqrt {a} \sqrt {b}} \]

[In]

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

[Out]

(ArcTan[(Sqrt[b]*x)/Sqrt[a]]*(I*n*ArcTan[(Sqrt[b]*x)/Sqrt[a]] + 2*n*Log[(2*I)/(I - (Sqrt[b]*x)/Sqrt[a])] + Log
[c*(a + b*x^2)^n]) + I*n*PolyLog[2, (I*Sqrt[a] + Sqrt[b]*x)/((-I)*Sqrt[a] + Sqrt[b]*x)])/(Sqrt[a]*Sqrt[b])

Maple [C] (warning: unable to verify)

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

Time = 1.25 (sec) , antiderivative size = 264, normalized size of antiderivative = 1.62

method	result	size
risch	\(\frac {\left (\ln \left (\left (b \,x^{2}+a \right )^{n}\right )-n \ln \left (b \,x^{2}+a \right )\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}+\frac {n \left (\munderset {\underline {\hspace {1.25 ex}}\alpha =\operatorname {RootOf}\left (b \,\textit {\_Z}^{2}+a \right )}{\sum }\frac {2 \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (b \,x^{2}+a \right )-b \left (\frac {\ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right )^{2}}{\underline {\hspace {1.25 ex}}\alpha b}+\frac {2 \underline {\hspace {1.25 ex}}\alpha \ln \left (x -\underline {\hspace {1.25 ex}}\alpha \right ) \ln \left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right )}{a}+\frac {2 \underline {\hspace {1.25 ex}}\alpha \operatorname {dilog}\left (\frac {x +\underline {\hspace {1.25 ex}}\alpha }{2 \underline {\hspace {1.25 ex}}\alpha }\right )}{a}\right )}{\underline {\hspace {1.25 ex}}\alpha }\right )}{4 b}+\frac {\left (\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{n}\right ) {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{n}\right )}^{2}}{2}-\frac {i \pi \,\operatorname {csgn}\left (i \left (b \,x^{2}+a \right )^{n}\right ) \operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{n}\right ) \operatorname {csgn}\left (i c \right )}{2}-\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{n}\right )}^{3}}{2}+\frac {i \pi {\operatorname {csgn}\left (i c \left (b \,x^{2}+a \right )^{n}\right )}^{2} \operatorname {csgn}\left (i c \right )}{2}+\ln \left (c \right )\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}\)	\(264\)

[In]

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

[Out]

(ln((b*x^2+a)^n)-n*ln(b*x^2+a))/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))+1/4*n/b*sum(1/_alpha*(2*ln(x-_alpha)*ln(b*
x^2+a)-b*(1/_alpha/b*ln(x-_alpha)^2+2*_alpha/a*ln(x-_alpha)*ln(1/2*(x+_alpha)/_alpha)+2*_alpha/a*dilog(1/2*(x+
_alpha)/_alpha))),_alpha=RootOf(_Z^2*b+a))+(1/2*I*Pi*csgn(I*(b*x^2+a)^n)*csgn(I*c*(b*x^2+a)^n)^2-1/2*I*Pi*csgn
(I*(b*x^2+a)^n)*csgn(I*c*(b*x^2+a)^n)*csgn(I*c)-1/2*I*Pi*csgn(I*c*(b*x^2+a)^n)^3+1/2*I*Pi*csgn(I*c*(b*x^2+a)^n
)^2*csgn(I*c)+ln(c))/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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