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

Optimal. Leaf size=163 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^n\right )}{\sqrt {a} \sqrt {b}}+\frac {i n \text {Li}_2\left (1-\frac {2 \sqrt {a}}{i \sqrt {b} x+\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}}+\frac {i n \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{\sqrt {a} \sqrt {b}}+\frac {2 n \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \]

[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)

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Rubi [A]  time = 0.15, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 7, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {205, 2470, 12, 4920, 4854, 2402, 2315} \[ \frac {i n \text {PolyLog}\left (2,1-\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right )}{\sqrt {a} \sqrt {b}}+\frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \log \left (c \left (a+b x^2\right )^n\right )}{\sqrt {a} \sqrt {b}}+\frac {i n \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )^2}{\sqrt {a} \sqrt {b}}+\frac {2 n \log \left (\frac {2 \sqrt {a}}{\sqrt {a}+i \sqrt {b} x}\right ) \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \]

Antiderivative was successfully verified.

[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 205

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

Rule 2315

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

Rule 2402

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 2470

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 4854

Int[((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)/((d_) + (e_.)*(x_)), x_Symbol] :> -Simp[((a + b*ArcTan[c*x])^p*Lo
g[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 4920

Int[(((a_.) + ArcTan[(c_.)*(x_)]*(b_.))^(p_.)*(x_))/((d_) + (e_.)*(x_)^2), x_Symbol] :> -Simp[(I*(a + b*ArcTan
[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*} \int \frac {\log \left (c \left (a+b x^2\right )^n\right )}{a+b x^2} \, 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}}-(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) \operatorname {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*}

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Mathematica [A]  time = 0.05, size = 128, normalized size = 0.79 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (\log \left (c \left (a+b x^2\right )^n\right )+2 n \log \left (\frac {2 i}{-\frac {\sqrt {b} x}{\sqrt {a}}+i}\right )+i n \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right )+i n \text {Li}_2\left (\frac {\sqrt {b} x+i \sqrt {a}}{\sqrt {b} x-i \sqrt {a}}\right )}{\sqrt {a} \sqrt {b}} \]

Antiderivative was successfully verified.

[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])

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fricas [F]  time = 0.56, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\log \left ({\left (b x^{2} + a\right )}^{n} c\right )}{b x^{2} + a}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (b x^{2} + a\right )}^{n} c\right )}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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maple [F]  time = 1.50, size = 0, normalized size = 0.00 \[ \int \frac {\ln \left (c \left (b \,x^{2}+a \right )^{n}\right )}{b \,x^{2}+a}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log \left ({\left (b x^{2} + a\right )}^{n} c\right )}{b x^{2} + a}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\ln \left (c\,{\left (b\,x^2+a\right )}^n\right )}{b\,x^2+a} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\log {\left (c \left (a + b x^{2}\right )^{n} \right )}}{a + b x^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[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)

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